Fundamentals.of.Physics_Halliday
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Fundamentals.of.Physics_Halliday

Course: PHYS 101-102, Spring 2008

School: Bilkent University

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curacy, should you need rescuing. Approximate values of time intervals are presented in Table 1.3. In addition to SI, another system of units, the British engineering system (sometimes called the conventional system), is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass, and 1.1 Standards of Length, Mass, and Time 7 TABLE 1.3...

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should curacy, you need rescuing. Approximate values of time intervals are presented in Table 1.3. In addition to SI, another system of units, the British engineering system (sometimes called the conventional system), is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass, and 1.1 Standards of Length, Mass, and Time 7 TABLE 1.3 Approximate Values of Some Time Intervals Interval (s) Age of the Universe Age of the Earth Average age of a college student One year One day (time for one rotation of the Earth about its axis) Time between normal heartbeats Period of audible sound waves Period of typical radio waves Period of vibration of an atom in a solid Period of visible light waves Duration of a nuclear collision Time for light to cross a proton 5 1.3 6.3 3.16 8.64 8 10 10 10 10 10 10 1017 1017 108 107 104 10 1 3 6 13 15 22 24 time are the foot (ft), slug, and second, respectively. In this text we shall use SI units because they are almost universally accepted in science and industry. We shall make some limited use of British engineering units in the study of classical mechanics. In addition to the basic SI units of meter, kilogram, and second, we can also use other units, such as millimeters and nanoseconds, where the prefixes milli- and nano- denote various powers of ten. Some of the most frequently used prefixes for the various powers of ten and their abbreviations are listed in Table 1.4. For TABLE 1.4 Prefixes for SI Units Power 10 24 10 21 10 18 10 15 10 12 10 9 10 6 10 3 10 2 10 1 101 103 106 109 1012 1015 1018 1021 1024 Prefix yocto zepto atto femto pico nano micro milli centi deci deka kilo mega giga tera peta exa zetta yotta Abbreviation y z a f p n m c d da k M G T P E Z Y 8 CHAPTER 1 Physics and Measurements example, 10 3 m is equivalent to 1 millimeter (mm), and 103 m corresponds to 1 kilometer (km). Likewise, 1 kg is 103 grams (g), and 1 megavolt (MV) is 106 volts (V). u u 1.2 Quark composition of a proton THE BUILDING BLOCKS OF MATTER d Proton Neutron Gold nucleus Nucleus Gold atoms Gold cube Figure 1.2 Levels of organization in matter. Ordinary matter consists of atoms, and at the center of each atom is a compact nucleus consisting of protons and neutrons. Protons and neutrons are composed of quarks. The quark composition of a proton is shown. A 1-kg cube of solid gold has a length of 3.73 cm on a side. Is this cube nothing but wall-to-wall gold, with no empty space? If the cube is cut in half, the two pieces still retain their chemical identity as solid gold. But what if the pieces are cut again and again, indefinitely? Will the smaller and smaller pieces always be gold? Questions such as these can be traced back to early Greek philosophers. Two of them -- Leucippus and his student Democritus -- could not accept the idea that such cuttings could go on forever. They speculated that the process ultimately must end when it produces a particle that can no longer be cut. In Greek, atomos means "not sliceable." From this comes our English word atom. Let us review briefly what is known about the structure of matter. All ordinary matter consists of atoms, and each atom is made up of electrons surrounding a central nucleus. Following the discovery of the nucleus in 1911, the question arose: Does it have structure? That is, is the nucleus a single particle or a collection of particles? The exact composition of the nucleus is not known completely even today, but by the early 1930s a model evolved that helped us understand how the nucleus behaves. Specifically, scientists determined that occupying the nucleus are two basic entities, protons and neutrons. The proton carries a positive charge, and a specific element is identified by the number of protons in its nucleus. This number is called the atomic number of the element. For instance, the nucleus of a hydrogen atom contains one proton (and so the atomic number of hydrogen is 1), the nucleus of a helium atom contains two protons (atomic number 2), and the nucleus of a uranium atom contains 92 protons (atomic number 92). In addition to atomic number, there is a second number characterizing atoms -- mass number, defined as the number of protons plus neutrons in a nucleus. As we shall see, the atomic number of an element never varies (i.e., the number of protons does not vary) but the mass number can vary (i.e., the number of neutrons varies). Two or more atoms of the same element having different mass numbers are isotopes of one another. The existence of neutrons was verified conclusively in 1932. A neutron has no charge and a mass that is about equal to that of a proton. One of its primary purposes is to act as a "glue" that holds the nucleus together. If neutrons were not present in the nucleus, the repulsive force between the positively charged particles would cause the nucleus to come apart. But is this where the breaking down stops? Protons, neutrons, and a host of other exotic particles are now known to be composed of six different varieties of particles called quarks, which have been given the names of up, down, strange, 2 charm, bottom, and top. The up, charm, and top quarks have charges of 3 that of the proton, whereas the down, strange, and bottom quarks have charges of 1 3 that of the proton. The proton consists of two up quarks and one down quark (Fig. 1.2), which you can easily show leads to the correct charge for the proton. Likewise, the neutron consists of two down quarks and one up quark, giving a net charge of zero. 1.3 Density 9 1.3 DENSITY A table of the letters in the Greek alphabet is provided on the back endsheet of this textbook. A property of any substance is its density (Greek letter rho), defined as the amount of mass contained in a unit volume, which we usually express as mass per unit volume: m V (1.1) For example, aluminum has a density of 2.70 g/cm3, and lead has a density of 11.3 g/cm3. Therefore, a piece of aluminum of volume 10.0 cm3 has a mass of 27.0 g, whereas an equivalent volume of lead has a mass of 113 g. A list of densities for various substances is given Table 1.5. The difference in density between aluminum and lead is due, in part, to their different atomic masses. The atomic mass of an element is the average mass of one atom in a sample of the element that contains all the element's isotopes, where the relative amounts of isotopes are the same as the relative amounts found in nature. The unit for atomic mass is the atomic mass unit (u), where 1 u 1.660 540 2 10 27 kg. The atomic mass of lead is 207 u, and that of aluminum is 27.0 u. However, the ratio of atomic masses, 207 u/27.0 u 7.67, does not correspond to the ratio of densities, (11.3 g/cm3)/(2.70 g/cm3) 4.19. The discrepancy is due to the difference in atomic separations and atomic arrangements in the crystal structure of these two substances. The mass of a nucleus is measured relative to the mass of the nucleus of the carbon-12 isotope, often written as 12C. (This isotope of carbon has six protons and six neutrons. Other carbon isotopes have six protons but different numbers of neutrons.) Practically all of the mass of an atom is contained within the nucleus. Because the atomic mass of 12C is defined to be exactly 12 u, the proton and neutron each have a mass of about 1 u. One mole (mol) of a substance is that amount of the substance that contains as many particles (atoms, molecules, or other particles) as there are atoms in 12 g of the carbon-12 isotope. One mole of substance A contains the same number of particles as there are in 1 mol of any other substance B. For example, 1 mol of aluminum contains the same number of atoms as 1 mol of lead. TABLE 1.5 Densities of Various Substances Substance Gold Uranium Lead Copper Iron Aluminum Magnesium Water Air Density (103 kg/m3) 19.3 18.7 11.3 8.92 7.86 2.70 1.75 1.00 0.0012 10 CHAPTER 1 Physics and Measurements Experiments have shown that this number, known as Avogadro's number, NA , is NA 6.022 137 10 23 particles/mol Avogadro's number is defined so that 1 mol of carbon-12 atoms has a mass of exactly 12 g. In general, the mass in 1 mol of any element is the element's atomic mass expressed in grams. For example, 1 mol of iron (atomic mass 55.85 u) has a mass of 55.85 g (we say its molar mass is 55.85 g/mol), and 1 mol of lead (atomic mass 207 u) has a mass of 207 g (its molar mass is 207 g/mol). Because there are 6.02 1023 particles in 1 mol of any element, the mass per atom for a given element is m atom molar mass NA (1.2) For example, the mass of an iron atom is m Fe 55.85 g/mol 6.02 10 23 atoms/mol 9.28 10 23 g/atom EXAMPLE 1.1 How Many Atoms in the Cube? minum (27 g) contains 6.02 1023 atoms: NA 27 g 6.02 (0.54 g)(6.02 27 g 10 23 atoms 27 g 10 23 atoms) N 0.54 g N 0.54 g 1.2 10 22 atoms A solid cube of aluminum (density 2.7 g/cm3) has a volume of 0.20 cm3. How many aluminum atoms are contained in the cube? Solution Since density equals mass per unit volume, the mass m of the cube is m V (2.7 g/cm3)(0.20 cm3) 0.54 g N To find the number of atoms N in this mass of aluminum, we can set up a proportion using the fact that one mole of alu- 1.4 DIMENSIONAL ANALYSIS The word dimension has a special meaning in physics. It usually denotes the physical nature of a quantity. Whether a distance is measured in the length unit feet or the length unit meters, it is still a distance. We say the dimension -- the physical nature -- of distance is length. The symbols we use in this book to specify length, mass, and time are L, M, and T, respectively. We shall often use brackets [ ] to denote the dimensions of a physical quantity. For example, the symbol we use for speed in this book is v, and in our notation the dimensions of speed are written [v] L/T. As another example, the dimensions of area, for which we use the symbol A, are [A] L2. The dimensions of area, volume, speed, and acceleration are listed in Table 1.6. In solving problems in physics, there is a useful and powerful procedure called dimensional analysis. This procedure, which should always be used, will help minimize the need for rote memorization of equations. Dimensional analysis makes use of the fact that dimensions can be treated as algebraic quantities. That is, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. 1.4 Dimensional Analysis 11 TABLE 1.6 Dimensions and Common Units of Area, Volume, Speed, and Acceleration System SI British engineering Area (L2) m2 ft2 Volume (L3) m3 ft3 Speed (L/T) m/s ft/s Acceleration (L/T 2) m/s2 ft/s2 By following these simple rules, you can use dimensional analysis to help determine whether an expression has the correct form. The relationship can be correct only if the dimensions are the same on both sides of the equation. To illustrate this procedure, suppose you wish to derive a formula for the distance x traveled by a car in a time t if the car starts from rest and moves with constant acceleration a. In Chapter 2, we shall find that the correct expression is x 1at 2. Let us use dimensional analysis to check the validity of this expression. 2 The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the dimensions for acceleration, L/T 2, and time, T, into the equation. That is, the dimensional form of the equation x 1at 2 is 2 L L T2 T2 L The units of time squared cancel as shown, leaving the unit of length. A more general procedure using dimensional analysis is to set up an expression of the form x a nt m where n and m are exponents that must be determined and the symbol indicates a proportionality. This relationship is correct only if the dimensions of both sides are the same. Because the dimension of the left side is length, the dimension of the right side must also be length. That is, [a nt m] L LT 0 Because the dimensions of acceleration are L/T 2 and the dimension of time is T, we have L T2 n Tm 2n L1 L1 Ln T m Because the exponents of L and T must be the same on both sides, the dimensional equation is balanced under the conditions m 2n 0, n 1, and m 2. Returning to our original expression x a nt mwe conclude that x , at 2This result . 1 2 differs by a factor of 2 from the correct expression, which is x 2at . Because the factor 1 is dimensionless, there is no way of determining it using dimensional 2 analysis. 12 CHAPTER 1 Physics and Measurements Quick Quiz 1.1 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. EXAMPLE 1.2 Analysis of an Equation The same table gives us L/T 2 for the dimensions of acceleration, and so the dimensions of at are [at] L (T) T2 L T Show that the expression v at is dimensionally correct, where v represents speed, a acceleration, and t a time interval. Solution For the speed term, we have from Table 1.6 [v] L T Therefore, the expression is dimensionally correct. (If the expression were given as v at 2, it would be dimensionally incorrect. Try it and see!) EXAMPLE 1.3 Analysis of a Power Law This dimensional equation is balanced under the conditions n Therefore n sion as m 1 and m 2 Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r n, and some power of v, say v m. How can we determine the values of n and m? 1, and we can write the acceleration expresv2 r Solution Let us take a to be a kr nv m a kr 1v 2 k where k is a dimensionless constant of proportionality. Knowing the dimensions of a, r, and v, we see that the dimensional equation must be L/T 2 Ln(L/T)m Ln m/T m When we discuss uniform circular motion later, we shall see that k 1 if a consistent set of units is used. The constant k would not equal 1 if, for example, v were in km/h and you wanted a in m/s2. QuickLab Estimate the weight (in pounds) of two large bottles of soda pop. Note that 1 L of water has a mass of about 1 kg. Use the fact that an object weighing 2.2 lb has a mass of 1 kg. Find some bathroom scales and check your estimate. 1.5 CONVERSION OF UNITS Sometimes it is necessary to convert units from one system to another. Conversion factors between the SI units and conventional units of length are as follows: 1 mi 1m 1 609 m 39.37 in. 1.609 km 3.281 ft 1 ft 1 in. 0.304 8 m 0.025 4 m 30.48 cm 2.54 cm (exactly) A more complete list of conversion factors can be found in Appendix A. Units can be treated as algebraic quantities that can cancel each other. For example, suppose we wish to convert 15.0 in. to centimeters. Because 1 in. is defined as exactly 2.54 cm, we find that 15.0 in. (15.0 in.)(2.54 cm/in.) 38.1 cm This works because multiplying by (2.54 cm) is the same as multiplying by 1, because 1 in. the numerator and denominator describe identical things. 1.6 Estimates and Order-of-Magnitude Calculations 13 (Left) This road sign near Raleigh, North Carolina, shows distances in miles and kilometers. How accurate are the conversions? (Billy E. Barnes/Stock Boston). (Right) This vehicle's speedometer gives speed readings in miles per hour and in kilometers per hour. Try confirming the conversion between the two sets of units for a few readings of the dial. (Paul Silverman/Fundamental Photographs) EXAMPLE 1.4 The Density of a Cube V Therefore, m, the m V 0.856 kg 1.53 10 4 The mass of a solid cube is 856 g, and each edge has a length of 5.35 cm. Determine the density of the cube in basic SI units. L3 (5.35 cm 10 6 10 m3 2 m/cm)3 10 4 (5.35)3 1.53 m3 Solution Because 1 g 10 3 kg and 1 cm mass m and volume V in basic SI units are m 856 g 10 3 10 2 kg/g 0.856 kg m3 5.59 10 3 kg/m3 1.6 ESTIMATES AND ORDER-OFMAGNITUDE CALCULATIONS It is often useful to compute an approximate answer to a physical problem even where little information is available. Such an approximate answer can then be used to determine whether a more accurate calculation is necessary. Approximations are usually based on certain assumptions, which must be modified if greater accuracy is needed. Thus, we shall sometimes refer to the order of magnitude of a certain quantity as the power of ten of the number that describes that quantity. If, for example, we say that a quantity increases in value by three orders of magnitude, this means that its value is increased by a factor of 103 1000. Also, if a quantity is given as 3 103, we say that the order of magnitude of that quantity is 103 (or in symbolic form, 3 103 103). Likewise, the quantity 8 107 108. The spirit of order-of-magnitude calculations, sometimes referred to as "guesstimates" or "ball-park figures," is given in the following quotation: "Make an estimate before every calculation, try a simple physical argument . . . before every derivation, guess the answer to every puzzle. Courage: no one else needs to 14 CHAPTER 1 Physics and Measurements know what the guess is." 4 Inaccuracies caused by guessing too low for one number are often canceled out by other guesses that are too high. You will find that with practice your guesstimates get better and better. Estimation problems can be fun to work as you freely drop digits, venture reasonable approximations for unknown numbers, make simplifying assumptions, and turn the question around into something you can answer in your head. EXAMPLE 1.5 Breaths in a Lifetime approximately 1 yr 400 days yr 25 h day 60 min h 6 105 min Estimate the number of breaths taken during an average life span. Solution We shall start by guessing that the typical life span is about 70 years. The only other estimate we must make in this example is the average number of breaths that a person takes in 1 min. This number varies, depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate of the average. (This is certainly closer to the true value than 1 breath per minute or 100 breaths per minute.) The number of minutes in a year is Notice how much simpler it is to multiply 400 25 than it is to work with the more accurate 365 24. These approximate values for the number of days in a year and the number of hours in a day are close enough for our purposes. Thus, in 70 years there will be (70 yr)(6 105 min/yr) 4 107 min. At a rate of 10 breaths/min, an individual would take 4 10 8 breaths in a lifetime. EXAMPLE 1.6 It's a Long Way to San Jose Now we switch to scientific notation so that we can do the calculation mentally: (3 10 3 mi)(2.5 10 3 steps/mi) 7.5 10 6 steps Estimate the number of steps a person would take walking from New York to Los Angeles. Solution Without looking up the distance between these two cities, you might remember from a geography class that they are about 3 000 mi apart. The next approximation we must make is the length of one step. Of course, this length depends on the person doing the walking, but we can estimate that each step covers about 2 ft. With our estimated step size, we can determine the number of steps in 1 mi. Because this is a rough calculation, we round 5 280 ft/mi to 5 000 ft/mi. (What percentage error does this introduce?) This conversion factor gives us 5 000 ft/mi 2 ft/step 2 500 steps/mi 10 7 steps So if we intend to walk across the United States, it will take us on the order of ten million steps. This estimate is almost certainly too small because we have not accounted for curving roads and going up and down hills and mountains. Nonetheless, it is probably within an order of magnitude of the correct answer. EXAMPLE 1.7 How Much Gas Do We Use? age distance each car travels per year is 10 000 mi. If we assume a gasoline consumption of 20 mi/gal or 0.05 gal/mi, then each car uses about 500 gal/yr. Multiplying this by the total number of cars in the United States gives an estimated total consumption of 5 1010 gal 10 11 gal. Estimate the number of gallons of gasoline used each year by all the cars in the United States. Solution There are about 270 million people in the United States, and so we estimate that the number of cars in the country is 100 million (guessing that there are between two and three people per car). We also estimate that the aver- 4 E. Taylor and J. A. Wheeler, Spacetime Physics, San Francisco, W. H. Freeman & Company, Publishers, 1966, p. 60. 1.7 Significant Figures 15 1.7 SIGNIFICANT FIGURES When physical quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. Suppose that we are asked to measure the area of a computer disk label using a meter stick as a measuring instrument. Let us assume that the accuracy to which we can measure with this stick is 0.1 cm. If the length of the label is measured to be 5.5 cm, we can claim only that its length lies somewhere between 5.4 cm and 5.6 cm. In this case, we say that the measured value has two significant figures. Likewise, if the label's width is measured to be 6.4 cm, the actual value lies between 6.3 cm and 6.5 cm. Note that the significant figures include the first estimated digit. Thus we could write the measured values as (5.5 0.1) cm and (6.4 0.1) cm. Now suppose we want to find the area of the label by multiplying the two measured values. If we were to claim the area is (5.5 cm)(6.4 cm) 35.2 cm2, our answer would be unjustifiable because it contains three significant figures, which is greater than the number of significant figures in either of the measured lengths. A good rule of thumb to use in determining the number of significant figures that can be claimed is as follows: When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the least accurate of the quantities being multiplied, where "least accurate" means "having the lowest number of significant figures." The same rule applies to division. Applying this rule to the multiplication example above, we see that the answer for the area can have only two significant figures because our measured lengths have only two significant figures. Thus, all we can claim is that the area is 35 cm2, realizing that the value can range between (5.4 cm)(6.3 cm) 34 cm2 and (5.6 cm)(6.5 cm) 36 cm2. Zeros may or may not be significant figures. Those used to position the decimal point in such numbers as 0.03 and 0.007 5 are not significant. Thus, there are one and two significant figures, respectively, in these two values. When the zeros come after other digits, however, there is the possibility of misinterpretation. For example, suppose the mass of an object is given as 1 500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point or whether they represent significant figures in the measurement. To remove this ambiguity, it is common to use scientific notation to indicate the number of significant figures. In this case, we would express the mass as 1.5 103 g if there are two significant figures in the measured value, 1.50 103 g if there are three significant figures, and 1.500 103 g if there are four. The same rule holds when the number is less than 1, so that 2.3 10 4 has two significant figures (and so could be written 0.000 23) and 2.30 10 4 has three significant figures (also written 0.000 230). In general, a significant figure is a reliably known digit (other than a zero used to locate the decimal point). For addition and subtraction, you must consider the number of decimal places when you are determining how many significant figures to report. QuickLab Determine the thickness of a page from this book. (Note that numbers that have no measurement errors -- like the count of a number of pages -- do not affect the significant figures in a calculation.) In terms of significant figures, why is it better to measure the thickness of as many pages as possible and then divide by the number of sheets? 16 CHAPTER 1 Physics and Measurements When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum. For example, if we wish to compute 123 5.35, the answer given to the correct number of significant figures is 128 and not 128.35. If we compute the sum 1.000 1 0.000 3 1.000 4, the result has five significant figures, even though one of the terms in the sum, 0.000 3, has only one significant figure. Likewise, if we perform the subtraction 1.002 0.998 0.004, the result has only one significant figure even though one term has four significant figures and the other has three. In this book, most of the numerical examples and end-of-chapter problems will yield answers having three significant figures. When carrying out estimates we shall typically work with a single significant figure. Quick Quiz 1.2 Suppose you measure the position of a chair with a meter stick and record that the center of the seat is 1.043 860 564 2 m from a wall. What would a reader conclude from this recorded measurement? EXAMPLE 1.8 The Area of a Rectangle (21.3 (209 9.80 21.3 0.1 0.2 9.80) cm2 A rectangular plate has a length of (21.3 0.2) cm and a width of (9.80 0.1) cm. Find the area of the plate and the uncertainty in the calculated area. 4) cm2 Solution Area w (21.3 0.2 cm) (9.80 0.1 cm) Because the input data were given to only three significant figures, we cannot claim any more in our result. Do you see why we did not need to multiply the uncertainties 0.2 cm and 0.1 cm? EXAMPLE 1.9 Installing a Carpet Note that in reducing 43.976 6 to three significant figures for our answer, we used a general rule for rounding off numbers that states that the last digit retained (the 9 in this example) is increased by 1 if the first digit dropped (here, the 7) is 5 or greater. (A technique for avoiding error accumulation is to delay rounding of numbers in a long calculation until you have the final result. Wait until you are ready to copy the answer from your calculator before rounding to the correct number of significant figures.) A carpet is to be installed in a room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room. Solution If you multiply 12.71 m by 3.46 m on your calculator, you will get an answer of 43.976 6 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of significant figures in the least accurate of the quantities being measured. In this example, we have only three significant figures in our least accurate measurement, so we should express our final answer as 44.0 m2. Problems 17 SUMMARY The three fundamental physical quantities of mechanics are length, mass, and time, which in the SI system have the units meters (m), kilograms (kg), and seconds (s), respectively. Prefixes indicating various powers of ten are used with these three basic units. The density of a substance is defined as its mass per unit volume. Different substances have different densities mainly because of differences in their atomic masses and atomic arrangements. The number of particles in one mole of any element or compound, called Avogadro's number, NA , is 6.02 1023. The method of dimensional analysis is very powerful in solving physics problems. Dimensions can be treated as algebraic quantities. By making estimates and making order-of-magnitude calculations, you should be able to approximate the answer to a problem when there is not enough information available to completely specify an exact solution. When you compute a result from several measured numbers, each of which has a certain accuracy, you should give the result with the correct number of significant figures. QUESTIONS 1. In this chapter we described how the Earth's daily rotation on its axis was once used to define the standard unit of time. What other types of natural phenomena could serve as alternative time standards? 2. Suppose that the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standarage, how many atoms were abraded from the ring during each second of your marriage? The molar mass of gold is 197 g/mol. 9. A small cube of iron is observed under a microscope. The edge of the cube is 5.00 10 6 cm long. Find (a) the mass of the cube and (b) the number of iron atoms in the cube. The molar mass of iron is 55.9 g/mol, and its density is 7.86 g/cm3. 10. A structural I-beam is made of steel. A view of its crosssection and its dimensions are shown in Figure P1.10. 15.0 cm WEB g where is the length of the pendulum and g is the freefall acceleration in units of length divided by the square of time. Show that this equation is dimensionally correct. 15. Which of the equations below are dimensionally correct? (a) v v 0 ax (b) y (2 m) cos(kx), where k 2 m 1 16. Newton's law of universal gravitation is represented by F GMm r2 1.00 cm 36.0 cm Here F is the gravitational force, M and m are masses, and r is a length. Force has the SI units kg m/s2. What are the SI units of the proportionality constant G ? 17. The consumption of natural gas by a company satisfies the empirical equation V 1.50t 0.008 00t 2, where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Put the proper units on the coefficients. Assume a month is 30.0 days. 1.00 cm Section 1.5 Conversion of Units Figure P1.10 18. Suppose your hair grows at the rate 1/32 in. per day. Find the rate at which it grows in nanometers per second. Since the distance between atoms in a molecule is Problems on the order of 0.1 nm, your answer suggests how rapidly layers of atoms are assembled in this protein synthesis. A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. An auditorium measures 40.0 m 20.0 m 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic feet and (b) the weight of air in the room in pounds? Assume that it takes 7.00 min to fill a 30.0-gal gasoline tank. (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time, in hours, required to fill a 1-cubic-meter volume at the same rate. (1 U.S. gal 231 in.3 ) A creature moves at a speed of 5.00 furlongs per fortnight (not a very common unit of speed). Given that 1 furlong 220 yards and 1 fortnight 14 days, determine the speed of the creature in meters per second. What kind of creature do you think it might be? A section of land has an area of 1 mi2 and contains 640 acres. Determine the number of square meters in 1 acre. A quart container of ice cream is to be made in the form of a cube. What should be the length of each edge in centimeters? (Use the conversion 1 gal 3.786 L.) A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3 ). An astronomical unit (AU) is defined as the average distance between the Earth and the Sun. (a) How many astronomical units are there in one lightyear? (b) Determine the distance from the Earth to the Andromeda galaxy in astronomical units. The mass of the Sun is 1.99 1030 kg, and the mass of an atom of hydrogen, of which the Sun is mostly composed, is 1.67 10 27 kg. How many atoms are there in the Sun? (a) Find a conversion factor to convert from miles per hour to kilometers per hour. (b) In the past, a federal law mandated that highway speed limits would be 55 mi/h. Use the conversion factor of part (a) to find this speed in kilometers per hour. (c) The maximum highway speed is now 65 mi/h in some places. In kilometers per hour, how much of an increase is this over the 55-mi/h limit? At the time of this book's printing, the U. S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000/s, how many years would it take to pay off a $6-trillion debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earth's equator, how many times would they encircle the Earth? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) 19 19. 20. WEB 21. 30. (a) How many seconds are there in a year? (b) If one micrometeorite (a sphere with a diameter of 1.00 10 6 m) strikes each square meter of the Moon each second, how many years will it take to cover the Moon to a depth of 1.00 m? (Hint: Consider a cubic box on the Moon 1.00 m on a side, and find how long it will take to fill the box.) 31. One gallon of paint (volume 3.78 10 3 m3 ) covers an area of 25.0 m2. What is the thickness of the paint on the wall? 32. A pyramid has a height of 481 ft, and its base covers an area of 13.0 acres (Fig. P1.32). If the volume of a pyramid is given by the expression V 1Bh, where B is the 3 area of the base and h is the height, find the volume of this pyramid in cubic meters. (1 acre 43 560 ft2 ) 22. 23. 24. 25. Figure P1.32 Problems 32 and 33. 26. 27. 28. 29. 33. The pyramid described in Problem 32 contains approximately two million stone blocks that average 2.50 tons each. Find the weight of this pyramid in pounds. 34. Assuming that 70% of the Earth's surface is covered with water at an average depth of 2.3 mi, estimate the mass of the water on the Earth in kilograms. 35. The amount of water in reservoirs is often measured in acre-feet. One acre-foot is a volume that covers an area of 1 acre to a depth of 1 ft. An acre is an area of 43 560 ft2. Find the volume in SI units of a reservoir containing 25.0 acre-ft of water. 36. A hydrogen atom has a diameter of approximately 1.06 10 10 m, as defined by the diameter of the spherical electron cloud around the nucleus. The hydrogen nucleus has a diameter of approximately 2.40 10 15 m. (a) For a scale model, represent the diameter of the hydrogen atom by the length of an American football field (100 yards 300 ft), and determine the diameter of the nucleus in millimeters. (b) The atom is how many times larger in volume than its nucleus? 37. The diameter of our disk-shaped galaxy, the Milky Way, is about 1.0 105 lightyears. The distance to Messier 31 -- which is Andromeda, the spiral galaxy nearest to the Milky Way -- is about 2.0 million lightyears. If a scale model represents the Milky Way and Andromeda galax- 20 CHAPTER 1 Physics and Measurements ifying examinations and for his own facility in making order-of-magnitude calculations. ies as dinner plates 25 cm in diameter, determine the distance between the two plates. 38. The mean radius of the Earth is 6.37 106 m, and that of the Moon is 1.74 108 cm. From these data calculate (a) the ratio of the Earth's surface area to that of the Moon and (b) the ratio of the Earth's volume to that of the Moon. Recall that the surface area of a sphere is 4 r 2 and that the volume of a sphere is 4 r 3. 3 WEB Section 1.7 Significant Figures 39. One cubic meter (1.00 m3 ) of aluminum has a mass of 2.70 103 kg, and 1.00 m3 of iron has a mass of 7.86 103 kg. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius 2.00 cm on an equal-arm balance. 40. Let A1 represent the density of aluminum and Fe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equalarm balance. Section 1.6 WEB Estimates and Order-ofMagnitude Calculations 41. Estimate the number of Ping-Pong balls that would fit into an average-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. 42. McDonald's sells about 250 million packages of French fries per year. If these fries were placed end to end, estimate how far they would reach. 43. An automobile tire is rated to last for 50 000 miles. Estimate the number of revolutions the tire will make in its lifetime. 44. Approximately how many raindrops fall on a 1.0-acre lot during a 1.0-in. rainfall? 45. Grass grows densely everywhere on a quarter-acre plot of land. What is the order of magnit mass 8-fold, and tripling the radius increases the mass 27-fold. Therefore, its mass is proportional to the cube of its radius. Because m r 3we can write m kr 3. Dimen, sional analysis shows that the proportionality constant k must have units kg/m3, but to determine its numerical value requires either experimental data or geometrical reasoning. 1.2 Reporting all these digits implies you have determined the location of the center of the chair's seat to the nearest 0.000 000 000 1 m. This roughly corresponds to being able to count the atoms in your meter stick because each of them is about that size! It would probably be better to record the measurement as 1.044 m: this indicates that you know the position to the nearest millimeter, assuming the meter stick has millimeter markings on its scale. P U Z Z L E R In a moment the arresting cable will be pulled taut, and the 140-mi/h landing of this F/A-18 Hornet on the aircraft carrier USS Nimitz will be brought to a sudden conclusion. The pilot cuts power to the engine, and the plane is stopped in less than 2 s. If the cable had not been successfully engaged, the pilot would have had to take off quickly before reaching the end of the flight deck. Can the motion of the plane be described quantitatively in a way that is useful to ship and aircraft designers and to pilots learning to land on a "postage stamp?" (Courtesy of the USS Nimitz/U.S. Navy) c h a p t e r Motion in One Dimension Chapter Outline 2.1 2.2 2.3 2.4 2.5 Displacement, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 (Optional) Kinematic Equations Derived from Calculus GOAL Problem-Solving Steps 23 24 CHAPTER 2 Motion in One Dimension s a first step in studying classical mechanics, we describe motion in terms of space and time while ignoring the agents that caused that motion. This portion of classical mechanics is called kinematics. (The word kinematics has the same root as cinema. Can you see why?) In this chapter we consider only motion in one dimension. We first define displacement, velocity, and acceleration. Then, using these concepts, we study the motion of objects traveling in one dimension with a constant acceleration. From everyday experience we recognize that motion represents a continuous change in the position of an object. In physics we are concerned with three types of motion: translational, rotational, and vibrational. A car moving down a highway is an example of translational motion, the Earth's spin on its axis is an example of rotational motion, and the back-and-forth movement of a pendulum is an example of vibrational motion. In this and the next few chapters, we are concerned only with translational motion. (Later in the book we shall discuss rotational and vibrational motions.) In our study of translational motion, we describe the moving object as a particle regardless of its size. In general, a particle is a point-like mass having infinitesimal size. For example, if we wish to describe the motion of the Earth around the Sun, we can treat the Earth as a particle and obtain reasonably accurate data about its orbit. This approximation is justified because the radius of the Earth's orbit is large compared with the dimensions of the Earth and the Sun. As an example on a much smaller scale, it is possible to explain the pressure exerted by a gas on the walls of a container by treating the gas molecules as particles. A 2.1 TABLE 2.1 Position of the Car at Various Times Position t(s) 0 10 20 30 40 50 x(m) 30 52 38 0 37 53 DISPLACEMENT, VELOCITY, AND SPEED The motion of a particle is completely known if the particle's position in space is known at all times. Consider a car moving back and forth along the x axis, as shown in Figure 2.1a. When we begin collecting position data, the car is 30 m to the right of a road sign. (Let us assume that all data in this example are known to two significant figures. To convey this information, we should report the initial position as 3..2 for convenience. The choice of which equation you use in a given situation depends on what you know beforehand. Sometimes it is necessary to use two of these equations to solve for two unknowns. For example, suppose initial velocity vxi and acceleration ax are given. You can then find (1) the velocity after an interval t has elapsed, using v x f v xi a x t, and (2) the displacement after an interval t has elapsed, using x f x i v xi t 1a x t 2. You should recognize that the quantities that vary dur2 ing the motion are velocity, displacement, and time. You will get a great deal of practice in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinematics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant. CONCEPTUAL EXAMPLE 2.5 The Velocity of Different Objects fined as x/ t.) There is one point at which the instantaneous velocity is zero -- at the top of the motion. (b) The car's average velocity cannot be evaluated unambiguously with the information given, but it must be some value between 0 and 100 m/s. Because the car will have every instantaneous velocity between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity. (c) Because the spacecraft's instantaneous velocity is constant, its instantaneous velocity at any time and its average velocity over any time interval are the same. Consider the following one-dimensional motions: (a) A ball thrown directly upward rises to a highest point and falls back into the thrower's hand. (b) A race car starts from rest and speeds up to 100 m/s. (c) A spacecraft drifts through space at constant velocity. Are there any points in the motion of these objects at which the instantaneous velocity is the same as the average velocity over the entire motion? If so, identify the point(s). Solution (a) The average velocity for the thrown ball is zero because the ball returns to the starting point; thus its displacement is zero. (Remember that average velocity is de- EXAMPLE 2.6 Entering the Traffic Flow of ax , but that value is hard to guess directly. The other three variables involved in kinematics are position, velocity, and time. Velocity is probably the easiest one to approximate. Let us assume a final velocity of 100 km/h, so that you can merge with traffic. We multiply this value by 1 000 to convert kilome- (a) Estimate your average acceleration as you drive up the entrance ramp to an interstate highway. Solution This problem involves more than our usual amount of estimating! We are trying to come up with a value 38 CHAPTER 2 Motion in One Dimension yields results that are not too different from those derived from careful measurements. (b) How far did you go during the first half of the time interval during which you accelerated? ters to meters and then divide by 3 600 to convert hours to seconds. These two calculations together are roughly equivalent to dividing by 3. In fact, let us just say that the final velocity is vx f 30 m/s. (Remember, you can get away with this type of approximation and with dropping digits when performing mental calculations. If you were starting with British units, you could approximate 1 mi/h as roughly 0.5 m/s and continue from there.) Now we assume that you started up the ramp at about onethird your final velocity, so that vxi 10 m/s. Finally, we assume that it takes about 10 s to get from vxi to vxf , basing this guess on our previous experience in automobiles. We can then find the acceleration, using Equation 2.8: ax vxf t vxi 30 m/s 10 m/s 10 s 2 m/s2 Solution We can calculate the distance traveled during the first 5 s from Equation 2.11: xf xi v xi t 50 m 1 2 2a x t (10 m/s)(5 s) 75 m 1 2 (2 m/s2)(5 s)2 25 m Granted, we made many approximations along the way, but this type of mental effort can be surprisingly useful and often This result indicates that if you had not accelerated, your initial velocity of 10 m/s would have resulted in a 50-m movement up the ramp during the first 5 s. The additional 25 m is the result of your increasing velocity during that interval. Do not be afraid to attempt making educated guesses and doing some fairly drastic number rounding to simplify mental calculations. Physicists engage in this type of thought analysis all the time. EXAMPLE 2.7 Carrier Landing (b) What is the displacement of the plane while it is stopping? A jet lands on an aircraft carrier at 140 mi/h ( 63 m/s). (a) What is its acceleration if it stops in 2.0 s? Solution We define our x axis as the direction of motion of the jet. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. We also note that we are not given the displacement of the jet while it is slowing down. Equation 2.8 is the only equation in Table 2.2 that does not involve displacement, and so we use it to find the acceleration: ax vx f t vxi 0 63 m/s 2.0 s 31 m/s2 Solution We can now use any of the other three equations in Table 2.2 to solve for the displacement. Let us choose Equation 2.10: xf xi 1 2 (vxi vx f )t 1 2 (63 m/s 0)(2.0 s) 63 m If the plane travels much farther than this, it might fall into the ocean. Although the idea of using arresting cables to enable planes to land safely on ships originated at about the time of the First World War, the cables are still a vital part of the operation of modern aircraft carriers. EXAMPLE 2.8 Watch Out for the Speed Limit! catch up to the car. While all this is going on, the car continues to move. We should therefore expect our result to be well over 15 s. A sketch (Fig. 2.12) helps clarify the sequence of events. First, we write expressions for the position of each vehicle as a function of time. It is convenient to choose the position of the billboard as the origin and to set t B 0 as the time the trooper begins moving. At that instant, the car has already traveled a distance of 45.0 m because it has traveled at a constant speed of vx 45.0 m/s for 1 s. Thus, the initial position of the speeding car is x B 45.0 m. Because the car moves with constant speed, its accelera- A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s2. How long does it take her to overtake the car? Solution A careful reading lets us categorize this as a constant-acceleration problem. We know that after the 1-s delay in starting, it will take the trooper 15 additional seconds to accelerate up to 45.0 m/s. Of course, she then has to continue to pick up speed (at a rate of 3.00 m/s per second) to 2.5 v x car = 45.0 m/s a x car = 0 ax trooper = 3.00 m/s 2 tA = 1.00 s tB = 0 One-Dimensional Motion with Constant Acceleration 39 The trooper starts from rest at t 0 and accelerates at 3.00 m/s2 away from the origin. Hence, her position after any time interval t can be found from Equation 2.11: tC = ? xf x trooper xi 0 v xi t 0t 1 2 2a x t 1 2 2 a xt 1 2 (3.00 m/s2)t 2 The trooper overtakes the car at the instant her position matches that of the car, which is position : x trooper 1 2 (3.00 x car 45.0 m (45.0 m/s)t m/s2)t 2 This gives the quadratic equation 1.50t 2 45.0t 45.0 0 31.0 s . Figure 2.12 A speeding car passes a hidden police officer. The positive solution of this equation is t tion is zero, and applying Equation 2.11 (with a x for the car's position at any time t: x car xB v x cart 45.0 m (45.0 m/s)t 0) gives (For help in solving quadratic equations, see Appendix B.2.) Note that in this 31.0-s time interval, the trooper travels a distance of about 1440 m. [This distance can be calculated from the car's constant speed: (45.0 m/s)(31 1) s 1 440 m.] A quick check shows that at t 0, this expression gives the car's correct initial position when the trooaccelerate at a maximum rate of 5.00 m/s2 as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time it needs before it can come to rest? (b) Can this plane land at a small tropical island airport where the runway is 0.800 km long? The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of 5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree? Help! One of our equations is missing! We describe constant-acceleration motion with the variables and parameters vxi , vx f , ax , t, and xf xi . Of the equations in Table 2.2, the first does not involve x f x i . The second does not contain ax , the third omits vx f , and the last 27. Section 2.5 One-Dimensional Motion with Constant Acceleration 21. Jules Verne in 1865 proposed sending people to the Moon by firing a space capsule from a 220-m-long cannon with a final velocity of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration, 9.80 m/s2. 22. A certain automobile manufacturer claims that its superdeluxe sports car will accelerate from rest to a speed of 42.0 m/s in 8.00 s. Under the (improbable) assumption that the acceleration is constant, (a) determine the acceleration of the car. (b) Find the distance the car travels in the first 8.00 s. (c) What is the speed of the car 10.0 s after it begins its motion, assuming it continues to move with the same acceleration? 23. A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration. 24. The minimum distance required to stop a car moving at 35.0 mi/h is 40.0 ft. What is the minimum stopping distance for the same car moving at 70.0 mi/h, assuming the same rate of acceleration? 25. A body moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is 5.00 cm, what is the magnitude of its acceleration? 26. Figure P2.26 represents part of the performance data of a car owned by a proud physics student. (a) Calculate from the graph the total distance traveled. (b) What distance does the car travel between the times t 10 s and t 40 s? (c) Draw a graph of its ac28. 29. 30. 31. 32. WEB 33. Problems 53 Figure P2.37 (Left) Col. John Stapp on rocket sled. (Courtesy of the U.S. Air Force) (Right) Col. Stapp's face is contorted by the stress of rapid negative acceleration. (Photri, Inc.) 34. 35. 36. 37. 38. 39. leaves out t. So to complete the set there should be an equation not involving vxi . Derive it from the others. Use it to solve Problem 32 in one step. An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the board with a speed of 420 m/s and emerges with a speed of 280 m/s. (a) What is the average acceleration of the bullet as it passes through the board? (b) What is the total time that the bullet is in contact with the board? (c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming the bullet's acceleration through all parts of the board is the same? A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described? A train is traveling down a straight track at 20.0 m/s when the engineer applies the brakes. This results in an acceleration of 1.00 m/s2 as long as the train is in motion. How far does the train move during a 40.0-s time interval starting at the instant the brakes are applied? For many years the world's land speed record was held by Colonel John P. Stapp, USAF (Fig. P2.37). On March 19, 1954, he rode a rocket-propelled sled that moved down the track at 632 mi/h. He and the sled were safely brought to rest in 1.40 s. Determine (a) the negative acceleration he experienced and (b) the distance he traveled during this negative acceleration. An electron in a cathode-ray tube (CRT) accelerates uniformly from 2.00 104 m/s to 6.00 106 m/s over 1.50 cm. (a) How long does the electron take to travel this 1.50 cm? (b) What is its acceleration? A ball starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 15.0 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball's speed 8.00 m along the second plane? 40. Speedy Sue, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sue applies her brakes but can accelerate only at 2.00 m/s2 because the road is wet. Will there be a collision? If so, determine how far into the tunnel and at what time the collision occurs. If not, determine the distance of closest approach between Sue's car and the van. Section 2.6 Freely Falling Objects Note: In all problems in this section, ignore the effects of air resistance. 41. A golf ball is released from rest from the top of a very tall building. Calculate (a) the position and (b) the velocity of the ball after 1.00 s, 2.00 s, and 3.00 s. 42. Every morning at seven o'clock There's twenty terriers drilling on the rock. The boss comes around and he says, "Keep still And bear down heavy on the cast-iron drill And drill, ye terriers, drill." And drill, ye terriers, drill. It's work all day for sugar in your tea . . . And drill, ye terriers, drill. One day a premature blast went off And a mile in the air went big Jim Goff. And drill . . . Then when next payday came around Jim Goff a dollar short was found. When he asked what for, came this reply: "You were docked for the time you were up in the sky." And drill . . . --American folksong What was Goff's hourly wage? State the assumptions you make in computing it. 54 CHAPTER 2 Motion in One Dimension WEB 43. A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? 44. A ball is thrown directly downward with an initial speed of 8.00 m/s from a height of 30.0 m. How many seconds later does the ball strike the ground? 45. Emily challenges her friend David to catch a dollar bill as follows: She holds the bill vertically, as in Figure P2.45, with the center of the bill between David's index finger and thumb. David must catch the bill after Emily releases it without moving his hand downward. If his reaction time is 0.2 s, will he succeed? Explain your reasoning. WEB 49. A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The speed of the horse is 10.0 m/s, and the distance from the limb to the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) How long is he in the air? 50. A ball thrown vertically upward is caught by the thrower after 20.0 s. Find (a) the initial velocity of the ball and (b) the maximum height it reaches. 51. A ball is thrown vertically upward from the ground with an initial speed of 15.0 m/s. (a) How long does it take the ball to reach its maximum altitude? (b) What is its maximum altitude? (c) Determine the velocity and acceleration of the ball at t 2.00 s. 52. The height of a helicopter above the ground is given by h 3.00t 3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? (Optional) 2.7 Kinematic Equations Derived from Calculus 53. Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax , velocity vx , and position x, given that its initial acceleration, speed, and position are ax i , vx i , and x i , respectively. (b) Show that a x2 a xi2 2J(v x v xi). 54. The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by the expression v ( 5.0 10 7)t 2 (3.0 10 5)t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel? 55. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared and is given (in SI units) by a 3.00v 2 for v 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value? Figure P2.45 (George Semple) 46. A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upward from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. 47. A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the maximum height it reaches. 48. A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Calculate (a) the speed of the woman just before she collided with the ventilator box, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box. ADDITIONAL PROBLEMS 56. A motorist is traveling at 18.0 m/s when he sees a deer in the road 38.0 m ahead. (a) If the maximum negative acceleration of the vehicle is 4.50 m/s2, what is the maximum reaction time t of the motorist that will allow him to avoid hitting the deer? (b) If his reaction time is actually 0.300 s, how fast will he be traveling when he hits the deer? Problems 57. Another scheme to catch the roadrunner has failed. A safe falls from rest from the top of a 25.0-m-high cliff toward Wile E. Coyote, who is standing at the base. Wile first notices the safe after it has fallen 15.0 m. How long does he have to get out of the way? 58. A dog's hair has been cut and is now getting longer by 1.04 mm each day. With winter coming on, this rate of hair growth is steadily increasing by 0.132 mm/day every week. By how much will the dog's hair grow during five weeks? 59. A test rocket is fired vertically upward from a well. A catapult gives it an initial velocity of 80.0 m/s at ground level. Subsequently, its engines fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1000 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of 9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (Hint: Consider the motion while the engine is operating separate from the free-fall motion.) 60. A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 2.00 m/s2 to overtake her. Assuming the officer maintains this acceleration, (a) determine the time it takes the police officer to reach the motorist. Also find (b) the speed and (c) the total displaccommunicate to the other honeybees how far it is to the flower and in what direction they must fly. This is exactly the kind of information that polar coordinates convey, as long as the origin of the coordinates is the beehive. 3.2 The resultant has magnitude A B when vector A is oriented in the same direction as vector B. The resultant vector is A B 0 when vector A is oriented in the direction opposite vector B and A B. 3.3 No. In two dimensions, a vector and its components form a right triangle. The vector is the hypotenuse and must be longer than either side. Problem 61 extends this concept to three dimensions. 3.4 No. The magnitude of a vector A is equal to Ay2 Az2. Therefore, if any component is nonzero, A cannot be zero. This generalization of the Pythagorean theorem is left for you to prove in Problem 61. 3.5 The fact that A B 0 tells you that A B. Therefore, the components of the two vectors must have oppoBy , Bx , Ay site signs and equal magnitudes: Ax Bz . and Az Ax2 P U Z Z L E R This airplane is used by NASA for astronaut training. When it flies along a certain curved path, anything inside the plane that is not strapped down begins to float. What causes this strange effect? (NASA) web For more information on microgravity in general and on this airplane, visit http://microgravity.msfc.nasa.gov/ and http://www.jsc.nasa.gov/coop/ kc135/kc135.html c h a p t e r Motion in Two Dimensions Chapter Outline 4.1 The Displacement, Velocity, and Acceleration Vectors 4.2 Two-Dimensional Motion with Constant Acceleration 4.4 Uniform Circular Motion 4.5 Tangential and Radial Acceleration 4.6 Relative Velocity and Relative Acceleration 4.3 Projectile Motion 76 4.1 The Displacement, Velocity, and Acceleration Vectors 77 I n this chapter we deal with the kinematics of a particle moving in two dimensions. Knowing the basics of two-dimensional motion will allow us to examine -- in future chapters -- a wide variety of motions, ranging from the motion of satellites in orbit to the motion of electrons in a uniform electric field. We begin by studying in greater detail the vector nature of displacement, velocity, and acceleration. As in the case of one-dimensional motion, we derive the kinematic equations for two-dimensional motion from the fundamental definitions of these three quantities. We then treat projectile motion and uniform circular motion as special cases of motion in two dimensions. We also discuss the concept of relative motion, which shows why observers in different frames of reference may measure different displacements, velocities, and accelerations for a given particle. y 4.1 THE DISPLACEMENT, VELOCITY, AND ACCELERATION VECTORS O ti ri rf r tf In Chapter 2 we found that the motion of a particle moving along a straight line is completely known if its position is known as a function of time. Now let us extend this idea to motion in the xy plane. We begin by describing the position of a particle by its position vector r, drawn from the origin of some coordinate system to the particle located in the xy plane, as in Figure 4.1. At time ti the particle is at point , and at some later time tf it is at point . The path from to is not necessarily a straight line. As the particle moves from to in the time interval t t f t i , its position vector changes from ri to rf . As we learned in Chapter 2, displacement is a vector, and the displacement of the particle is the difference between its final position and its initial position. We now formally define the displacement vector r for the particle of Figure 4.1 as being the difference between its final position vector and its initial position vector: r rf ri (4.1) The direction of r is indicated in Figure 4.1. As we see from the figure, the magnitude of r is less than the distance traveled along the curved path followed by the particle. As we saw in Chapter 2, it is often useful to quantify motion by looking at the ratio of a displacement divided by the time interval during which that displacement occurred. In two-dimensional (or three-dimensional) kinematics,alue of the ratio v/ t as t approaches zero: a lim t:0 v t dv dt (4.5) Instantaneous acceleration 3.5 In other words, the instantaneous acceleration equals the derivative of the velocity vector with respect to time. It is important to recognize that various changes can occur when a particle accelerates. First, the magnitude of the velocity vector (the speed) may change with time as in straight-line (one-dimensional) motion. Second, the direction of the velocity vector may change with time even if its magnitude (speed) remains constant, as in curved-path (two-dimensional) motion. Finally, both the magnitude and the direction of the velocity vector may change simultaneously. Quick Quiz 4.1 The gas pedal in an automobile is called the accelerator. (a) Are there any other controls in an automobile that can be considered accelerators? (b) When is the gas pedal not an accelerator? 4.2 TWO-DIMENSIONAL MOTION WITH CONSTANT ACCELERATION Let us consider two-dimensional motion during which the acceleration remains constant in both magnitude and direction. The position vector for a particle moving in the xy plane can be written r xi yj (4.6) where x, y, and r change with time as the particle moves while i and j remain constant. If the position vector is known, the velocity of the particle can be obtained from Equations 4.3 and 4.6, which give v vxi vy j (4.7) 80 CHAPTER 4 Motion in Two Dimensions Because a is assumed constant, its components ax and ay also are constants. Therefore, we can apply the equations of kinematics to the x and y components of the velocity vector. Substituting vx f vxi a x t and vy f vyi a y t into Equation 4.7 to determine the final velocity at any time t, we obtain vf Velocity vector as a function of time vf (v xi a x t)i (v yi a y t)j (v xi i v yi j) (a x i a y j)t vi at (4.8) This result states that the velocity of a particle at some time t equals the vector sum of its initial velocity vi and the additional velocity at acquired in the time t as a result of constant acceleration. Similarly, from Equation 2.11 we know that the x and y coordinates of a particle moving with constant acceleration are xf xi v xit 1 2 2 a xt yf yi v yit 1 2 2 a yt Position vector as a function of time Substituting these expressions into Equation 4.6 (and labeling the final position vector rf ) gives rf (x i v xit 1a xt 2)i (y i v yit 1a yt 2)j 2 2 (x i i y i j) (v xi i v yi j)t 1(a x i a y j)t 2 2 (4.9) rf ri vit 1at 2 2 This equation tells us that the displacement vector r rf ri is the vector sum of a displacement vi t arising from the initial velocity of the particle and a displacement 1at 2 resulting from the uniform acceleration of the particle. 2 Graphical representations of Equations 4.8 and 4.9 are shown in Figure 4.4. For simplicity in drawing the figure, we have taken ri 0 in Figure 4.4a. That is, we assume the particle is at the origin at t t i 0. Note from Figure 4.4a that rf is generally not along the direction of either vi or a because the relationship between these quantities is a vector expression. For the same reason, from Figure 4.4b we see that vf is generally not along the direction of vi or a. Finally, note that vf and rf are generally not in the same direction. y y ayt 1 a t2 2 y vf yf rf vyf 1 at 2 2 at vyi vyit vit vxit xf (a) (b) 1 a t2 2 x vi x vxi axt x vxf Figure 4.4 Vector representations and components of (a) the displacement and (b) the velocity of a particle moving with a uniform acceleration a. To simplify the drawing, we have set ri 0. 4.2 Two-Dimensional Motion with Constant Acceleration 81 Because Equations 4.8 and 4.9 are vector expressions, we may write them in component form: vf rf vi ri at vit 1 2 2 at v xf v yf xf yf v xi v yi xi yi a xt a yt v xit v yit 1a t 2 2 x 1 2 2 a yt (4.8a) (4.9a) These components are illustrated in Figure 4.4. The component form of the equations for vf and rf show us that two-dimensional motion at constant acceleration is equivalent to two independent motions -- one in the x direction and one in the y direction -- ensional free fall) resistance is neglected, we know that a y and that a x 0. Furthermore, let us assume that at t 0, the projectile leaves the origin (x i y i 0) with speed vi , as shown in Figure 4.6. The vector vi makes an angle i with the horizontal, where i is the angle at which the projectile leaves the origin. From the definitions of the cosine and sine functions we have 3.5 cos i v xi /v i sin i v yi /v i Therefore, the initial x and y components of velocity are v xi v i cos i v yi v i sin i Horizontal position component Substituting the x component into Equation 4.9a with xi that x f v xit (v i cos i)t Repeating with the y component and using yi 0 and ay 1 2 2 gt 1 2 2 a yt 0 and ax 0, we find (4.10) g, we obtain (4.11) Vertical position component yf v yit (v i sin i)t i) Next, we solve Equation 4.10 for t for t into Equation 4.11; this gives y 1 xf /(vi cos g and substitute this expression (tan i)x 2v i cos2 2 x2 i (4.12) This assumption is reasonable as long as the range of motion is small compared with the radius of the Earth (6.4 106 m). In effect, this assumption is equivalent to assuming that the Earth is flat over the range of motion considered. 2 This assumption is generally not justified, especially at high velocities. In addition, any spin imparted to a projectile, such as that applied when a pitcher throws a curve ball, can give rise to some very interesting effects associated with aerodynamic forces, which will be discussed in Chapter 15. 4.3 Projectile Motion 83 y vy vi vyi v vy = 0 vx i g vx i vx i vy v i vxi vx i x i vyi v Figure 4.6 The parabolic path of a projectile that leaves the origin with a velocity vi . The velocity vector v changes with time in both magnitude and direction. This change is the result of acceleration in the negative y direction. The x component of velocity remains constant in time because there is no acceleration along the horizontal direction. The y component of velocity is zero at the peak of the path. A welder cuts holes through a heavy metal construction beam with a hot torch. The sparks generated in the process follow parabolic paths. This equation is valid for launch angles in the range 0 /2. We have left i the subscripts off the x and y because the equation is valid for any point (x, y) along the path of the projectile. The equation is of the form y ax bx 2, which is the equation of a parabola that passes through the origin. Thus, we have shown that the trajectory of a projectile is a parabola. Note that the trajectory is completely specified if both the initial speed vi and the launch angle i are known. The vector expression for the position vector of the projectile as a function of time follows directly from Equation 4.9, with ri 0 and a g: r vit 1 2 2 gt QuickLab Place two tennis balls at the edge of a tabletop. Sharply snap one ball horizontally off the table with one hand while gently tapping the second ball off with your other hand. Compare how long it takes the two to reach the floor. Explain your results. This expression is plotted in Figure 4.7. y 1 2 gt 2 vit r (x, y) O x Figure 4.7 The position vector r of a projectile whose initial velocity at the origin is vi . The vector vi t would be the displacement of the projectile if gravity were absent, and the vector 1 gt 2 is its 2 vertical displacement due to its downward gravitational acceleration. 84 CHAPTER 4 Motion in Two Dimensions Multiflash exposure of a tennis player executing a forehand swing. Note that the ball follows a parabolic path characteristic of a projectile. Such photographs can be used to study the quality of sports equipment and the performance of an athlete. It is interesting to realize that the motion of a particle can be considered the superposition of the term vi t, the displacement if no acceleration were present, and the term 1 gt 2, which arises from the acceleration due to gravity. In other 2 words, if there were no gravitational acceleration, the particle would continue to move along a straight path in the direction of vi . Therefore, the vertical distance 1 2 2 gt through which the particle "falls" off the straight-line path is the same distance that a freely falling body would fall during the same time interval. We conclude that projectile motion is the superposition of two motions: (1) constant-velocity motion in the horizontal direction and (2) free-fall motion in the vertical direction. Except for t, the time of flight, the horizontal and vertical components of a projectile's motion are completely independent of each other. EXAMPLE 4.2 Approximating Projectile Motion A ball is thrown in such a way that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s, respectively. Estimate the total time of flight and the distance the ball is from its starting point when it lands. Solution We start by remembering that the two velocity components are independent of each other. By considering the vertical motion first, we can determine how long the ball remains in the air. Then, we can use the time of flight to estimate the horizontal distance covered. A motion diagram like Figure 4.8 helps us organize what we know about the problem. The acceleration vectors are all the same, pointing downward with a magnitude of nearly 10 m/s2. The velocity vectors change direction. Their hori- Figure 4.8 Motion diagram for a projectile. 4.3 zontal components are all the same: 20 m/s. Because the vertical motion is free fall, the vertical components of the velocity vectors change, second by second, from 40 m/s to roughly 30, 20, and 10 m/s in the upward direction, and then to 0 m/s. Subsequently, its velocity becomes 10, 20, 30, and 40 m/s in the downward direction. Thus it takes the ball Projectile Motion 85 about 4 s to go up and another 4 s to come back down, for a total time of flight of approximately 8 s. Because the horizontal component of velocity is 20 m/s, and because the ball travels at this speed for 8 s, it ends up approximately 160 m from its starting point. Horizontal Range and Maximum Height of a Projectile Let us assume that a projectile is fired from the origin at ti 0 with a positive vyi component, as shown in Figure 4.9. Two points are especially interesting to analyze: the peak point , which has cartesian coordinates (R/2, h), and the point , which has coordinates (R, 0). The distance R is called the horizontal range of the projectile, and the distance h is its maximum height. Let us find h and R in terms of vi , i , and g. We can determine h by noting that at the peak, vy A 0. Therefore, we can use Equation 4.8a to determine the time t A it takes the projectile to reach the peak: vyf 0 tA v yi v i sin v i sin g a yt i i y vy A = 0 vi h i O x R gt A Substituting this expression for t A into the y part of Equation 4.9a and replacing y f y A with h, we obtain an expression for h in terms of the magnitude and direction of the initial velocity vector: h h (v i sin i) v i2 sin2 2g i Figure 4.9 A projectile fired from the origin at ti 0 with an initial velocity vi . The maximum height of the projectile is h, and the horizontal range is R. At , the peak of the trajectory, the particle has coordinates (R/2, h). v i sin g i 1 2g v i sin g i 2 (4.13) Maximum height of projectile The range R is the horizontal distance that the projectile travels in twice the time it takes to reach its peak, that is, in a time t B 2t A . Using the x part of Equation 4.9a, noting that vxi vx B vi cos i , and setting R x B at t 2t A , we find that R v xit B (v i cos i)2t A 2v i sin g cos i (v i cos i) Using the identity sin 2 more compact form 2 sin 2v i2 sin i cos g i (see Appendix B.4), we write R in the i R v i2 sin 2 g (4.14) Range of projectile Keep in mind that Equations 4.13 and 4.14 are useful for calculating h and R only if vi and i are known (which means that only vi has to be specified) and if the projectile lands at the same height from which it started, as it does in Figure 4.9. The maximum value of R from Equation 4.14 is R max v i2/g. This result follows from the fact that the maximum valuits vertical acceleration, you can calibrate the distances depicted in the photograph. Then you can find the horizontal speed of the yellow ball.) Figure 4.15 This multiflash photograph of two balls released simultaneously illustrates both free fall (red ball) and projectile motion (yellow ball). The yellow ball was projected horizontally, while the red ball was released from rest. (Richard Megna/Fundamental Pho- tographs) side each other; that is, they have the same trajectory. An astronaut can release a piece of equipment and it will float freely alongside her hand. The same thing happens in the space shuttle. The craft and everything in it are falling as they orbit the Earth. 4.4 3.6 UNIFORM CIRCULAR MOTION Figure 4.16a shows a car moving in a circular path with constant linear speed v. Such motion is called uniform circular motion. Because the car's direction of motion changes, the car has an acceleration, as we learned in Section 4.1. For any motion, the velocity vector is tangent to the path. Consequently, when an object moves in a circular path, its velocity vector is perpendicular to the radius of the circle. We now show that the acceleration vector in uniform circular motion is always perpendicular to the path and always points toward the center of the circle. An acvi r v r O O (a) (b) (c) r r vf vi vf v Figure 4.16 (a) A car moving along a circular path at constant speed experiences uniform circular motion. (b) As a particle moves from to , its velocity vector changes from vi to vf . (c) The construction for determining the direction of the change in velocity v, which is toward the center of the circle for small r. 92 CHAPTER 4 Motion in Two Dimensions celeration of this nature is called a centripetal (center-seeking) acceleration, and its magnitude is ar v2 r (4.15) where r is the radius of the circle and the notation ar is used to indicate that the centripetal acceleration is along the radial direction. To derive Equation 4.15, consider Figure 4.16b, which shows a particle first at point and then at point . The particle is at at time ti , and its velocity at that time is vi . It is at at some later time tf , and its velocity at that time is vf . Let us assume here that vi and vf differ only in direction; their magnitudes (speeds) are the same (that is, v i v f v). To calculate the acceleration of the particle, let us begin with the defining equation for average acceleration (Eq. 4.4): a vf tf vi ti v t This equation indicates that we must subtract vi from vf , being sure to treat them as vectors, where v vf vi is the change in the velocity. Because vi v vf , we can find the vector v, using the vector triangle in Figure 4.16c. Now consider the triangle in Figure 4.16b, which has sides r and r. This triangle and the one in Figure 4.16c, which has sides v and v, are similar. This fact enables us to write a relationship between the lengths of the sides: v v r r This equation can be solved for v and the expression so obtained substituted into a v/ t (Eq. 4.4) to give a v r r t Now imagine that points and in Figure 4.16b are extremely close together. In this case v points toward the center of the circular path, and because the acceleration is in the direction of v, it too points toward the center. Furthermore, as and approach each other, t approaches zero, and the ratio r/ t approaches the speed v. Hence, in the limit t : 0, the magnitude of the acceleration is ar v2 r Thus, we conclude that in uniform circular motion, the acceleration is directed toward the center of the circle and has a magnitude given by v 2/r, where v is the speed of the particle and r is the radius of the circle. You should be able to show that the dimensions of a r are L/T 2. We shall return to the discussion of circular motion in Section 6.1. 4.5 3.6 TANGENTIAL AND RADIAL ACCELERATION Now let us consider a particle moving along a curved path where the velocity changes both in direction and in magnitude, as shown in Figure 4.17. As is always the case, the velocity vector is tangent to the path, but now the direction oe initial speed. Firing at what other projectile angle results in the same range if the initial speed is the same in both cases? Neglect air resistance. A projectile is fired on the Earth with some initial velocity. Another projectile is fired on the Moon with the same initial velocity. If air resistance is neglected, which projectile has the greater range? Which reaches the greater altitude? (Note that the free-fall acceleration on the Moon is about 1.6 m/s2.) As a projectile moves through its parabolic trajectory, which of these quantities, if any, remain constant: (a) speed, (b) acceleration, (c) horizontal component of velocity, (d) vertical component of velocity? A passenger on a train that is moving with constant velocity drops a spoon. What is the acceleration of the spoon relative to (a) the train and (b) the Earth? 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. Problems 101 PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 4.1 Vectors WEB The Displacement, Velocity, and Acceleration 1. A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Use a coordinate system in which east is the positive x axis. 2. Suppose that the position vector for a particle is given as r x i y j, with x at b and y ct 2 d, where a 1.00 m/s, b 1.00 m, c 0.125 m/s2, and d 1.00 m. (a) Calculate the average velocity during the time interval from t 2.00 s to t 4.00 s. (b) Determine the velocity and the speed at t 2.00 s. 3. A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates versus time are given by the following expressions: x and y (4.00 m/s)t (4.90 m/s2)t 2 (18.0 m/s)t 6. The vector position of a particle varies in time according to the expression r (3.00i 6.00t 2 j) m. (a) Find expressions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at t 1.00 s. 7. A fish swimming in a horizontal plane has velocity vi (4.00i 1.00j) m/s at a point in the ocean whose displacement from a certain rock is ri (10.0i 4.00j) m. After the fish swims with constant acceleration for 20.0 s, its velocity is v (20.0i 5.00j) m/s. (a) What are the components of the acceleration? (b) What is the direction of the acceleration with respect to the unit vector i? (c) Where is the fish at t 25.0 s if it maintains its original acceleration and in what direction is it moving? 8. A particle initially located at the origin has an acceleration of a 3.00j m/s2 and an initial velocity of vi 5.00i m/s. Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t 2.00 s. Section 4.3 Projectile Motion (a) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. By taking derivatives of your results, write expressions for (b) the velocity vector as a function of time and (c) the acceleration vector as a function of time. Now use unit vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the ball, all at t 3.00 s. 4. The coordinates of an object moving in the xy plane vary with time according to the equations x and y (4.00 m) (5.00 m)cos t where t is in seconds and has units of seconds 1. (a) Determine the components of velocity and components of acceleration at t 0. (b) Write expressions for the position vector, the velocity vector, and the acceleration vector at any time t 0. (c) Describe the path of the object on an xy graph. (5.00 m) sin t (Neglect air resistance in all problems and take g 9.80 m/s2.) WEB 9. In a local bar, a customer slides an empty beer mug dip makes it very difficult to cause any sudden changes in its motion? (Courtesy of Edward E. Ogden) web For more information about the airship, visit http://www.goodyear.com/us/blimp/ index.html c h a p t e r The Laws of Motion Chapter Outline 5.1 The Concept of Force 5.2 Newton's First Law and Inertial Frames 5.3 Mass 5.4 Newton's Second Law 5.5 The Force of Gravity and Weight 5.6 Newton's Third Law 5.7 Some Applications of Newton's Laws 5.8 Forces of Friction 110 5.1 The Concept of Force 111 I n Chapters 2 and 4, we described motion in terms of displacement, velocity, and acceleration without considering what might cause that motion. What might cause one particle to remain at rest and another particle to accelerate? In this chapter, we investigate what causes changes in motion. The two main factors we need to consider are the forces acting on an object and the mass of the object. We discuss the three basic laws of motion, which deal with forces and masses and were formulated more than three centuries ago by Isaac Newton. Once we understand these laws, we can answer such questions as "What mechanism changes motion?" and "Why do some objects accelerate more than others?" 5.1 THE CONCEPT OF FORCE Everyone has a basic understanding of the concept of force from everyday experience. When you push your empty dinner plate away, you exert a force on it. Similarly, you exert a force on a ball when you throw or kick it. In these examples, the word force is associated with muscular activity and some change in the velocity of an object. Forces do not always cause motion, however. For example, as you sit reading this book, the force of gravity acts on your body and yet you remain stationary. As a second example, you can push (in other words, exert a force) on a large boulder and not be able to move it. What force (if any) causes the Moon to orbit the Earth? Newton answered this and related questions by stating that forces are what cause any change in the velocity of an object. Therefore, if an object moves with uniform motion (constant velocity), no force is required for the motion to be maintained. The Moon's velocity is not constant because it moves in a nearly circular orbit around the Earth. We now know that this change in velocity is caused by the force exerted on the Moon by the Earth. Because only a force can cause a change in velocity, we can think of force as that which causes a body to accelerate. In this chapter, we are concerned with the relationship between the force exerted on an object and the acceleration of that object. What happens when several forces act simultaneously on an object? In this case, the object accelerates only if the net force acting on it is not equal to zero. The net force acting on an object is defined as the vector sum of all forces acting on the object. (We sometimes refer to the net force as the total force, the resultant force, or the unbalanced force.) If the net force exerted on an object is zero, then the acceleration of the object is zero and its velocity remains constant. That is, if the net force acting on the object is zero, then the object either remains at rest or continues to move with constant velocity. When the velocity of an object is constant (including the case in which the object remains at rest), the object is said to be in equilibrium. When a coiled spring is pulled, as in Figure 5.1a, the spring stretches. When a stationary cart is pulled sufficently hard that friction is overcome, as in Figure 5.1b, the cart moves. When a football is kicked, as in Figure 5.1c, it is both deformed and set in motion. These situations are all examples of a class of forces called contact forces. That is, they involve physical contact between two objects. Other examples of contact forces are the force exerted by gas molecules on the walls of a container and the force exerted by your feet on the floor. Another class of forces, known as field forces, do not involve physical contact between two objects but instead act through empty space. The force of gravitational attraction between that the air rushing out strikes the ball. Try a variety of configurations: Blow in opposite directions against the ball, blow in the same direction, blow at right angles to each other, and so forth. Can you verify the vector nature of the forces? 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 F2 F1 F1 F1 F F2 (a) (b) (c) F2 (d) The vector nature of a force is tested with a spring scale. (a) A downward force F1 elongates the spring 1 cm. (b) A downward force F2 elongates the spring 2 cm. (c) When F1 and F2 are applied simultaneously, the spring elongates by 3 cm. (d) When F1 is downward and F2 is 2 2 cm horizontal, the combination of the two forces elongates the spring 1 2 5 cm. Figure 5.2 114 CHAPTER 5 The Laws of Motion 5.2 4.2 NEWTON'S FIRST LAW AND INERTIAL FRAMES QuickLab Use a drinking straw to impart a strong, short-duration burst of air against a tennis ball as it rolls along a tabletop. Make the force perpendicular to the ball's path. What happens to the ball's motion? What is different if you apply a continuous force (constant magnitude and direction) that is directed along the direction of motion? Before we state Newton's first law, consider the following simple experiment. Suppose a book is lying on a table. Obviously, the book remains at rest. Now imagine that you push the book with a horizontal force great enough to overcome the force of friction between book and table. (This force you exert, the force of friction, and any other forces exerted on the book by other objects are referred to as external forces.) You can keep the book in motion with constant velocity by applying a force that is just equal in magnitude to the force of friction and acts in the opposite direction. If you then push harder so that the magnitude of your applied force exceeds the magnitude of the force of friction, the book accelerates. If you stop pushing, the book stops after moving a short distance because the force of friction retards its motion. Suppose you now push the book across a smooth, highly waxed floor. The book again comes to rest after you stop pushing but not as quickly as before. Now imagine a floor so highly polished that friction is absent; in this case, the book, once set in motion, moves until it hits a wall. Before about 1600, scientists felt that the natural state of matter was the state of rest. Galileo was the first to take a different approach to motion and the natural state of matter. He devised thought experiments, such as the one we just discussed for a book on a frictionless surface, and concluded that it is not the nature of an object to stop once set in motion: rather, it is its nature to resist changes in its motion. In his words, "Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed." This new approach to motion was later formalized by Newton in a form that has come to be known as Newton's first law of motion: In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line). In simpler terms, we can say that when no force acts on an object, the acceleration of the object is zero. If nothing acts to change the object's motion, then its velocity does not change. From the first law, we conclude that any isolated object (one that does not interact with its environment) is either at rest or moving with constant velocity. The tendency of an object to resist any attempt to change its velocity is called the inertia of the object. Figure 5.3 shows one dramatic example of a consequence of Newton's first law. Another example of uniform (constant-velocity) motion on a nearly frictionless surface is the motion of a light disk on a film of air (the lubricant), as shown in Figure 5.4. If the disk is given an initial velocity, it coasts a great distance before stopping. Finally, consider a spaceship traveling in space and far removed from any planets or other matter. The spaceship requires some propulsion hich ball is more likely to keep moving when you try to catch it? Which ball has the greater tendency to remain motionless when you try to throw it? Because the bowling ball is more resistant to changes in its velocity, we say it has greater inertia than the basketball. As noted in the preceding section, inertia is a measure of how an object responds to an external force. Mass is that property of an object that specifies how much inertia the object has, and as we learned in Section 1.1, the SI unit of mass is the kilogram. The greater the mass of an object, the less that object accelerates under the action of an applied force. For example, if a given force acting on a 3-kg mass produces an acceleration of 4 m/s2, then the same force applied to a 6-kg mass produces an acceleration of 2 m/s2. To describe mass quantitatively, we begin by comparing the accelerations a given force produces on different objects. Suppose a force acting on an object of mass m1 produces an acceleration a1 , and the same force acting on an object of mass m 2 produces an acceleration a2 . The ratio of the two masses is defined as the inverse ratio of the magnitudes of the accelerations produced by the force: m1 m2 a2 a1 (5.1) Mass and weight are different quantities If one object has a known mass, the mass of the other object can be obtained from acceleration measurements. Mass is an inherent property of an object and is independent of the object's surroundings and of the method used to measure it. Also, mass is a scalar quantity and thus obeys the rules of ordinary arithmetic. That is, several masses can be combined in simple numerical fashion. For example, if you combine a 3-kg mass with a 5-kg mass, their total mass is 8 kg. We can verify this result experimentally by comparing the acceleration that a known force gives to several objects separately with the acceleration that the same force gives to the same objects combined as one unit. Mass should not be confused with weight. Mass and weight are two different quantities. As we see later in this chapter, the weight of an object is equal to the magnitude of the gravitational force exerted on the object and varies with location. For example, a person who weighs 180 lb on the Earth weighs only about 30 lb on the Moon. On the other hand, the mass of a body is the same everywhere: an object having a mass of 2 kg on the Earth also has a mass of 2 kg on the Moon. 5.4 4.4 NEWTON'S SECOND LAW Newton's first law explains what happens to an object when no forces act on it. It either remains at rest or moves in a straight line with constant speed. Newton's second law answers the question of what happens to an object that has a nonzero resultant force acting on it. 5.4 Newton's Second Law 117 Imagine pushing a block of ice across a frictionless horizontal surface. When you exert some horizontal force F, the block moves with some acceleration a. If you apply a force twice as great, the acceleration doubles. If you increase the applied force to 3F, the acceleration triples, and so on. From such observations, we conclude that the acceleration of an object is directly proportional to the resultant force acting on it. The acceleration of an object also depends on its mass, as stated in the preceding section. We can understand this by considering the following experiment. If you apply a force F to a block of ice on a frictionless surface, then the block undergoes some acceleration a. If the mass of the block is doubled, then the same applied force produces an acceleration a/2. If the mass is tripled, then the same applied force produces an acceleration a/3, and so on. According to this observation, we conclude that the magnitude of the acceleration of an object is inversely proportional to its mass. These observations are summarized in Newton's second law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Thus, we can relate mass and force through the following mathematical statement of Newton's second law:1 F ma (5.2) Newton's seock 1 on block 2 is less than the applied force F. This is consistent with the fact that the force required to accelerate block 2 alone must be less than the force required to produce the same acceleration for the two-block system. It is instructive to check this expression for P by considering the forces acting on block 1, shown in Figure 5.13b. The horizontal forces acting on this block are the applied force F to the right and the contact force P to the left (the force exerted by block 2 on block 1). From Newton's third law, P is the reaction to P, so that P P . Applying Newton's second law to block 1 produces (4) Fx F P F P m 1a x Figure 5.13 128 CHAPTER 5 The Laws of Motion Substituting into (4) the value of ax from (1), we obtain P F m 1a x F m 1F m1 m 2 m2 m1 m2 F Exercise If m1 4.00 kg, m 2 3.00 kg, and F 9.00 N, find the magnitude of the acceleration of the system and the magnitude of the contact force. ax 1.29 m/s2; P 3.86 N. This agrees with (3), as it must. Answer EXAMPLE 5.8 Weighing a Fish in an Elevator If the elevator moves upward with an acceleration a relative to an observer standing outside the elevator in an inertial frame (see Fig. 5.14a), Newton's second law applied to the fish gives the net force on the fish: (1) Fy T mg ma y A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator, as illustrated in Figure 5.14. Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from the weight of the fish. Solution The external forces acting on the fish are the downward force of gravity Fg mg and the force T exerted by the scale. By Newton's third law, the tension T is also the reading of the scale. If the elevator is either at rest or moving at constant velocity, the fish is not accelerating, and so F y T mg 0 or T mg (remember that the scalar mg is the weight of the fish). a where we have chosen upward as the positive direction. Thus, we conclude from (1) that the scale reading T is greater than the weight mg if a is upward, so that ay is positive, and that the reading is less than mg if a is downward, so that ay is negative. For example, if the weight of the fish is 40.0 N and a is upward, so that ay 2.00 m/s2, the scale reading from (1) is a T T mg (a) (b) mg Observer in inertial frame Figure 5.14 Apparent weight versus true weight. (a) When the elevator accelerates upward, the spring scale reads a value greater than the weight of the fish. (b) When the elevator accelerates downward, the spring scale reads a value less than the weight of the fish. 5.7 ay g Some Applications of Newton's Laws 129 (2) T ma y mg mg 1 1 (40.0 N) 48.2 N If a is downward so that ay T mg ay g 1 2.00 m/s2 9.80 m/s2 Hence, if you buy a fish by weight in an elevator, make sure the fish is weighed while the elevator is either at rest or accelerating downward! Furthermore, note that from the information given here one cannot determine the direction of motion of the elevator. 2.00 m/s2, then (2) gives us (40.0 N) 2.00 m/s2 9.80 m/s2 1 31.8 N Special Cases If the elevator cable breaks, the elevator falls freely and ay g. We see from (2) that the scale reading T is zero in this case; that is, the fish appears to be weightless. If the elevator accelerates downward with an acceleration greater than g, the fish (along with the person in the elevator) eventually hits the ceiling because the acceleration of fish and person is still that of a freely falling object relative to an outside observer. EXAMPLE 5.9 Atwood's Machine vice is sometimes used in the laboratory to measure the freefall acceleration. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord. When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass, as shown in Figure 5.15a, the arrangement is called an Atwood machine. The de- a m1 m2 a (a) T T m1 m2 m1g m2g (b) Solution If we were to define our system as being made up of both objects, as we did in Example 5.7, we would have to determine an internal force (tension in the cord). We must define two systems here -- one for each object -- and apply Newton's second law to each. The free-body diagrams for the two objects are shown in Figure 5.15b. Two forces act on each object: the upward force T exerted by the cord and the downward force of gravity. We need to be very careful with signs in problems such as this, in which a string or rope passes over a pulley or some other structure that causes the string or rope to bend. In Figure 5.15a, notice that if object 1 accelerates upward, then object 2 accelerates downward. Thus, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. With this sign convention applied to the forces, the y component of the net force exerted on object 1 is T m1g, and the y component of the net force exerted on object 2 is m 2g T. Because the objects are connected by a cord, their accelerations must be equal in magnitude. (Otherwise the cord would stretch or break as the distance between the objects increased.) If we assume m 2 m1 , then object 1 must accelerate upward and object 2 downward. When Newton's second law is applied to object 1, we obtain (1) Fy T m 1g m 1a y Similarly, for object 2 we find (2) Fy m 2g T m 2a y Figure 5.15 Atwood's machine. (a) Two objects (m 2 m1 ) connected by a cord of negligible mass strung over a frictionless pulley. (b) Free-body diagrams for the two objects. 130 CHAPTER 5 The Laws of Motion the ratio of the unbalanced force on the system (m 2g m 1g) to the total mass of the system (m 1 m 2), as expected from Newton's second law. When (2) is added to (1), T drops out and we get m 1g (3) m 2g ay m 1a y m2 m1 m 2a y m1 g m2 When (3) is substituted into (1), we obtain (4) T 2m 1m 2 g m1 m2 Special Cases When m 1 m 2 , then ay 0 and T m1 g, as we would expect for this balanced case. If m 2 m1 , then ay g (a freely falling body) and T 2m1 g. Exercise Find the magnitude of the acceleration and the string tension for an Atwood machine in which m1 2.00 kg and m 2 4.00 kg. ay 3.27 m/s2, T 26.1 N. The result for the acceleration in (3) can be interpreted as Answer EXAMPLE 5.10 Acceleration of Two Objects Connected by a Cord rection. Applying Newton's second law in component form to the block gives (3) (4) Fx Fy m 2g sin n m 2g cos T m 2a x 0 m 2a A ball of mass m1 and a block of mass m 2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass, as shown in Figure 5.16a. The block lies on a frictionless incline of angle . Find the magnitude of the acceleration of the two objects and the tension in the cord. Solution Because the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude. The free-body diagrams are shown in Figures 5.16b and 5.16c. Applying Newton's second law in component form to the ball, with the choice of the upward direction as positive, yields (1) (2) Fx Fy 0 T m 1g m 1a y m 1a In (3) we have replaced ax with a because that is the acceleration's only component. In other words, the two objects have accelerations of the same magnitude a, which is what we are trying to find. Equations (1) and (4) provide no information regarding the acceleration. However, if we solve (2) for T and then substitute this value for T into (3) and solve for a, we obtain m 2g sin m1 m 1g m2 (5) a Note that in order for the ball to accelerate upward, it is necessary that T m1 g. In (2) we have replaced ay with a because the acceleration has only a y component. For the block it is convenient to choose the positive x axis along the incline, as shown in Figure 5.16c. Here we choose the positive direction to be down the incline, in the x di- When this value for a is substituted into (2), we find 1) m 1m 2g(sin m1 m2 (6) T y y a m2 T T m1 x m 2g cos m 1g (b) m 2g (c) m2g sin nent of kinetic friction between the puck and ice. Solution The forces acting on the puck after it is in motion are shown in Figure 5.20. If we assume that the force of kinetic friction fk remains constant, then this force produces a uniform acceleration of the puck in the direction opposite its velocity, causing the puck to slow down. First, we find this acceleration in terms of the coefficient of kinetic friction, using Newton's second law. Knowing the acceleration of the puck and the distance it travels, we can then use the equations of kinematics to find the coefficient of kinetic friction. n Motion But fk k n, and from (2) we see that n (1) becomes kn kmg kg ma x ax The negative sign means the acceleration is to the left; this means that the puck is slowing down. The acceleration is independent of the mass of the puck and is constant because we assume that k remains constant. Because the acceleration is constant, we can use Equation 2.12, v xf 2 v xi2 2a x(x f x i), with xi 0 and vxf 0: v xi2 2ax f k v xi2 v xi 2gx f 2 2 k gx f 0 fk mg k (20.0 m/s)2 2(9.80 m/s2)(115 m) 0.177 Figure 5.20 After the puck is given an initial velocity to the right, the only external forces acting on it are the force of gravity mg, the normal force n, and the force of kinetic friction fk . Note that k is dimensionless. EXAMPLE 5.14 Acceleration of Two Connected Objects When Friction Is Present Motion of block: (1) (2) Motion of ball: (3) Fx Fy Fx Fy F cos m 1a n m 1a y m 2a x T F sin 0 0 m 2g m 2a y m 2a m 1g fk T m 1a x A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m 2 by a lightweight cord over a lightweight, frictionless pulley, as shown in Figure 5.21a. A force of magnitude F at an angle with the horizontal is applied to the block as shown. The coefficient of kinetic friction between the block and surface is k . Determine the magnitude of the acceleration of the two objects. We start by drawing free-body diagrams for the two objects, as shown in Figures 5.21b and 5.21c. (Are you beginning to see the similarities in all these examples?) Next, we apply Newton's second law in component form to each object and use Equation 5.9, f k kn. Then we can solve for the acceleration in terms of the parameters given. The applied force F has x and y components F cos and F sin , respectively. Applying Newton's second law to both objects and assuming the motion of the block is to the right, we obtain Solution Note that because the two objects are connected, we can equate the magnitudes of the x component of the acceleration of the block and the y component of the acceleration of the ball. From Equation 5.9 we know that fk k n, and from (2) we know that n m1 g F sin (note that in this case n is not equal to m1 g); therefore, (4) fk k(m 1g F sin ) That is, the frictional force is reduced because of the positive 136 CHAPTER 5 The Laws of Motion Note that the acceleration of the block can be either to the right or to the left,6 depending on the sign of the numerator in (5). If the motion is to the left, then we must reverse the sign of fk in (1) because the force of kinetic friction must oppose the motion. In this case, the value of a is the same as in (5), with k replaced by k. y component of F. Substituting (4) and the value of T from (3) into (1) gives F cos k(m 1g F sin ) m 2(a g) m 1a Solving for a, we obtain F(cos sin ) g(m 2 m1 m2 km 1) (5) a k y a m1 n F x T T fk F sin F F cos m2 m 1g (c) a m2 (a) m 2g (b) Figure 5.21 (a) The external force F applied as shown can cause the block to accelerate to the right. (b) and (c) The free-body diagrams, under the assumption that the block accelerates to the right and the ball accelerates upward. The magnitude of the force of kinetic friction in this case is given by fk F sin ). kn k(m1g APPLICATION Automobile Antilock Braking Systems (ABS) have developed antilock braking systems (ABS) that very briefly release the brakes when a wheel is just about to stop turning. This maintains rolling contact between the tire and the paing? 16. Suppose a large and spirited Freshman team is beating the Sophomores in a tug-of-war contest. The center of the 17. 18. 19. 20. PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Sections 5.1 through 5.6 1. A force F applied to an object of mass m1 produces an acceleration of 3.00 m/s2. The same force applied to a second object of mass m 2 produces an acceleration of 1.00 m/s2. (a) What is the value of the ratio m1 /m 2 ? (b) If m1 and m 2 are combined, find their acceleration under the action of the force F. 2. A force of 10.0 N acts on a body of mass 2.00 kg. What are (a) the body's acceleration, (b) its weight in newtons, and (c) its acceleration if the force is doubled? 3. A 3.00-kg mass undergoes an acceleration given by a (2.00i 5.00j) m/s2. Find the resultant force F and its magnitude. 4. A heavy freight train has a mass of 15 000 metric tons. If the locomotive can pull with a force of 750 000 N, how long does it take to increase the speed from 0 to 80.0 km/h? 5. A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. The expanding gases behind it exert what force on the bullet while it is traveling down the barrel of the rifle, 0.820 m long? Assume constant acceleration and negligible friction. 6. After uniformly accelerating his arm for 0.090 0 s, a pitcher releases a baseball of weight 1.40 N with a veloc- ity of 32.0 m/s horizontally forward. If the ball starts from rest, (a) through what distance does the ball accelerate before its release? (b) What force does the pitcher exert on the ball? 7. After uniformly accelerating his arm for a time t, a pitcher releases a baseball of weight Fg j with a velocity vi. If the ball starts from rest, (a) through what distance does the ball accelerate before its release? (b) What force does the pitcher exert on the ball? 8. Define one pound as the weight of an object of mass 0.453 592 37 kg at a location where the acceleration due to gravity is 32.174 0 ft/s2. Express the pound as one quantity with one SI unit. WEB 9. A 4.00-kg object has a velocity of 3.00i m/s at one instant. Eight seconds later, its velocity has increased to (8.00i 10.0j) m/s. Assuming the object was subject to a constant total force, find (a) the components of the force and (b) its magnitude. 10. The average speed of a nitrogen molecule in air is about 6.70 102 m/s, and its mass is 4.68 10 26 kg. (a) If it takes 3.00 10 13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in the opposite direction, what is the average acceleration of the molecule during this time interval? (b) What average force does the molecule exert on the wall? Problems 11. An electron of mass 9.11 10 31 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to 7.00 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we neglected. 12. A woman weighs 120 lb. Determine (a) her weight in newtons and (b) her mass in kilograms. 13. If a man weighs 900 N on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 m/s2? 14. The distinction between mass and weight was discovered after Jean Richer transported pendulum clocks from Paris to French Guiana in 1671. He found that they ran slower there quite systematically. The effect was reversed when the clocks returned to Paris. How much weight would you personally lose in traveling from Paris, where g 9.809 5 m/s2, to Cayenne, where g 9.780 8 m/s2 ? (We shall consider how the free-fall acceleration influences the period of a pendulum in Section 13.4.) 15. Two forces F1 and F2 act on a 5.00-kg mass. If F1 20.0 N and F2 15.0 N, find the accelerations in (a) and (b) of Figure P5.15. F2 F, but once her parachute opens, her downward velocity will be greatly reduced. Why does she slow down rapidly when her chute opens, enabling her to fall safely to the ground? If the chute does not function properly, the sky diver will almost certainly be seriously injured. What force exerted on her limits her maximum speed? (Guy Savage/Photo Researchers, Inc.) c h a p t e r Circular Motion and Other Applications of Newton's Laws Chapter Outline 6.1 Newton's Second Law Applied to Uniform Circular Motion 6.4 (Optional) Motion in the Presence of Resistive Forces 6.2 Nonuniform Circular Motion 6.3 (Optional) Motion in Accelerated Frames 6.5 (Optional) Numerical Modeling in Particle Dynamics 151 152 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws I n the preceding chapter we introduced Newton's laws of motion and applied them to situations involving linear motion. Now we discuss motion that is slightly more complicated. For example, we shall apply Newton's laws to objects traveling in circular paths. Also, we shall discuss motion observed from an accelerating frame of reference and motion in a viscous medium. For the most part, this chapter is a series of examples selected to illustrate the application of Newton's laws to a wide variety of circumstances. 6.1 NEWTON'S SECOND LAW APPLIED TO UNIFORM CIRCULAR MOTION In Section 4.4 we found that a particle moving with uniform speed v in a circular path of radius r experiences an acceleration ar that has a magnitude ar v2 r 4.7 The acceleration is called the centripetal acceleration because ar is directed toward the center of the circle. Furthermore, ar is always perpendicular to v. (If there were a component of acceleration parallel to v, the particle's speed would be changing.) Consider a ball of mass m that is tied to a string of length r and is being whirled at constant speed in a horizontal circular path, as illustrated in Figure 6.1. Its weight is supported by a low-friction table. Why does the ball move in a circle? Because of its inertia, the tendency of the ball is to move in a straight line; however, the string prevents motion along a straight line by exerting on the ball a force that makes it follow the circular path. This force is directed along the string toward the center of the circle, as shown in Figure 6.1. This force can be any one of our familiar forces causing an object to follow a circular path. If we apply Newton's second law along the radial direction, we find that the value of the net force causing the centripetal acceleration can be evaluated: Fr mar m v2 r (6.1) Force causing centripetal acceleration m Fr r Fr Figure 6.1 Overhead view of a ball moving in a circular path in a horizontal plane. A force Fr directed toward the center of the circle keeps the ball moving in its circular path. 6.1 Newton's Second Law Applied to Uniform Circular Motion 153 Figure 6.2 When the string breaks, the ball moves in the direction tangent to the circle. r A force causing a centripetal acceleration acts toward the center of the circular path and causes a change in the direction of the velocity vector. If that force should vanish, the object would no longer move in its circular path; instead, it would move along a straight-line path tangent to the circle. This idea is illustrated in Figure 6.2 for the ball whirling at the end of a string. If the string breaks at some instant, the ball moves along the straight-line path tangent to the circle at the point where the string broke. Quick Quiz 6.1 Is it possible for a car to move in a circular path in such a way that it has a tangential acceleration but no centripetal acceleration? An athlete in the process of throwing the hammer at the 1996 Olympic Games in Atlanta, Georgia. The force exerted by the chain is the force causing the circular motion. Only when the athlete releases the hammer will it move along a straight-line path tangent to the circle. CONCEPTUAL EXAMPLE 6.1 Forces That Cause Centripetal Acceleration Consider some examples. For the motion of the Earth around the Sun, the centripetal force is gravity. For an object sitting on a rotating turntable, the centripetal force is friction. For a rock whirled on the end of a string, the centripetal force is the force of tension in the string. For an amusementpark patron pressed against the inner wall of a rapidly rotating circular room, the centripetal force is the normal force exerted by the wall. What's more, the centripetal force could be a combination of two or more forces. For example, as a Ferris-wheel rider passes through the lowest point, the centripetal force on her is the difference between the normal force exerted by the seat and her weight. The force causing centripetal acceleration is sometimes called a centripetal force. We are familiar with a variety of forces in nature -- friction, gravity, normal forces, tension, and so forth. Should we add centripetal force to this list? Solution No; centripetal force should not be added to this list. This is a pitfall for many students. Giving the force causing circular motion a name -- centripetal force -- leads many students to consider it a new kind of force rather than a new role for force. A common mistake in force diagrams is to draw all the usual forces and then to add another vector for the centripetal force. But it is not a separate force -- it is simply one of our familiar forces acting in the role of a force that causes a circular motion. 154 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws (a) (b) (c) (d) Figure 6.3 A ball that had been moving in a circular path is acted on by various external forces that change its path. Quick Quiz 6.2 QuickLab Tie a string to a tennis ball, swing it in a circle, and then, while it is swinging, let go of the string to verify your answer to the last part of Quick Quiz 6.2. A ball is following the dotted circular path shown in Figure 6.3 under the influence of a force. At a certain instant of time, the force on the ball changes abruptly to a new force, and the ball follows the paths indicated by the solid line with an arrowhead in each of the four parts of the figure. For each part of the figure, describe the magnitude and direction of the force required to make the ball move in the solid path. If the dotted line represents the path of a ball being whirled on the end of a string, which path does the ball follow if the string breaks? Let us consider some examples of uniform circular motion. In each case, be sure to recognize the external force (or forces) that causes the body to move in its circular path. EXAMPLE 6.2 How Fast Can It Spin? Solving for v, we have v A ball of mass 0.500 kg is attached to the end of a cord 1.50 m long. The ball is whirled in a horizontal circle as was shown in Figure 6.1. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Assume that the string remains horizontal during the motion. Tr m Solution It is difficult to know what might be a reasonable value for the answer. Nonetheless, we know that it cannot be too large, say 100 m/s, because a person cannot make a ball move so quickly. It makes sense that the stronger the cord, the faster the ball can twirl before the cord breaks. Also, we expect a more massive ball to break the cord at a lower speed. (Imagine whirling a bowling ball!) Because the force causing the centripetal acceleration in this case is the force T exerted by the cord on the ball, Equation 6.1 yields for Fr mar v2 T m r This shows that v increases with T and decreases with larger m, as we expect to see -- for a given v, a large mass requires a large tension and a small mass needs only a small tension. The maximum speed the ball can have corresponds to the maximum tension. Hence, we find vmax Tmaxr m (50.0 N)(1.50 m) 0.500 kg 12.2 m/s Exercise Answer Calculate the tension in the cord if the speed of the ball is 5.00 m/s. 8.33 N. EXAMPLE 6.3 The Conical Pendulum Let us choose to represent the angle between string and vertical. In the free-body diagram shown in Figure 6.4, the force T exerted by the string is resolved into a vertical component T cos and a horizontal component T sin acting toward the center of revolution. Because the object does A small object of mass m is suspended from a string of length L. The object revolves with constant speed v in a horizontal circle of radius r, as shown in Figure 6.4. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for v. Solution 6.1 Newton's Second Law Applied to Uniform Circular Motion 155 not accelerate in the vertical direction, Fy may 0, and the upward vertical component of T must balance the downward force of gravity. Therefore, (1) T cos mg Because the force providing the centripetal acceleration in this example is the component T sin , we can use Newton's second law and Equation 6.1 to obtain (2) Fr T sin ma r mv 2 r /cos Dividing (2) by (1) and remembering that sin tan , we eliminate T and find that L T cos tan v T sin T r v2 rg rg tan L sin ; From the geometry in Figure 6.4, we note that r therefore, v mg mg The conical pendulum and its free-body diagram. Lg sin tan Figure 6.4 Note that the speed is independent of the mass of the object. EXAMPLE 6.4 What Is the Maximum Speed of the Car? and dry pavement is 0.500, find the maximum speed the car can have and still make the turn successfully. A 1 500-kg car moving on a flat, horizontal road negotiates a curve, as illustrated in Figure 6.5. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires fs Solution From experience, we should expect a maximum speed less than 50 m/s. (A convenient mental conversion is that 1 m/s is roughly 2 mi/h.) In this case, the force that enables the car to remain in its circular path is the force of static friction. (Because no slipping occurs at the point of contact between road and tires, the acting force is a force of static friction directed toward the center of the curve. If this force of static friction were zero -- for example, if the car were on an icy road -- the car would continue in a straight line and slide off the road.) Hence, from Equation 6.1 we have (a) (1) fs m v2 r n fs mg (b) The maximum speed the car can have around the curve is the speed at which it is on the verge of skidding outward. At this point, the friction force has its maximum value fs,max sn. Because the car is on a horizontal road, the magnitude of the normal force equals the weight (n mg) and thus fs,max smg. Substituting this value for fs into (1), we find that the maximum speed is vmax Figure 6.5 (a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. (b) The freebody diagram for the car. fs,maxr m smgr m s gr (0.500)(9.80 m/s2)(35.0 m) 13.1 m/s 156 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws Note that the maximum speed does not depend on the mass of the car. That is why curved highways do not need multiple speed limit signs to cover the various masses of vehicles using the road. Exercise Answer On a wet day, the car begins to skid on the curve when its speed reaches 8.00 m/s. What is the coefficient of static friction in this case? 0.187. EXAMPLE 6.5 The Banked Exit Ramp n sin pointing toward the center of the curve. Because the ramp is to be designed so that the force of static friction is zero, only the component n sin causes the centripetal acceleration. Hence, Newton's second law written for the radial direction gives (1) Fr n sin mv2 r A civil engineer wishes to design a curved exit ramp for a highway in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a ramp is usually banked; this means the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 13.4 m/s (30.0 mi/h) and the radius of the curve is 50.0 m. At what angle should the curvom of the loop the normal and gravitational forces act in opposite directions, whereas at the top of the loop these two forces act in the same direction. It is the vector sum of these two forces that gives the force of constant magnitude that keeps the pilot moving in a circular path. To yield net force vectors with the same magnitude, the normal force at the bottom (where the normal and gravitational forces are in opposite directions) must be greater than that at the top (where the normal and gravitational forces are in the same direction). (a) The free-body diagram for the pilot at the bottom of the loop is shown in Figure 6.8b. The only forces acting on him are the downward force of gravity Fg mg and the upward force n bot exerted by the seat. Because the net upward force that provides the centripetal ac- Substituting the values given for the speed and radius gives nbot mg 1 (2.70 (225 m/s)2 103 m)(9.80 m/s2) 2.91mg Hence, the magnitude of the force n bot exerted by the seat on the pilot is greater than the weight of the pilot by a factor of 2.91. This means that the pilot experiences an apparent weight that is greater than his true weight by a factor of 2.91. (b) The free-body diagram for the pilot at the top of the loop is shown in Figure 6.8c. As we noted earlier, both the gravitational force exerted by the Earth and the force n top exerted by the seat on the pilot act downward, and so the net downward force that provides the centripetal acceleration has 158 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws n bot Top Figure 6.8 (a) An aircraft executes a loop-the-loop maneuver as it moves in a vertical circle at constant speed. (b) Free-body diagram for the pilot at the bottom of the loop. In this position the pilot experiences an apparent weight greater than his true weight. (c) Free-body diagram for the pilot at the top of the loop. A ntop mg (b) mg (c) Bottom (a) a magnitude n top Fr ntop ntop ntop m mg v2 r mg mg mg. Applying Newton's second law yields m v2 r v2 rg 1 1 0.913mg In this case, the magnitude of the force exerted by the seat on the pilot is less than his true weight by a factor of 0.913, and the pilot feels lighter. Exercise mg Determine the magnitude of the radially directed force exerted on the pilot by the seat when the aircraft is at point A in Figure 6.8a, midway up the loop. nA 1.913mg directed to the right. (2.70 (225 m/s)2 103 m)(9.80 m/s2) Answer Quick Quiz 6.3 A bead slides freely along a curved wire at constant speed, as shown in the overhead view of Figure 6.9. At each of the points , , and , draw the vector representing the force that the wire exerts on the bead in order to cause it to follow the path of the wire at that point. QuickLab Hold a shoe by the end of its lace and spin it in a vertical circle. Can you feel the difference in the tension in the lace when the shoe is at top of the circle compared with when the shoe is at the bottom? Figure 6.9 6.2 NONUNIFORM CIRCULAR MOTION In Chapter 4 we found that if a particle moves with varying speed in a circular path, there is, in addition to the centripetal (radial) component of acceleration, a tangential component having magnitude dv/dt. Therefore, the force acting on the 6.2 Nonuniform Circular Motion 159 Some examples of forces acting during circular motion. (Left) As these speed skaters round a curve, the force exerted by the ice on their skates provides the centripetal acceleration. (Right) Passengers on a "corkscrew" roller coaster. What are the origins of the forces in this example? Figure 6.10 When the force acting on a particle moving in a circular path has a tangential component Ft , the particle's speed changes. The total force exerted on the particle in this case is the vector sum of the radial force and the tangential force. That is, F Fr Ft . F Fr Ft particle must also have a tangential and a radial component. Because the total acceleration is a ar at , the total force exerted on the particle is F Fr Ft , as shown in Figure 6.10. The vector Fr is directed toward the center oe passenger tends to move along the original straightline path. This is in accordance with Newton's first law: The natural tendency of a body is to continue moving in a straight line. However, if a sufficiently large force (toward the center of curvature) acts on the passenger, as in Figure 6.12c, she will move in a curved path along with the car. The origin of this force is the force of friction between her and the car seat. If this frictional force is not large enough, she will slide to the right as the car turns to the left under her. Eventually, she encounters the door, which provides a force large enough to enable her to follow the same curved path as the car. She slides toward the door not because of some mysterious outward force but because the force of friction is not sufficiently great to allow her to travel along the circular path followed by the car. In general, if a particle moves with an acceleration a relative to an observer in an inertial frame, that observer may use Newton's second law and correctly claim that F ma. If another observer in an accelerated frame tries to apply Newton's second law to the motion of the particle, the person must introduce fictitious forces to make Newton's second law work. These forces "invented" by the observer in the accelerating frame appear to be real. However, we emphasize that these fictitious forces do not exist when the motion is observed in an inertial frame. Fictitious forces are used only in an accelerating frame and do not represent "real" forces acting on the particle. (By real forces, we mean the interaction of the particle with its environment.) If the fictitious forces are properly defined in the accelerating frame, the description of motion in this frame is equivalent to the description given by an inertial observer who considers only real forces. Usually, we analyze motions using inertial reference frames, but there are cases in which it is more convenient to use an accelerating frame. QuickLab Use a string, a small weight, and a protractor to measure your acceleration as you start sprinting from a standing position. Fictitious forces (a) (b) Figure 6.12 (a) A car approaching a curved exit ramp. What causes a front-seat passenger to move toward the right-hand door? (b) From the frame of reference of the passenger, a (fictitious) force pushes her toward the right door. (c) Relative to the reference frame of the Earth, the car seat applies a leftward force to the passenger, causing her to change direction along with the rest of the car. (c) 162 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws EXAMPLE 6.9 Fictitious Forces in Linear Motion Because the deflection of the cord from the vertical serves as a measure of acceleration, a simple pendulum can be used as an accelerometer. According to the noninertial observer riding in the car (Fig. 6.13b), the cord still makes an angle with the vertical; however, to her the sphere is at rest and so its acceleration is zero. Therefore, she introduces a fictitious force to balance the horizontal component of T and claims that the net force on the sphere is zero! In this noninertial frame of reference, Newton's second law in component form yields Noninertial observer 0 Fx Fy T sin T cos Ffictitious mg 0 0 A small sphere of mass m is hung by a cord from the ceiling of a boxcar that is accelerating to the right, as shown in Figure 6.13. According to the inertial observer at rest (Fig. 6.13a), the forces on the sphere are the force T exerted by the cord and the force of gravity. The inertial observer concludes that the acceleration of the sphere is the same as that of the boxcar and that this acceleration is provided by the horizontal component of T. Also, the vertical component of T balances the force of gravity. Therefore, she writes Newton's second law as F T m g ma, which in component form becomes Inertial observer (1) (2) Fx Fy T sin T cos ma mg Thus, by solving (1) and (2) simultaneously for a, the inertial observer can determine the magnitude of the car's acceleration through the relationshie time it takes to reach 0.63vt . Figure 6.15 Resistive Force Proportional to Object Speed If we assume that the resistive force acting on an object moving through a liquid or gas is proportional to the object's speed, then the magnitude of the resistive force can be expressed as R bv (6.2) where v is the speed of the object and b is a constant whose value depends on the properties of the medium and on the shape and dimensions of the object. If the object is a sphere of radius r, then b is proportional to r. Consider a small sphere of mass m released from rest in a liquid, as in Figure 6.15a. Assuming that the only forces acting on the sphere are the resistive force bv and the force of gravity Fg , let us describe its motion.1 Applying Newton's second law to the vertical motion, choosing the downward direction to be positive, and noting that Fy mg bv, we obtain mg bv ma m dv dt (6.3) where the acceleration dv/dt is downward. Solving this expression for the acceleration gives dv dt g b v m (6.4) Terminal speed This equation is called a differential equation, and the methods of solving it may not be familiar to you as yet. However, note that initially, when v 0, the resistive force bv is also zero and the acceleration dv/dt is simply g. As t increases, the resistive force increases and the acceleration decreases. Eventually, the acceleration becomes zero when the magnitude of the resistive force equals the sphere's weight. At this point, the sphere reaches its terminal speed vt , and from then on There is also a buoyant force acting on the submerged object. This force is constant, and its magnitude is equal to the weight of the displaced liquid. This force changes the apparent weight of the sphere by a constant factor, so we will ignore the force here. We discuss buoyant forces in Chapter 15. 1 6.4 Motion in the Presence of Resistive Forces 165 it continues to move at this speed with zero acceleration, as shown in Figure 6.15b. We can obtain the terminal speed from Equation 6.3 by setting a dv/dt 0. This gives mg bvt 0 or vt mg/b The expression for v that satisfies Equation 6.4 with v v mg (1 b e bt/m) 0 at t ) 0 is (6.5) vt (1 e t/ This function is plotted in Figure 6.15c. The time constant m/b (Greek letter tau) is the time it takes the sphere to reach 63.2% ( 1 1/e) of its terminal speed. This can be seen by noting that when t , Equation 6.5 yields v 0.632vt . We can check that Equation 6.5 is a solution to Equation 6.4 by direct differentiation: mg d mg bt/m dv d mg e e bt/m ge bt/m dt dt b b b dt (See Appendix Table B.4 for the derivative of e raised to some power.) Substituting into Equation 6.4 both this expression for dv/dt and the expression for v given by Equation 6.5 shows that our solution satisfies the differential equation. Aerodynamic car. A streamlined body reduces air drag and increases fuel efficiency. EXAMPLE 6.11 Sphere Falling in Oil 0.900vt 1 e e t/ t/ A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the time constant and the time it takes the sphere to reach 90% of its terminal speed. vt(1 0.900 0.100 e t/ ) t t ln(0.100) 2.30 2.30 10 3 Solution vt b Because the terminal mg/b, the coefficient b is mg vt speed is given by 2.30(5.10 s) 11.7 10 3 s (2.00 g)(980 cm/s2) 5.00 cm/s 392 g/s 11.7 ms Thus, the sphere reaches 90% of its terminal (maximum) speed in a very short time. Therefore, the time constant is m b 2.00 g 392 g/s 5.10 10 3 s The speed of the sphere as a function of time is given by Equation 6.5. To find the time t it takes the sphere to reach a speed of 0.900vt , we set v 0.900vt in Equation 6.5 and solve for t: Exercise What is the sphere's speed through the oil at t 11.7 ms? Compare this value with the speed the sphere would have if it were falling in a vacuum and so were influenced only by gravity. Answer 4.50 cm/s in oil versus 11.5 cm/s in free fall. Air Drag at High Speeds For objects moving at high speeds through air, such as airplanes, sky divers, cars, and baseballs, the resistive force is approximately proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as R 1 2D Av2 (6.6) 166 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws R v R mg vt where is the density of air, A is the cross-sectional area of the falling object measured in a plane perpendicular to its motion, and D is a dimensionless empirical quantity called the drag coefficient. The drag coefficient has a value of about 0.5 for spherical objects but can have a value as great as 2 for irregularly shaped objects. Let us analyze the motion of an object in free fall subject to an upward air resistive force of magnitude R 1 D Av2. Suppose an object of mass m is re2 leased from rest. As Figure 6.16 shows, the object experiences two external forces: the downward force of gravity Fg mg and the upward resistive force R. (There is also an upward buoyant force that we neglect.) Hence, the magnitude of the net force is F mg 1 2D Av2 (6.7) mg where we have taken downward to be the positive vertical direction. Substituting F ma into Equation 6.7, we find that the object has a downward acceleration of magnitude D A a g v2 (6.8) 2m We can calculate the terminal speed vt by using the fact that when the force of gravity is balanced by the resistive force, the net force on the object is zero and therefore its acceleration is zero. Setting a 0 in Equation 6.8 gives g D A 2m vt2 vt 0 Figure 6.16 An object falling through air experiences a resistive force R and a gravitational force Fg mg. The object reaches terminal speed (on the right) when the net force acting on it is zero, that is, when R Fg or R mg. Before this occurs, the acceleration varies with speed according to Equation 6.8. 2mg D A (6.9) Using this expression, we can determine how the terminal speed depends on the dimensions of the object. Suppose the object is a sphere of radius r. In this case, A r2 (from A r 2 ) and m r3 (because the mass is proportional to the volume of the sphere, which is V 4 r3). Therefore, vt r. 3 Table 6.1 lists the terminal speeds for several objects falling through air. The high cost of fuel has prompted many truck owners to install wind deflectors on their cabs to reduce drag. 6.4 Motion in the Presence of Resistive Forces 167 TABLE 6.1 Terminal Speed for Various Objects Falling Through Air Object Sky diver Baseball (radius 3.7 cm) Golf ball (radius 2.1 cm) Hailstone (radius 0.50 cm) Raindrop (radius 0.20 cm) Mass (kg) 75 0.145 0.046 4.8 10 3.4 10 Cross-Sectional Area (m2) 4.2 1.4 7.9 1.3 0.70 10 10 10 10 3 3 5 5 vt (m/s) 60 43 44 14 9.0 4 5 CONCEPTUAL EXAMPLE 6.12 Consider a sky surfer who jumps from a plane with her feet attached firmly to her surfboard, does some tricks, and then opens her parachute. Describe the forces acting on her during these maneuvers. Solution When the surfer first steps out of the plane, she has no vertical velocity. The downward force of gravity causes her to accelerate toward the ground. As her downward speed increases, so does the upward resistive force exerted by the air on her body and the board. This upward force reduces their acceleration, and so their speed increases more slowly. Eventually, they are going so fast that the upward resistive force matches the downward force of gravity. Now the net force is zero and they no longer accelerate, but reach their terminal speed. At some point after reaching terminal speed, she opens her parachute, resulting in a drastic increase in the upward resistive force. The net force (and thus the acceleration) is now upward, in the direction opposite the direction of the velocity. This causes the downward velocity to decrease rapidly; this means the resistive force on the chute also decreases. Eventually the upward resistive force and the downward force of gravity balance each other and a much smaller terminal speed is reached, permitting a safe landing. (Contrary to popular belief, the velocity vector of a sky diver never points upward. You may have seen a videotape in which a sky diver appeared to "rocket" upward once the chute opened. In fact, what happened is that the diver slowed down while the person holding the camera continued falling at high speed.) A sky surfer takes advantage of the upward force of the air on her board. ( EXAMPLE 6.13 Falling Coffee Filters presents data for these coffee filters as they fall through the air. The time constant is small, so that a dropped filter quickly reaches terminal speed. Each filter has a mass of 1.64 g. When the filters are nested together, they stack in The dependence of resistive force on speed is an empirical relationship. In other words, it is based on observation rather than on a theoretical model. A series of stacked filters is dropped, and the terminal speeds are measured. Table 6.2 168 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws Two filters nested together experience 0.032 2 N of resistive force, and so forth. A graph of the resistive force on the filters as a function of terminal speed is shown in Figure 6.17a. A straight line would not be a good fit, indicating that the resistive force is not proportional to the speed. The curved line is for a second-order polynomial, indicating a proportionality of the resistive force to the square of the speed. This proportionality is more clearly seen in Figure 6.17b, in which the resistive force is plotted as a function of the square of the terminal speed. such a way that the front-facing surface area does not increase. Determine the relationship between the resistive force exerted by the air and the speed of the falling filters. Solution At terminal speed, the upward resistive force balances the downward force of gravity. So, a single filter falling at its terminal speed experiences a resistive force of R mg 1.64 g 1000 g/kg (9.80 m/s2) 0.016 1 N TABLE 6.2 Terminal Speed for Stacked Coffee Filters Number of Filters 1 2 3 4 5 6 7 8 9 10 a vt (m/s)a 1.01 1.40 1.63 2.00 2.25 2.40 2.57 2.80 3.05 3.22 Pleated coffee filters can be nested together so that the force of air resistance can be studied. ( All values of vt are approximate. 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 Resistive force (N) Resistive force (N) 0 1 2 Terminal speed (m/s) (a) 3 4 0 2 4 6 (b) 8 10 12 Terminal speed squared (m/s)2 Figure 6.17 (a) Relationship between the resistive force acting on falling coffee filters and their terminal speed. The curved line is a second-order polynomial fit. (b) Graph relating the resistive force to the square of the terminal speed. The fit of the straight line to the data points indicates that the resistive force is proportional to the terminal speed squared. Can you find the proportionality constant? 6.5 Numerical Modeling in Particle Dynamics 169 EXAMPLE 6.14 Resistive Force Exerted on a Baseball D 2 mg vt2 A 0.284 2(0.145 kg)(9.80 m/s2) (43 m/s)2 (1.29 kg/m3)(4.2 10 3 A pitcher hurls a 0.145-kg baseball past a batter at 40.2 m/s ( 90 mi/h). Find the resistive force acting on the ball at this speed. m2) We do not expect the air to exert a huge force on the ball, and so the resistive force we calculate from Equation 6.6 should not be more than a few newtons. First, we must determine the drag coefficient D. We do this by imagining that we drop the baseball and allow it to reach terminal speed. We solve Equation 6.9 for D and substitute the appropriate values for m, vt , and A from Table 6.1. Taking the density of air as 1.29 kg/m3, we obtain Solution This number has no dimensions. We have kept an extra digit beyond the two that are significant and will drop it at the end of our calculation. We can now use this value for D in Equation 6.6 to find the magnitude of the resistive force: R 1 2 2 D Av 1 2 (0.284)(1.29 kg/m3)(4.2 10 3 m2)(40.2 m/s)2 1.2 N Optional Section 6.5 NUMERICAL MODELING IN PARTICLE DYNAMICS 2 As we have seen in this and the preceding chapter, the study of the dynamicst v(t) t) of the particle at the end of the time interval t is apThen the speed v(t proximately equal to the speed v(t) at the beginning of the time interval plus the magnitude of the acceleration during the interval multiplied by t: v(t t) v(t) a(t) t (6.10) Because the acceleration is a function of time, this estimate of v(t t) is accurate only if the time interval t is short enough that the change in acceleration during it is very small (as is discussed later). Of course, Equation 6.10 is exact if the acceleration is constant. 6.5 Numerical Modeling in Particle Dynamics 171 The position x(t t) of the particle at the end of the interval found in the same manner: v(t) x(t x t t) x(t x(t) 1 2 t can be t) t v(t) t t)2 x(t) (6.11) You may be tempted to add the term a( to this result to make it look like the familiar kinematics equation, but this term is not included in the Euler method because t is assumed to be so small that t 2 is nearly zero. If the acceleration at any instant t is known, the particle's velocity and position at a time t t can be calculated from Equations 6.10 and 6.11. The calculation then proceeds in a series of finite steps to determine the velocity and position at any later time. The acceleration is determined from the net force acting on the particle, and this force may depend on position, velocity, or time: a(x, v, t) F(x, v, t) m (6.12) It is convenient to set up the numerical solution to this kind of problem by numbering the steps and entering the calculations in a table, a procedure that is illustrated in Table 6.3. The equations in the table can be entered into a spreadsheet and the calculations performed row by row to determine the velocity, position, and acceleration as functions of time. The calculations can also be carried out by using a program written in either BASIC, C , or FORTRAN or by using commercially available mathematics packages for personal computers. Many small increments can be taken, and accurate results can usually be obtained with the help of a computer. Graphs of velocity versus time or position versus time can be displayed to help you visualize the motion. One advantage of the Euler method is that the dynamics is not obscured -- the fundamental relationships between acceleration and force, velocity and acceleration, and position and velocity are clearly evident. Indeed, these relationships form the heart of the calculations. There is no need to use advanced mathematics, and the basic physics governs the dynamics. The Euler method is completely reliable for infinitesimally small time increments, but for practical reasons a finite increment size must be chosen. For the finite difference approximation of Equation 6.10 to be valid, the time increment must be small enough that the acceleration can be approximated as being constant during the increment. We can determine an appropriate size for the time in- See the spreadsheet file "Baseball with Drag" on the Student Web site (address below) for an example of how this technique can be applied to find the initial speed of the baseball described in Example 6.14. We cannot use our regular approach because our kinematics equations assume constant acceleration. Euler's method provides a way to circumvent this difficulty. A detailed solution to Problem 41 involving iterative integration appears in the Student Solutions Manual and Study Guide and is posted on the Web at http:/ www.saunderscollege.com/physics TABLE 6.3 The Euler Method for Solving Dynamics Problems Step 0 1 2 3 n Time t0 t1 t2 t3 tn t0 t1 t2 t t t Position x0 x1 x2 x3 xn x0 x1 x2 v0 t v1 t v2 t Velocity v0 v1 v2 v3 vn v0 v1 v2 a0 t a1 t a2 t Acceleration a0 a1 a2 a3 an F(x 0 , v0 , t 0)/m F(x 1 , v 1 , t 1)/m F(x 2 , v 2 , t 2)/m F(x 3 , v 3 , t 3)/m 172 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws crement by examining the particular problem being investigated. The criterion for the size of the time increment may need to be changed during the course of the motion. In practice, however, we usually choose a time increment appropriate to the initial conditions and use the same value throughout the calculations. The size of the time increment influences the accuracy of the result, but unfortunately it is not easy to determine the accuracy of an Euler-method solution without a knowledge of the correct analytical solution. One method of determining the accuracy of the numerical solution is to repeat the calculations with a smaller time increment and compare results. If the two calculations agree to a certain number of significant figures, you can assume that the results are correct to that precision. SUMMARY Newton's second law applied to a particle moving in uniform circular motion states that the net force causing the particle to undergo a centripetal acceleration is Fr mar mv2 r (6.1) You should be able to use this formula in situations where the force providing the centripetal acceleration could be the force of gravity, a force of friction, a force of string tension, or a normal force. A particle moving in nonuniform circular motion has both a centripetal component of acceleration and a nonzero tangential component of acceleration. In the case of a particle rotating in a vertical circle, the force of gravity provides the tangential component of acceleration and part or all of the centripetal component of acceleration. Be sure you understand the directions and magnitudes of the velocity and acceleration vectors for nonuniform circular motion. An observer in a noninertial (accelerating) frame of reference must introduce fictitious forces when applying Newton's second law in that frame. If these fictitious forces are properly defined, the description of motion in the noninertial frame is equivalent to that made by an observer in an inertial frame. However, the observers in the two frames do not agree on the causes of the motion. You should be able to distinguish between inertial and noninertial frames and identify the fictitious forces acting in a noninertial frame. A body moving through a liquid or gas experiences a resistive force that is speed-dependent. This resistive force, which opposes the motion, generally increases with speed. The magnitude of the resistive force depends on the shape of the body and on the properties of the medium through which the body is moving. In the limiting case for a falling body, when the magnitude of the resistive force equals the body's weight, the body reaches its terminal speed. You should be able to apply Newton's laws to analyze the motion of objects moving under the influence of resistive forces. You may need to apply Euler's method if the force depends on velocity, as it does for air drag. QUESTIONS 1. Because the Earth rotates about its axis and revolves around the Sun, it is a noninertial frame of reference. Assuming the Earth is a uniform sphere, why would the apparent weight of an object be greater at the poles than at the equator? 2. Explain why the Earth bulges at the equator. Problems 3. Why is it that an astronaut in a space capsule orbiting the Earth experiences a feeling of weightlessness? 4. Why does mud fly off a rapidly turning automobile tire? 5. Imagine that you attach a heavy object to one end of a spring and then whirl the spring and object in a horizontal circle (by holding the free end of the spring). Does the spring stretch? If so, why? Discuss this in terms of the force causing the circular motion. 6. It has been suggested that rotating cylinders about 10 mi in length and 5 mi in diameter be placed in space and used as colonies. The purpose of the rotation is to simulate gravity for the inhabitants. Explain this concept for producing an effective gravity. 7. Why does a pilot tend to black out when pulling out of a steep dive? 173 8. Describe a situation in which a car driver can have a centripetal acceleration but no tangential acceleration. 9. Describe the path of a moving object if its acceleration is constant in magnitude at all times and (a) perpendicular to the velocity; (b) parallel to the velocity. 10. Analyze the motion of a rock falling through water in terms of its speed and acceleration as it falls. Assume that the resistive force acting on the rock increases as the speed increases. 11. Consider a small raindrop and a large raindrop falling through the atmosphere. Compare their terminal speeds. What are their accelerations when they reach terminal speed? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 6.1 Newton's Second Law Applied to Uniform Circular Motion 1. A toy car moving at constant speed completes one lap around a circular track (a distance of 200 m) in 25.0 s. (a) What is its average speed? (b) If the mass of the car is 1.50 kg, what is the magnitude of the force that keeps it in a circle? 2. A 55.0-kg ice skater is moving at 4.00 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.800 m around the pole. (a) Determine the force exerted by the rope on her arms. (b) Compare this force with her weight. 3. A light string can support a stationary hanging load of 25.0 kg before breaking. A 3.00-kg mass attached to the string rotates on a horizontal, frictionless table in a circle of radius 0.800 m. What range of speeds can the mass have before the string breaks? 4. In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.20 106 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.530 10 10 m and (b) the centripetal acceleration of the electron. 5. In a cyclotron (one type of particle accelerator), a deuteron (of atomic mass 2.00 u) reaches a final speed of 10.0% of the speed of light while moving in a circular path of radius 0.480 m. The deuteron is maintained in the circular path by a magnetic force. What magnitude of force is required? 6. A satellite of mass 300 kg is in a circular orbit around the Earth at an altitude equal to the Earth's mean radius (see Example 6.6). Find (a) the satellite's orbital 7. 8. 9. 10. 11. speed, (b) the period of its revolution, and (c) the gravitational force acting on it. Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 100 km above the surface of the Moon. If the mass of the Moon is 7.40 1022 kg and its radius is 1.70 106 m, determine (a) the orbiting astronaut's acceleration, (b) his orbital speed, and (c) the period of the orbit. The speed of the tip of the minute hand on a town clock is 1.75 10 3 m/s. (a) What is the speed of the tip of the second hand of the same length? (b) What is the centripetal acceleration of the tip of the second hand? A coin placed 30.0 cm from the center of a rotating, horizontal turntable slips when its speed is 50.0 cm/s. (a) What provides the force in the radial direction when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between coin and turntable? The cornering performance of an automobile is evaluated on a skid pad, where the maximum speed that a car can maintain around a circular path on a dry, flat surface is measured. The centripetal acceleration, also called the lateral acceleration, is then calculated as a multiple of the free-fall acceleration g. The main factors affecting the performance are the tire characteristics and the suspension system of the car. A Dodge Viper GTS can negotiate a skid pad of radius 61.0 m at 86.5 km/h. Calculate its maximum lateral acceleration. A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates an unbanked 174 CHAPTER 6 Circular Motion and Other Applications of Newton's Laws hump? (b) What must be the speed of the car over the hump if she is to experience weightlessness? (That is, if her apparent weight is zero.) 15. Tarzan (m 85.0 kg) tries to cross a river by swinging from a vine. The vinthe puck? 52. An air puck of mass m1 is tied to a string and allowed to revolve in a circle of radius R on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass m 2 is tied to it (Fig. P6.51). The suspended mass remains in equilibrium while the puck on the tabletop revolves. What are (a) the tension in the string? (b) the central force exerted on the puck? (c) the speed of the puck? m L Figure P6.55 56. The pilot of an airplane executes a constant-speed loopthe-loop maneuver. His path is a vertical circle. The speed of the airplane is 300 mi/h, and the radius of the circle is 1 200 ft. (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) Describe how the pilot could experience apparent weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the force that the seat exerts on his body.) 57. For a satellite to move in a stable circular orbit at a constant speed, its centripetal acceleration must be inversely proportional to the square of the radius r of the orbit. (a) Show that the tangential speed of a satellite is proportional to r 1/2. (b) Show that the time required to complete one orbit is proportional to r 3/2. 58. A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk (Fig. P6.58). If the coeffi- Figure P6.51 Problems 51 and 52. WEB 53. Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a point at one of the poles experiences no centripetal acceleration. (a) Show that at the equator the gravitational force acting on an object (the true weight) must exceed the object's apparent weight. (b) What is the apparent weight at the equator and at the poles of a person having a mass of 75.0 kg? (Assume the Earth is a uniform sphere and take g 9.800 m/s2.) 54. A string under a tension of 50.0 N is used to whirl a rock in a horizontal circle of radius 2.50 m at a speed of 20.4 m/s. The string is pulled in and the speed of the rock increases. When the string is 1.00 m long and the speed of the rock is 51.0 m/s, the string breaks. What is the breaking strength (in newtons) of the string? 55. A child's toy consists of a small wedge that has an acute angle (Fig. P6.55). The sloping side of the wedge is frictionless, and a mass m on it remains at constant height if the wedge is spun at a certain constant speed. The wedge is spun by rotating a vertical rod that is firmly attached to the wedge at the bottom end. Show Disk Penny 12.0 cm Block Figure P6.58 Problems cients of friction between block and disk are 0.750 (static) and 0.640 (kinetic) while those for the penny and block are 0.450 (kinetic) and 0.520 (static), what is the maximum rate of rotation (in revolutions per minute) that the disk can have before either the block or the penny starts to slip? 59. Figure P6.59 shows a Ferris wheel that rotates four times each minute and has a diameter of 18.0 m. (a) What is the centripetal acceleration of a rider? What force does the seat exert on a 40.0-kg rider (b) at the lowest point of the ride and (c) at the highest point of the ride? (d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom? 179 8.00 m 2.50 m Figure P6.61 63. An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (Fig. P6.63). The coefficient of static friction between person and wall is s , and the radius of the cylinder is R. (a) Show that the maximum period of revolution necessary to keep the person from falling is T (4 2R s /g)1/2. (b) Obtain a numerical value for T Figure P6.59 (Color Box/FPG) 60. A space station, in the form of a large wheel 120 m in diameter, rotates to provide an "artificial gravity" of 3.00 m/s2 for persons situated at the outer rim.f 2 vf 0.15 m2/s2 kg)vf 2 0 0.39 m/s 0.40 J 1.6 kg 0.25 m2/s2 0.50 m/s As expected, this value is somewhat less than the 0.50 m/s we found in part (a). If the frictional force were greater, then the value we obtained as our answer would have been even smaller. (b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 4.0 N retards its motion from the moment it is released. 7.5 5.8 POWER Imagine two identical models of an automobile: one with a base-priced four-cylinder engine; and the other with the highest-priced optional engine, a mighty eightcylinder powerplant. Despite the differences in engines, the two cars have the same mass. Both cars climb a roadway up a hill, but the car with the optional engine takes much less time to reach the top. Both cars have done the same amount of work against gravity, but in different time periods. From a practical viewpoint, it is interesting to know not only the work done by the vehicles but also the rate at which it is done. In taking the ratio of the amount of work done to the time taken to do it, we have a way of quantifying this concept. The time rate of doing work is called power. If an external force is applied to an object (which we assume acts as a particle), and if the work done by this force in the time interval t is W, then the average power expended during this interval is defined as W t The work done on the object contributes to the increase in the energy of the object. Therefore, a more general definition of power is the time rate of energy transfer. In a manner similar to how we approached the definition of velocity and acceleraAverage power 200 CHAPTER 7 Work and Kinetic Energy tion, we can define the instantaneous power age power as t approaches zero: lim t:0 as the limiting value of the averdW dt W t where we have represented the increment of work done by dW. We find from Equation 7.2, letting the displacement be expressed as ds, that dW F ds. Therefore, the instantaneous power can be written Instantaneous power dW dt F ds dt F v (7.18) where we use the fact that v ds/dt. The SI unit of power is joules per second (J/s), also called the watt (W) (after James Watt, the inventor of the steam engine): The watt 1W 1 J/s 1 kg m2/s3 The symbol W (not italic) for watt should not be confused with the symbol W (italic) for work. A unit of power in the British engineering system is the horsepower (hp): 1 hp 746 W A unit of energy (or work) can now be defined in terms of the unit of power. One kilowatt hour (kWh) is the energy converted or consumed in 1 h at the constant rate of 1 kW 1 000 J/s. The numerical value of 1 kWh is The kilowatt hour is a unit of energy 1 kWh (103 W)(3 600 s) 3.60 106 J It is important to realize that a kilowatt hour is a unit of energy, not power. When you pay your electric bill, you pay the power company for the total electrical energy you used during the billing period. This energy is the power used multiplied by the time during which it was used. For example, a 300-W lightbulb run for 12 h would convert (0.300 kW)(12 h) 3.6 kWh of electrical energy. Quick Quiz 7.6 Suppose that an old truck and a sports car do the same amount of work as they climb a hill but that the truck takes much longer to accomplish this work. How would graphs of versus t compare for the two vehicles? EXAMPLE 7.12 Power Delivered by an Elevator Motor a free-body diagram in Figure 7.18b and have arbitrarily specified that the upward direction is positive. From Newton's second law we obtain Fy T f Mg 0 An elevator car has a mass of 1 000 kg and is carrying passengers having a combined mass of 800 kg. A constant frictional force of 4 000 N retards its motion upward, as shown in Figure 7.18a. (a) What must be the minimum power delivered by the motor to lift the elevator car at a constant speed of 3.00 m/s? where M is the total mass of the system (car plus passengers), equal to 1 800 kg. Therefore, T f Mg 103 N 104 N (1.80 103 kg)(9.80 m/s2) 4.00 2.16 Solution The motor must supply about 0.016. For a 1 450-kg car, the weight is 14 200 N and the force of rolling friction has a magnitude of n mg 227 N. As the speed of the car increases, a small reduction in the normal force occurs as a result of a decrease in atmospheric pressure as air flows over the top of the car. (This phenomenon is discussed in Chapter 15.) This reduction in the normal force causes a slight reduction in the force of rolling friction fr with increasing speed, as the data in Table 7.2 indicate. Now let us consider the effect of the resistive force that results from the movement of air past the car. For large objects, the resistive force fa associated with air friction is proportional to the square of the speed (in meters per second; see Section 6.4) and is given by Equation 6.6: fa 1 2D Av2 where D is the drag coefficient, is the density of air, and A is the cross-sectional area of the moving object. We can use this expression to calculate the fa values in Table 7.2, using D 0.50, 1.293 kg/m3, and A 2 m2. The magnitude of the total frictional force ft is the sum of the rolling frictional force and the air resistive force: ft fr fa At low speeds, road friction is the predominant resistive force, but at high speeds air drag predominates, as shown in Table 7.2. Road friction can be decreased by a reduction in tire flexing (for example, by an increase in the air pres- TABLE 7.2 Frictional Forces and Power Requirements for a Typical Car a v (m/s) 0 8.9 17.8 26.8 35.9 44.8 n (N) 14 200 14 100 13 900 13 600 13 200 12 600 fr (N) 227 226 222 218 211 202 fa (N) 0 51 204 465 830 1 293 ft (N) 227 277 426 683 1 041 1 495 ft v (kW) 0 2.5 7.6 18.3 37.3 67.0 is a In this table, n is the normal force, f is road friction, f is air friction, f is total friction, and r a t the power delivered to the wheels. 7.6 Energy and the Automobile 203 sure slightly above recommended values) and by the use of radial tires. Air drag can be reduced through the use of a smaller cross-sectional area and by streamlining the car. Although driving a car with the windows open increases air drag and thus results in a 3% decrease in mileage, driving with the windows closed and the air conditioner running results in a 12% decrease in mileage. The total power needed to maintain a constant speed v is ft v, and it is this power that must be delivered to the wheels. For example, from Table 7.2 we see that at v 26.8 m/s (60 mi/h) the required power is ft v (683 N) 26.8 m s 18.3 kW This power can be broken down into two parts: (1) the power fr v needed to compensate for road friction, and (2) the power fa v needed to compensate for air drag. At v 26.8 m/s, we obtain the values r fr v fa v (218 N) 26.8 (465 N) 26.8 m s m s 5.84 kW 12.5 kW a Note that r a. On the other hand, at v 44.8 m/s (100 mi/h), r 9.05 kW, a 57.9 kW, and 67.0 kW. This shows the importance of air drag at high speeds. EXAMPLE 7.14 Gas Consumed by a Compact Car would supply 1.3 108 J of energy. Because the engine is only 18% efficient, each gallon delivers only (0.18)(1.3 108 J) 2.3 107 J. Hence, the number of gallons used to accelerate the car is Number of gallons 2.9 2.3 105 J 107 J/gal 0.013 gal A compact car has a mass of 800 kg, and its efficiency is rated at 18%. (That is, 18% of the available fuel energy is delivered to the wheels.) Find the amount of gasoline used to accelerate the car from rest to 27 m/s (60 mi/h). Use the fact that the energy equivalent of 1 gal of gasoline is 1.3 108 J. The energy required to accelerate the car from rest to a speed v is its final kinetic energy 1 mv 2: 2 K 1 2 2 mv 1 2 (800 Solution kg)(27 m/s)2 2.9 105 J At cruising speed, this much gasoline is sufficient to propel the car nearly 0.5 mi. This demonstrates the extreme energy requirements of stop-and-start driving. If the engine were 100% efficient, each gallon of gasoline EXAMPLE 7.15 Power Delivered to Wheels 2.2 3.6 108 J 103 s 62 kW Suppose the compact car in Example 7.14 gets 35 mi/gal at 60 mi/h. How much power is delivered to the wheels? Solution By simply canceling units, we dn must equal the maximum potential energy stored in the fully compressed spring: EA KA 1 2 2 mvA fk kn kmg 0.50(0.80 kg)(9.80 m/s2) 3.92 N Therefore, the change in the block's mechanical energy due to friction as the block is displaced from the equilibrium position of the spring (where we have set our origin) to x B is E fk xB 1 2 2 k xB ) 3.92xB Substituting this into Equation 8.15 gives E Ef Ei (0 (1mvA2 2 3.92xB 0 0) fkxB EC KC 0 Us C 1 2 2 kxm Us A 0 xm 1 2 2 (50)xB 1 2 2 (0.80)(1.2) m v k A 25xB2 0.80 kg (1.2 m/s) 50 N/m 3.92xB 0.576 Solving the quadratic equation for x B gives xB 0.092 m and xB 0.25 m. The physically meaningful root is xB 0.092 m. The negative root does not apply to this situation because the block must be to the right of the origin (positive value of x) when it comes to rest. Note that 0.092 m is less than the distance obtained in the frictionless case of part (a). This result is what we expect because friction retards the motion of the system. 0.15 m Note that we have not included Ug terms because no change in vertical position occurred. (b) Suppose a constant force of kinetic friction acts between the block and the surface, with k 0.50. If the speed 230 CHAPTER 8 Potential Energy and Conservation of Energy EXAMPLE 8.9 Connected Blocks in Motion where Ug Ug f Ug i is the change in the system's gravitational potential energy and Us Usf Usi is the change in the system's elastic potential energy. As the hanging block falls a distance h, the horizontally moving block moves the same distance h to the right. Therefore, using Equation 8.15, we find that the loss in energy due to friction between the horizontally sliding block and the surface is (2) E fk h km1gh Two blocks are connected by a light string that passes over a frictionless pulley, as shown in Figure 8.13. The block of mass m1 lies on a horizontal surface and is connected to a spring of force constant k. The system is released from rest when the spring is unstretched. If the hanging block of mass m 2 falls a distance h before coming to rest, calculate the coefficient of kinetic friction between the block of mass m1 and the surface. Solution The key word rest appears twice in the problem statement, telling us that the initial and final velocities and kinetic energies are zero. (Also note that because we are concerned only with the beginning and ending points of the motion, we do not need to label events with circled letters as we did in the previous two examples. Simply using i and f is sufficient to keep track of the situation.) In this situation, the system consists of the two blocks, the spring, and the Earth. We need to consider two forms of potential energy: gravitational and elastic. Because the initial and final kinetic energies of the system are zero, K 0, and we can write (1) E Ug Us The change in the gravitational potential energy of the system is associated with only the falling block because the vertical coordinate of the horizontally sliding block does not change. Therefore, we obtain (3) Ug Ug f Ugi 0 m2 gh where the coordinates have been measured from the lowest position of the falling block. The change in the elastic potential energy stored in the spring is (4) Us Us f Usi 1 2 2 kh 0 Substituting Equations (2), (3), and (4) into Equation (1) gives m2gh 1kh2 km1gh 2 k m1 k m2g m1g 1 2 kh m2 h Figure 8.13 As the hanging block moves from its highest elevation to its lowest, the system loses gravitational potential energy but gains elastic potential energy in the spring. Some mechanical energy is lost because of friction between the sliding block and the surface. This setup represents a way of measuring the coefficient of kinetic friction between an object and some surface. As you can see from the problem, sometimes it is easier to work with the changes in the various types of energy rather than the actual values. For example, if we wanted to calculate the numerical value of the gravitational potential energy associated with the horizontally sliding block, we would need to specify the height o between the two is given by Einstein's most famous formula: ER mc 2 (8.17) where c is the speed of light and E R is the energy equivalent of a mass m. The subscript R on the energy refers to the rest energy of an object of mass m -- that is, the energy of the object when its speed is v 0. The rest energy associated with even a small amount of matter is enormous. For example, the rest energy of 1 kg of any substance is ER mc 2 (1 kg)(3 10 8 m/s)2 9 10 16 J This is equivalent to the energy content of about 15 million barrels of crude oil -- about one day's consumption in the United States! If this energy could easily be released as useful work, our energy resources would be unlimited. In reality, only a small fraction of the energy contained in a material sample can be released through chemical or nuclear processes. The effects are greatest in nuclear reactions, in which fractional changes in energy, and hence mass, of approximately 10 3 are routinely observed. A good example is the enormous amount of energy released when the uranium-235 nucleus splits into two smaller nuclei. This happens because the sum of the masses of the product nuclei is slightly less than the mass of the original 235 U nucleus. The awesome nature of the energy released in such reactions is vividly demonstrated in the explosion of a nuclear weapon. Equation 8.17 indicates that energy has mass. Whenever the energy of an object changes in any way, its mass changes as well. If E is the change in energy of an object, then its change in mass is m E c2 (8.18) Anytime energy E in any form is supplied to an object, the change in the mass of the object is m E/c 2. However, because c 2 is so large, the changes in mass in any ordinary mechanical experiment or chemical reaction are too small to be detected. 8.10 Quantization of Energy 237 EXAMPLE 8.12 Here Comes the Sun The Sun radiates uniformly in all directions, and so only a very tiny fraction of its total output is collected by the Earth. Nonetheless this amount is sufficient to supply energy to nearly everything on the Earth. (Nuclear and geothermal energy are the only alternatives.) Plants absorb solar energy and convert it to chemical potential energy (energy stored in the plant's molecules). When an animal eats the plant, this chemical potential energy can be turned into kinetic and other forms of energy. You are reading this book with solarpowered eyes! The Sun converts an enormous amount of matter to energy. Each second, 4.19 109 kg -- approximately the capacity of 400 average-sized cargo ships -- is changed to energy. What is the power output of the Sun? Solution We find the energy liberated per second by means of a straightforward conversion: ER (4.19 109 kg)(3.00 108 m/s)2 3.77 1026 J We then apply the definition of power: 3.77 1026 J 1.00 s 3.77 1026 W Optional Section 8.10 QUANTIZATION OF ENERGY Certain physical quantities such as electric charge are quantized; that is, the quantities have discrete values rather than continuous values. The quantized nature of energy is especially important in the atomic and subatomic world. As an example, let us consider the energy levels of the hydrogen atom (which consists of an electron orbiting around a proton). The atom can occupy only certain energy levels, called quantum states, as shown in Figure 8.18a. The atom cannot have any energy values lying between these quantum states. The lowest energy level E 1 is called the E E4 E3 E2 Energy (arbitrary units) E1 Energy Hydrogen atom (a) Earth satellite (b) Figure 8.18 Energy-level diagrams: (a) Quantum states of the hydrogen atom. The lowest state E 1 is the ground state. (b) The energy levels of an Earth satellite are also quantized but are so close together that they cannot be distinguished from one another. 238 CHAPTER 8 Potential Energy and Conservation of Energy ground state of the atom. The ground state corresponds to the state that an isolated atom usually occupies. The atom can move to higher energy states by absorbing energy from some external source or by colliding with othere conservative: the force of gravity and the spring force. There are two forms of potential energy: (1) gravitational potential energy and (2) elastic potential energy stored in the spring. 8.3 The first and third balls speed up after they are thrown, while the second ball initially slows down but then speeds up after reaching its peak. The paths of all three balls are parabolas, and the balls take different times to reach the ground because they have different initial velocities. However, all three balls have the same speed at the moment they hit the ground because all start with the same kinetic energy and undergo the same change in gravitational potential energy. In other words, E total 1mv 2 mgh is the same for all three balls at the 2 start of the motion. 8.4 Designate one object as No. 1 and the other as No. 2. The external force does work Wapp on the system. If Wapp 0, then the system energy increases. If Wapp 0, then the system energy decreases. The effect of friction is to decrease the total system energy. Equation 8.15 then becomes E W app K [K 1f [(Ug1f U K 2f ) Ug 2f (K 1i Usf) K 2i )] (Ug1i Ug 2i Usi)] E friction You may find it easier to think of this equation with its terms in a different order, saying total initial energy K 1i K 2i Ug1i net change Ug2i Usi K 1f K 2f total final energy W app f kd Ug1f Ug 2f Usf 8.5 The slope of a U(x)-versus-x graph is by definition dU(x)/dx. From Equation 8.16, we see that this expression is equal to the negative of the x component of the conservative force acting on an object that is part of the system. P U Z Z L E R Airbags have saved countless lives by reducing the forces exerted on vehicle occupants during collisions. How can airbags change the force needed to bring a person from a high speed to a complete stop? Why are they usually safer than seat belts alone? (Courtesy of Saab) c h a p t e r Linear Momentum and Collisions Chapter Outline 9.1 Linear Momentum and Its Conservation 9.2 Impulse and Momentum 9.3 Collisions 9.4 Elastic and Inelastic Collisions in One Dimension 9.5 9.6 9.7 9.8 Two-Dimensional Collisions The Center of Mass Motion of a System of Particles (Optional) Rocket Propulsion 251 252 CHAPTER 9 Linear Momentum and Collisions C onsider what happens when a golf ball is struck by a club. The ball is given a very large initial velocity as a result of the collision; consequently, it is able to travel more than 100 m through the air. The ball experiences a large acceleration. Furthermore, because the ball experiences this acceleration over a very short time interval, the average force exerted on it during the collision is very great. According to Newton's third law, the ball exerts on the club a reaction force that is equal in magnitude to and opposite in direction to the force exerted by the club on the ball. This reaction force causes the club to accelerate. Because the club is much more massive than the ball, however, the acceleration of the club is much less than the acceleration of the ball. One of the main objectives of this chapter is to enable you to understand and analyze such events. As a first step, we introduce the concept of momentum, which is useful for describing objects in motion and as an alternate and more general means of applying Newton's laws. For example, a very massive football player is often said to have a great deal of momentum as he runs down the field. A much less massive player, such as a halfback, can have equal or greater momentum if his speed is greater than that of the more massive player. This follows from the fact that momentum is defined as the product of mass and velocity. The concept of momentum leads us to a second conservation law, that of conservation of momentum. This law is especially useful for treating problems that involve collisions between objects and for analyzing rocket propulsion. The concept of the center of mass of a system of particles also is introduced, and we shall see that the motion of a system of particles can be described by the motion of one representative particle located at thntum of particle 2 is p2 m 2v2 . Note that F12 F21 . The total momentum of the system ptot is equal to the vector sum p1 p2 . this chapter, the terms momentum and linear momentum have the same meaning. Later, in Chapter 11, we shall use the term angular momentum when dealing with rotational motion. 254 CHAPTER 9 Linear Momentum and Collisions Because the time derivative of the total momentum ptot p1 p2 is zero, we conclude that the total momentum of the system must remain constant: ptot system p p1 p2 constant (9.4) or, equivalently, p1i p2i p1f p2f (9.5) where pli and p2i are the initial values and p1f and p2f the final values of the momentum during the time interval dt over which the reaction pair interacts. Equation 9.5 in component form demonstrates that the total momenta in the x, y, and z directions are all independently conserved: pix system system pf x system piy system pf y system piz system pf z (9.6) This result, known as the law of conservation of linear momentum, can be extended to any number of particles in an isolated system. It is considered one of the most important laws of mechanics. We can state it as follows: Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. This law tells us that the total momentum of an isolated system at all times equals its initial momentum. Notice that we have made no statement concerning the nature of the forces acting on the particles of the system. The only requirement is that the forces must be internal to the system. Conservation of momentum Quick Quiz 9.2 Your physical education teacher throws a baseball to you at a certain speed, and you catch it. The teacher is next going to throw you a medicine ball whose mass is ten times the mass of the baseball. You are given the following choices: You can have the medicine ball thrown with (a) the same speed as the baseball, (b) the same momentum, or (c) the same kinetic energy. Rank these choices from easiest to hardest to catch. EXAMPLE 9.1 The Floating Astronaut A SkyLab astronaut discovered that while concentrating on writing some notes, he had gradually floated to the middle of an open area in the spacecraft. Not wanting to wait until he floated to the opposite side, he asked his colleagues for a push. Laughing at his predicament, they decided not to help, and so he had to take off his uniform and throw it in one direction so that he would be propelled in the opposite direction. Estimate his resulting velocity. v1f v2f Solution We begin by making some reasonable guesses of relevant data. Let us assume we have a 70-kg astronaut who threw his 1-kg uniform at a speed of 20 m/s. For conve- Figure 9.2 somewhere. A hapless astronaut has discarded his uniform to get 9.2 nience, we set the positive direction of the x axis to be the direction of the throw (Fig. 9.2). Let us also assume that the x axis is tangent to the circular path of the spacecraft. We take the system to consist of the astronaut and the uniform. Because of the gravitational force (which keeps the astronaut, his uniform, and the entire spacecraft in orbit), the system is not really isolated. However, this force is directed perpendicular to the motion of the system. Therefore, momentum is constant in the x direction because there are no external forces in this direction. The total momentum of the system before the throw is zero (m1v1i m2v2i 0). Therefore, the total momentum after the throw must be zero; that is, m1v1f m2v2f 0 Impulse and Momentum 255 With m1 70 kg, v2f 20i m/s, and m2 1 kg, solving for v1f , we find the recoil velocity of the astronaut to be v1f m2 v m1 2f 1 kg (20i m/s) 70 kg 0.3i m/s The negative sign for v1f indicates that the astronaut is moving to the left after the throw, in the direction opposite the direction of motion of the uniform, in accordance with Newton's third law. Because the astronaut is much more massive than his uniform, his acceleration and consequent velocity are much smaller than the acceleration and velocity of the uniform. EXAMPLEof time for the two colliding particles described in Figure F21. 9.7a. Note that F12 260 CHAPTER 9 Linear Momentum and Collisions Momentum is conserved for any collision Therefore, we conclude that the total momentum of an isolated system just before a collision equals the total momentum of the system just after the collision. EXAMPLE 9.5 Carry Collision Insurance! the entangled cars is pf (m1 m2)vf (2 700 kg)vf A car of mass 1800 kg stopped at a traffic light is struck from the rear by a 900-kg car, and the two become entangled. If the smaller car was moving at 20.0 m/s before the collision, what is the velocity of the entangled cars after the collision? Solution We can guess that the final speed is less than 20.0 m/s, the initial speed of the smaller car. The total momentum of the system (the two cars) before the collision must equal the total momentum immediately after the collision because momentum is conserved in any type of collision. The magnitude of the total momentum before the collision is equal to that of the smaller car because the larger car is initially at rest: pi m1v1i (900 kg)(20.0 m/s) 1.80 104 kg m/s After the collision, the magnitude of the momentum of Equating the momentum before to the momentum after and solving for vf , the final velocity of the entangled cars, we have vf pi m1 m2 1.80 104 kg m/s 2 700 kg 6.67 m/s The direction of the final velocity is the same as the velocity of the initially moving car. Exercise What would be the final speed if the two cars each had a mass of 900 kg? Answer 10.0 m/s. Quick Quiz 9.5 As a ball falls toward the Earth, the ball's momentum increases because its speed increases. Does this mean that momentum is not conserved in this situation? Quick Quiz 9.6 A skater is using very low-friction rollerblades. A friend throws a Frisbee straight at her. In which case does the Frisbee impart the greatest impulse to the skater: (a) she catches the Frisbee and holds it, (b) she catches it momentarily but drops it, (c) she catches it and at once throws it back to her friend? When the bowling ball and pin collide, part of the ball's momentum is transferred to the pin. Consequently, the pin acquires momentum and kinetic energy, and the ball loses momentum and kinetic energy. However, the total momentum of the system (ball and pin) remains constant. 9.4 ELASTIC AND INELASTIC COLLISIONS IN ONE DIMENSION Elastic collision As we have seen, momentum is conserved in any collision in which external forces are negligible. In contrast, kinetic energy may or may not be constant, depending on the type of collision. In fact, whether or not kinetic energy is the same before and after the collision is used to classify collisions as being either elastic or inelastic. An elastic collision between two objects is one in which total kinetic energy (as well as total momentum) is the same before and after the collision. Billiard-ball collisions and the collisions of air molecules with the walls of a container at ordinary temperatures are approximately elastic. Truly elastic collisions do occur, however, between atomic and subatomic particles. Collisions between certain objects in the macroscopic world, such as billiard-ball collisions, are only approximately elastic because some deformation and loss of kinetic energy take place. 9.4 Elastic and Inelastic Collisions in One Dimension 261 An inelastic collision is one in which total kinetic energy is not the same before and after the collision (even though momentum is constant). Inelastic collisions are of two types. When the colliding objects stick together after the collision, as happens when a meteorite collides with the Earth, the collision is called perfectly inelastic. When the colliding objects do not stick together, but some kinetic energy is lost, as in the case of a rubber ball colliding with a hard surface, the collision is called inelastic (with no modifying adverb). For example, when a rubber ball collides with a hard surface, the collision is inelastic because some of the kinetic energy of the ball is lost when the ball is deformed while it is in contact with the surface. In most collisions, kinetic energy is not the same before and after the collision because some of it is converted to internal energy, to elastic potential energy when the objects are deformed, and to rotational energy. Elastic and perfectly inelastic collisions are limiting cases; most collisions fall somewhere between them. In the remainder of this section, we treat collisions in one dimension and consider the two extreme cases -- perfectly inelastic and elastic collisions. The important distinction between these two types of collisions is that momentum is constant in all collisions, but kinetic energy is constant only in elastic collisions. Inelastic collision QuickLab Hold a Ping-Pong ball or tennis ball on top of a basketball. Drop them both at the same time so that the basketball hits the floor, bounces up, and hits the smaller falling ball. What happens and why? Perfectly Inelastic Collisions Consider two particles of masses m1 and m 2 moving with initial velocities v1i and v2i along a straight line, as shown in Figure 9.9. The two particles collide head-on, stick together, and then move with some common velocity vf after the collision. Because momentum is conserved in any collision, we can say that the total momentum before the collision equals the total momentum of the composite system after the collision: m1v1i m2v2i vf (m1 m1v1i m1 m2)vf m2v2i m2 (9.13) (9.14) m1 Before collision m2 v1i (a) v2i Quick Quiz 9.7 Which is worse, crashing into a brick wall at 40 mi/h or crashing head-on into an oncoming car that is identical to yours and also moving at 40 mi/h? After collision vf m1 + m2 (b) Elastic Collisions 6.6 Now consider two particles that undergo an elastic head-on collision (Fig. 9.10). In this case, both momentum and kinetic energy are conserved; therefore, we have m1v1i 1 2 2 m1v1i Figure 9.9 Schematic representation of a perfectly inelastic head-on collision between two particles: (a) before collision and (b) after collision. m2v2i 1 2 2 m2v2i m1v1f 1 2 2 m1v1f m2v2f 1 2 2 m2v2f (9.15) (9.16) Because all velocities in Figure 9.10 are either to the left or the right, they can be represented by the corresponding speeds along with algebraic signs indicating direction. We shall indicate v as positive if a particle moves to the right and negative 262 Before collision m1 v1i (a) v2i m2 CHAPTER 9 Linear Momentum and Collisions After collision v1f (b) v2f if it moves to the left. As has been seen in earlier chapters, it is common practice to call these values "speed" even though this term technically refers to the magnitude of the velocity vector, which does not have an algebraic sign. In a typical problem involving elastic collisions, there are two unknown quantities, and Equations 9.15 and 9.16 can be solved simultaneously to find these. An alternative approach, however -- one that involves a little mathematical manipulation of Equation 9.16 -- often simplifies this process. To see how, let us cancel the factor 1 in Equation 9.16 and rewrite it as 2 m1(v1i2 and then factor both sides: m1(v1i v1f )(v1i v1f ) m2(v2f v2i )(v2f v2i ) (9.17) v1f 2 ) m2(v2f 2 v2i2 ) Schematic representation of an elastic head-on collision between two particles: (a) before collision and (b) after collision. Figure 9.10 Next, let us separate the terms containing m1 and m 2 in Equation 9.15 to get m1(v1i v1f ) m2(v2f v2i ) (9.18) To obtain our final result, we divide Equation 9.17 by Equation 9.18 and get v1i v1i v1f v2i v2f (v1f v2i v2f ) (9.19) This equation, in combination with Equation 9.15, can be used to solve problems dealing with elastic collisions. According to Equation 9.19, the relative speed of the two particles before the collision v1i v2i equals the negative of their relative speed after the collision, (v1f v2f ). Suppose that the masses and initial velocities of both particles are known. Equations 9.15 and 9.19 can be solved for the final speeds in terms of the initial speeds because there are two equations and two unknowns: v1f Elasticpt to a system of many particles in three dimensions. The x coordinate of the center of mass of n particles is defined to be xCM m1x1 m2 x2 m1 m2 m3x3 m3 mn xn mn i mi xi i mi (9.28) where xi is the x coordinate of the ith particle. For convenience, we express the total mass as M mi , where the sum runs over all n particles. The y and z coordii nates of the center of mass are similarly defined by the equations yCM i mi yi M and zCM i mi zi M (9.29) The center of mass can also be located by its position vector, rCM . The cartesian coordinates of this vector are x CM , y CM , and z C M , defined in Equations 9.28 and 9.29. Therefore, rCM xCMi yCM j zCMk mi xi i mi yi j mi zi k i i i M Vector position of the center of mass for a system of particles rCM i mi ri M (9.30) where ri is the position vector of the ith particle, defined by y mi CM ri rCM x ri xi i yi j zi k z Although locating the center of mass for an extended object is somewhat more cumbersome than locating the center of mass of a system of particles, the basic ideas we have discussed still apply. We can think of an extended object as a system containing a large number of particles (Fig. 9.19). The particle separation is very small, and so the object can be considered to have a continuous mass distribution. By dividing the object into elements of mass mi , with coordinates xi , yi , zi , we see that the x coordinate of the center of mass is approximately xCM i Figure 9.19 An extended object can be considered a distribution of small elements of mass mi . The center of mass is located at the vector position rCM , which has coordinates x CM , y CM , and z CM . xi mi M with similar expressions for y CM and z CM . If we let the number of elements n approach infinity, then x CM is given precisely. In this limit, we replace the sum by an 9.6 The Center of Mass 271 integral and mi by the differential element dm: xCM lim i xi mi M mi :0 1 M x dm (9.31) Likewise, for y CM and z CM we obtain yCM 1 M y dm and zCM 1 M C z dm (9.32) A We can express the vector position of the center of mass of an extended object in the form rCM 1 M r dm (9.33) B C which is equivalent to the three expressions given by Equations 9.31 and 9.32. The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry.4 For example, the center of mass of a rod lies in the rod, midway between its ends. The center of mass of a sphere or a cube lies at its geometric center. One can determine the center of mass of an irregularly shaped object by suspending the object first from one point and then from another. In Figure 9.20, a wrench is hung from point A, and a vertical line AB (which can be established with a plumb bob) is drawn when the wrench has stopped swinging. The wrench is then hung from point C, and a second vertical line CD is drawn. The center of mass is halfway through the thickness of the wrench, under the intersection of these two lines. In general, if the wrench is hung freely from any point, the vertical line through this point must pass through the center of mass. Because an extended object is a continuous distribution of mass, each small mass element is acted upon by the force of gravity. The net effect of all these forces is equivalent to the effect of a single force, Mg, acting through a special point, called the center of gravity. If g is constant over the mass distribution, then the center of gravity coincides with the center of mass. If an extended object is pivoted at its center of gravity, it balances in any orientation. A Center of mass D B Figure 9.20 An experimental technique for determining the center of mass of a wrench. The wrench is hung freely first from point A and then from point C. The intersection of the two lines AB and CD locates the center of mass. Quick Quiz 9.9 If a baseball bat is cut at the location of its center of mass as shown in Figure 9.21, do the two pieces have the same mass? QuickLab Cut a triangle from a piece of cardboard and draw a set of adjacent strips inside it, parallel to one of the sides. Put a dot at the approximate location of the center of mass of each strip and then draw a straight line through the dots and into the angle opposite your starting side. The center of mass for the triangle must lie on this bisector of the angle. Repeat these steps for the other two sides. The three angle bisectors you have drawn will intersect at the center of mass of the triangle. If you poke a hole anywhere in the triangle and hang the cardboard from a string attached at that hole, the center of mass will be vertically aligned with the hole. Figure 9.21 A baseball bat cut at the location of its center of mass. 4This statement is valid only for objects that have a uniform mass per unit volume. 272 CHAPTER 9 Linear Momentum and Collisions EXAMPLE 9.12 The Center of Mass of Three Particles A system consists of three particles located as shown in Figure 9.22a. Find the center of mass of the system. y(m) 3 Solution We set up the problem by labeling the masses of the particles as shown in the figure, with m1 m2 1.0 kg and m3 2.0 kg. Using the basic defining equations for the coordinates of the center of mass and noting that zCM 0, we obtain xCM m1x1 m2x2 m3x3 i M m1 m2 m3 (1.0 kg)(1.0 m) (1.0 kg)(2.0 m) (2.0 kg)(0 m) 1.0 kg 1.0 kg 2.0 kg 3.0 kg m 0.75 m 4.0 kg miyi my my my i 1 1 2 2 3 3 2 mixi m3 1 rCM yCM m1 0 1 (a) m2 2 3 x(m) M m1 m2 m3 (1.0 kg)(0) (1.0 kg)(0) (2.0 kg)(2.0 m) 4.0 kg 4.0 kg m 1.0 m 4.0 kg The position vector to the center of mass measured from the origin is therefore rCM xCM i yCM j 0.75i m 1.0 j m MrCM m3r3 rCM We can verify this result graphically by adding together m1r1 m2r2 m3r3 and dividing the vector sum by M, the total mass. This is shown in Figure 9.22b. Figure 9.22 (a) Two 1-kg masses and a single 2-kg mass are located as shown. The vector indicates the location of the system's center of mass. (b) The vector sum of m i ri . m1r1 m2r2 (b) EXAMPLE 9.13 The Center of Mass of a Rod Because M/L, this reduces to xCM L2 2M M L L 2 (a) Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length. Solution The rod is shown aligned along the x axis in Figure 9.23, so that yCM zCM 0. Furthermore, if we call the mass per unit length (this quantity is called the linear mass density), then M/L for the uniform rod we assume here. If we divide the rod into elements of length dx, then the mass of each element is dm dx. For an arbitrary element located a distance x from the origin, Equation 9.31 gives xCM 1 M x dm 1 M L One can also use symmetry arguments to obtain the same result. (b) Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to the expression x, where is a constant. Find the x coordinate of the center of mass as a fraction of L. x dx 0 x2 M 2 L 0 L2 2M Solution In this case, we replace dm by dx where constant. Therefore, x CM is is not 9.7 L L Motion of a System of Particles 273 xCM 1 M M 0 x dm L 1 M x dx 0 L3 1 M x x dx 0 x2 dx 3M y dm = dx L O x dx x We can eliminate by noting that the total mass of the rod is related to through the relationship L L M dm 0 dx 0 x dx L2 2 Substituting this into the expression for x CM gives xCM L3 3 L2/2 2 L 3 Figure 9.23 cated at x CM The center of mass of a uniform rod of length L is loL/2. EXAMPLE 9.14 The Center of Mass of a Right Triangle With this substitution, x CM becomes xCM 2 ab a a An object of mass M is in the shape of a right triangle whose dimensions are shown in Figure 9.24. Locate the coordinates of the center of mass, assuming the object has a uniform mass per unit area. By inspection we can estimate that the x coordinate of the center of mass must be past the center of the base, that is, greater than a/2, because the largest part of the triangle lies beyond that point. A similar argument indicates that its y coordinate must be less than b/2. To evaluate the x coordinate, we divide the triangle into narrow strips of width dx and height y as in Figure 9.24. The mass dm of each strip is dm total mass of object total area of object M (y dx) 1/2ab area of strip x 0 b x dx a 2 a2 x2 dx 0 2 a2 x3 3 a 0 Solution 2 a 3 By a similar calculation, we get for the y coordinate of the center of mass yCM 1 b 3 These values fit our original estimates. 2M y dx ab y dm Therefore, the x coordinate of the center of mass is xCM 1 M x dm 1 M a x 0 2M y dx ab 2 ab a c xy dx 0 b y dx x To evaluate this integral, we must express y in terms of x. From similar triangles in Figure 9.24, we see that y x b a or y b x a O x a Figure 9.24 9.7 6.8 MOTION OF A SYSTEM OF PARTICLES We can begin to understand the physical significance and utility of the center of mass concept by taking the time derivative of the position vector given by Equation 9.30. From Section 4.1 we know that the time derivative of a position vector is by 274 CHAPTER 9 Linear Momentum and Collisions Velocity of the center of mass definition a velocity. Assuming M remains constant for a system of particles, that is, no particles enter or leave the system, we get the following expression for the velocity of the center of mass of the system: mivi drCM 1 dr i (9.34) vCM mi i dt M i dt M where vi is the velocity of the ith particle. Rearranging Equation 9.34 gives Total momentum of a system of particles MvCM i mi vi i pi ptot (9.35) Therefore, we conclude that the total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass. In other words, the total linear momentum of the system is equal to that of a single particle of mass M moving with a velocity vCM . If we now differentiate Equation 9.34 with respect to time, we get the acceleration of the center of mass of the system: Acceleration of the center of mass aCM d vCM dt 1 M mi i dvi dt 1 M mi a i i (9.36) Rearranging this expression and using Newton's second law, we obtain MaCM i mi a i i Fi (9.37) where Fi is the net force on particle i. The forces on any particle in the system may include both external forces (from outside the system) and internal forces (from within the system). However, by Newton's third law, the internal force exerted by particle 1 on particle 2, for example, is equal in magnitude and opposite in direction to the internal force exerted by particle 2 on particle 1. Thus, when we sum over all internal forces in Equation 9.37, they cancel in pairs and the net force on the system is caused only by external forces. Thus, we can write Equation 9.37 in the form Newton's second law for a system of particles Fext MaCM dptot dt (9.38) That is, the resultant external force on a system of particles equals the total mass of the system multiplied by the acceleration of the center of mass. If we compare this with Newton's second law for a single particle, we see that The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the resultant external force on the system. Finally, we see that if the resultant external force is zero, then from Equation 9.38 it follows that dptot dt MaCM 0 9.7 Motion of a System of Particles 275 Figure 9.25 Multiflash photograph showing an overhead view of a wrench moving on a horizontal surface. The center of mass of the wrench moves in a straight line as the wrench rotates about this point, shown by the white dots. so that ptot MvCM constant (when Fext 0) (9.39) That is, the total linear momentum of a system of particles is conserved if no net external force is acting on the system. It follows that for an isolated system of particles, both the total momentum and the velocity of the center of mass are constant in time, as shown in Figure 9.25. This is a generalization to a many-particle system of the law of conservation of momentum discussed in Section 9.1 for a two-particle system. Suppose an isolated system consisting of two or more members is at rest. The center of mass of such a system remains at rest unless acted upon by an external force. For example, consider a system made up of a swimmer standing on a raft, with the system initially at rest. When the swimmer dives horizontally off the raft, the center of mass of the system remains at rest (if we neglect friction between raft and water). Furthermore, the linear momentum of the diver is equal in magnitude to that of the raft but opposite in direction. As another example, suppose an unstable atom initially at rest suddenly breaks up into two fragments of masses MA and MB , with velocities vA and vB , respectively. Because the total momentum of the system before the breakup is zero, the total momentum of the system after the breakup must also be zero. Therefore, MAvA MBvB 0. If the velocity of one of the fragments is known, the recoil velocity of the other fragment can be calculated. EXAMPLE 9.15 The Sliding Bear noting your location. Take off your spiked shoes and pull on the rope hand over hand. Both you and the bear will slide over the ice until you meet. From the tape, observe how far you have slid, xp , and how far the bear has slid, xb . The point where you meet the bear is the constant location of the center of mass of the system (bear plus you), and so you can determine the mass of the bear from m b x b m p x p . (Unfortunately, you cannot get back to your spiked shoes and so are in big trouble if the bear wakes up!) Suppose you tranquilize a polar bear on a smooth glacier as part of a research effort. How might you estimate the bear's mass using a measuring tape, a rope, and knowledge of your own mass? Solution Tie one end of the rope around the bear, and then lay out the tape measure on the ice with one end at the bear's original position, as shown in Figure 9.26. Grab hold of the free end of the rope and position yourself as shown, 276 CHAPTER 9 Linear Momentum and Collisions xp CM xb Figure 9.26 The center of mass of an isolated system remains at rest unless acted on by an external force. How can you determine the mass of the polar bear? CONCEPTUAL EXAMPLE 9.16 Exploding Projectile the center of mass of the system made up of all the fragments after the explosion? A projectile fired into the air suddenly explodes into several fragments (Fig. 9.27). What can be said about the motion of Motion of center of mass Solution Neglecting air resistance, the only external force on the projectile is the gravitational force. Thus, if the projectile did not explode, it would continue to move along the parabolic path indicated by the broken line in Figure 9.27. Because the forces caused by the explosion are internal, they do not affect the motion of the center of mass. Thus, after the explosion the center of mass of the system (the fragments) follows the same parabolic path the projectile would have followed if there had been no explosion. Figure 9.27 When a projectile explodes into several fragments, the center of mass of the system made up of all the fragments follows the same parabolic path the projectile would have taken had there been no explosion. EXAMPLE 9.17 The Exploding Rocket A rocket is fired vertically upward. At the instant it reaches an altitude of 1 000 m and a speed of 300 m/s, it explodes into three equal fragments. One fragment continues to move upward with a speed of 450 m/s following the explosion. The second fragment has a speed of 240 m/s and is moving east right after the explosion. What is the velocity of the third fragment right after the explosion? Solution Let us call the total mass of the rocket M; hence, the mass of each fragment is M/3. Because the forces of the explosion are internal to the system and cannot affect its total momentum, the total momentum pi of the rocket just before the explosion must equal the total momentum pf of the fragments right after the explosion. 9.8 Before the explosion: pi After the explosion: pf M (240 i) m/s 3 M (450 j) m/s 3 M v 3 f M vi M(300 j) m/s Rocket Propulsion 277 vf ( 240i 450j) m/s What does the sum of the momentum vectors for all the fragments look like? Exercise where vf is the unknown velocity of the third fragment. Equating these two expressions (because pi pf ) gives M v 3 f M(80 i) m/s M(150 j) m/s M(300 j) m/s Find the position of the center of mass of the system of fragments relative to the ground 3.00 s after the explosion. Assume the rocket engine is nonoperative after the explosion. The x coordinate does not change; yCM 1.86 km. Answer Optional Section 9.8 ROCKET PROPULSION When ordinary vehicles, such as automobiles and locomotives, are propelled, the driving force for the motion is friction. In the case of the automobile, the driving force is the force exerted by the road on the car. A locomotive "pushes" against the tracks; hence, the driving force is the force exerted by the tracks on the locomotive. However, a rocket moving in space has no road or tracks to push against. Therefore, the source of the propulsion of a rocket must be something other than friction. Figure 9.28 is a dramatic photograph of a spacecraft at liftoff. The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus its ejected fuel. Rocket propulsion can be understood by first considering the mechanical system consisting of a machine gun mounted on a cart on wheels. As the gun is fired, Figure 9.28 Liftoff of the space shuttle Columbia. Enormous thrust is generated by the shuttle's liquid-fuel engines, aided by the two solid-fuel boosters. Many physical principles from mechanics, thermodynamics, and electricity and magnetism are involved in such a launch. 278 CHAPTER 9 Linear Momentum and Collisions The force from a nitrogen-propelled, hand-controlled device allows an astronaut to move about freely in space without restrictive tethers. v M + m each bullet receives a momentum mv in some direction, where v is measured with respect to a stationary Earth frame. The momentum of the system made up of cart, gun, and bullets must be conserved. Hence, for each bullet fired, the gun and cart must receive a compensating momentum in the opposite direction. That is, the reaction force exerted by the bullet on the gun accelerates the cart and gun, and the cart moves in the direction opposite that of the bullets. If n is the number of bullets fired each second, then the average force exerted on the gun is Fav nmv. In a similar manner, as a rocket moves in free space, its linear momentum changes when some of its mass is released in the form of ejected gases. Because the gases are given momentum when they are ejected out of the engine, the rocket receives a compensating momentum in the opposite direction. Therefore, the rocket is accelerated as a result of the "push," or thrust, from the exhaust gases. In free space, the center of mass of the system (rocket plus expelled gases) moves uniformly, independent of the propulsion process.5 Suppose that at some time t, the magnitude of the momentum of a rocket plus its fuel is (M m)v, where v is the speed of the rocket relative to the Earth (Fig. 9.29a). Over a short time interval t, the rocket ejects fuel of mass m, and so at the end of this interval the rocket's speed is v v, where v is the change in speed of the rocket (Fig. 9.29b). If the fuel is ejected with a speed ve relative to the rocket (the subscript "e" stands for exhaust, and ve is usually called the exhaust speed), the velocity of the fuel relative to a stationary frame of reference is v ve . Thus, if we equate the total initial momentum of the system to the total final momentum, we obtain (M m)v M(v v) m(v ve) pi = (M + m)v (a) where M represents the mass of the rocket and its remaining fuel after an amount of fuel having mass m has been ejected. Simplifying this expression gives M v ve m M m v + v (b) Figure 9.29 Rocket propulsion. (a) The initial mass of the rocket plus all its fuel is M m at a time t, and its speed is v. (b) At a time t t, the rocket's mass has been reduced to M and an amount of fuel m has been ejected. The rocket's speed increases by an amount v. We also could have arrived at this result by considering the system in the center-of-mass frame of reference, which is a frame having the same velocity as the center of mass of the system. In this frame, the total momentum of the system is zero; therefore, if the rocket gains a momentum M v by ejecting some fuel, the exhausted fuel obtains a momentum ve m in the opposite direction, so that M v ve m 0. If we now take the limit as t goes to zero, we get v : dv and m : dm. Futhermore, the increase in the exhaust mass dm corresponds to an equal decrease in the rocket mass, so that dm dM. Note that dM is given a negative sign because it represents a decrease in mass. Using this fact, we obtain M dv ve dm ve dM (9.40) Integrating this equation and taking the initial mass of the rocket plus fuel to be Mi and the final mass of the rocket plus its remaining fuel to be Mf , we obtain vf Mf dv vi ve Mi dM M (9.41) Expression for rocket propulsion vf vi ve ln Mi Mf 5It is interesting to note that the rocket and machine gun represent cases of the reverse of a perfectly inelastic collision: Momentum is conserved, but the kinetic energy of the system increases (at the expense of chemical potential energy in the fuel). 9.8 Rocket Propulsion 279 This is the basic expression of rocket propulsion. First, it tells us that the increase in rocket speed is proportional to the exhaust speed of the ejected gases, ve . Therefore, the exhaust speed should be very high. Second, the increase in rocket speed is proportional to the natural logarithm of the ratio Mi /Mf . Therefore, this ratio should be as large as possible, which means that the mass of the rocket without its fuel should be as small as possible and the rocket should carry as much fuel as possible. The thrust on the rocket is the force exerted on it by the ejected exhaust gases. We can obtain an expression for the thrust from Equation 9.40: Thrust M dv dt ve dM dt (9.42) This expression shows us that the thrust increases as the exhaust speed increases and as the rate of change of mass (called the burn rate) increases. EXAMPLE 9.18 A Rocket in Space 3.0 6.5 103 m/s 103 m/s (5.0 103 m/s)ln Mi 0.5 Mi A rocket moving in free space has a speed of 3.0 103 m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket's motion at a speed of 5.0 103 m/s relative to the rocket. (a) What is the speed of the rocket relative to the Earth once the rocket's mass is reduced to one-half its mass before ignition? Solution We can guess that the speed we are looking for must be greater than the original speed because the rocket is accelerating. Applying Equation 9.41, we obtain vf vi Mi ve ln Mf (b) What is the thrust on the rocket if it burns fuel at the rate of 50 kg/s? Solution Thrust ve dM dt (5.0 105 N 103 m/s)(50 kg/s) 2.5 EXAMPLE 9.19 Fighting a Fire their hands, the movement of the hose due to the thrust it receives from the rapidly exiting water could injure the firefighters. Two firefighters must apply a total force of 600 N to steady a hose that is discharging water at 3 600 L/min. Estimate the speed of the water as it exits the nozzle. Solution The water is exiting at 3 600 L/min, which is 60 L/s. Knowing that 1 L of water has a mass of 1 kg, we can say that about 60 kg of water leaves the nozzle every second. As the water leaves the hose, it exerts on the hose a thrust that must be counteracted by the 600-N force exerted on the hose by the firefighters. So, applying Equation 9.42 gives Thrust 600 N ve ve dM dt ve(60 kg/s) 10 m/s Firefighters attack a burning house with a hose line. Firefighting is dangerous work. If the nozzle should slip from 280 CHAPTER 9 Linear Momentum and Collisions SUMMARY The linear momentum p of a particle of mass m moving with a velocity v is p mv (9.1) The law of conservation of linear momentum indicates that the total momentum of an isolated system is conserved. If two particles form an isolated system, their total momentum is conserved regardless of the nature of the force between them. Therefore, the total momentum of the system at all times equals its initial ens to the momentum of the mud? Is momentum conserved? Explain. 17. Early in this century, Robert Goddard proposed sending a rocket to the Moon. Critics took the position that in a vacuum, such as exists between the Earth and the Moon, the gases emitted by the rocket would have nothing to push against to propel the rocket. According to Scientific American ( January 1975), Goddard placed a gun in a vacuum and fired a blank cartridge from it. (A blank cartridge fires only the wadding and hot gases of the burning gunpowder.) What happened when the gun was fired? 18. A pole-vaulter falls from a height of 6.0 m onto a foam rubber pad. Can you calculate his speed just before he reaches the pad? Can you estimate the force exerted on him due to the collision? Explain. 19. Explain how you would use a balloon to demonstrate the mechanism responsible for rocket propulsion. 20. Does the center of mass of a rocket in free space accelerate? Explain. Can the speed of a rocket exceed the exhaust speed of the fuel? Explain. 21. A ball is dropped from a tall building. Identify the system for which linear momentum is conserved. 22. A bomb, initially at rest, explodes into several pieces. (a) Is linear momentum conserved? (b) Is kinetic energy conserved? Explain. 23. NASA often uses the gravity of a planet to "slingshot" a probe on its way to a more distant planet. This is actually a collision where the two objects do not touch. How can the probe have its speed increased in this manner? 24. The Moon revolves around the Earth. Is the Moon's linear momentum conserved? Is its kinetic energy conserved? Assume that the Moon's orbit is circular. 25. A raw egg dropped to the floor breaks apart upon impact. However, a raw egg dropped onto a thick foam rubber cushion from a height of about 1 m rebounds without breaking. Why is this possible? (If you try this experiment, be sure to catch the egg after the first bounce.) 26. On the subject of the following positions, state your own view and argue to support it: (a) The best theory of motion is that force causes acceleration. (b) The true measure of a force's effectiveness is the work it does, and the best theory of motion is that work on an object changes its energy. (c) The true measure of a force's effect is impulse, and the best theory of motion is that impulse injected into an object changes its momentum. 282 CHAPTER 9 Linear Momentum and Collisions PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 9.1 Linear Momentum and Its Conservation 1. A 3.00-kg particle has a velocity of (3.00i 4.00j) m/s. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum. 2. A 0.100-kg ball is thrown straight up into the air with an initial speed of 15.0 m/s. Find the momentum of the ball (a) at its maximum height and (b) halfway up to its maximum height. 3. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction between child and ice, find the recoil velocity of the child. 4. A pitcher claims he can throw a baseball with as much momentum as a 3.00-g bullet moving with a speed of 1 500 m/s. A baseball has a mass of 0.145 kg. What must be its speed if the pitcher's claim is valid? 5. How fast can you set the Earth moving? In particular, when you jump straight up as high as you can, you give the Earth a maximum recoil speed of what order of magnitude? Model the Earth as a perfectly solid object. In your solution, state the physical quantities you take as data and the values you measure or estimate for them. 6. Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.6). A cord initially holding the blocks together ison? (b) How much mechanical energy is lost in the collision? Account for this loss in energy. 22. A railroad car of mass 2.50 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much energy is lost in the collision? Figure P9.13 14. A professional diver performs a dive from a platform 10 m above the water surface. Estimate the order of magnitude of the average impact force she experiences in her collision with the water. State the quantities you take as data and their values. Section 9.3 Collisions Section 9.4 Elastic and Inelastic Collisions in One Dimension 15. High-speed stroboscopic photographs show that the head of a golf club of mass 200 g is traveling at 55.0 m/s just before it strikes a 46.0-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact. 16. A 75.0-kg ice skater, moving at 10.0 m/s, crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 5.00 m/s. Suppose the average force a skater can experience without breaking a bone is 4 500 N. If the impact time is 0.100 s, does a bone break? 17. A 10.0-g bullet is fired into a stationary block of wood (m 5.00 kg). The relative motion of the bullet stops 284 25.0 m/s CHAPTER 9 20.0 m/s BIG Linear Momentum and Collisions 18.0 m/s v BIG Joes IRISH BEER Joes IRISH BEER Before After Figure P9.21 23. Four railroad cars, each of mass 2.50 104 kg, are coupled together and coasting along horizontal tracks at a speed of vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving toward the south, now at 2.00 m/s. (a) Find the initial speed of the cars. (b) How much work did the actor do? (c) State the relationship between the process described here and the process in Problem 22. 24. A 7.00-kg bowling ball collides head-on with a 2.00-kg bowling pin. The pin flies forward with a speed of 3.00 m/s. If the ball continues forward with a speed of 1.80 m/s, what was the initial speed of the ball? Ignore rotation of the ball. 25. A neutron in a reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is 1.60 10 13 J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is about 12.0 times greater than the mass of the neutron.) 26. Consider a frictionless track ABC as shown in Figure P9.26. A block of mass m 1 5.00 kg is released from A. It makes a head-on elastic collision at B with a block of mass m 2 10.0 kg that is initially at rest. Calculate the maximum height to which m 1 rises after the collision. 0.650, what was the speed of the bullet immediately before impact? 28. A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, would penetrate the block to a depth of 8.00 cm. This block of wood is placed on a frictionless horizontal surface, and a 7.00-g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case? Section 9.5 Two-Dimensional Collisions WEB 29. A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. If the collision is perfectly inelastic, (a) calculate the speed and direction of the players just after the tackle and (b) determine the energy lost as a result of the collision. Account for the missing energy. 30. The mass of the blue puck in Figure P9.30 is 20.0% greater than the mass of the green one. Before colliding, the pung it back. Since this is when the skater imparts the greatest impulse to the Frisbee, then this also is when the Frisbee imparts the greatest impulse to her. 9.7 Both are equally bad. Imagine watching the collision from a safer location alongside the road. As the "crush zones" of the two cars are compressed, you will see that 291 the actual point of contact is stationary. You would see the same thing if your car were to collide with a solid wall. 9.8 No, such movement can never occur if we assume the collisions are elastic. The momentum of the system before the collision is mv, where m is the mass of ball 1 and v is its speed just before the collision. After the collision, we would have two balls, each of mass m and moving with a speed of v/2. Thus, the total momentum of the system after the collision would be m(v/2) m(v/2) mv. Thus, momentum is conserved. However, the kinetic energy just before the collision is K i 1 mv 2, and that 2 after the collision is K f 1 m(v/2)2 1 m(v/2)2 1mv 2. 2 2 4 Thus, kinetic energy is not conserved. Both momentum and kinetic energy are conserved only when one ball moves out when one ball is released, two balls move out when two are released, and so on. 9.9 No they will not! The piece with the handle will have less mass than the piece made up of the end of the bat. To see why this is so, take the origin of coordinates as the center of mass before the bat was cut. Replace each cut piece by a small sphere located at the center of mass for each piece. The sphere representing the handle piece is farther from the origin, but the product of lesser mass and greater distance balances the product of greater mass and lesser distance for the end piece: P U Z Z L E R Did you know that the CD inside this player spins at different speeds, depending on which song is playing? Why would such a strange characteristic be incorporated into the design of every CD player? (George Semple) c h a p t e r Rotation of a Rigid Object About a Fixed Axis Chapter Outline 10.1 Angular Displacement, Velocity, and Acceleration 10.5 Calculation of Moments of Inertia 10.2 Rotational Kinematics: Rotational Motion with Constant Angular Acceleration 10.6 Torque 10.7 Relationship Between Torque and Angular Acceleration 10.3 Angular and Linear Quantities 10.4 Rotational Energy 292 10.8 Work, Power, and Energy in Rotational Motion 10.1 Angular Displacement, Velocity, and Acceleration 293 W hen an extended object, such as a wheel, rotates about its axis, the motion cannot be analyzed by treating the object as a particle because at any given time different parts of the object have different linear velocities and linear accelerations. For this reason, it is convenient to consider an extended object as a large number of particles, each of which has its own linear velocity and linear acceleration. In dealing with a rotating object, analysis is greatly simplified by assuming that the object is rigid. A rigid object is one that is nondeformable -- that is, it is an object in which the separations between all pairs of particles remain constant. All real bodies are deformable to some extent; however, our rigid-object model is useful in many situations in which deformation is negligible. In this chapter, we treat the rotation of a rigid object about a fixed axis, which is commonly referred to as pure rotational motion. Rigid object 10.1 ANGULAR DISPLACEMENT, VELOCITY, AND ACCELERATION y Figure 10.1 illustrates a planar (flat), rigid object of arbitrary shape confined to the xy plane and rotating about a fixed axis through O. The axis is perpendicular to the plane of the figure, and O is the origin of an xy coordinate system. Let us look at the motion of only one of the millions of "particles" making up this object. A particle at P is at a fixed distance r from the origin and rotates about it in a circle of radius r. (In fact, every particle on the object undergoes circular motion about O.) It is convenient to represent the position of P with its polar coordinates (r, ), where r is the distance from the origin to P and ear speed and acceleration of an arbitrary point in the object. To do so, we must keep in mind that when a rigid object rotates about a fixed axis, as in Figure 10.4, every particle of the object moves in a circle whose center is the axis of rotation. We can relate the angular speed of the rotating object to the tangential speed of a point P on the object. Because point P moves in a circle, the linear velocity vector v is always tangent to the circular path and hence is called tangential velocity. The magnitude of the tangential velocity of the point P is by definition the tangential speed v ds/dt, where s is the distance traveled by this point measured along the circular path. Recalling that s r (Eq. 10.1a) and noting that r is constant, we obtain v Because d /dt ds dt r d dt O x Figure 10.4 As a rigid object rotates about the fixed axis through O, the point P has a linear velocity v that is always tangent to the circular path of radius r. (see Eq. 10.4), we can say v r (10.10) Relationship between linear and angular speed That is, the tangential speed of a point on a rotating rigid object equals the perpendicular distance of that point from the axis of rotation multiplied by the angular speed. Therefore, although every point on the rigid object has the same angular speed, not every point has the same linear speed because r is not the same for all points on the object. Equation 10.10 shows that the linear speed of a point on the rotating object increases as one moves outward from the center of rotation, as we would intuitively expect. The outer end of a swinging baseball bat moves much faster than the handle. We can relate the angular acceleration of the rotating rigid object to the tangential acceleration of the point P by taking the time derivative of v: at at dv dt r r d dt (10.11) QuickLab Spin a tennis ball or basketball and watch it gradually slow down and stop. Estimate and at as accurately as you can. Relationship between linear and angular acceleration That is, the tangential component of the linear acceleration of a point on a rotating rigid object equals the point's distance from the axis of rotation multiplied by the angular acceleration. 298 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis y at a ar x P In Section 4.4 we found that a point rotating in a circular path undergoes a centripetal, or radial, acceleration ar of magnitude v 2/r directed toward the center of rotation (Fig. 10.5). Because v r for a point P on a rotating object, we can express the radial acceleration of that point as ar v2 r r 2 (10.12) O The total linear acceleration vector of the point is a at ar . (at describes the change in how fast the point is moving, and ar represents the change in its direction of travel.) Because a is a vector having a radial and a tangential component, the magnitude of a for the point P on the rotating rigid object is a a t 2 a r2 r 2 2 r2 4 r 2 4 (10.13) As a rigid object rotates about a fixed axis through O, the point P experiences a tangential component of linear acceleration at and a radial component of linear acceleration ar . The total linear acceleration of this point is a a t ar . Figure 10.5 Quick Quiz 10.2 When a wheel of radius R rotates about a fixed axis, do all points on the wheel have (a) the same angular speed and (b) the same linear speed? If the angular speed is constant and equal to , describe the linear speeds and linear accelerations of the points located at (c) r 0, (d) r R/2, and (e) r R, all measured from the center of the wheel. EXAMPLE 10.2 CD Player 1 (56.5 rad/s) 2 rev/rad (60 s/min) 5.4 10 2 rev/min On a compact disc, audio information is stored in a series of pits and flat areas on the surface of the disc. The information is stored digitally, and the alternations between pits and flat areas on the surface represent binary ones and zeroes to be read by the compact disc player and converted back to sound waves. The pits and flat areas are detected by a system consisting of a laser and lenses. The length of a certain number of ones and zeit is common to every particle. Figure 10.7 A rigid object rotating about a z axis with angular speed . The kinetic energy of the particle of mass mi is 1 m iv i 2. 2 The total kinetic energy of the object is called its rotational kinetic energy. 300 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis We simplify this expression by defining the quantity in parentheses as the moment of inertia I: Moment of inertia I i mi ri 2 (10.15) From the definition of moment of inertia, we see that it has dimensions of ML2 (kg m2 in SI units).1 With this notation, Equation 10.14 becomes Rotational kinetic energy KR 1 2 2I (10.16) 1 Although we commonly refer to the quantity 2I 2 as rotational kinetic energy, it is not a new form of energy. It is ordinary kinetic energy because it is derived from a sum over individual kinetic energies of the particles contained in the rigid object. However, the mathematical form of the kinetic energy given by Equation 10.16 is a convenient one when we are dealing with rotational motion, provided we know how to calculate I. It is important that you recognize the analogy between kinetic energy associated with linear motion 1mv 2 and rotational kinetic energy 1 I 2. The quantities I 2 2 and in rotational motion are analogous to m and v in linear motion, respectively. (In fact, I takes the place of m every time we compare a linear-motion equation with its rotational counterpart.) The moment of inertia is a measure of the resistance of an object to changes in its rotational motion, just as mass is a measure of the tendency of an object to resist changes in its linear motion. Note, however, that mass is an intrinsic property of an object, whereas I depends on the physical arrangement of that mass. Can you think of a situation in which an object's moment of inertia changes even though its mass does not? EXAMPLE 10.3 The Oxygen Molecule 1.95 10 46 Consider an oxygen molecule (O2 ) rotating in the xy plane about the z axis. The axis passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is 2.66 10 26 kg, and at room temperature the average separation between the two atoms is d 1.21 10 10 m (the atoms are treated as point masses). (a) Calculate the moment of inertia of the molecule about the z axis. kg m2 This is a very small number, consistent with the minuscule masses and distances involved. (b) If the angular speed of the molecule about the z axis is 4.60 1012 rad/s, what is its rotational kinetic energy? Solution This is a straightforward application of the definition of I. Because each atom is a distance d/2 from the z axis, the moment of inertia about the axis is I i 1 2 (2.66 Solution We apply the result we just calculated for the moment of inertia in the formula for K R : KR 1 2 2I 1 2 (1.95 mi ri 2 m 10 d 2 26 2 m d 2 2 1 2 2 md 10 10 10 46 21 kg m2)(4.60 J 1012 rad/s)2 kg)(1.21 10 m)2 2.06 1 Civil engineers use moment of inertia to characterize the elastic properties (rigidity) of such structures as loaded beams. Hence, it is often useful even in a nonrotational context. 10.5 Calculation of Moments of Inertia 301 EXAMPLE 10.4 Four Rotating Masses Therefore, the rotational kinetic energy about the y axis is KR 1 2 Iy 2 1 2 2 (2Ma ) 2 Four tiny spheres are fastened to the corners of a frame of negligible mass lying in the xy plane (Fig. 10.8). We shall assume that the spheres' radii are small compared with the dimensions of the frame. (a) If the system rotates about the y axis with an angular speed , find the moment of inertia and the rotational kinetic energy about this axis. Ma 2 2 Solution First, note that the two spheres of mass m, which lie on the y axis, do not contribute to Iy (that is, ri 0 for these spheres about this axis). Applying Equation 10.15, we obtain Iy i The fact that the two spheres of mass m do not enter into this result makes sense because they have no motion about the axis of rotation; hence, they have no rotational kinetic energy. By similar logic, we expect the moment of inertia about the x axis to be Ix 2mb 2 with a rotational kinetic energy about that axis of K R mb 2 2. (b) Suppose the system rotates in the xy plane about an axis through O (the z axis). Calculate the moment of inertia and rotational kinetic energy about this axis. m i ri 2 Ma 2 Ma 2 2Ma 2 y m Solution Iz i Because ri in Equation 10.15 is the perpendicular distance to the axis of rotation, we obtain mi ri 2 1 2 Iz 2 Ma 2 1 2 2 (2Ma Ma 2 mb 2 2 mb 2 (Ma 2 2Ma 2 mb 2) 2 2mb 2 b M a O b a M KR x 2mb 2) m Figure 10.8 The four spheres are at a fixed separation as shown. The moment of inertia of the system depends on the axis about which it is evaluated. Comparing the results for parts (a) and (b), we conclude that the moment of inertia and therefore the rotational kinetic energy associated with a given angular speed depend on the axis of rotation. In part (b), we expect the result to include all four spheres and distances because all four spheres are rotating in the xy plane. Furthermore, the fact that the rotational kinetic energy in part (a) is smaller than that in part (b) indicates that it would take less effort (work) to set the system into rotation about the y axis than about the z axis. 10.5 7.5 CALCULATION OF MOMENTS OF INERTIA We can evaluate the moment of inertia of an extended rigid object by imagining the object divided into many small volume elements, each of which has mass m. r i 2 m i and take the limit of this sum as m : 0. In We use the definition I i this limit, the sum becomes an integral over the whole object: I lim ri 2 mi r 2 dm (10.17) m i :0 i It is usually easier to calculate moments of inertia in terms of the volume of the elements rather than their mass, and we can easily make that change by using m/V, where is the density of the object and V is its volume. We Equation 1.1, want this expression in its differential form dm/dV because the volumes we are dealing with are very small. Solving for dm dV and substituting the result 302 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis into Equation 10.17 gives I r 2 dV If the object is homogeneous, then is constant and the integral can be evaluated for a known geometry. If is not constant, then its variation with position must be known to complete the integration. The density given by m/V sometimes is referred to as volume density for the obvious reason that it relates to volume. Often we use other ways of expressing density. For instance, when dealing with a sheet of uniform thickness t, we can define a surface density t, which signifies mass per unit area. Finally, when mass is distributed along a uniform rod of cross-sectional area A, we sometimes use linear density M/L A, which is the mass per unit length. EXAMPLE 10.5 Uniform Hoop y Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center (Fig. 10.9). dm Solution All mass elements dm are the same distance r R from the axis, and so, applying Equation 10.17, we obtain for the moment of inertia about the z axis through O: Iz r 2 dm R2 dm MR 2 O x R Note that this moment of inertia is the same as that of a single particle of mass M located a distance R from the axis of rotation. Figure 10.9 The mass elements dm of a uniform hoop are all the same distance from O. Quick Quiz 10.3 (a) Based on what you have learned from Example 10.5, what do you expect to find for the moment of inertia of two particles, each of mass M/2, located anywhere on a circle of radius R around the axis of rotation? (b) How about the moment of inertia of four particles, each of mass M/4, again located a distance R from the rotation axis? EXAMPLE 10.6 Uniform Rigid Rod Substituting this expression for dm into Equation 10.17, with r x, we obtain L/2 Calculate the moment of inertia of a uniform rigid rod of length L and mass M (Fig. 10.10) about an axis perpendicular to the rod (the y axis) and passing through its center of mass. Iy r 2 dm L/2 x2 L/2 L/2 M dx L M L L/2 x 2 dx L/2 Solution The shaded length element dx has a mass dm equal to the mass per unit length multiplied by dx : dm dx M dx L M L x3 3 1 2 12 ML 10.5 Calculation of Moments of Inertia y y 303 dx x O x Figure 10.10 L A uniform rigid rod of length L. The moment of inertia about the y axis is less than that about the y axis. The latter axis is examined in Example 10.8. EXAMPLE 10.7 Uniform Solid Cylinder cylindrical shells, each of which has radius r, thickness dr, and length L, as shown in Figure 10.11. The volume dV of each shell is its cross-sectional area multiplied by its length: dV dA L (2 r dr)L. If the mass per unit volume is , then the mass of this differential volume element is dm dV 2 rL dr. Substituting this expression for dm into Equation 10.17, we obtain R A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis in Fig. 10.11). Solution It is convenient to divide the cylinder into many z dr r R L Iz r 2 dm 2 L 0 r 3 dr 1 2 LR 4 Because the total volume of the cylinder is R 2L, we see that M/V M/ R 2L. Substituting this value for into the above result gives (1) Iz 1 2 2 MR Figure 10.11 cylinder. Calculating I about the z axis for a uniform solid Note that this result does not depend on L, the length of the cylinder. In other words, it applies equally well to a long cylinder and a flat disc. Also note that this is exactly half the value we would expect were all the mass concentrated at the outer edge of the cylinder or disc. (See Example 10.5.) Table 10.2 gives the moments of inertia for a number of bodies about specific axes. The moments of inertia of rigid bodies with simple geometry (high symmetry) are relatively easy to calculate provided the rotation axis coincides with an axis of symmetry. The calculation of moments of inertia about an arbitrary axis can be cumbersome, however, even for a highly symmetric object. Fortunately, use of an important theorem, called the parallel-axis theorem, often simplifies the calculation. Suppose the moment of inertia about an axis through the center of mass of an object is ICM . The parallel-axis theorem states that the moment of inertia about any axis parallel to and a distance D away from this axis is I ICM MD 2 (10.18) Parallel-axis theorem 304 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis TABLE 10.2 Moments of Inertia of Homogeneous Rigid Bodies with Different Geometries Hoop or cylindrical shell I CM = MR 2 Hollow cylinder R I CM = 1 M(R 12 + R 22) 2 R1 R2 Solid cylinder or disk I CM = 1 MR 2 2 R Rectangular plate I CM = 1 M(a 2 + b 2) 12 b a Long thin rod with rotation axis through center I CM = 1 ML 2 12 L Long thin rod with rotation axis through end I = 1 ML 2 3 L Solid sphere I CM = 2 MR 2 5 R Thin spherical shell I CM = 2 MR 2 3 R Proof of the Parallel-Axis Theorem (Optional). Suppose that an object rotates in the xy plane about the z axis, as shown in Figure 10.12, and that the coordinates of the center of mass are x CM , y CM . Let the mass element dm have coordinates x, y. Because this element is a distance r x 2 y 2 from the z axis, the moment of inertia about the z axis is I r 2 dm (x 2 y 2) dm However, we can relate the coordinates x, y of the mass element dm to the coordinates of this same element located in a coordinate system having the object's center of mass as its origin. If the coordinates of the center of mass are x CM , y CM in the original coordinate system centered on O, then from Figure 10.12a we see that the relationships between the unprimed and primed coordinates are x x x CM 10.5 Calculation of Moments of Inertia 305 y dm x, y z Axis through CM y CM xCM, yCM yCM O xCM x x O D x x CM y y r Rotation axis (a) (b) (a) The parallel-axis theorem: If the moment of inertia about an axis perpendicular to the figure through the center of mass is ICM , then the moment of inertia about the z axis is Iz I CM MD 2. (b) Perspective drawing showing the z axis (the axis of rotation) and the parallel axis through the CM. Figure 10.12 and y I y [(x [(x )2 y CM . Therefore, x CM)2 (y y CM)2] dm 2x CM x dm 2y CM y dm (x CM2 y CM 2) dm (y )2] dm The first integral is, by definition, the moment of inertia about an axis that is parallel to the z axis and passes through the center of mass. The second two integrals are zero because, by definition of the center of mass, x dm y dm 0. The last integral is simply MD 2 because dm M and D 2 x CM2 y CM2. Therefore, we conclude that I ICM MD 2 EXAMPLE 10.8 Applying the Parallel-Axis Theorem I ICM MD 2 1 12 Consider once again the uniform rigid rod of mass M and length L shown in Figure 10.10. Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the y axis in Fig. 10.10). ML 2 M L 2 2 1 3 ML 2 Solution Intuitively, we expect the moment of inertia to 1 be greater than ICM 12ML 2 because it should be more difficult to change the rotational motion of a rod spinning about an axis at one end than one that is spinning about its center. Because the distance between the center-of-mass axis and the y axis is D L/2, the parallel-axis theorem gives So, it is four times more difficult to change the rotation of a rod spinning about its end than it is to change the motion of one spinning about its center. Exercise Answer Calculate the moment of inertia of the rod about a perpendicular axis through the point x L/4. I 7 48 ML 2. 306 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis 10.6 7.6 TORQUE Why are a door's doorknob and hinges placed near opposite edges of the door? This question actually has an answer based on common sense ideas. The harder we push against the door and the farther we are from the hinges, the more likely we are to open or close the door. When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque (tau). Consider the wrench pivoted on the axis through O in Figure 10.13. The applied force F acts at an angle to the horizontal. We define the magnitude of the torque associated with the force F by the expression r F sin Fd (10.19) Definition of torque Moment arm where r is the distance between the pivot point and the point of application of F and d is the perpendicular distance from the pivot point to the line of action of F. (The line of action of a force is an imaginary line extending out both ends of the vector representing the force. The dashed line extending from the tail of F in Figure 10.13 is part of the line of action of F.) From the right triangle in Figure 10.13 that has the wrench as its hypotenuse, we see that d r sin . This quantity d is called the moment arm (or lever arm) of F. It is very important that you recognize that torque is defined only when a reference axis is specified. Torque is the product of a force and the moment arm of that force, and moment arm is defined only in terms of an axis of rotation. In Figure 10.13, the only component of F that tends to cause rotation is F sin , the component perpendicular to r. The horizontal component F cos , because it passes through O, has no tendency to produce rotation. From the definition of torque, we see that the rotating tendency increases as F increases and as d increases. This explains the observation that it is easier to close a door if we push at the doorknob rather than at a point close to the hinge. We also want to apply our push as close to perpendicular to the door as we can. Pushing sideways on the doorknob will not cause the door to rotate. If two or more forces are acting on a rigid object, as shown in Figure 10.14, each tends to produce rotation about the pivot at O. In this example, F2 tends to F1 F sin r F d1 O d F cos Line of action O d2 Figure 10.13 The force F has a greater rotating tendency about O as F increases and as the moment arm d increases. It is the component F sin that tends to rotate the wrench about O. F2 Figure 10.14 The force F1 tends to rotate the object counterclockwise about O, and F2 tends to rotate it clockwise. 10.7 Relationship Between Torque and Angular Acceleration 307 rotate the object clockwise, and F1 tends to rotate it counterclockwise. We use the convention that the sign of the torque resulting from a force is positive if the turning tendency of the force is counterclockwise and is negative if the turning tendency is clockwise. For example, in Figure 10.14, the torque resulting from F1 , which has a moment arm d 1 , is positive and equal to F1 d 1 ; the torque from F2 is negative and equal to F2 d 2 . Hence, the net torque about O is 1 2 F 1d 1 F 2d 2 Torque should not be confused with force. Forces can cause a change in linear motion, as described by Newton's second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the forces and the moment arms of the forces, in the combination that we call torque. Torque has units of force times length -- newton meters in SI units -- and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. EXAMPLE 10.9 The Net Torque on a Cylinder A one-piece cylinder is shaped as shown in Figure 10.15, with a core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, which has radius R 1 , exerts a force F1 to the right on the cylinder. A rope wrapped around the core, which has radius R 2 , exerts a force F2 downward on the cylinder. (a) What is the net torque acting on the cylinder about the rotation axis (which is the z axis in Figure 10.15)? y Solution The torque due to F1 is R 1 F1 (the sign is negative because the torque tends to produce clockwise rotation). The torque due to F2 is R 2 F 2 (the sign is positive because the torque tends to produce counterclockwise rotation). Therefore, the net torque about the rotation axis is 1 2 R 1F 1 R 2F 2 F1 We can make a quick check by noting that if the two forces are of equal magnitude, the net torque is negative because R 1 R 2 . Starting from rest with both forces acting on it, the cylinder would rotate clockwise because F 1 would be more effective at turning it than would F 2 . (b) Suppose F 1 5.0 N, R 1 1.0 m, F 2 15.0 N, and R 2 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate from rest? (5.0 N)(1.0 m) (15.0 N)(0.50 m) 2.5 N m R1 R2 O x z F2 Figure 10.15 A solid cylinder pivoted about the z axis through O. The moment arm of F1 is R 1 , and the moment arm of F2 is R 2 . Because the net torque is positive, if the cylinder starts from rest, it will commence rotating counterclockwise with increasing angular velocity. (If the cylinder's initial rotation is clockwise, it will slow to a stop and then rotate counterclockwise with increasing angular speed.) 10.7 RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION 7.6 In this section we show that the angular acceleration of a rigid object rotating about a fixed axis is proportional to the net torque acting about that axis. Before discussing the more complex case of rigid-body rotation, however, it is instructive 308 CHAPTER 10 Rotation of a Rigid Object About a Fixed Axis first to discuss the case of a particle rotating about some fixed point under the influence of an external force. Consider a particle of mass m rotating in a circle of radius r under the influence of a tangential force Ft and a radial force Fr , as shown in Figure 10.16. (As we learned in Chapter 6, the radial force must be present to keep the particle moving in its circular path.) The tangential force provides a tangential acceleration at , and Ft ma t The torque about the center of the circle due to Ft is Figure 10.16 A particle rotating in a circle under the influence of a tangential force Ft . A force Fr in the radial direction also must be present to maintain the circular motion. Ft r (ma t )r Because the tangential acceleration is related to the angular acceleration through the relationship at r (see Eq. 10.11), the torque can be expressed as (mr )r (mr 2) Recall from Equation 10.15 that mr 2 is the moment of inertia of the rotating particle about the z axis passing through the origin, so that Relationship between torque and angular acceleration I (10.20) y d Ft dm That is, the torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is the moment of inertia. It is important to note that I is the rotational analog of Newton's second law of motion, F ma. Now let us extend this discussion to a rigid object of arbitrary shape rotating about a fixed axis, as shown in Figure 10.17. The object can be regarded as an infinite number of mass elements dm of infinitesimal size. If we impose a cartesian coordinate system on the object, then each mass element rotates in a circle about the origin, and each has a tangential acceleration at produced by an external tangential force dFt . For any given element, we know from Newton's second law that dFt (dm)a t (r dm)a t (r 2 dm) The torque d associated with the force dFt acts about the origin and is given by x r O d Because at r dF t r , the expression for d becomes d (r dm)r A rigid object rotating about an axis through O. Each mass element dm rotates about O with the same angular acceleration , and the net torque on the object is proportional to . Figure 10.17 It is important to recognize that although each mass element of the rigid object may have a different linear acceleration at , they all have the same angular acceleration . With this in mind, we can integrate the above expression to obtain the net torque about O due to the external forces: (r 2 dm) r 2 dm where can be taken outside the integral because it is common to all mass elements. From Equation 10.17, we know that r 2 dm is the moment of inertia of the object about the rotation axis through O, and so the expression for becomes Torque is proportional to angular acceleration I (10.21) Note that this is the same relationship we found for a particle rotating in a circle (see Eq. 10.20). So, again we see that the net torque about the rotation axis is pro- 10.7 Relationship Between Torque and Angular Acceleration 309 portional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the axis of rotation and upon the size and shape of the object. In view of the complex nature of the system, it is interesting to note that the relationship I is strikingly simple and in complete agreement with experimental observations. The simplicity of the result lies in the manner in which the motion is described. Although each point on a rigid object rotating about a fixed axis may not experience the same force, linear acceleration, or linear speed, each point experiences the same angular acceleration and angular speed at any instant. Therefore, at any instant the rotating rigid object as a whole is characterized by specific values for angular acceleration, net torque, and angular speed. Every point has the same and QuickLab Tip over a child's tall tower of blocks. Try this several times. Does the tower "break" at the same place each time? What affects where the tower comes apart as it tips? If the tower were made of toy bricks that snap together, what would happen? (Refer to Conceptual Example 10.11.) Finally, note that the result I also applies when the forces acting on the mass elements have radial components as well as tangential components. This is because the line of action of all radial components must pass through the axis of rotation, and hence all radial components produce zero torque about that axis. EXAMPLE 10.10 Rotating Rod compute the torque on the rod, we can assume that the gravitational force acts at the center of mass of the rod, as shown in Figure 10.18. The torque due to this force about an axis through the pivot is g With I , and I Table 10.2), we obtain 1 2 3 ML A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is freis Fd (10.19) where d is the moment arm of the force, which is the perpendicular distance from some origin to the line of action of the force. Torque is a measure of the tendency of the force to change the rotation of the object about some axis. If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration , where I (10.21) The rate at which work is done by an external force in rotating a rigid object about a fixed axis, or the power delivered, is (10.23) The net work done by external forces in rotating a rigid object about a fixed axis equals the change in the rotational kinetic energy of the object: W 1 2 2I f 1 2 2I i (10.24) QUESTIONS 1. What is the angular speed of the second hand of a clock? What is the direction of as you view a clock hanging vertically? What is the magnitude of the angular acceleration vector of the second hand? 2. A wheel rotates counterclockwise in the xy plane. What is the direction of ? What is the direction of if the angular velocity is decreasing in time? 3. Are the kinematic expressions for , , and valid when the angular displacement is measured in degrees instead of in radians? 4. A turntable rotates at a constant rate of 45 rev/min. What is its angular speed in radians per second? What is the magnitude of its angular acceleration? 5. Suppose a b and M m for the system of particles described in Figure 10.8. About what axis (x, y, or z) does the moment of inertia have the smallest value? the largest value? 6. Suppose the rod in Figure 10.10 has a nonuniform mass distribution. In general, would the moment of inertia about the y axis still equal ML2/12? If not, could the moment of inertia be calculated without knowledge of the manner in which the mass is distributed? 7. Suppose that only two external forces act on a rigid body, and the two forces are equal in magnitude but opposite in direction. Under what condition does the body rotate? 8. Explain how you might use the apparatus described in Example 10.12 to determine the moment of inertia of the wheel. (If the wheel does not have a uniform mass density, the moment of inertia is not necessarily equal to 1 2 2 MR .) Problems 9. Using the results from Example 10.12, how would you calculate the angular speed of the wheel and the linear speed of the suspended mass at t 2 s, if the system is released from rest at t 0? Is the expression v R valid in this situation? 10. If a small sphere of mass M were placed at the end of the rod in Figure 10.23, would the result for be greater than, less than, or equal to the value obtained in Example 10.14? 11. Explain why changing the axis of rotation of an object changes its moment of inertia. 12. Is it possible to change the translational kinetic energy of an object without changing its rotational energy? 13. Two cylinders having the same dimensions are set into rotation about their long axes with the same angular speed. 317 14. 15. 16. 17. One is hollow, and the other is filled with water. Which cylinder will be easier to stop rotating? Explain your answer. Must an object be rotating to have a nonzero moment of inertia? If you see an object rotating, is there necessarily a net torque acting on it? Can a (momentarily) stationary object have a nonzero angular acceleration? The polar diameter of the Earth is slightly less than the equatorial diameter. How would the moment of inertia of the Earth change if some mass from near the equator were removed and transferred to the polar regions to make the Earth a perfect sphere? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 10.2 Rotational Kinematics: Rotational Motion with Constant Angular Acceleration 1. A wheel starts from rest and rotates with constant angular acceleration and reaches an angular speed of 12.0 rad/s in 3.00 s.e disk in the dashed position? (c) Repeat part (a), using a uniform hoop. M R m1 3.00 m M = 3.00 kg R = 10.0 cm m1 = 15.0 kg m2 = 10.0 kg m2 Figure P10.43 Problems 43 and 44. 45. A weight of 50.0 N is attached to the free end of a light string wrapped around a reel with a radius of 0.250 m and a mass of 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center. The weight is released 6.00 m above the floor. (a) Determine the tension in the string, the acceleration of the mass, and the speed with which the weight hits the floor. (b) Find the speed calculated in part (a), using the principle of conservation of energy. 46. A constant torque of 25.0 N m is applied to a grindstone whose moment of inertia is 0.130 kg m2. Using energy principles, find the angular speed after the grindstone has made 15.0 revolutions. (Neglect friction.) 47. This problem describes one experimental method of determining the moment of inertia of an irregularly shaped object such as the payload for a satellite. Figure P10.47 shows a mass m suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. When the mass is released from rest, it descends through a distance h, acquiring a speed Pivot R g Figure P10.49 50. A horizontal 800-N merry-go-round is a solid disk of radius 1.50 m and is started from rest by a constant horizontal force of 50.0 N applied tangentially to the cylinder. Find the kinetic energy of the solid cylinder after 3.00 s. ADDITIONAL PROBLEMS 51. Toppling chimneys often break apart in mid-fall (Fig. P10.51) because the mortar between the bricks cannot Problems withstand much shear stress. As the chimney begins to fall, shear forces must act on the topmost sections to accelerate them tangentially so that they can keep up with the rotation of the lower part of the stack. For simplicity, let us model the chimney as a uniform rod of length pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the bottom) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin , where is the angle the chimney makes with the vertical? 323 exerts on the wheel. (a) How long does the wheel take to reach its final rotational speed of 1 200 rev/min? (b) Through how many revolutions does it turn while accelerating? 54. The density of the Earth, at any distance r from its center, is approximately [14.2 11.6 r/R] 10 3 kg/m3 where R is the radius of the Earth. Show that this density leads to a moment of inertia I 0.330MR 2 about an axis through the center, where M is the mass of the Earth. 55. A 4.00-m length of light nylon cord is wound around a uniform cylindrical spool of radius 0.500 m and mass 1.00 kg. The spool is mounted on a frictionless axle and is initially at rest. The cord is pulled from the spool with a constant acceleration of magnitude 2.50 m/s2. (a) How much work has been done on the spool when it reaches an angular speed of 8.00 rad/s? (b) Assuming that there is enough cord on the spool, how long does it take the spool to reach this angular speed? (c) Is there enough cord on the spool? 56. A flywheel in the form of a heavy circular disk of diameter 0.600 m and mass 200 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1 000 rev/min. (a) What is the moment of inertia of the flywheel? (b) How much work is done on it during this acceleration? (c) When the angular speed reaches 1 000 rev/min, the motor is disengaged. A friction brake is used to slow the rotational rate to 500 rev/min. How much energy is dissipated as internal energy in the friction brake? 57. A shaft is turning at 65.0 rad/s at time zero. Thereafter, its angular acceleration is given by 10 rad/s2 5t rad/s3 Figure P10.51 A building demolition site in Baltimore, MD. At the left is a chimney, mostly concealed by the building, that has broken apart on its way down. Compare with Figuays a key role in rotational dynamics. In analogy to the conservation of linear momentum, we find that the angular momentum of a rigid object is always conserved if no external torques act on the object. Like the law of conservation of linear momentum, the law of conservation of angular momentum is a fundamental law of physics, equally valid for relativistic and quantum systems. 11.1 7.7 ROLLING MOTION OF A RIGID OBJECT In this section we treat the motion of a rigid object rotating about a moving axis. In general, such motion is very complex. However, we can simplify matters by restricting our discussion to a homogeneous rigid object having a high degree of symmetry, such as a cylinder, sphere, or hoop. Furthermore, we assume that the object undergoes rolling motion along a flat surface. We shall see that if an object such as a cylinder rolls without slipping on the surface (we call this pure rolling motion), a simple relationship exists between its rotational and translational motions. Suppose a cylinder is rolling on a straight path. As Figure 11.1 shows, the center of mass moves in a straight line, but a point on the rim moves in a more complex path called a cycloid. This means that the axis of rotation remains parallel to its initial orientation in space. Consider a uniform cylinder of radius R rolling without slipping on a horizontal surface (Fig. 11.2). As the cylinder rotates through an angle , its center of mass moves a linear distance s R (see Eq. 10.1a). Therefore, the linear speed of the center of mass for pure rolling motion is given by v CM ds dt R d dt R (11.1) where is the angular velocity of the cylinder. Equation 11.1 holds whenever a cylinder or sphere rolls without slipping and is the condition for pure rolling Figure 11.1 One light source at the center of a rolling cylinder and another at one point on the rim illustrate the different paths these two points take. The center moves in a straight line (green line), whereas the point on the rim moves in the path called a cycloid (red curve). (Henry Leap and Jim Lehman) 11.1 Rolling Motion of a Rigid Object 329 R s s = R Figure 11.2 In pure rolling motion, as the cylinder rotates through an angle , its center of mass moves a linear distance s R . motion. The magnitude of the linear acceleration of the center of mass for pure rolling motion is a CM dv CM dt R d dt R (11.2) where is the angular acceleration of the cylinder. The linear velocities of the center of mass and of various points on and within the cylinder are illustrated in Figure 11.3. A short time after the moment shown in the drawing, the rim point labeled P will have rotated from the six o'clock position to, say, the seven o'clock position, the point Q will have rotated from the ten o'clock position to the eleven o'clock position, and so on. Note that the linear velocity of any point is in a direction perpendicular to the line from that point to the contact point P. At any instant, the part of the rim that is at point P is at rest relative to the surface because slipping does not occur. All points on the cylinder have the same angular speed. Therefore, because the distance from P to P is twice the distance from P to the center of mass, P has a speed 2v CM 2R . To see why this is so, let us model the rolling motion of the cylinder in Figure 11.4 as a combination of translational (linear) motion and rotational motion. For the pure translational motion shown in Figure 11.4a, imagine that the cylinder does not rotate, so that each point on it moves to the right with speed v CM . For the pure rotational motion shown in Figure 11.4b, imagine that a rotation axis through the center of mass is stationary, so that each point on the cylinder has the same rotational speed . The combination of these two motions represents the rolling motion shown in Figure 11.4c. Note in Figure 11.4c that the top of the cylinder has linear speed v CM R v CM v CM 2v CM , which is greater than the linear speed of any other point on the cylinder. As noted earlier, the center of mass moves with linear speed v CM while the contact point between the surface and cylinder has a linear speed of zero. We can express the total kinetic energy of the rolling cylinder as 7.2 P 2vCM Q CM vCM P Figure 11.3 All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity vCM , and the point P moves with a velocity 2vCM . K 1 2I P 2 (11.3) where IP is the moment of inertia about a rotation axis through P. Applying the parallel-axis theorem, we can substitute I P I CM MR 2 into Equation 11.3 to obtain K 1 2 2 I CM 1 2 2 2 MR 330 CHAPTER 11 Rolling Motion and Angular Momentum P P v CM v = R CM v CM CM v=0 P v CM v = R P (b) Pure rotation (a) Pure translation P v = v CM + R = 2v CM CM v = v CM v=0 P (c) Combination of translation and rotation Figure 11.4 The motion of a rolling object can be modeled as a combination of pure translation and pure rotation. or, because v CM Total kinetic energy of a rolling body R , K 1 2 2 I CM 1 2 2 Mv CM (11.4) M R x h vCM Figure 11.5 A sphere rolling down an incline. Mechanical energy is conserved if no slipping occurs. The term 1 I CM 2 represents the rotational kinetic energy of the cylinder about its 2 center of mass, and the term 1Mv CM2 represents the kinetic energy the cylinder 2 would have if it were just translating through space without rotating. Thus, we can say that the total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass. We can use energy methods to treat a class of problems concerning the rolling motion of a sphere down a rough incline. (The analysis that follows also applies to the rolling motion of a cylinder or hoop.) We assume that the sphere in Figure 11.5 rolls without slipping and is released from rest at the top of the incline. Note that accelerated rolling motion is possible only if a frictional force is present between the sphere and the incline to produce a net torque about the center of mass. Despite the presence of friction, no loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant. On the other hand, if the sphere were to slip, mechanical energy would be lost as motion progressed. Using the fact that v CM R for pure rolling motion, we can express Equation 11.4 as 2 v 1 2 K 1 I CM CM 2 2 Mv CM R K 1 2 I CM R2 M v CM2 (11.5) 11.1 Rolling Motion of a Rigid Object 331 By the time the sphere reaches the bottom of the incline, work equal to Mgh has been done on it by the gravitational field, where h is the height of the incline. Because the sphere starts from rest at the top, its kinetic energy at the bottom, given by Equation 11.5, must equal this work done. Therefore, the speed of the center of mass at the bottom can be obtained by equating these two quantities: 1 2 I CM R2 M v CM2 v CM Mgh 2gh I CM/MR 2 1/2 1 (11.6) Quick Quiz 11.1 Imagine that you slide your textbook across a gymnasium floor with a certain initial speed. It quickly stops moving because of friction between it and the floor. Yet, if you were to start a basketball rolling with the same initial speed, it would probably keep rolling from one end of the gym to the other. Why does a basketball roll so far? Doesn't friction affect its motion? EXAMPLE 11.1 Sphere Rolling Down an Incline x sin . Hence, after squaring both sides, we can express the equation above as v CM2 10 gx sin 7 Comparing this with the expression from kinematics, v CM2 2a CMx (see Eq. 2.12), we see that the acceleration of the center of mass is a CM 5 7 For the solid sphere shown in Figure 11.5, calculate the linear speed of the center of mass at the bottom of the incline and the magnitude of the linear acceleration of the center of mass. The sphere starts from the top of the incline with potential energy U g Mgh and kinetic energy K 0. As we have seen before, if it fell vertically ar momentum about O that has magnitude mvr. The vector L r p points out of the diagram. A car of mass 1 500 kg moves with a linear speed of 40 m/s on a circular race track of radius 50 m. What is the magnitude of its angular momentum relative to the center of the track? 3.0 106 kg m2/s Answer 11.4 ANGULAR MOMENTUM OF A ROTATING RIGID OBJECT Consider a rigid object rotating about a fixed axis that coincides with the z axis of a coordinate system, as shown in Figure 11.12. Let us determine the angular momentum of this object. Each particle of the object rotates in the xy plane about the z axis with an angular speed . The magnitude of the angular momentum of a particle of mass mi about the origin O is m i vi ri . Because vi ri , we can express the magnitude of the angular momentum of this particle as Li m ir i 2 . The vector Li is directed along the z axis, as is the vector 338 z CHAPTER 11 Rolling Motion and Angular Momentum We can now find the angular momentum (which in this situation has only a z component) of the whole object by taking the sum of Li over all particles: Lz i L m ir i 2 i m ir i 2 (11.21) Lz r mi x vi y I where I is the moment of inertia of the object about the z axis. Now let us differentiate Equation 11.21 with respect to time, noting that I is constant for a rigid body: dL z dt I d dt I (11.22) When a rigid body rotates about an axis, the angular momentum L is in the same direction as the angular velocity , according to the expression L I . Figure 11.12 where is the angular acceleration relative to the axis of rotation. Because dL z /dt is equal to the net external torque (see Eq. 11.20), we can express Equation 11.22 as dL z dt I (11.23) ext That is, the net external torque acting on a rigid object rotating about a fixed axis equals the moment of inertia about the rotation axis multiplied by the object's angular acceleration relative to that axis. Equation 11.23 also is valid for a rigid object rotating about a moving axis provided the moving axis (1) passes through the center of mass and (2) is a symmetry axis. You should note that if a symmetrical object rotates about a fixed axis passing through its center of mass, you can write Equation 11.21 in vector form as L I , where L is the total angular momentum of the object measured with respect to the axis of rotation. Furthermore, the expression is valid for any object, regardless of its symmetry, if L stands for the component of angular momentum along the axis of rotation.2 EXAMPLE 11.5 Bowling Ball solid sphere. A typical bowling ball might have a mass of 6 kg and a radius of 12 cm. The moment of inertia of a solid sphere about an axis through its center is, from Table 10.2, I 2 2 5 MR 2 5 (6 Estimate the magnitude of the angular momentum of a bowling ball spinning at 10 rev/s, as shown in Figure 11.13. We start by making some estimates of the relevant physical parameters and model the ball as a uniform Solution kg)(0.12 m)2 0.035 kg m2 Therefore, the magnitude of the angular momentum is 2 In general, the expression L I is not always valid. If a rigid object rotates about an arbitrary axis, L and may point in different directions. In this case, the moment of inertia cannot be treated as a scalar. Strictly speaking, L I applies only to rigid objects of any shape that rotate about one of three mutually perpendicular axes (called principal axes) through the center of mass. This is discussed in more advanced texts on mechanics. 11.4 Angular Momentum of a Rotating Rigid Object L I (0.035 kg m2)(10 rev/s)(2 rad/rev) z L 339 2.2 kg m2/s Because of the roughness of our estimates, we probably want to keep only one significant figure, and so L 2 kg m2/s. y Figure 11.13 A bowling ball that rotates about the z axis in the direction shown has an angular momentum L in the positive z direction. If the direction of rotation is reversed, L points in the negative z direction. x EXAMPLE 11.6 Rotating Rod y m2 A rigid rod of mass M and length is pivoted without friction at its center (Fig. 11.14). Two particles of masses m 1 and m 2 are connected to its ends. The combination rotates in a vertical plane with an angular speed . (a) Find an expression for the magnitude of the angular momentum of the system. This is different from the last example in that we now must account for the motion of more than one object. The moment of inertia of the system equals the sum of the moments of inertia of the three components: the rod and the two particles. Referring to Table 10.2 to obtain the expression for the moment of inertia of the rod, and using the expression I mr 2 for each particle, we find that the total moment of inertia about the z axis through O is I 1 M 12 2 2 2 2 Solution O x m 2g m1 m 1g m1 m1 2 m2 m2 2 4 M 3 Figure 11.14 Because gravitational forces act on the rotating rod, there is in general a net nonzero torque about O when m 1 m 2 . This net torque produces an angular acceleration given by ext I. The torque due to the force m 2 g about the pivot is 2 Therefore, the magnitude of the angular momentum is 2 L I 4 M 3 m 2g m1 m2 ext 1 ext 2 cos ( 2 into page) Hence, the net torque exerted on the system about O is 2 1 2 (m 1 (b) Find an expression for the magnitude of the angular acceleration of the system when the rod makes an angle with the horizontal. m 2)g cos Solution If the masses of the two particles are equal, then the system has no angular acceleration because the net torque on the system is zero when m 1 m 2 . If the initial angle is exactly /2 or /2 (vertical position), then the rod will be in equilibrium. To find the angular acceleration of the system at any angle , we first calculate the net torque on the system and then use ext I to obtain an expression for . The torque due to the force m 1 g about the pivot is 1 The direction of into the page if m 2 To find , we use is out of the page if m1 m 2 and is m1 . I , where I was obtained in part (a): ext 2(m 1 (M/3 m 2)g cos m1 m 2) ext I /2 (vertical position) Note that is zero when is /2 or and is a maximum when is 0 or (horizontal position). m 1g 2 cos ( 1 out of page) Exercise Answer If m 2 m 1, at what value of /2. is a maximum? 340 CHAPTER 11 Rolling Motion and Angular Momentum EXAMPLE 11.7 Two Connected Masses Now let us evaluate the total external torque acting on the system about the pulley axle. Because it has a moment arm of zero, the force exerted by the axle on the pulley does not contribute to the torque. Furthermore, the normal force acting on the block is balanced by the force of gravity m 2 g, and so these forces do not contribute to the torque. The force of gravity m 1 g acting on the sphere produces a torque about the axle equal in magnitude to m 1 gR, where R is the moment arm of the force about the axle. (Note that in this situation, the tension is not equal to m 1 g.) This is the total external torque about the pulley axle; that is, ext m 1 gR. Using this result, together with Equation (1) and Equation 11.23, we find ext A sphere of mass m 1 and a block of mass m 2 are connected by a light cord that passes over a pulley, as shown in Figure 11.15. The radius of the pulley is R, and the moment of inertia about its axle is I. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque. Solution We need to determine the angular momentum of the system, which consists of the two objects and the pulley. Let us calculate the angular momentum about an axis that coincides with the axle of the pulley. At the instant the sphere and block have a common speed v, the angular momentum of the sphere is m 1 vR , and that of the block is m 2 vR . At the same instant, the angular momentum of the pulley is I Iv/R. Hence, the total angular momentum of the system is (1) L m 1vR m 2vR I v R dL dt d (m 1 dt (m 1 m 2)Rv dv dt I v R m 1gR (2) Because dv/dt m 1gR m 2)R I dv R dt v m2 R a, we can solve this for a to obtain a m 1g m 2) (m 1 I/R 2 v m1 Figure 11.15 You may wonder why we did not include the forces thatg on the system about the vertical axis. Therefore, the angular momentum of the system is conserved. Initially, we have L system Li L wheel (upward) After the wheel is inverted, we have Linverted wheel L i . For angular momentum to be conserved, some other part of the system has to start rotating so that the total angular momentum remains the initial angular momentum L i . That other part of the system is the student plus the stool she is sitting on. So, we can now state that Lf Li L student 2L i stool Li Figure 11.17 The wheel is initially spinning when the student is at rest. What happens when the wheel is inverted? L student stool 344 CHAPTER 11 Rolling Motion and Angular Momentum EXAMPLE 11.11 Disk and Stick We used the fact that radians are dimensionless to ensure consistent units for each term. Finally, the elastic nature of the collision reminds us that kinetic energy is conserved; in this case, the kinetic energy consists of translational and rotational forms: Ki 1 2 2 m dv di 1 2 (2.0 A 2.0-kg disk traveling at 3.0 m/s strikes a 1.0-kg stick that is lying flat on nearly frictionless ice, as shown in Figure 11.18. Assume that the collision is elastic. Find the translational speed of the disk, the translational speed of the stick, and the rotational speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.33 kg m2. Because the disk and stick form an isolated system, we can assume that total energy, linear momentum, and angular momentum are all conserved. We have three unknowns, and so we need three equations to solve simultaneously. The first comes from the law of the conservation of linear momentum: pi pf m dv di (2.0 kg)(3.0 m/s) (1) 6.0 kg m/s (2.0 kg)v d f m dv d f m sv s (1.0 kg)v s Kf 1 1 1 2 2 2 2 m dv d f 2 m sv s 2I 1 1 2 2 2 (2.0 kg)v d f 2 (1.0 kg)v s 1 2 2 2 (1.33 kg m /s) Solution kg)(3.0 m/s)2 54 m2/s2 (3) 6.0v d f 2 3.0v s2 (4.0 m2) 2 (2.0 kg)v d f (1.0 kg)v s Now we apply the law of conservation of angular momentum, using the initial position of the center of the stick as our reference point. We know that the component of angular momentum of the disk along the axis perpendicular to the plane of the ice is negative (the right-hand rule shows that Ld points into the ice). Li Lf rm dv di (2.0 m)(2.0 kg)(3.0 m/s) 12 kg m2/s (2) 9.0 rad/s (3.0 rad/m)v d f rm dv d f I In solving Equations (1), (2), and (3) simultaneously, we find that vd f 2.3 m/s, vs 1.3 m/s, and 2.0 rad/s. These values seem reasonable. The disk is moving more slowly than it was before the collision, and the stick has a small translational speed. Table 11.1 summarizes the initial and final values of variables for the disk and the stick and verifies the conservation of linear momentum, angular momentum, and kinetic energy. Exercise Verify the values in Table 11.1. After Before vdi = 3.0 m/s 2.0 m vs vdf (2.0 m)(2.0 kg)v d f (1.33 kg m2) (4.0 kg m)v d f (1.33 kg m2) Figure 11.18 Overhead view of a disk striking a stick in an elastic collision, which causes the stick to rotate. TABLE 11.1 Comparison of Values in Example 11.11 Before and After the Collisiona v (m/s) Before Disk Stick Total After Disk Stick Total a Notice (rad/s) p (kg m/s) L (kg m2/s) Ktrans ( J) Krot ( J) 3.0 0 -- 2.3 1.3 -- -- 0 -- -- 2.0 -- 6.0 0 6.0 4.7 1.3 6.0 12 0 12 9.3 2.7 12 9.0 0 9.0 5.4 0.9 6.3 -- 0 0 -- 2.7 2.7 that linear momentum, angular momentum, and total kinetic energy are conserved. 11.6 The Motion of Gyroscopes and Tops 345 Optional Section 11.6 THE MOTION OF GYROSCOPES AND TOPS A very unusual and fascinating type of motion you probably have observed is that of a top spinning about its axis of symmetry, as shown in Figure 11.19a. If the top spins very rapidly, the axis rotates about the z axis, sweeping out a cone (see Fig. 11.19b). The motion of the axis about the vertical -- known as precessional motion -- is usually slow relative to the spin motion of the top. It is quite natural to wonder why the top does not fall over. Because the center of mass is not directly above the pivot point O, a net torque is clearly acting on the top about O -- a torque resulting from the force of gravity Mg. The top would certainly fall over if it were not spinning. Because it is spinning, however, it has an angular momentum L directed along its symmetry axis. As we shall show, the motion of this symmetry axis about the z axis (the precessional motion) occurs because the torque produces a change in the direction of the symmetry axis. This is an excellent example of the importance of the directional nature of angular momentum. The two forces acting on the top are the downward force of gravity Mg and the normal force n acting upward at the pivot point O. The normal force produces no torque about the pivot because its moment arm through that point is zero. However, the force of gravity produces a torque r Mg about O, where the direction of is perpendicular to the plane formed by r and Mg. By necessity, the vector lies in a horizontal xy plane perpendicular to the angular momentum vector. The net torque and angular momentum of the top are related through Equation 11.19: dL dt From this expression, we see that the nonzero torque produces a change in angular momentum d L -- a change that is in the same direction as . Therefore, like the torque vector, d L must also be at right angles to L. Figure 11.19b illustrates the resulting precessional motion of the symmetry axis of the top. In a time t, the change in angular momentum is L L f L i t. Because L is perpendicular to L, the magnitude of L does not change ( L i L f ). Rather, what is changing is the direction of L. Because the change in angular momentum L is in the direction of , which lies in the xy plane, the top undergoes precessional motion. The essential features of precessional motion can be illustrated by considering the simple gyroscope shown in Figure 11.20a. This device consists of a wheel free to spin about an axle that is pivoted at a distance h from the center of mass of the wheel. When given an angular velocity about the axle, the wheel has an angular momentum L I directed along the axle as shown. Let us consider the torque acting on the wheel about the pivot O. Again, the force n exerted by the support on the axle produces no torque about O, and the force of gravity Mg produces a torque of magnitude Mgh about O, where the axle is perpendicular to the support. The direction of this torque is perpendicular to the axle (and perpendicular to L), as shown in Figure 11.20a. This torque causes the angular momentum to change in the direction perpendicular to the axle. Hence, the axle moves in the direction of the torque -- that is, in the horizontal plane. To simplify the description of the system, we must make an assumption: The total angular momentum of the precessing wheel is the sum of the angular momentum I due to the spinning and the angular momentum due to the motion of Precessional motion z L CM n Mg r O x (a) y L Li Lf (b) Precessional motion of a top spinning about its symmetry axis. (a) The only external forces acting on the top are the normal force n and the force of gravity Mg. The direction of the angular momentum L is along the axis of symmetry. The right-hand r F rule indicates that r Mg is in the xy plane. (b). The direction of L is parallel to that of in part (a). The fact that Lf L L i indicates that the top precesses about the z axis. Figure 11.19 346 CHAPTER 11 Rolling Motion and Angular Momentum h n O Li Lf Mg dL (a) (b) Li d Lf Figure 11.20 (a) The motion of a simple gyroscope pivoted a distance h from its center of mass. The force of gravity Mg produces a torque about the pivot, and this torque is perpendicular to the axle. (b) This torque results in a change in angular momentum d L in a direction perpendicular to the axle. The axle sweeps out an angle d in a time dt. L r Mg n This toy gyroscope undergoes precessional motion about the vertical axis as it spins about its axis of symmetry. The only forces acting on it are the force of gravity Mg and the upward force of the pivot n. The direcStrictly classical models were unsuccessful in describing many properties of the hydrogen atom. Bohr postulated that the electron could occupy only those circular orbits about the proton for which the orbital angular momentum was equal to n , where n is an integer. That is, he made the bold assumption that orbital angular momentum is quantized. From this simple model, the rotational frequencies of the electron in the various orbits can be estimated (see Problem 43). SUMMARY The total kinetic energy of a rigid object rolling on a rough surface without slipping equals the rotational kinetic energy about its center of mass, 1 I CM 2, plus the 2 1 translational kinetic energy of the center of mass, 2 Mv CM2: K The torque to be 1 2 2 I CM 1 2 2 Mv CM (11.4) due to a force F about an origin in an inertial frame is defined r F (11.7) B is a vector C having a (11.9) Given two vectors A and B, the cross product A magnitude C AB sin where is the angle between A and B. The direction of the vector C A B is perpendicular to the plane formed by A and B, and this direction is determined by the right-hand rule. The angular momentum L of a particle having linear momentum p mv is L r p (11.15) where r is the vector position of the particle relative to an origin in an inertial frame. The net external torque acting on a particle or rigid object is equal to the time rate of change of its angular momentum: ext dL dt (11.20) The z component of angular momentum of a rigid object rotating about a fixed z axis is Lz I (11.21) Questions 349 where I is the moment of inertia of the object about the axis of rotation and is its angular speed. The net external torque acting on a rigid object equals the product of its moment of inertia about the axis of rotation and its angular acceleration: ext I (11.23) If the net external torque acting on a system is zero, then the total angular momentum of the system is constant. Applying this law of conservation of angular momentum to a system whose moment of inertia changes gives Ii i If f constant (11.27) QUESTIONS 1. Is it possible to calculate the torque acting on a rigid body without specifying a center of rotation? Is the torque independent of the location of the center of rotation? 2. Is the triple product defined by A (B C) a scalar or a vector quantity? Explain why the operation (A B) C has no meaning. 3. In some motorcycle races, the riders drive over small hills, and the motorcycles become airborne for a short time. If a motorcycle racer keeps the throttle open while leaving the hill and going into the air, the motorcycle tends to nose upward. Why does this happen? 4. If the torque acting on a particle about a certain origin is zero, what can you say about its angular momentum about that origin? 5. Suppose that the velocity vector of a particle is completely specified. What can you conclude about the direction of its angular momentum vector with respect to the direction of motion? 6. If a single force acts on an object, and the torque caused by that force is nonzero about some point, is there any other point about which the torque is zero? 7. If a system of particles is in motion, is it possible for the total angular momentum to be zero about some origin? Explain. 8. A ball is thrown in such a way that it does not spin about its own axis. Does this mean that the angular momentum is zero about an arbitrary origin? Explain. 9. In a tape recorder, the tape is pulled past the read-andwrite heads at a constant speed by the drive mechanism. Consider the reel from which the tape is pulled -- as the tape is pulled off it, the radius of the roll of remaining tape decreases. How does the torque on the reel change with time? How does the angular speed of the reel change with time? If the tape mechanism is suddenly turned on so that the tape is quickly pulled with a great force, is the tape more likely to break when pulled from a nearly full reel or a nearly empty reel? 10. A scientist at a hotel sought assistance from a bellhop to carry a mysterious suitcase. When the unaware bellhop rounded a corner carrying the suitcase, it suddenly moved away from him for some unknown reason. At this point, the alarmed bellhop dropped the suitcase and ran off. What do you suppose might have been in the suitcase? 11. When a cylinder rolls on a horizontal surface as in Figure 11.3, do any points on the cylinder have only a vertical component of velocity at some instant? If so, where are they? 12. Three objects of uniform density -- a solid sphere, a solid cylinder, and a hollow cylinder -- are placed at the top of an incline (Fig. Q11.12). If they all are released from rest at the same elevation and roll without slipping, which object reaches the bottom first? Which reaches it last? You should try this at home and note that the result is independent of the masses and the radii of the objects. Figure Q11.12 Which object wins the race? 13. A mouse is initially at rest on a horizontal turntable mounted on a frictionless vertical axle. If the mouse begins to walk around the perimeter, what happens to the turntable? Explain. 14. Stars originate as large bodies of slowly rotating gas. Because of gravity, these regions of gas slowly decrease in size. What happens to the angular speed of a star as it shrinks? Explain. 15. Often, when a high diver wants to execute a flip in midair, she draws her legs up against her chest. Why does this make her rotate faster? What should she do when she wants to come out of her flip? 16. As a tether ball winds around a thin pole, what happens to its angular speed? Explain. 350 CHAPTER 11 Rolling Motion and Angular Momentum 20. Two balls have the same size and mass. One is hollow, whereas the other is solid. How would you determine which is which without breaking them apart? 21. A particle is moving in a circle with constant speed. Locate one point about which the particle's angular momentum is constant and another about which it changes with time. 22. If global warming occurs over the next century, it is likely that some polar ice will melt and the water will be distributed closer to the equator. How would this change the moment of inertia of the Earth? Would the length of the day (one revolution) increase or decrease? 17. Two solid spheres -- a large, massive sphere and a small sphere with low mass -- are rolled down a hill. Which sphere reaches the bottom of the hill first? Next, a large, low-density sphere and a small, high-density sphere having the same mass are rolled down the hill. Which one reaches the bottom first in this case? 18. Suppose you are designing a car for a coasting race -- the cars in this race have no engines; they simply coast down a hill. Do you want to use large wheels or small wheels? Do you want to use solid, disk-like wheels or hoop-like wheels? Should the wheels be heavy or light? 19. Why do tightrope walkers carry a long pole to help themselves keep their balance? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 11.1 Rolling Motion of a Rigid Object WEB v 1. A cylinder of mass 10.0 kg rolls without slipping on a horizontal surface. At the instant its center of mass has a speed of 10.0 m/s, determine (a) the translational kinetic energy of its center of mass, (b) the rotational energy about its center of mass, and (c) its total energy. 2. A bowling ball has a mass of 4.00 kg, a moment of inertia of 1.60 10 2 kg m2, and a radius of 0.100 m. If it rolls down the lane without slipping at a linear speed of 4.00 m/s, what is its total energy? 3. A bowling ball has a mass M, a radius R, and a moment of inertia 2MR 2. If it starts from rest, how much work 5 must be done on it to set it rolling without slipping at a linear speed v? Express the work in terms of M and v. 4. A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from rest and roll withouotal torque about the point O be zero. Will the total torque change if F3 is applied not at B, but rather at any other point along BC ? WEB x 4.00 kg v Figure P11.19 20. A 1.50-kg particle moves in the xy plane with a velocity of v (4.20i 3.60j) m/s. Determine the particle's angular momentum when its position vector is r (1.50i 2.20j) m. 21. The position vector of a particle of mass 2.00 kg is given as a function of time by r (6.00i 5.00t j) m. Determine the angular momentum of the particle about the origin as a function of time. 22. A conical pendulum consists of a bob of mass m in motion in a circular path in a horizontal plane, as shown in Figure P11.22. During the motion, the supporting wire of length maintains the constant angle with the vertical. Show that the magnitude of the angular momen- WEB 1. 00 m 352 CHAPTER 11 tum of the mass about the center of the circle is L (m 2g 3 Rolling Motion and Angular Momentum v1 = vxi i sin4 /cos )1/2 vi O R v2 Figure P11.25 cle about the origin when the particle is (a) at the origin, (b) at the highest point of its trajectory, and (c) just about to hit the ground. (d) What torque causes its angular momentum to change? 26. Heading straight toward the summit of Pike's Peak, an airplane of mass 12 000 kg flies over the plains of Kansas at a nearly constant altitude of 4.30 km and with a constant velocity of 175 m/s westward. (a) What is the airplane's vector angular momentum relative to a wheat farmer on the ground directly below the airplane? (b) Does this value change as the airplane continues its motion along a straight line? (c) What is its angular momentum relative to the summit of Pike's Peak? 27. A ball of mass m is fastened at the end of a flagpole connected to the side of a tall building at point P, as shown in Figure P11.27. The length of the flagpole is , and is the angle the flagpole makes with the horizontal. Suppose that the ball becomes loose and starts to fall. Determine the angular momentum (as a function of time) of the ball about point P. Neglect air resistance. m Figure P11.22 23. A particle of mass m moves in a circle of radius R at a constant speed v, as shown in Figure P11.23. If the motion begins at point Q, determine the angular momentum of the particle about point P as a function of time. y v m R P Q x m Figure P11.23 24. A 4.00-kg mass is attached to a light cord that is wound around a pulley (see Fig. 10.20). The pulley is a uniform solid cylinder with a radius of 8.00 cm and a mass of 2.00 kg. (a) What is the net torque on the system about the point O? (b) When the mass has a speed v, the pulley has an angular speed v/R. Determine the total angular momentum of the system about O. (c) Using the fact that d L/dt and your result from part (b), calculate the acceleration of the mass. 25. A particle of mass m is shot with an initial velocity vi and makes an angle with the horizontal, as shown in Figure P11.25. The particle moves in the gravitational field of the Earth. Find the angular momentum of the parti- P Figure P11.27 28. A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. The hose passes between the fireman's feet, which are 1.30 m vertically below the nozzle. Choose the origin to be inside the hose between the fireman's Problems feet. What torque must the fireman exert on the hose? That is, what is the rate of change of angular momentum of the water? 353 34. Section 11.4 Angular Momentum of a Rotating Rigid Object 29. A uniform solid sphere with a radius of 0.500 m and a mass of 15.0 kg turns counterclockwise about a vertical axis through its center. Find its vector angular momentum when its angular speed is 3.00 rad/s. 30. A uniform solid disk with a mass of 3.00 kg and a radius of 0.200 m rotates about a fixed axis perpendicular to its face. If the angular speed is 6.00 rad/s, calculate the angular momentum of the disk when the axis of rotation (a) passes through its center of mass and (b) passes through a point midway between the center and the rim. 31. A particle with a mass of 0.400 kg is attached to the 100-cm mark of a meter stick with a mass of 0.100 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 4.00 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the 50.0-cm mark and (b) perpendicular to the table through the 0-cm mark. 32. The hour and minute hands of Big Ben, the famous Parliament Building tower clock in London, are 2.70 m and 4.50 m long and have masses of 60.0 kg and 100 kg, respectively. Calculate the total angular momentum of these hands about the center point. Treat the hands as long thin rods. 35. 36. Section 11.5 Conservation of Angular Momentum 33. A cylinder with a moment of inertia of I1 rotates about a vertical, frictionless axle with angular velocity i . A second cylinder that has a moment of inertia of I 2 and initially is not rotating drops onto the first cylinder (Fig. P11.33). Because of friction between the surfaces, the two eventually reach the same angular speed f . (a) Calculate f . (b) Show that the kinetic energy of the system decreases in this interaction and calculate WEB 37. 38. the ratio of the final rotational energy to the initial rotational energy. A playground merry-go-round of radius R 2.00 m has a moment of inertia of I 250 kg m2 and is rotating at 10.0 rev/min about a frictionless vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? A student sits on a freely rotating stool holding two weights, each of which has a mass of 3.00 kg. When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kg m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. (a) Find the new angular speed of the student. (b) Find the kinetic energy of the rotating system before and after he pulls the weights inward. A uniform rod with a mass of 100 g and a length of 50.0 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin passing through its center. Two small beads, each having a mass 30.0 g, are mounted on the rod so that they are able to slide without friction along its length. Initially, the beads are held by catches at positions 10.0 cm on each side of center; at this time, the system rotates at an angular speed of 20.0 rad/s. Suddenly, the catches are released, and the small beads slide outward along the rod. Find (a) the angular speed of the system at the instant the beads reach the ends of the rod and (b) the angular speed of the rod after the beads fly off the rod's ends. A 60.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 500 kg m2 and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.50 m/s relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion? A puck with a mass of 80.0 g and a radius of 4.00 cm slides along an air table at a speed of 1.50 m/s, as shown in Figure P11.38a. It makes a glancing collision I2 1.50 m/s i I1 f Before After (a) (b) Figure P11.33 Figure P11.38 354 CHAPTER 11 Rolling Motion and Angular Momentum with a second puck having a radius of 6.00 cm and a mass of 120 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision (Fig. P11.38b). (a) What is the angular momentum of the system relative to the center of mass? the speed of the topmost point on the ball's surface relative to the center of mass. Find the ratio v CM /vr . 54. A projectile of mass m moves to the right with speed vi (Fig. P11.54a). The projectile strikes and sticks to the end of a stationary rod of mass M and length d that is pivoted about a frictionless axle through its center (Fig. P11.54b). (a) Find the angular speed of the system right after the collision. (b) Determine the fractional loss in mechanical energy due to the collision. Problems 357 m vi O d O (a) (b) Figure P11.54 WEB cal grape at the top of his bald head, which itself has the shape of a sphere. After all of the children have had time to giggle, the grape starts from rest and rolls down your uncle's head without slipping. It loses contact with your uncle's scalp when the radial line joining it to the center of curvature makes an angle with the vertical. What is the measure of angle ? 58. A thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) Determine the speed of its center of mass just before it hits the horizontal surface. (b) Now suppose that the rod has a fixed pivot at its lower end. Determine the speed of the rod's center of mass just before it hits the surface. 59. Two astronauts (Fig. P11.59), each having a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. (c) What is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronaut in shortening the rope? 60. Two astronauts (see Fig. P11.59), each having a mass M, are connected by a rope of length d having negligible mass. They are isolated in space, orbiting their center of mass at speeds v. Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, one of the astronauts shortens the distance between them to d/2. (c) What is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronaut in shortening the rope? 55. A mass m is attached to a cord passing through a small hole in a frictionless, horizontal surface (Fig. P11.55). The mass is initially orbiting with speed vi in a circle of radius ri . The cord is then slowly pulled from below, and the radius of the circle decreases to r. (a) What is the speed of the mass when the radius is r ? (b) Find the tension in the cord as a function of r. (c) How much work W is done in moving m from ri to r ? (Note: The tension depends on r.) (d) Obtain numerical values for v, T, and W when r 0.100 m, m 50.0 g, ri 0.300 m, and vi 1.50 m/s. ri m vi Figure P11.55 CM 56. A bowler releases a bowling ball with no spin, sending it sliding straight down the alley toward the pins. The ball continues to slide for some distance before its motion becomes rolling without slipping; of what order of magnitude is this distance? State the quantities you take as data, the values you measure or estimate for them, and your reasoning. 57. Following Thanksgiving dinner, your uncle falls into a deep sleep while sitting straight up and facing the television set. A naughty grandchild balances a small spheri- d Figure P11.59 Problems 59 and 60. 61. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to ro- 358 CHAPTER 11 Rolling Motion and Angular Momentum tate about an axis AB (Fig. P11.61). A bullet of mass m and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becomes embedded in the cube. Find the minimum value of v required to tip the cube so that it falls on face ABCD. Assume m V M. C 2a v 4a/3 D B A Figure P11.64 Problems 64 and 65. Figure P11.61 62. A large, cylindrical roll of paper of initial radius R lies on a long, horizontal surface with the open end of the paper nailed to the surface. The roll is given a slight shove (vi 0) and begins to unroll. (a) Determine the speed of the center of mass of the roll when its radius has diminished to r. (b) Calculate a numerical value for this speed at r 1.00 mm, assuming R 6.00 m. (c) What happens to the energy of the system when the paper is completely unrolled? (Hint: Assume that the roll has a uniform density and apply energy methods.) 63. A spool of wire of mass M and radius R is unwound under a constant force F (Fig. P11.63). Assuming that the spool is a uniform solid cylinder that does not slip, show that (a) the acceleration of the center of mass is 4F/3M and that (b) the force of friction is to the right and is equal in magnitude to F/3. (c) If the cylinder starts from rest and rolls without slipping, what is the speed of its center of mass after it has rolled through a distance d ? F a horizontal surface and released, as shown in Figure P11.64. (a) What is the angular speed of the disk once pure rolling takes place? (b) Find the fractional loss in kinetic energy from the time the disk is released until the time pure rolling occurs. (Hint: Consider torques about the center of mass.) 65. Suppose a solid disk of radius R is given an angular speed i about an axis through its center and is then lowered to a horizontal surface and released, as shown in Problem 64 (see Fig. P11.64). Furthermore, assume that the coefficient of friction between the disk and the surface is . (a) Show that the time it takes for pure rolling motion to occur is R i /3 g. (b) Show that the distance the disk travels before pure rolling occurs is R 2 i 2/18 g. 66. A solid cube of side 2a and mass M is sliding on a frictionless surface with uniform velocity v, as shown in Figure P11.66a. It hits a small obstacle at the end of the table; this causes the cube to tilt, as shown in Figure M 2a M R Mg v (b) Figure P11.63 64. A uniform solid disk is set into rotation with an angular speed i about an axis through its center. While still rotating at this speed, the disk is placed into contact with (a) Figure P11.66 Problems P11.66b. Find the minimum value of v such that the cube falls off the table. Note that the moment of inertia of the cube about an axis along one of its edges is 8Ma 2/3. (Hint: The cube undergoes an inelastic collision at the edge.) 67. A plank with a mass M 6.00 kg rides on top of two identical solid cylindrical rollers that have R 5.00 cm and m 2.00 kg (Fig. P11.67). The plank is pulled by a constant horizontal force of magnitude F 6.00 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. Also, no slipping occurs between the cylinders and the plank. (a) Find the acceleration of the plank and that of the rollers. (b) What frictional forces are acting? 359 that the critical angle for which the spool does not slip and remains stationary is cos c r R M F m R m R (Hint: At the critical angle, the line of action of the applied force passes through the contact point.) 70. In a demonstration that employs a ballistics cart, a ball is projected vertically upward from a cart moving with constant velocity along the horizontal direction. The ball lands in the catching cup of the cart because both the cart and the ball have the same horizontal component of velocity. Now consider a ballistics cart on an incline making an angle with the horizontal, as shown in Figure P11.70. The cart (including its wheels) has a mass M, and the moment of inertia of each of the two wheels is mR 2/2. (a) Using conservation of energy considerations (assuming that there is no friction between the cart and the axles) and assuming pure rolling motion (that is, no slipping), show that the acceleration of the cart along the incline is ax M M 2m g sin Figure P11.67 68. A spool of wire rests on a horizontal surface as in Figure P11.68. As the wire is pulled, the spool does not slip at the contact point P. On separate trials, each one of the forces F1 , F2 , F3 , and F4 is applied to the spool. For each one of these forces, determine the direction in which the spool will roll. Note that the line of action of F2 passes through P. (b) Note that the x component of acceleration of the ball released by the cart is g sin . Thus, the x component of the cart's acceleration is smaller than that of the ball by the factor M/(M 2m). Use this fact and kinematic equations to show that the ball overshoots the cart by an amount x, where x 4m M 2m sin cos2 v yi 2 g F3 F2 and vyi is the initial speed of the ball imparted to it by the spring in the cart. (c) Show that the distance d that the ball travels measured along the incline is d 2vyi 2 g sin cos2 F4 r R c F1 y x P Figure P11.68 Problems 68 and 69. x 69. The spool of wire shown in Figure P11.68 has an inner radius r and an outer radius R. The angle between the applied force and the horizontal can be varied. Show Figure P11.70 360 CHAPTER 11 Rolling Motion and Angular Momentum ANSWERS TO QUICK QUIZZES 11.1 There is very little resistance to motion that can reduce the kinetic energy of the rolling ball. Even though there is friction between the ball and the floor (if there were not, then no rotation would occur, and the ball would slide), there is no relative motion of the two surfaces (by the definition of "rolling"), and so kinetic friction cannot reduce K. (Air drag and friction associated with deformation of the ball eventually stop the ball.) 11.2 The box. Because none of the box's initial potential energy is converted to rotational kinetic energy, at any time t 0 its translational kinetic energy is greater than that of the rolling ball. 11.3 Zero. If she were moving directly toward the pole, r and p would be antiparallel to each other, and the sine of the angle between them is zero; therefore, L 0. 11.4 Both (a) and (b) are false. The net force is not necessarily zero. If the line of action of the net force passes through the point, then the net torque about an axis passing through that point is zero even though the net force is not zero. Because the net force is not necessarily zero, you cannot conclude that the particle's velocity is constant. 11.5 The student does work as he walks from the rim of the platform toward its center. 11.6 Because it is perpendicular to the precessional motion of the top, the force of gravity does no work. This is a situation in which a force causes motion but does no work. 2.2 This is the Nearest One Head 361 P U Z Z L E R This one-bottle wine holder is an interesting example of a mechanical system that seems to defy gravity. The system (holder plus bottle) is balanced when its center of gravity is directly over the lowest support point. What two conditions are necessary for an object to exhibit this kind of stability? (Charles D. Winters) c h a p t e r Static Equilibrium and Elasticity Chapter Outline 12.1 The Conditions for Equilibrium 12.2 More on the Center of Gravity 12.3 Examples of Rigid Objects in Static Equilibrium 12.4 Elastic Properties of Solids 361 362 CHAPTER 12 Static Equilibrium and Elasticity I n Chapters 10 and 11 we studied the dynamics of rigid objects -- that is, objects whose parts remain at a fixed separation with respect to each other when subjected to external forces. Part of this chapter addresses the conditions under which a rigid object is in equilibrium. The term equilibrium implies either that the object is at rest or that its center of mass moves with constant velocity. We deal here only with the former case, in which the object is described as being in static equilibrium. Static equilibrium represents a common situation in engineering practice, and the principles it involves are of special interest to civil engineers, architects, and mechanical engineers. If you are an engineering student you will undoubtedly take an advanced course in statics in the future. The last section of this chapter deals with how objects deform under load conditions. Such deformations are usually elastic and do not affect the conditions for equilibrium. An elastic object returns to its original shape when the deforming forces are removed. Several elastic constants are defined, each corresponding to a different type of deformation. 12.1 THE CONDITIONS FOR EQUILIBRIUM F P In Chapter 5 we stated that one necessary condition for equilibrium is that the net force acting on an object be zero. If the object is treated as a particle, then this is the only condition that must be satisfied for equilibrium. The situation with real (extended) objects is more complex, however, because these objects cannot be treated as particles. For an extended object to be in static equilibrium, a second condition must be satisfied. This second condition involves the net torque acting on the extended object. Note that equilibrium does not require the absence of motion. For example, a rotating object can have constant angular velocity and still be in equilibrium. Consider a single force F acting on a rigid object, as shown in Figure 12.1. The effect of the force depends on its point of application P. If r is the position vector of this point relative to O, the torque associated with the force F about O is given by Equation 11.7: r F d r O Figure 12.1 A single force F acts on a rigid object at the point P. Recall from the discussion of the vector product in Section 11.2 that the vector is perpendicular to the plane formed by r and F. You can use the right-hand rule to determine the direction of : Curl the fingers of your right hand in the direction of rotation that F tends to cause about an axis through O : your thumb then points in the direction of . Hence, in Figure 12.1 is directed toward you out of the page. As you can see from Figure 12.1, the tendency of F to rotate the object about an axis through O depends on the moment arm d, as well as on the magnitude of F. Recall that the magnitude of is Fd (see Eq. 10.19). Now suppose a rigid object is acted on first by force F1 and later by force F2 . If the two forces have the same magnitude, they will produce the same effect on the object only if they have the same direction and the same line of action. In other words, two forces F1 and F2 are equivalent if and only if F1 two produce the same torque about any axis. F2 and if and only if the Equivalent forces The two forces shown in Figure 12.2 are equal in magnitude and opposite in direction. They are not equivalent. The force directed to the right tends to rotate 12.1 The Conditions for Equilibrium 363 the object clockwise about an axis perpendicular to the diagram through O, whereas the force directed to the left tends to rotate it counterclockwise about that axis. Suppose an object is pivoted about an axis through its center of mass, as shown in Figure 12.3. Two forces of equal magnitude act in opposite directions along parallel lines of action. A pair of forces acting in this manner form what is called a couple. (The two forces shown in Figure 12.2 also form a couple.) Do not make the mistake of thinking that the forces in a couple are a result of Newton's third law. They cannot be third-law forces because they act on the same object. Third-law force pairs act on different objects. Because each force produces the same torque Fd, the net torque has a magnitude of 2Fd. Clearly, the object rotates clockwise and undergoes an angular acceleration about the axis. With respect to rotational motion, this is a nonequilibrium situation. The net torque on the obaccording to the relationship ject gives rise to an angular acceleration 2Fd I (see Eq. 10.21). In general, an object is in rotational equilibrium only if its angular acceleration 0. Because I for rotation about a fixed axis, our second necessary condition for equilibrium is that the net torque about any axis must be zero. We now have two necessary condieinforce the concrete, as illustrated in Figure 12.17b. Because concrete is much stronger under compression (squeezing) than under tension (stretching) or shear, vertical columns of concrete can support very heavy loads, whereas horizontal beams of concrete tend to sag and crack. However, a significant increase in shear strength is achieved if the reinforced concrete is prestressed, as shown in Figure 12.17c. As the concrete is being poured, the steel rods are held under tension by external forces. The external Load force Concrete Cracks Steel reinforcing rod Steel rod under tension (a) (b) (c) Figure 12.17 (a) A concrete slab with no reinforcement tends to crack under a heavy load. (b) The strength of the concrete is increased by using steel reinforcement rods. (c) The concrete is further strengthened by prestressing it with steel rods under tension. 376 CHAPTER 12 Static Equilibrium and Elasticity forces are released after the concrete cures; this results in a permanent tension in the steel and hence a compressive stress on the concrete. This enables the concrete slab to support a much heavier load. EXAMPLE 12.6 Stage Design The radius of the wire can be found from A r r 2: 3 Recall Example 8.10, in which we analyzed a cable used to support an actor as he swung onto the stage. The tension in the cable was 940 N. What diameter should a 10-m-long steel wire have if we do not want it to stretch more than 0.5 cm under these conditions? From the definition of Young's modulus, we can solve for the required cross-sectional area. Assuming that the cross section is circular, we can determine the diameter of the wire. From Equation 12.6, we have Y A F/A L/L i FL i Y L (20 (940 N)(10 m) 10 10 N/m2)(0.005 m) 9.4 10 6 ! ! A 2r 9.4 10 6 m2 1.7 10 m 1.7 mm d 2(1.7 mm) 3.4 mm Solution To provide a large margin of safety, we would probably use a flexible cable made up of many smaller wires having a total cross-sectional area substantially greater than our calculated value. m2 EXAMPLE 12.7 Squeezing a Brass Sphere V Vi P B A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 105 N/m2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth at which the pressure is 2.0 107 N/m2. The volume of the sphere in air is 0.50 m3. By how much does this volume change once the sphere is submerged? Because the final pressure is so much greater than the initial pressure, we can neglect the initial pressure and state that P Pf Pi Pf 2.0 10 7 N/m2. Therefore, V (0.50 m3)(2.0 10 7 N/m2) 6.1 10 10 N/m2 1.6 10 4 Solution From the definition of bulk modulus, we have B P V/Vi m3 The negative sign indicates a decrease in volume. SUMMARY A rigid object is in equilibrium if and only if the resultant external force acting on it is zero and the resultant external torque on it is zero about any axis: F 0 0 (12.1) (12.2) The first condition is the condition for translational equilibrium, and the second is the condition for rotational equilibrium. These two equations allow you to analyze a great variety of problems. Make sure you can identify forces unambiguously, create a free-body diagram, and then apply Equations 12.1 and 12.2 and solve for the unknowns. Problems 377 The force of gravity exerted on an object can be considered as acting at a single point called the center of gravity. The center of gravity of an object coincides with its center of mass if the object is in a uniform gravitational field. We can describe the elastic properties of a substance using the concepts of stress and strain. Stress is a quantity proportional to the force producing a deformation; strain is a measure of the degree of deformation. Strain is proportional to stress, and the constant of proportionality is the elastic modulus: Elastic modulus stress strain (12.5) Three common types of deformation are (1) the resistance of a solid to elongation under a load, characterized by Young's modulus Y ; (2) the resistance of a solid to the motion of internal planes sliding past each other, characterized by the shear modulus S ; and (3) the resistance of a solid or fluid to a volume change, characterized by the bulk modulus B. QUESTIONS 1. Can a body be in equilibrium if only one external force acts on it? Explain. 2. Can a body be in equilibrium if it is in motion? Explain. 3. Locate the center of gravity for the following uniform objects: (a) sphere, (b) cube, (c) right circular cylinder. 4. The center of gravity of an object may be located outside the object. Give a few examples for which this is the case. 5. You are given an arbitrarily shaped piece of plywood, together with a hammer, nail, and plumb bob. How could you use these items to locate the center of gravity of the plywood? (Hint: Use the nail to suspend the plywood.) 6. For a chair to be balanced on one leg, where must the center of gravity of the chair be located? 7. Can an object be in equilibrium if the only torques acting on it produce clockwise rotation? 8. A tall crate and a short crate of equal mass are placed side by side on an incline (without touching each other). As the incline angle is increased, which crate will topple first? Explain. 9. When lifting a heavy object, why is it recommended to keep the back as vertical as possible, lifting from the knees, rather than bending over and lifting from the waist? Give a few examples in which several forces are acting on a system in such a way that their sum is zero but the system is not in equilibrium. If you measure the net torque and the net force on a system to be zero, (a) could the system still be rotating with respect to you? (b) Could it be translating with respect to you? A ladder is resting inclined against a wall. Would you feel safer climbing up the ladder if you were told that the ground is frictionless but the wall is rough or that the wall is frictionless but the ground is rough? Justify your answer. What kind of deformation does a cube of Jell-O exhibit when it "jiggles"? Ruins of ancient Greek temples often have intact vertical columns, but few horizontal slabs of stone are still in place. Can you think of a reason why this is so? 10. 11. 12. 13. 14. PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 12.1 The Conditions for Equilibrium 1. A baseball player holds a 36-oz bat (weight 10.0 N) with one hand at the point O (Fig. P12.1). The bat is in equilibrium. The weight of the bat acts along a line 60.0 cm to the right of O. Determine the force and the torque exerted on the bat by the player. O 60.0 cm mg Figure P12.1 378 CHAPTER 12 Static Equilibrium and Elasticity 2. Write the necessary conditions of equilibrium for the body shown in Figure P12.2. Take the origin of the torque equation at the point O. Section 12.2 More on the Center of Gravity 5. A 3.00-kg particle is located on the x axis at x 5.00 m, and a 4.00-kg particle is located on the x axis at x 3.00 m. Find the center of gravity of this twoparticle system. 6. A circular pizza of radius R has a circular piece of radius R/2 removed from one side, as shown in Figure P12.6. Clearly, the center of gravity has moved from C to C along the x axis. Show that the distance from C to C is R/6. (Assume that the thickness and density of the pizza are uniform throughout.) Fy Fx Ry Rx O Fg Figure P12.2 WEB C C 3. A uniform beam of mass m b and length supports blocks of masses m 1 and m 2 at two positions, as shown in Figure P12.3. The beam rests on two points. For what value of x will the beam be balanced at P such that the normal force at O is zero? Figure P12.6 7. A carpenter's square has the shape of an L, as shown in Figure P12.7. Locate its center of gravity. 2 m1 O CG d P m2 4.0 cm x Figure P12.3 4. A student gets his car stuck in a snow drift. Not at a loss, having studied physics, he attaches one end of a stout rope to the vehicle and the gravity? (b) The couple in the bottom left corner moves 10.0 m to the right. Where is the new center of gravity? (c) What was the average velocity of the center of gravity if it took that couple 8.00 s to change position? 61. A shelf bracket is mounted on a vertical wall by a single screw, as shown in Figure P12.61. Neglecting the weight of the bracket, find the horizontal component of the force that the screw exerts on the bracket when an 80.0-N vertical force is applied as shown. (Hint: Imagine that the bracket is slightly loose.) static friction between the cylinder and all surfaces is 0.500. In terms of Fg , find the maximum force P that can be applied that does not cause the cylinder to rotate. (Hint: When the cylinder is on the verge of slipping, both friction forces are at their maximum values. Why?) 63. Review Problem. A wire of length L i , Young's modulus Y, and cross-sectional area A is stretched elastically by an amount L. According to Hooke's law, the restoring force is k L. (a) Show that k YA/L i . (b) Show that the work done in stretching the wire by an amount L is W YA( L)2/2L i . 64. Two racquetballs are placed in a glass jar, as shown in Figure P12.64. Their centers and the point A lie on a straight line. (a) Assuming that the walls are frictionless, determine P1 , P2 , and P3 . (b) Determine the magnitude of the force exerted on the right ball by the left ball. Assume each ball has a mass of 170 g. 80.0 N 5.00 cm 3.00 cm P1 P3 P2 A 6.00 cm Figure P12.61 62. Figure P12.62 shows a vertical force applied tangentially to a uniform cylinder of weight Fg . The coefficient of Figure P12.64 65. In Figure P12.65, the scales read Fg1 380 N and Fg 2 320 N. Neglecting the weight of the supporting plank, Problems 2.00 m 387 Fg1 Fg 2 Figure P12.65 how far from the woman's feet is her center of mass, given that her height is 2.00 m? A steel cable 3.00 cm2 in cross-sectional area has a mass of 2.40 kg per meter of length. If 500 m of the cable is hung over a vertical cliff, how much does the cable stretch under its own weight? (For Young's modulus for steel, refer to Table 12.1.) (a) Estimate the force with which a karate master strikes a board if the hand's speed at time of impact is 10.0 m/s and decreases to 1.00 m/s during a 0.002 00-s time-of-contact with the board. The mass of coordinated hand-and-arm is 1.00 kg. (b) Estimate the shear stress if this force is exerted on a 1.00-cm-thick pine board that is 10.0 cm wide. (c) If the maximum shear stress a pine board can receive before breaking is 3.60 106 N/m2, will the board break? A bucket is made from thin sheet metal. The bottom and top of the bucket have radii of 25.0 cm and 35.0 cm, respectively. The bucket is 30.0 cm high and filled with water. Where is the center of gravity? (Ignore the weight of the bucket itself.) Review Problem. A trailer with a loaded weight of Fg is being pulled by a vehicle with a force P, as illustrated in Figure P12.69. The trailer is loaded such that its center of mass is located as shown. Neglect the force of rolling friction and let a represent the x component of the acceleration of the trailer. (a) Find the vertical component of P in terms of the given parameters. (b) If a 2.00 m/s2 and h 1.50 m, what must be the value of d 66. so that Py 0 (that is, no vertical load on the vehicle)? (c) Find the values of Px and Py given that Fg 1 500 N, d 0.800 m, L 3.00 m, h 1.50 m, and a 2.00 m/s2. 70. Review Problem. An aluminum wire is 0.850 m long and has a circular cross section of diameter 0.780 mm. Fixed at the top end, the wire supports a 1.20-kg mass that swings in a horizontal circle. Determine the angular velocity required to produce strain 1.00 10 3. 71. A 200-m-long bridge truss extends across a river (Fig. P12.71). Calculate the force of tension or compression in each structural component when a 1 360-kg car is at the center of the bridge. Assume that the structure is free to slide horizontally to permit thermal expansion and contraction, that the structural components are connected by pin joints, and that the massinusoidal function of time with no loss in mechanical energy. In real mechanical systems, damping (frictional) forces are often present. These forces are considered in optional Section 13.6 at the end of this chapter. A 13.1 8.10 SIMPLE HARMONIC MOTION Fs (a) x Fs = 0 m x=0 m x x=0 x m x=0 (b) x Consider a physical system that consists of a block of mass m attached to the end of a spring, with the block free to move on a horizontal, frictionless surface (Fig. 13.1). When the spring is neither stretched nor compressed, the block is at the position x 0, called the equilibrium position of the system. We know from experience that such a system oscillates back and forth if disturbed from its equilibrium position. We can understand the motion in Figure 13.1 qualitatively by first recalling that when the block is displaced a small distance x from equilibrium, the spring exerts on the block a force that is proportional to the displacement and given by Hooke's law (see Section 7.3): Fs kx (13.1) Fs (c) x Figure 13.1 A block attached to a spring moving on a frictionless surface. (a) When the block is displaced to the right of equilibrium (x 0), the force exerted by the spring acts to the left. (b) When the block is at its equilibrium position (x 0), the force exerted by the spring is zero. (c) When the block is displaced to the left of equilibrium (x 0), the force exerted by the spring acts to the right. We call this a restoring force because it is is always directed toward the equilibrium position and therefore opposite the displacement. That is, when the block is displaced to the right of x 0 in Figure 13.1, then the displacement is positive and the restoring force is directed to the left. When the block is displaced to the left of x 0, then the displacement is negative and the restoring force is directed to the right. Applying Newton's second law to the motion of the block, together with Equation 13.1, we obtain Fs a kx k x m ma (13.2) That is, the acceleration is proportional to the displacement of the block, and its direction is opposite the direction of the displacement. Systems that behave in this way are said to exhibit simple harmonic motion. An object moves with simple harmonic motion whenever its acceleration is proportional to its displacement from some equilibrium position and is oppositely directed. 13.1 Simple Harmonic Motion 391 m Motion of paper Figure 13.2 An experimental apparatus for demonstrating simple harmonic motion. A pen attached to the oscillating mass traces out a wavelike pattern on the moving chart paper. 8.2 & 8.3 An experimental arrangement that exhibits simple harmonic motion is illustrated in Figure 13.2. A mass oscillating vertically on a spring has a pen attached to it. While the mass is oscillating, a sheet of paper is moved perpendicular to the direction of motion of the spring, and the pen traces out a wavelike pattern. In general, a particle moving along the x axis exhibits simple harmonic motion when x, the particle's displacement from equilibrium, varies in time according to the relationship x A cos( t ) (13.3) Displacement versus time for simple harmonic motion where A, , and are constants. To give physical significance to these constants, we have labeled a plot of x as a function of t in Figure 13.3a. This is just the pattern that is observed with the experimental apparatus shown in Figure 13.2. The amplitude A of the motion is the maximum displacement of the particle in either the positive or negative x direction. The constant is called the angular frequency of the motion and has units of radians per second. (We shall discuss the geometric significance of in Section 13.2.) The constant angle , called the phase constant (or phase angle), is determined by the initial displacement and velocity of the particle. If the particle is at its maximum position x A at t 0, then 0 and the curve of x versus t is as shown in Figure 13.3b. If the particle is at some other position at t 0, the constants and A tell us what the position was at time t 0. The quantity ( t ) is called the physical pendulum. Consider a rigid body pivoted at a point O that is a distance d from the center of mass (Fig. 13.13). The force of gravity provides a torque about an axis through O, and the magnitude of that torque is mgd sin , where is as shown in Figure I , where I is the moment of inertia about 13.13. Using the law of motion 13.4 The Pendulum 405 the axis through O, we obtain mgd sin I d2 dt 2 Pivot O The minus sign indicates that the torque about O tends to decrease . That is, the force of gravity produces a restoring torque. Because this equation gives us the angular acceleration d 2 /dt 2 of the pivoted body, we can consider it the equation of motion for the system. If we again assume that is small, the approximation sin is valid, and the equation of motion reduces to d2 dt 2 mgd I 2 d d sin CM (13.27) mg Because this equation is of the same form as Equation 13.17, the motion is simple harmonic motion. That is, the solution of Equation 13.27 is ), max cos( t where max is the maximum angular displacement and Figure 13.13 lum. A physical pendu- The period is T 2 2 mgd I I mgd (13.28) Period of motion for a physical pendulum One can use this result to measure the moment of inertia of a flat rigid body. If the location of the center of mass -- and hence the value of d -- are known, the moment of inertia can be obtained by measuring the period. Finally, note that Equation 13.28 reduces to the period of a simple pendulum (Eq. 13.26) when I md 2 -- that is, when all the mass is concentrated at the center of mass. EXAMPLE 13.6 A Swinging Rod A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane (Fig. 13.14). Find the period of oscillation if the amplitude of the motion is small. In Chapter 10 we found that the moment of inertia of a uniform rod about an axis through one end is 1 2 3 ML . The distance d from the pivot to the center of mass is L/2. Substituting these quantities into Equation 13.28 gives T 2 Exercise Answer Calculate the period of a meter stick that is pivoted about one end and is oscillating in a vertical plane. 1.64 s. O Pivot Solution ML 2 L Mg 2 1 3 2 2L 3g L CM Comment In one of the Moon landings, an astronaut walking on the Moon's surface had a belt hanging from his space suit, and the belt oscillated as a physical pendulum. A scientist on the Earth observed this motion on television and used it to estimate the free-fall acceleration on the Moon. How did the scientist make this calculation? Mg Figure 13.14 A rigid rod oscillating about a pivot through one end is a physical pendulum with d L/2 and, from Table 10.2, I 1 ML 2. 3 406 CHAPTER 13 Oscillatory Motion Balance wheel O max P Figure 13.16 The balance wheel of this antique pocket watch is a torsional pendulum and regulates the time-keeping mechanism. Figure 13.15 A torsional pendulum consists of a rigid body suspended by a wire attached to a rigid support. The body oscillates about the line OP with an amplitude max . Torsional Pendulum Figure 13.15 shows a rigid body suspended by a wire attached at the top to a fixed support. When the body is twisted through some small angle , the twisted wire exerts on the body a restoring torque that is proportional to the angular displacement. That is, where (kappa) is called the torsion constant of the support wire. The value of can be obtained by applying a known torque to twist the wire through a measurable angle . Applying Newton's second law for rotational motion, we find I d2 dt 2 d2 dt 2 (13.29) I Again, this is the equation of motion for a simple harmonic oscillator, with /I and a period Period of motion for a torsional pendulum T 2 I (13.30) This system is called a torsional pendulum. There is no small-angle restriction in this situation as long as the elastic limit of the wire is not exceeded. Figure 13.16 shows the balance wheel of a watch oscillating as a torsional pendulum, energized by the mainspring. 13.5 COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR MOTION 8.8 We can bettery in the absence of a retarding force (the undamped oscillator) and is called the natural frequency of the system. When the magnitude of the maximum retarding force R max bv max kA, the system is said to be underdamped. As the value of R approaches kA, the amplitudes of the oscillations decrease more and more rapidly. This motion is represented by the blue curve in Figure 13.20. When b reaches a critical value bc such that bc /2m 0 , the system does not oscillate and is said to be critically damped. In this case the system, once released from rest at some nonequilibrium position, returns to equilibrium and then stays there. The graph of displacement versus time for this case is the red curve in Figure 13.20. If the medium is so viscous that the retarding force is greater than the restoring force -- that is, if R max bv max kA and b/2m 0 --the system is overdamped. Again, the displaced system, when free to move, does not oscillate but simply returns to its equilibrium position. As the damping increases, the time it takes the system to approach equilibrium also increases, as indicated by the black curve in Figure 13.20. In any case in which friction is present, whether the system is overdamped or underdamped, the energy of the oscillator eventually falls to zero. The lost mechanical energy dissipates into internal energy in the retarding medium. m (b) Figure 13.19 (a) Graph of displacement versus time for a damped oscillator. Note the decrease in amplitude with time. (b) One example of a damped oscillator is a mass attached to a spring and submersed in a viscous liquid. x b a c t Figure 13.20 Graphs of displacement versus time for (a) an underdamped oscillator, (b) a critically damped oscillator, and (c) an overdamped oscillator. 410 CHAPTER 13 Oscillatory Motion Oil or other viscous fluid Coil spring Shock absorber Piston with holes (a) (b) Figure 13.21 (a) A shock absorber consists of a piston oscillating in a chamber filled with oil. As the piston oscillates, the oil is squeezed through holes between the piston and the chamber, causing a damping of the piston's oscillations. (b) One type of automotive suspension system, in which a shock absorber is placed inside a coil spring at each wheel. web To learn more about shock absorbers, visit http://www.hdridecontrol.com Quick Quiz 13.6 An automotive suspension system consists of a combination of springs and shock absorbers, as shown in Figure 13.21. If you were an automotive engineer, would you design a suspension system that was underdamped, critically damped, or overdamped? Discuss each case. Optional Section 13.7 FORCED OSCILLATIONS It is possible to compensate for energy loss in a damped system by applying an external force that does positive work on the system. At any instant, energy can be put into the system by an applied force that acts in the direction of motion of the oscillator. For example, a child on a swing can be kept in motion by appropriately timed pushes. The amplitude of motion remains constant if the energy input per cycle exactly equals the energy lost as a result of damping. Any motion of this type is called forced oscillation. A common example of a forced oscillator is a damped oscillator driven by an external force that varies periodically, such as F F ext cos t, where is the angular frequency of the periodic force and Fext is a constant. Adding this driving force to the left side of Equation 13.32 gives F ext cos t kx b dx dt m d 2x dt 2 (13.35) (As earlier, we present the solution of this equation without proof.) After a sufficiently long period of time, when the energy input per cycle equals the energy lost per cycle, a steady-state condition is reached in which the oscillations proceed with constant amplitude. At this time, when the system is in a steady state, the solution of Equation 13.35 is x A cos( t ) (13.36) 13.7 Forced Oscillations 411 where A F ext/m ( 2 0 2)2 b m 2 (13.37) and where 0 k/m is the angular frequency of the undamped oscillator (b 0). One could argue that in steady state the oscillator must physically have the same frequency as the driving force, and thus the solution given by Equation 13.36 is expected. In fact, when this solution is substituted into Equation 13.35, one finds that it is indeed a solution, provided the amplitude is given by Equation 13.37. Equation 13.37 shows that, because an external force is driving it, the motion of the forced oscillator is not damped. The external agent provides the necessary energy to overcome the losses due to the retarding force. Note that the system oscillates at the angular frequency of the driving force. For small damping, the amplitude becomes very large when the frequency of the driving force is near the natural frequency of oscillation. The dramatic increase in amplitude near the natural frequency 0 is called resonance, and for this reason 0 is sometimes called the resonance frequency of the system. The reason for large-amplitude oscillations at the resonance frequency is that energy is being transferred to the system under the most favorable conditions. We can better understand this by taking the first time derivative of x in Equation 13.36, which gives an expression for the velocity of the oscillator. We find that v is proportional to sin( t ). When the applied force F is in phase with the velocity, the rate at which work is done on the oscillator by F equals the dot product F v. Remember that "rate at which work is done" is the definition of power. Because the product F v is a maximum when F and v are in phase, we conclude that at resonance the applied force is in phase with the velocity and that the power transferred to the oscillator is a maximum. Figure 13.22 is a graph of amplitude as a function of frequency for a forced oscillator with and without damping. Note that the amplitude increases with decreasing damping (b : 0) and that the resonance curve broadens as the damping increases. Under steady-state conditions and at any driving frequency, the energy transferred into the system equals the energy lost because of the damping force; hence, the average total energy of the oscillator remains constant. In the absence of a damping force (b 0), we see from Equation 13.37 that the steady-state amplitude approaches infinity as : 0 . In other words, if there are no losses in the system and if we continue to drive an initially motionless oscillator with a periodic force that is in phase with the velocity, the amplitude of motion builds without limit (see the red curve in Fig. 13.22). This limitless building does not occur in practice because some damping is always present. The behavior of a driven oscillating system after the driving force is removed depends on b and on how close was to 0 . This behavior is sometimes quantified by a parameter called the quality factor Q. The closer a system is to being undamped, the greater its Q. The amplitude of oscillation drops by a factor of e ( 2.718 . . . ) in Q/ cycles. Later in this book we shall see that resonance appears in other areas of physics. For example, certain electrical circuits have natural frequencies. A bridge has natural frequencies that can be set into resonance by an appropriate driving force. A dramatic example of such resonance occurred in 1940, when the Tacoma Narrows Bridge in the state of Washington was destroyed by resonant vibrations. Although the winds were not particularly strong on that occasion, the bridge ultimately collapsed (Fig. 13.23) because the bridge design had no built-in safety features. A b=0 Undamped Small b Large b 0 0 Figure 13.22 Graph of amplitude versus frequency for a damped oscillator when a periodic driving force is present. When the frequency of the driving force equals the natural frequency 0 , resonance occurs. Note that the shape of the resonance curve depends on the size of the damping coefficient b. QuickLab Tie several objects to strings and suspend them from a horizontal string, as illustrated in the figure. Make two of the hanging strings approximately the same length. If one of this pair, such as P, is set into sideways motion, all the others betational force that exists between any two masses. Newton's law of universal gravitation, together with his development of the laws of motion, provides the basis for a full mathematical solution to the motion of planets and satellites. a c F1 F2 b Figure 14.5 Plot of an ellipse. The semimajor axis has a length a, and the semiminor axis has a length b. The focal points are located at a distance c from the cenb2 c 2. ter, where a 2 14.5 THE LAW OF GRAVITY AND THE MOTION OF PLANETS In formulating his law of gravity, Newton used the following reasoning, which supports the assumption that the gravitational force is proportional to the inverse square of the separation between the two interacting bodies. He compared the acceleration of the Moon in its orbit with the acceleration of an object falling near the Earth's surface, such as the legendary apple (Fig. 14.6). Assuming that both accelerations had the same cause -- namely, the gravitational attraction of the Earth -- Newton used the inverse-square law to reason that the acceleration of the Moon toward the Earth (centripetal acceleration) should be proportional to 1/rM2, where rM is the distance between the centers of the Earth and the Moon. Furthermore, the acceleration of the apple toward the Earth should be proportional to 1/R E 2, where R E is the radius of the Earth, or the distance between the centers of the Earth and the apple. Using the values r M 3.84 10 8 m and 14.5 The Law of Gravity and the Motion of Planets 431 Moon aM v g rM Earth RE Figure 14.6 As it revolves around the Earth, the Moon experiences a centripetal acceleration aM directed toward the Earth. An object near the Earth's surface, such as the apple shown here, experiences an acceleration g. (Dimensions are not to scale.) R E 6.37 10 6 m, Newton predicted that the ratio of the Moon's acceleration aM to the apple's acceleration g would be aM g (1/r M)2 (1/R E)2 RE rM 2 6.37 3.84 m/s2) 10 6 m 10 8 m 2 2.75 10 4 Therefore, the centripetal acceleration of the Moon is aM (2.75 10 4)(9.80 2.70 10 3 m/s2 Acceleration of the Moon Newton also calculated the centripetal acceleration of the Moon from a knowledge of its mean distance from the Earth and its orbital period, T 27.32 days 2.36 106 s. In a time T, the Moon travels a distance 2 rM , which equals the circumference of its orbit. Therefore, its orbital speed is 2 rM /T and its centripetal acceleration is aM v2 rM 2.72 (2 r M/T)2 rM 10 3 4 2r M 4 2(3.84 T 2 (2.36 10 8 m) 10 6 s)2 m/s2 9.80 m/s2 60 2 In other words, because the Moon is roughly 60 Earth radii away, the gravitational acceleration at that distance should be about 1/602 of its value at the Earth's surface. This is just the acceleration needed to account for the circular motion of the Moon around the Earth. The nearly perfect agreement between this value and the value Newton obtained using g provides strong evidence of the inverse-square nature of the gravitational force law. Although these results must have been very encouraging to Newton, he was deeply troubled by an assumption he made in the analysis. To evaluate the acceleration of an object at the Earth's surface, Newton treated the Earth as if its mass were all concentrated at its center. That is, he assumed that the Earth acted as a particle as far as its influence on an exterior object was concerned. Several years later, in 1687, on the basis of his pioneering work in the development of calculus, Newton proved that this assumption was valid and was a natural consequence of the law of universal gravitation. 432 v Mp r MS CHAPTER 14 The Law of Gravity Kepler's Third Law It is informative to show that Kepler's third law can be predicted from the inversesquare law for circular orbits.2 Consider a planet of mass Mp moving around the Sun of mass MS in a circular orbit, as shown in Figure 14.7. Because the gravitational force exerted by the Sun on the planet is a radially directed force that keeps the planet moving in a circle, we can apply Newton's second law ( F ma) to the planet: GM S M p M pv 2 2 r r Because the orbital speed v of the planet is simply 2 r/T, where T is its period of revolution, the preceding expression becomes Figure 14.7 A planet of mass Mp moving in a circular orbit around the Sun. The orbits of all planets except Mercury and Pluto are nearly circular. Kepler's third law GM S r2 T2 where K S is a constant given by KS 4 2 GM S (2 r/T)2 r 4 2 GM S r3 K Sr 3 (14.7) 2.97 10 19 s2/m3 Equation 14.7 is Kepler's third law. It can be shown that the law is also valid for elliptical orbits if we replace r with the length of the semimajor axis a. Note that the constant of proportionality K S is independent of the mass of the planet. Therefore, Equation 14.7 is valid for any planet.3 Table 14.2 contains a collection of useful planetary data. The last column verifies that T 2/r 3 is a constant. The small variations in the values in this column reflect uncertainties in the measured values of the periods and semimajor axes of the planets. If we were to consider the orbit around the Earth of a satellite such as the Moon, then the proportionality constant would have a different value, with the Sun's mass replaced by the Earth's mass. EXAMPLE 14.4 The Mass of the Sun 1.99 10 30 kg Calculate the mass of the Sun using the fact that the period of the Earth's orbit around the Sun is 3.156 107 s and its distance from the Sun is 1.496 1011 m. Solution MS 4 GT 2 Using Equation 14.7, we find that (6.67 4 10 2(1.496 11 2r 3 10 11 m)3 10 7 s)2 In Example 14.3, an understanding of gravitational forces enabled us to find out something about the density of the Earth's core, and now we have used this understanding to determine the mass of the Sun. N m2/kg 2)(3.156 2 The orbits of all planets except Mercury and Pluto are very close to being circular; hence, we do not introduce much error with this assumption. For example, the ratio of the semiminor axis to the semimajor axis for the Earth's orbit is b/a 0.999 86. 3 Equation 14.7 is indeed a proportion because the ratio of the two quantities T 2 and r 3 is a constant. The variables in a proportion are not required to be limited to the first power only. 14.5 The Law of Gravity and the Motion of Planets 433 TABLE 14.2 Useful Planetary Data Mean Radius (m) 2.43 6.06 6.37 3.37 6.99 5.85 2.33 2.21 1.5 1.74 6.96 106 106 106 106 107 107 107 107 106 106 108 Body Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Moon Sun Mass (kg) 3.18 4.88 5.98 6.42 1.90 5.68 8.68 1.03 1.4 7.36 1.991 1023 1024 1024 1023 1027 1026 1025 1026 1022 1022 1030 Period of Revolution (s) 7.60 1.94 3.156 5.94 3.74 9.35 2.64 5.22 7.82 -- -- 106 107 107 107 108 108 109 109 109 Mean Distance from Sun (m) 5.79 1.08 1.496 2.28 7.78 1.43 2.87 4.50 5.91 -- -- 1010 1011 1011 1011 1011 1012 1012 1012 1012 T2 2 3 (s /m ) r3 2.97 2.99 2.97 2.98 2.97 2.99 2.95 2.99 2.96 10 10 10 10 10 10 10 10 10 -- -- 19 19 19 19 19 19 19 19 19 Kepler's Second Law and Conservation of Angular Momentum Consider a planet of mass Mp moving around the Sun in an elliptical orbit (Fig. 14.8). The gravitational force acting on the planet is always along the radius vector, directed toward the Sun, as shown in Figure 14.9a. When a force is directed toward or away from a fixed point and is a function of r only, it is called a central force. The torque acting on the planet due to this force is clearly zero; that is, because F is parallel to r, r F r F^ r 0 Sun S C A B D (You may want to revisit Section 11.2 to refresh your memory on the vector product.) Recall from Equation 11.19, however, that torque equals the time rate of d L/dt. Therefore, because the gravitational change of angular momentum: Figure 14.8 Kepler's second law is called the law of equal areas. When the time interval required for a planet to travel from A to B is equal to the time interval required for it to go from C to D, the two areas swept out by the planet's radius vector are equal. Note that in order for this to be true, the planet must be moving faster between C and D than between A and B. Separate views of Jupiter and of Periodic Comgravitational interaction in a different way. As described in Section 5.1, this alternative approach uses the concept of a gravitational field that exists at every point in space. When a particle of mass m is placed at a point where the gravitational field is g, the particle experiences a force Fg mg. In other words, the field exerts a force on the particle. Hence, the gravitational field g is defined as g Fg m (14.10) Gravitational field That is, the gravitational field at a point in space equals the gravitational force experienced by a test particle placed at that point divided by the mass of the test particle. Notice that the presence of the test particle is not necessary for the field to exist --the Earth creates the gravitational field. We call the object creating the field the source particle (although the Earth is clearly not a particle; we shall discuss shortly the fact that we can approximate the Earth as a particle for the purpose of finding the gravitational field that it creates). We can detect the presence of the field and measure its strength by placing a test particle in the field and noting the force exerted on it. Although the gravitational force is inherently an interaction between two objects, the concept of a gravitational field allows us to "factor out" the mass of one of the objects. In essence, we are describing the "effect" that any object (in this case, the Earth) has on the empty space around itself in terms of the force that would be present if a second object were somewhere in that space.4 As an example of how the field concept works, consider an object of mass m near the Earth's surface. Because the gravitational force acting on the object has a magnitude GME m/r 2 (see Eq. 14.4), the field g at a distance r from the center of the Earth is g Fg m GM E ^ r r2 (14.11) ^ where r is a unit vector pointing radially outward from the Earth and the minus 4 We shall return to this idea of mass affecting the space around it when we discuss Einstein's theory of gravitation in Chapter 39. 436 CHAPTER 14 The Law of Gravity (a) (b) Figure 14.11 (a) The gravitational field vectors in the vicinity of a uniform spherical mass such as the Earth vary in both direction and magnitude. The vectors point in the direction of the acceleration a particle would experience if it were placed in the field. The magnitude of the field vector at any location is the magnitude of the free-fall acceleration at that location. (b) The gravitational field vectors in a small region near the Earth's surface are uniform in both direction and magnitude. sign indicates that the field points toward the center of the Earth, as illustrated in Figure 14.11a. Note that the field vectors at different points surrounding the Earth vary in both direction and magnitude. In a small region near the Earth's surface, the downward field g is approximately constant and uniform, as indicated in Figure 14.11b. Equation 14.11 is valid at all points outside the Earth's surface, assuming that the Earth is spherical. At the Earth's surface, where r R E , g has a magnitude of 9.80 N/kg. 14.7 GRAVITATIONAL POTENTIAL ENERGY In Chapter 8 we introduced the concept of gravitational potential energy, which is the energy associated with the position of a particle. We emphasized that the gravitational potential energy function U mgy is valid only when the particle is near the Earth's surface, where the gravitational force is constant. Because the gravitational force between two particles varies as 1/r 2, we expect that a more general potential energy function -- one that is valid without the restriction of having to be near the Earth's surface -- will be significantly different from U mgy. Before we calculate this general form for the gravitational potential energy function, let us first verify that the gravitational force is conservative. (Recall from Section 8.2 that a force is conservative if the work it does on an object moving between any two points is independent of the path taken by the object.) To do this, we first note that the gravitational 2/kg 2)(5.98 2 1 6.65 10 6 m 3 T2 KE 3 (86 400 9.89 10 14 s2/m3 s)2 4.23 10 7 m 4.23 Rf 1.19 10 7 m 10 10 J This is a little more than 26 000 mi above the Earth's surface. 14.8 Energy Considerations in Planetary and Satellite Motion This is the energy equivalent of 89 gal of gasoline. NASA engineers must account for the changing mass of the spacecraft as it ejects burned fuel, something we have not done here. Would you expect the calculation that includes the effect of this changing mass to yield a greater or lesser amount of energy required from the engine? If we wish to determine how the energy is distributed after the engine is fired, we find from Equation 14.18 that the change in kinetic energy is K (GM E m/2)(1/R f 1/R i ) 1.19 10 10 J (a decrease), 441 and the corresponding change in potential energy is U GM E m(1/R f 1/R i ) 2.38 10 10 J (an increase). Thus, the change in mechanical energy of the system is E K U 1.19 10 10 J, as we already calculated. The firing of the engine results in an increase in the total mechanical energy of the system. Because an increase in potential energy is accompanied by a decrease in kinetic energy, we conclude that the speed of an orbiting satellite decreases as its altitude increases. Escape Speed Suppose an object of mass m is projected vertically upward from the Earth's surface with an initial speed vi , as illustrated in Figure 14.17. We can use energy considerations to find the minimum value of the initial speed needed to allow the object to escape the Earth's gravitational field. Equation 14.17 gives the total energy of the object at any point. At the surface of the Earth, v v i and r r i R E . When the object reaches its maximum altitude, v v f 0 and r r f r max . Because the total energy of the system is constant, substituting these conditions into Equation 14.20 gives 1 2 2 mv i vf = 0 h rmax vi m RE GM E m RE GM E m r max Solving for v i 2 gives v i2 2GM E 1 RE 1 r max (14.21) ME Therefore, if the initial speed is known, this expression can be used to calculate the maximum altitude h because we know that h r max RE Figure 14.17 We are now in a position to calculate escape speed, which is the minimum speed the object must have at the Earth's surface in order to escape from the influence of the Earth's gravitational field. Traveling at this minimum speed, the object continues to move farther and farther away from the Earth as its speed asymptotically approaches zero. Letting r max : in Equation 14.21 and taking v i v esc , we obtain v esc An object of mass m projected upward from the Earth's surface with an initial speed vi reaches a maximum altitude h. 2GM E RE (14.22) Escape speed Note that this expression for vesc is independent of the mass of the object. In other words, a spacecraft has the same escape speed as a molecule. Furthermore, the result is independent of the direction of the velocity and ignores air resistance. If the object is given an initial speed equal to vesc , its total energy is equal to zero. This can be seen by noting that when r : , the object's kinetic energy and its potential energy are both zero. If vi is greater than vesc , the total energy is greater than zero and the object has some residual kinetic energy as r : . 442 CHAPTER 14 The Law of Gravity EXAMPLE 14.8 Escape Speed of a Rocket 1.12 10 4 m/s Calculate the escape speed from the Earth for a 5 000-kg spacecraft, and determine the kinetic energy it must have at the Earth's surface in order to escape the Earth's gravitational field. Solution v esc This corresponds to about 25 000 mi/h. The kinetic energy of the spacecraft is K 1 2 2 mv esc 1 2 (5.00 Using Equation 14.22 gives 2GM E RE 2(6.67 10 11 10 3 kg)(1.12 10 4 m/s)2 3.14 N m2/kg 2)(5.98 6.37 10 6 m 10 24 kg) 10 11 J This is equivalent to about 2 300 gal of gasoline. Equations 14.21 and 14.22 can be applied to objects projected from any planet. That is, in general, the escape speed from the surface of any planet of mass M and radius R is TABLE 14.3 Escape Speeds from the Surfaces of the Planets, Moon, and Sun Body Mercury Venus Earth Moon Mars Jupiter Saturn Uranus Neptune Pluto Sun v esc 2GM R vesc (km/s) 4.3 10.3 11.2 2.3 5.0 60 36 22 24 1.1 618 Escape speeds for the planets, the Moon, and the Sun are provided in Table 14.3. Note that the values vary from 1.1 km/s for Pluto to about 618 km/s for the Sun. These results, together with some ideas from the kinetic theory of gases (see Chapter 21), explain why some planets have atmospheres and others do not. As we shall see later, a gas molecule has an average kinetic energy that depends on the temperature of the gas. Hence, lighter molecules, such as hydrogen and helium, have a higher average speed than heavier species at the same temperature. When the average speed of the lighter molecules is not much less than the escape speed of a planet, a significant fraction of them have a chance to escape from the planet. This mechanism also explains why the Earth does not retain hydrogen molecules and helium atoms in its atmosphere but does retain heavier molecules, such as oxygen and nitrogen. On the other hand, the very large escape speed for Jupiter enables that planet to retain hydrogen, the primary constituent of its atmosphere. Quick Quiz 14.2 If you were a space prospector and discovered gold on an asteroid, it probably would not be a good idea to jump up and down in excitement over your find. Why? Quick Quiz 14.3 Figure 14.18 is a drawing by Newton showing the path of a stone thrown from a mountaintop. He shows the stone landing farther and farther away when thrown at higher and higher speeds (at points D, E, F, and G), until finally it is thrown all the way around the Earth. Why didn't Newton show the stone landing at B and A before it was going fast enough to complete an orbit? 14.9 The Gravitational Force Between an Extended Object and a Particle 443 "The greater the velocity . . . with which [a stone] is projected, the farther it goes before it falls to the Earth. We may therefore suppose the velocity to be so increased, that it would describe an arc of 1, 2, 5, 10, 100, 1000 miles before it arrived at the Earth, till at last, exceeding the limits of the Earth, it should pass into space without touching." Sir Isaac Newton, System of the World. Figure 14.18 Optional Section 14.9 THE GRAVITATIONAL FORCE BETWEEN AN EXTENDED OBJECT AND A PARTICLE M Mi ^ r We have emphasized that the law of universal gravitation given by Equation 14.3 is valid only if the interacting objects are treated as particles. In view of this, how can we calculate the force between a particle and an object having finite dimensions? This is accomplished by treating the extended object as a collection of particles and making use of integral calculus. We first evaluate the potential energy function, and then calculate the gravitational force from that function. We obtain the potential energy associated with a system consisting of a particle of mass m and an extended object of mass M by dividing the object into many elements, each having a mass Mi (Fig. 14.19). The potential energy associated with the system consisting of any one element and the particle is U Gm M i /r i , where ri is the distance from the particle to the element Mi . The total potential energy of the overall system is obtained by taking the sum over all elements as Mi : 0. In this limit, we can express U in integral form as U Gm dM r (14.23) ri m Once U has been evaluated, we obtain the force exerted by the extended object on the particle by taking the negative derivative of this scalar function (see Section 8.6). If the extended object has spherical symmetry, the function U depends only on r, and the force is given by dU/dr. We treat this situation in Section 14.10. In principle, one can evaluate U for any geometry; however, the integration can be cumbersome. An alternative approach to evaluating the gravitational force between a particle and an extended object is to perform a vector sum over all mass elements of the object. Using the procedure outlined in evaluating U and the law of universal gravitation in the form shown in Equation 14.3, we obtain, for the total force exerted on the particle Fg Gm dM ^ r r2 (14.24) Figure 14.19 A particle of mass m interacting with an extended object of mass M. The total gravitational force exerted by the object on the particle can be obtained by dividing the object into numerous elements, each having a mass Mi , and then taking a vector sum over the forces exerted by all elements. Total force exerted on a particle by an extended object ^ where r is a unit vector directed from the element dM toward the particle (see Fig. 14.19) and the minus sign indicates that the direction of the force is opposite that ^ of r. This procedure is not always recommended because working with a vector function is more difficult than working with the scalar potential energy function. However, if the geometry is simple, as in the following example, the evaluation of F can be straightforward. 444 CHAPTER 14 The Law of Gravity EXAMPLE 14.9 Gravitational Force Between a Particle and a Bar of lengths dx/L, and so dM (M/L) dx. In this problem, the variable r in Equation 14.24 is the distance x shown in Figure ^ ^ 14.20, the unit vector r is r i, and the force acting on the particle is to the right; therefore, Equation 14.24 gives us h L The left end of a homogeneous bar of length L and mass M is at a distance h from a particle of mass m (Fig. 14.20). Calculate the total gravitational force exerted by the bar on the particle. The arbitrary segment of the bar of length dx has a mass dM. Because the mass per unit length is constant, it follows that the ratio of masses dM/M is equal to the ratio y h L dx m O x x Solution Fg Gm h Mdx 1 ( i) L x2 h L Gm M L h L h dx i x2 Fg GmM L 1 x i h GmM i h(h L) Figure 14.20 The gravitational force exerted by the bar on the particle is directed to the right. Note that the bar is not equivalent to a particle of mass M located at the center of mass of the bar. We see that the force exerted on the particle is in the positive x direction, which is what we expect because the gravitational force is attractive. Note that in the limit L : 0, the force varies as 1/h 2, which is what we expect for the force between two point masses. Furthermore, if h W L, the force also varies as 1/h 2. This can be seen by noting that the denominator of the expression for Fg can be expressed in the form h 2(1 L/h), which is approximately equal to h2 when h W L . Thus, when bodies are separated by distances that are great relative to their characteristic dimensions, they behave like particles. Optional Section 14.10 THE GRAVITATIONAL FORCE BETWEEN A PARTICLE AND A SPHERICAL MASS We have already stated that a large sphere attracts a particle outside it as if the total mass of the sphere were concentrated at its center. We now describe the force acting on a particle when the extended object is either a spherical shell or a solid sphere, and then apply these facts to some interesting systems. Spherical Shell Case 1. If a particle of mass m is located outside a spherical shell of mass M at, for instance, point P in Figure 14.21a, the shell attracts the particle as though the mass of the shell were concentrated at its center. We can show this, as Newton did, with integral calculus. Thus, as far as the gravitational force acting on a particle outside the shell is concerned, a spherical shell acts no differently from the solid spherical distributions of mass we have seen. Case 2. If the particle is located inside the shell (at point P in Fig. 14.21b), the gravitational force acting on it can be shown to be zero. We can express these two important results in the following way: Force on a particle due to a spherical shell Fg Fg 0 GMm ^ r r2 for r for r R R (14.25a) (14.25b) The gravitational force as a function of the distance r is plotted in Figure 14.21c. 14.10 The Gravitational Force Between a Particle and a Spherical Mass M Q FQP P m FQP 445 Q (a) M P FTop, P m FBottom, P (b) Fg O R r (c) Figure 14.21 (a) The nonradial components of the gravitational forces exerted on a particle of mass m located at point P outside a spherical shell of mass M cancel out. (b) The spherical shell can be broken into rings. Even though point P is closer to the top ring than to the bottom ring, the bottom ring is larger, and the gravitational forces exerted on the particle at P by the matter in the two rings cancel each other. Thus, for a particle located at any point P inside the shell, there is no gravitational force exerted on the particle by the mass M of the shell. (c) The magnitude of the gravitational force versus the radial distance r from the center of the shell. The shell does not act as a gravitational shield, which means that a particle inside a shell may experience forces exerted by bodies outside the shell. Solid Sphere Case 1. If a particle of mass m is located outside a homogeneous solid sphere of mass M (at point P in Fig. 14.22), the sphere attracts the particle as though the 446 CHAPTER 14 The Law of Gravity mass of the sphere were concentrated at its center. We have used this notion at several places in this chapter already, and we can argue it from Equation 14.25a. A solid sphere can be considered to be a collection of concentric spherical shells. The masses of all of the shells can be interpreted as being concentrated at their common center, and the gravitational force is equivalent to that due to a particle of mass M located at that center. Case 2. If a particle of mass m is located inside a homogeneous solid sphere of mass M (at point Q in Fig. 14.22), the gravitational force acting on it is due only to the mass M contained within the sphere of radius r R, shown in Figure 14.22. In other words, Force on a particle due to a solid sphere Fg Fg GmM ^ r r2 GmM ^ r r2 for r for r R R (14.26a) (14.26b) This also follows from spherical-shell Case 1 because the part of the sphere that is m P M R M Fg Q r Fg r O R Figure 14.22 The gravitational force acting on a particle when it is outside a uniform solid sphere is GMm/r 2 and is directed toward the center of the sphere. The gravitational force acting on the particle when it is inside such a sphere is proportional to r and goes to zero at the center. 14.10 The Gravitational Force Between a Particle and a Spherical Mass 447 farther from the center than Q can be treated as a series of concentric spherical shells that do not exert a net force on the particle because the particle is inside them. Because the sphere is assumed to have a uniform density, it follows that the ratio of masses M /M is equal to the ratio of volumes V /V, where V is the total volume of the sphere and V is the volume within the sphere of radius r only: M M V V 4 3 4 3 r3 R3 r3 R3 Solving this equation for M and substituting the value obtained into Equation 14.26b, we have Fg GmM ^ rr R3 for r R (14.27) This equation tells us that at the center of the solid sphere, where r 0, the gravitational force goes to zero, as we intuitively expect. The force as a function of r is plotted in Figure 14.22. Case 3. If a particle is located inside a solid sphere having a density that is spherically symmetric but not uniform, then M in Equation 14.26b is given by an integral of the form M dV, where the integration is taken over the volume contained within the sphere of radius r in Figure 14.22. We can evaluate this integral if the radial variation of is given. In this case, we take the volume element dV as the volume of a spherical shell of radius r and thickness dr, and thus dV 4 r 2 dr. For example, if Ar, where A is a constant, it is left to a problem (Problem 63) to show that M Ar 4. 2 in this case and is Hence, we see from Equation 14.26b that F is proportional to r zero at the center. Quick Quiz 14.4 A particle is projected through a small hole into the interior of a spherical shell. Describe EXAMPLE 14.10 A Free Ride, Thanks to Gravity The y component of the gravitational force on the object is balanced by the normal force exerted by the tunnel wall, and the x component is Fx GmM E r cos R E3 r cos , we can An object of mass m moves in a smooth, straight tunnel dug between two points on the Earth's surface (Fig. 14.23). Show that the object moves with simple harmonic motion, and find the period of its motion. Assume that the Earth's density is uniform. Solution The gravitational force exerted on the object acts toward the Earth's center and is given by Equation 14.27: Fg GmM r^ r R3 Because the x coordinate of the object is x write Fx GmM E x R E3 We receive our first indication that this force should result in simple harmonic motion by comparing it to Hooke's law, first seen in Section 7.3. Because the gravitational force on the object is linearly proportional to the displacement, the object experiences a Hooke's law force. Applying Newton's second law to the motion along the x direction gives Fx GmM E x R E3 ma x 448 y CHAPTER 14 The Law of Gravity which we have derived for the acceleration of our object in the tunnel, is the acceleration equation for simple harmonic motion at angular speed with x O Fg m r x T 2 2 GM E R E3 Thus, the object in the tunnel moves in the same way as a block hanging from a spring! The period of oscillation is 2 An object moves along a tunnel dug through the Earth. The component of the gravitational force Fg along the x axis is the driving force for the motion. Note that this component always acts toward O. Figure 14.23 R E3 GM E 10 11 (6.67 10 3 s (6.37 10 6 m)3 N m2/kg 2)(5.98 10 24 kg) 5.06 84.3 min Solving for ax , we obtain ax If we use the symbol 2 -- we see that (1) ax 2 GM E x R E3 for the coefficient of x -- GME /R E3 2x an expression that matches the mathematical form of Equation 13.9, which gives the acceleration of a particle in simple 2x. Therefore, Equation (1), harmonic motion: a x This period is the same as that of a satellite traveling in a circular orbit just above the Earth's surface (ignoring any trees, buildings, or other objects in the way). Note that the result is independent of the length of the tunnel. A proposal has been made to operate a mass-transit system between any two cities, using the principle described in this example. A one-way trip would take about 42 min. A more precise calculation of the motion must account for the fact that the Earth's density is not uniform. More important, there are many practical problems to consider. For instance, it would be impossible to achieve a frictionless tunnel, and so some auxiliary power source would be required. Can you think of other problems? the motion of the particle inside the shell. SUMMARY Newton's law of universal gravitation states that the gravitational force of attraction between any two particles of masses m 1 and m 2 separated by a distance r has the magnitude Fg G m 1m 2 r2 (14.1) where G 6.673 10 11 N m2/kg 2 is the universal gravitational constant. This equation enables us to calculate the force of attraction between masses under a wide variety of circumstances. An object at a distance h above the Earth's surface experiences a gravitational force of magnitude mg , where g is the free-fall acceleration at that elevation: g GM E r2 GM E (R E h)2 (14.6) Summary 449 In this expression, ME is the mass of the Earth and RE is its radius. Thus, the weight of an object decreases as the object moves away from the Earth's surface. Kepler's laws of planetary motion state that 1. All planets move in elliptical orbits with the Sun at one focal point. 2. The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. 3. The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. Kepler's third law can be expressed as T2 4 2 3 r GM S (14.7) where MS is the mass of the Sun and r is the orbital radius. For elliptical orbits, Equation 14.7 is valid if r is replaced by the semimajor axis a. Most planets have nearly circular orbits around the Sun. The gravitational field at a point in space equals the gravitational force experienced by any test particle located at that point divided by the mass of the test particle: g Fg m circular orbit just above the surface of a planet, which is assumed to offer no air resistance. Show that its orbital speed v and the escape speed from the planet are related by the expression v esc 2v. 40. The planet Uranus has a mass about 14 times the Earth's mass, and its radius is equal to about 3.7 Earth 454 CHAPTER 14 The Law of Gravity d L m L m radii. (a) By setting up ratios with the corresponding Earth values, find the acceleration due to gravity at the cloud tops of Uranus. (b) Ignoring the rotation of the planet, find the minimum escape speed from Uranus. 41. Determine the escape velocity for a rocket on the far side of Ganymede, the largest of Jupiter's moons. The radius of Ganymede is 2.64 106 m, and its mass is 1.495 1023 kg. The mass of Jupiter is 1.90 1027 kg, and the distance between Jupiter and Ganymede is 1.071 109 m. Be sure to include the gravitational effect due to Jupiter, but you may ignore the motions of Jupiter and Ganymede as they revolve about their center of mass (Fig. P14.41). Figure P14.44 M R m Figure P14.45 (Optional) v Section 14.10 The Gravitational Force Between a Particle and a Spherical Mass 46. (a) Show that the period calculated in Example 14.10 can be written as T 2 Ganymede Jupiter RE g Figure P14.41 42. In Robert Heinlein's The Moon is a Harsh Mistress, the colonial inhabitants of the Moon threaten to launch rocks down onto the Earth if they are not given independence (or at least representation). Assuming that a rail gun could launch a rock of mass m at twice the lunar escape speed, calculate the speed of the rock as it enters the Earth's atmosphere. (By lunar escape speed we mean the speed required to escape entirely from a stationary Moon alone in the Universe.) 43. Derive an expression for the work required to move an Earth satellite of mass m from a circular orbit of radius 2R E to one of radius 3R E . (Optional) where g is the free-fall acceleration on the surface of the Earth. (b) What would this period be if tunnels were made through the Moon? (c) What practical problem regarding these tunnels on Earth would be removed if they were built on the Moon? 47. A 500-kg uniform solid sphere has a radius of 0.400 m. Find the magnitude of the gravitational force exerted by the sphere on a 50.0-g particle located (a) 1.50 m from the center of the sphere, (b) at the surface of the sphere, and (c) 0.200 m from the center of the sphere. 48. A uniform solid sphere of mass m1 and radius R1 is inside and concentric with a spherical shell of mass m 2 and radius R 2 (Fig. P14.48). Find the gravitational force exerted by the spheres on a particle of mass m located at (a) r a, (b) r b, and (c) r c, where r is measured from the center of the spheres. Section 14.9 The Gravitational Force Between an Extended Object and a Particle 44. Consider two identical uniform rods of length L and mass m lying along the same line and having their closest points separated by a distance d (Fig. P14.44). Show that the mutual gravitational force between these rods has a magnitude F Gm 2 L2 ln (L d(2L d )2 d) m2 m1 R1 R2 a b c 45. A uniform rod of mass M is in the shape of a semicircle of radius R (Fig. P14.45). Calculate the force on a point mass m placed at the center of the semicircle. Figure P14.48 Problems 455 ADDITIONAL PROBLEMS 49. Let g M represent the difference in the gravitational fields produced by the Moon at the points on the Earth's surface nearest to and farthest from the Moon. Find the fraction gM /g, where g is the Earth's gravitational field. (This difference is responsible for the occurrence of the lunar tides on the Earth.) 50. Two spheres having masses M and 2M and radii R and 3R, respectively, are released from rest when the distance between their centers is 12R. How fast will each sphere be moving when they collide? Assume that the two spheres interact only with each other. 51. In Larry Niven's science-fiction novel Ringworld, a rigid ring of material rotates about a star (Fig. P14.51). The rotational speed of the ring is 1.25 106 m/s, and its radius is 1.53 1011 m. (a) Show that the centripetal acceleration of the inhabitants is 10.2 m/s2. (b) The inhabitants of this ring world experience a normal contact force n. Acting alone, this normal force would produce an inward acceleration of 9.90 m/s2. Additionally, the star at the center of the ring exerts a gravitational force on the ring and its inhabitants. The difference between the total acceleration and the acceleration provided by the normal force is due to the gravitational attraction of the central star. Show that the mass of the star is approximately 1032 kg. 53. 54. 55. 56. WEB 57. Star 58. Fg n Figure P14.51 52. (a) Show that the rate of change of the free-fall acceleration with distance above the Earth's surface is dg dr 2GM E R E3 This rate of change over distance is called a gradient. (b) If h is small compared to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is g 2GM E h R E3 59. 60. (c) Evaluate this difference for h 6.00 m, a typical height for a two-story building. A particle of mass m is located inside a uniform solid sphere of radius R and mass M, at a distance r from its center. (a) Show that the gravitational potential energy of the system is U (GmM/2R 3)r 2 3GmM/2R. (b) Write an expression for the amount of work done by the gravitational force in bringing the particle from the surface of the sphere to its center. Voyagers 1 and 2 surveyed the surface of Jupiter's moon Io and photographed active volcanoes spewing liquid sulfur to heights of 70 km above the surface of this moon. Find the speed with which the liquid sulfur left the volcano. Io's mass is 8.9 1022 kg, and its radius is 1 820 km. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always straight ahead, and find yourself returning to your spacecraft from the opposite side after completing a lap of 25.0 km. You hold a hammer and a falcon feather at a height of 1.40 m, release them, and observe that they fall together to the surface in 29.2 s. Determine the mass of the planet. A cylindrical habitat in space, 6.00 km in diameter and 30 km long, was proposed by G. K. O'Neill in 1974. Such a habitat would have cities, land, and lakes on the inside surface and air and clouds in the center. All of these would be held in place by the rotation of the cylinder about its long axis. How fast would the cylinder have to rotate to imitate the Earth's gravitational field at the walls of the cylinder? In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.50 kg and 15.0 g whose centers are separated by about 4.50 cm. Calculate the gravitational force between these spheres, treating each as a point mass located at the center of the sphere. Newton's law of universal gravitation is valid for distances covering an enormous range, but it is thought to fail for very small distances, where the structure of space itself is uncertain. The crossover distance, far less than the diameter of an atomic nucleus, is called the Planck length. It is determined by a combination of the constants G, c, and h, where c is the speed of light in vacuum and h is Planck's constant (introduced briefly in Chapter 11 and discussed in greater detail in Chapter 40) with units of angular momentum. (a) Use dimensional analysis to find a combination of these three universal constants that has units of length. (b) Determine the order of magnitude of the Planck length. (Hint: You will need to consider noninteger powers of the constants.) Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet. (a) Suppose that the Earth (or another object) has density (r), which can vary with radius but is spherically 456 CHAPTER 14 The Law of Gravity of which the Sun is typical. What is the order of magnitude of the number of stars in the Milky Way? 67. The oldest artificial satellite in orbit is Vanguard I, launched March 3, 1958. Its mass is 1.60 kg. In its initial orbit, its minimum distance from the center of the Earth was 7.02 Mm, and its speed at this perigee point was 8.23 km/s. (a) Find its total energy. (b) Find the magnitude of its angular momentum. (c) Find its speed at apogee and its maximum (apogee) distance from the center of the Earth. (d) Find the semimajor axis of its orbit. (e) Determine its period. 68. A rocket is given an initial speed vertically upward of v i 2Rg at the surface of the Earth, which has radius R and surface free-fall acceleration g. The rocket motors are quickly cut off, and thereafter the rocket coasts under the action of gravitational forces only. (Ignore atmospheric friction and the Earth's rotation.) Derive an expression for the subsequent speed v as a function of the distance r from the center of the Earth in terms of g, R, and r. 69. Two stars of masses M and m, separated by a distance d, revolve in circular orbits about their center of mass (Fig. P14.69). Show that each star has a period given by T2 4 2d 3 G(M m) WEB 61. 62. 63. 64. 65. 66. symmetric. Show that at any particular radius r inside the Earth, the gravitational field strength g(r) will increase as r increases, if and only if the density there exceeds 2/3 the average density of the portion of the Earth inside the radius r. (b) The Earth as a whole has an average density of 5.5 g/cm3, while the density at the surface is 1.0 g/cm3 on the oceans and about 3 g/cm3 on land. What can you infer from this? Two hypothetical planets of masses m1 and m 2 and radii r 1 and r 2 , respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. (a) When their center-to-center separation is d, find expressions for the speed of each planet and their relative velocity. (b) Find the kinetic energy of each planet just before they collide, if m 1 2.00 1024 kg, m 2 8.00 1024 kg, r 1 3.00 106 m, and r 2 5.00 106 m. (Hint: Both energy and momentum are conserved.) The maximum distance from the Earth to the Sun (at our aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelion) is 1.471 1011 m. If the Earth's orbital speed at perihelion is 30.27 km/s, determine (a) the Earth's orbital speed at aphelion, (b) the kinetic and potential energies at perihelion, and (c) the kinetic and potential energies at aphelion. Is the total energy constant? (Neglect the effect of the Moon and other planets.) A sphere of mass M and radius R has a nonuniform density that varies with r, the distance from its center, according to the expression Ar, for 0 r R. (a) What is the constant A in terms of M and R ? (b) Determine an expression for the force exerted on a particle of mass m placed outside the sphere. (c) Determine an expression for the force exerted on the particle if it is inside the sphere. (Hint: See Section 14.10 and note that the distribution is spherically symmetric.) (a) Determine the amount of work (in joules) that must be done on a 100-kg payload to elevate it to a height of 1 000 km above the Earth's surface. (b) Determine the amount of additional work that is required to put the payload into circular orbit at this elevation. X-ray pulses from Cygnus X-1, a celestial x-ray source, have been recorded during high-altitude rocket flights. The signals can be interpreted as originating when a blob of ionized matter orbits a black hole with a period of 5.0 ms. If the blob is in a circular orbit about a black hole whose mass is 20MSun , what is the orbital radius? Studies of the relationship of the Sun to its galaxy -- the Milky Way -- have revealed that the Sun is located near the outer edge of the galactic disk, about 30 000 lightyears from the center. Furthermore, it has been found that the Sun has an orbital speed of approximately 250 km/s around the galactic center. (a) What is the period of the Sun's galactic motion? (b) What is the order of magnitude of the mass of the Milky Way galaxy? Suppose that the galaxy is made mostly of stars, (Hint: Apply Newton's second law to each star, and note that the center-of-mass condition requires that Mr 2 mr 1 , where r 1 r 2 d.) m CM v1 v2 M r1 d r2 Figure P14.69 70. (a) A 5.00-kg mass is released 1.20 107 m from the center of the Earth. It moves with what acceleration relative to the Earth? (b) A 2.00 1024 kg mass is released 1.20 107 m from the center of the Earth. It moves with what acceleration relative to the Earth? Assume that the objects behave as pairs of particles, isolated from the rest of the Universe. 71. The acceleration of an object moving in the gravitational field of the Earth is a GM E r r3 Answers to Quick Quizzes where r is the position vector directed from the center of the Earth to the object. Choosing the origin at the center of the Earth and assuming that the small object is moving in the xy plane, we find that the rectangular (cartesian) components of its acceleration are ax GM E x (x 2 y 2)3/2 ay GM E y (x 2 y 2)3/2 457 diction of the motion of the object, according to Euler's method. Assume that the initial position of the object is x 0 and y 2R E , where RE is the radius of the Earth. Give the object an initial velocity of 5 000 m/s in the x direction. The time increment should be made as small as practical. Try 5 s. Plot the x and y coordinates of the object as time goes on. Does the object hit the Earth? Vary the initial velocity until you find a circular orbit. Use a computer to set up and carry out a numerical pre- ANSWERS TO QUICK QUIZZES 14.1 Kepler's third law (Eq. 14.7), which applies to all the planets, tells us that the period of a planet is proportional to r 3/2. Because Saturn and Jupiter are farther from the Sun than the Earth is, they have longer periods. The Sun's gravitational field is much weaker at Saturn and Jupiter than it is at the Earth. Thus, these planets experience much less centripetal acceleration than the Earth does, and they have correspondingly longer periods. 14.2 The mass of the asteroid might be so small that you would be able to exceed escape velocity by leg power alone. You would jump up, but you would never come back down! 14.3 Kepler's first law applies not only to planets orbiting the Sun but also to any relatively small object orbiting another under the influence of gravity. Any elliptical path that does not touch the Earth before reaching point G will continue around the other side to point V in a complete orbit (see figure in next column). 14.4 The gravitational force is zero inside the shell (Eq. 14.25b). Because the force on it is zero, the particle moves with constant velocity in the direction of its original motion outside the shell until it hits the wall opposite the entry hole. Its path thereafter depends on the nature of the collision and on the particle's original direction. P U Z Z L E R Have you ever wondered why a tennis ball is fuzzy and why a golf ball has dimples? A "spitball" is an illegal baseball pitch because it makes the ball act too much like the fuzzy tennis ball or the dimpled golf ball. What principles of physics govern the behavior of these three pieces of sporting equipment (and also keep airplanes in the sky)? (George Semple) c h a p t e r Fluid Mechanics Chapter Outline 15.1 15.2 15.3 15.4 Pressure Variation of Pressure with Depth Pressure Measurements Buoyant Forces and Archimedes's Principle 15.6 Streamlines and the Equation of Continuity 15.7 Bernoulli's Equation 15.8 (Optional) Other Applications of Bernoulli's Equation 15.5 Fluid Dynamics 458 15.1 Pressure 459 atter is normally classified as being in one of three states: solid, liquid, or gas. From everyday experience, we know that a solid has a definite volume and shape. A brick maintains its familiar shape and size day in and day out. We also know that a liquid has a definite volume but no definite shape. Finally, we know that an unconfined gas has neither a definite volume nor a definite shape. These definitions help us picture the states of matter, but they are somewhat artificial. For example, asphalt and plastics are normally considered solids, but over long periods of time they tend to flow like liquids. Likewise, most substances can be a solid, a liquid, or a gas (or a combination of any of these), depending on the temperature and pressure. In general, the time it takes a particular substance to change its shape in response to an external force determines whether we treat the substance as a solid, as a liquid, or as a gas. A fluid is a collection of molecules that are randomly arranged and held together by weak cohesive forces and by forces exerted by the walls of a container. Both liquids and gases are fluids. In our treatment of the mechanics of fluids, we shall see that we do not need to learn any new physical principles to explain such effects as the buoyant force acting on a submerged object and the dynamic lift acting on an airplane wing. First, we consider the mechanics of a fluid at rest -- that is, fluid statics -- and derive an expression for the pressure exerted by a fluid as a function of its density and depth. We then treat the mechanics of fluids in motion -- that is, fluid dynamics. We can describe a fluid in motion by using a model in which we make certain simplifying assumptions. We use this model to analyze some situations of practical importance. An analysis leading to Bernoulli's equation enables us to determine relationships between the pressure, density, and velocity at every point in a fluid. M 15.1 PRESSURE Fluids do not sustain shearing stresses or tensile stresses; thus, the only stress that can be exerted on an object submerged in a fluid is one that tends to compress the object. In other words, the force exerted by a fluid on an object is always perpendicular to the surfaces of the object, as shown in Figure 15.1. The pressure in a fluid can be measured with the device pictured in Figure 15.2. The device consists of an evacuated cylinder that encloses a light piston connected to a spring. As the device is submerged in a fluid, the fluid presses on the top of the piston and compresses the spring until the inward force exerted by the fluid is balanced by the outward force exerted by the spring. The fluid pressure can be measured directly if the spring is calibrated in advance. If F is the magnitude of the force exerted on the piston and A is the surface area of the piston, A F Vacuum Figure 15.1 At any point on the surface of a submerged object, the force exerted by the fluid is perpendicular to the surface of the object. The force exerted by the fluid on the walls of the container is perpendicular to the walls at all points. Figure 15.2 by a fluid. A simple device for measuring the pressure exerted 460 CHAPTER 15 Fluid Mechanics then the pressure P of the fluid at the level to which the device has been submerged is defined as the ratio F/A: Definition of pressure P F A (15.1) Note that pressure is a scalar quantity because it is proportional to the magnitude of the force on the piston. To define the pressure at a specific point, we consider a fluid acting on the device shown in Figure 15.2. If the force exerted by the fluid over an infinitesimal surface element of area dA containing the point in question is dF, then the pressure at that point is P dF dA (15.2) Snowshoes keep you from sinking into soft snow because they spread the downward force you exert on the snow over a large area, reducing the pressure on the snow's surface. As we shall see in the next section, the pressure exerted by a fluid varies with depth. Therefore, to calculate the total force exerted on a flat wall of a container, we must integrate Equation 15.2 over the surface area of the wall. Because pressure is force per unit area, it has units of newtons per square meter (N/m2 ) in the SI system. Another name for the SI unit of pressure is pascal (Pa): 1 Pa 1 N/m2 (15.3) Quick Quiz 15.1 Suppose you are standing directly behind someone who steps back and accidentally stomps on your foot with the heel of one shoe. Would you be better off if that person were a professional basketball player wearing sneakers or a petie at the top by an amount gh, where h is the length of any side of the cube. The pressure difference P between the bottom and top faces of the cube is equal to the buoyant force per unit area of those faces -- that is, P B/A. Therefore, B ( P)A ( gh)A gV, where V is the volume of the cube. Because the mass of the fluid in the cube is M V, we see that B Fg Vg Mg (15.5) where Mg is the weight of the fluid in the cube. Thus, the buoyant force is a result of the pressure differential on a submerged or partly submerged object. Before we proceed with a few examples, it is instructive for us to compare the forces acting on a totally submerged object with those acting on a floating (partly submerged) object. Case 1: Totally Submerged Object When an object is totally submerged in a fluid of density f , the magnitude of the upward buoyant force is B fVo g, where Vo is the volume of the object. If the object has a mass M and density o , its weight is equal to F g Mg Fg ( f oVo g, and the net force on it is B o)Vo g. Hence, if the density of the object is less than the density of the fluid, then the downward force of gravity is less than the buoyant force, and the unconstrained object accelerates upward (Fig. 15.10a). If the density of the object is greater than the density of the fluid, then the upward buoyant force is less than the downward force of gravity, and the unsupported object sinks (Fig. 15.10b). Case 2: Floating Object Now consider an object of volume Vo in static equilibrium floating on a fluid -- that is, an object that is only partially submerged. In this h Fg B Figure 15.9 The external forces acting on the cube of liquid are the force of gravity Fg and the buoyant force B. Under equilibrium conditions, B Fg. 15.4 Buoyant Forces and Archimedes's Principle 467 a B Fg a B Fg Figure 15.10 (a) (b) (a) A totally submerged object that is less dense than the fluid in which it is submerged experiences a net upward force. (b) A totally submerged object that is denser than the fluid sinks. case, the upward buoyant force is balanced by the downward gravitational force acting on the object. If Vf is the volume of the fluid displaced by the object (this volume is the same as the volume of that part of the object that is beneath the fluid level), the buoyant force has a magnitude B fVf g. Because the weight of the object is F g Mg Vo g, and because F g B, we see that fVf g o oVo g, or o f Vf Vo (15.6) Under normal conditions, the average density of a fish is slightly greater than the density of water. It follows that the fish would sink if it did not have some mechanism for adjusting its density. The fish accomplishes this by internally regulating the size of its air-filled swim bladder to balance the change in the magnitude of the buoyant force acting on it. In this manner, fish are able to swim to various depths. Unlike a fish, a scuba diver cannot achieve neutral buoyancy (at which the buoyant force just balances the weight) by adjusting the magnitude of the buoyant force B. Instead, the diver adjusts Fg by manipulating lead weights. Hot-air balloons. Because hot air is less dense than cold air, a net upward force acts on the balloons. Quick Quiz 15.6 Steel is much denser than water. In view of this fact, how do steel ships float? Quick Quiz 15.7 A glass of water contains a single floating ice cube (Fig. 15.11). When the ice melts, does the water level go up, go down, or remain the same? Quick Quiz 15.8 When a person in a rowboat in a small pond throws an anchor overboard, does the water level of the pond go up, go down, or remain the same? Figure 15.11 EXAMPLE 15.5 Eureka! scale read 7.84 N in air and 6.86 N in water. What should Archimedes have told the king? Archimedes supposedly was asked to determine whether a crown made for the king consisted of pure gold. Legend has it that he solved this problem by weighing the crown first in air and then in water, as shown in Figure 15.12. Suppose the Solution When the crown is suspended in air, the scale 468 CHAPTER 15 Fluid Mechanics he had been cheated. Either the crown was hollow, or it was not made of pure gold. reads the true weight T1 F g (neglecting the buoyancy of air). When it is immersed in water, the buoyant force B reduces the scale reading to an apparent weight of T2 F g B. Hence, the buoyant force exerted on the crown is the difference between its weight in air and its weight in water: B Fg T2 7.84 N 6.86 N 0.98 N Because this buoyant force is equal in magnitude to the weight of the displaced water, we have w gVw 0.98 N, where Vw is the volume of the displaced water and w is its density. Also, the volume of the crown Vc is equal to the volume of the displaced water because the crown is completely submerged. Therefore, Vc Vw 1.0 0.98 N g w 10 4 B T2 T1 Fg 0.98 N (9.8 m/s2)(1 000 kg/m3) m3 Fg Finally, the density of the crown is c mc Vc 8.0 mc g Vc g (1.0 10 7.84 N 4 m3)(9.8 m/s2) (a) (b) 10 3 kg/m3 Figure 15.12 From Table 15.1 we see that the density of gold is 19.3 103 kg/m3. Thus, Archimedes should have told the king that (a) When the crown is suspended in air, the scale reads its true weight T1 F g (the buoyancy of air is negligible). (b) When the crown is immersed in water, the buoyant force B reduces the scale reading to the apparent weight T2 F g B. EXAMPLE 15.6 A Titanic Surprise ward buoyant force equals the weight of the displaced water: B wVw g, where Vw , the volume of the displaced water, is equal to the volume of the ice beneath the water (the shaded region in Fig. 15.13b) and w is the density of seawater, 1 030 kg/m3. Because iVi g w wVw g, the fraction of ice beneath the water's surface is f Vw Vi i w An iceberg floating in seawater, as shown in Figure 15.13a, is extremely dangerous because much of the ice is below the surface. This hidden ice can damage a ship that is still a considerable distance from the visible ice. What fraction of the iceberg lies below the water level? Solution This problem corresponds to Case 2. The weight of the iceberg is F g i 917 kg/m3 and Vi is iVi g, where i the volume of the whole iceberg. The magnitude of the up- 917 kg/m3 1 030 kg/m3 0.890 or 89.0% Figure 15.13 (a) (b) (a) Much of the volume of this iceberg is beneath the water. (b) A ship can be damaged even when it is not near the exposed ice. 15.5 Fluid Dynamics 469 15.5 FLUID DYNAMICS Thus far, our study of fluids has been restricted to fluids at rest. We now turn our attention to fluids in motion. Instead of trying to study the motion of each particle of the fluid as a function of time, we describe the properties of a moving fluid at each point as a function of time. Flow Characteristics When fluid is in motion, its flow can be characterized as being one of two main types. The flow is said to be steady, or laminar, if each particle of the fluid follows a smooth path, such that the paths of different particles never cross each other, as shown in Figure 15.14. In steady flow, the velocity of the fluid at any point remains constant in time. Above a certain critical speed, fluid flow becomes turbulent; turbulent flow is irregular flow characterized by small whirlpool-like regions, as shown in Figure 15.15. The term viscosity is commonly used in the description of fluid flow to characterize the degree of internal friction in the fluid. This internal friction, or viscous force, is associated with the resistance that two adjacent layers of fluid have to moving relative to each other. Viscosity causes part of the kinetic energy of a fluid to be converted to internal energy. This mechanism is similar to the one by which an object sliding on a rough horizontal surface loses kinetic energy. Because the motion of real fluids is very complex and not fully understood, we make some simplifying assumptions in our approach. In our model of an ideal fluid, we make the following four assumptions: 1. The fluid is nonviscous. In a nonviscous fluid, internal friction is neglected. An object moving through the fluid experiences no viscous force. 2. The flow is steady. In steady (laminar) flow, the velocity of the fluid at each point remains constant. Properties of an ideal fluid Figure 15.15 Figure 15.14 Laminar flow around an automobile in a test wind tunnel. Hot gases from a cigarette made visible by smoke particles. The smoke first moves in laminar flow at the bottom and then in turbulent flow above. 470 CHAPTER 15 Fluid Mechanics 3. The fluid is incompressible. The density of an incompressible fluid is constant. 4. The flow is irrotational. In irrotational flow, the fluid has no angular momentum about any point. If a small paddle wheel placed anywhere in the fluid does not rotate about the wheel's center of mass, then the flow is irrotational. 15.6 STREAMLINES AND THE EQUATION OF CONTINUITY v P Figure 15.16 A particle in laminar flow follows a streamline, and at each point along its path the particle's velocity is tangent to the streamline. Equation of continuity The path taken by a fluid particle under steady flow is called a streamline. The velocity of the particle is always tangent to the streamline, as shown in Figure 15.16. A set of streamlines like the ones shown in Figure 15.16 form a tube of flow. Note that fluid particles cannot flow into or out of the sides of this tube; if they could, then the streamlines would cross each other. Consider an ideal fluid flowing through a pipe of nonuniform size, as illustrated in Figure 15.17. The particles in the fluid move along streamlines in steady flow. In a time t, the fluid at the bottom end of the pipe moves a distance x 1 v 1t. If A1 is the cross-sectional area in this region, then the mass of fluid contained in the left shaded region in Figure 15.17 is m 1 A1 x 1 A1v 1t, where is the (nonchanging) density of the ideal fluid. Similarly, the fluid that moves through the upper end of the pipe in the time t has a mass m 2 A2v 2t. However, because mass is conserved and because the flow is steady, the mass that crosses A1 in a time t must equal the mass that crosses A2 in the time t. That is, m 1 m 2, or A1v 1t A2v 2t ; this means that A1v 1 A2v 2 constant (15.7) This expression is called the equation of continuity. It states that the product of the area and the fluid speed at all points along the pipe is a constant for an incompressible fluid. This equation tells us that the speed is high where the tube is constricted (small A) and low where the tube is wide (large A). The product Av, which has the dimensions of volume per unit time, is called either the volume flux or the flow rate. The condition Av constant is equivalent to the statement that the volume of fluid that enters one end of a tube in a given time interval equals the volume leaving the other end of the tube in the same time interval if no leaks are present. v2 A2 v1 A1 x 1 x 2 Figure 15.17 A fluid moving with steady flow through a pipe of varying cross-sectional area. The volume of fluid flowing through area A1 in a time interval t must equal the volume flowing through area A2 in the same time inA 2v 2 . terval. Therefore, A 1v 1 Quick Quiz 15.9 Figure 15.18 As water flows from a faucet, as shown in Figure 15.18, why does the stream of water become narrower as it descends? 15.7 Bernoulli's Equation 471 EXAMPLE 15.7 Niagara Falls Note that we have kept only one significant figure because our value for the depth has only one significant figure. Each second, 5 525 m3 of water flows over the 670-m-wide cliff of the Horseshoe Falls portion of Niagara Falls. The water is approximately 2 m deep as it reaches the cliff. What is its speed at that instant? Exercise Solution The cross-sectional area of the water as it reaches the edge of the cliff is A (670 m)(2 m) 1 340 m2. The flow rate of 5 525 m3/s is equal to Av. This gives v 5 525 m3/s A 5 525 m3/s 1 340 m2 4 m/s A barrel floating along in the river plunges over the Falls. How far from the base of the cliff is the barrel when it reaches the water 49 m below? 13 m 10 m. Answer 15.7 BERNOULLI'S EQUATION When you press your thumb over the end of a garden hose so that the opening becomes a small slit, the water comes out at high speed, as shown in Figure 15.19. Is theuation 15.8 to points 1 and 2 gives (1) P1 1 2 y 2 , and ap- v 12 P2 1 2 v 22 P1 P2 (a) Pressure P1 is greater than pressure P2 because v 1 v 2 . This device can be used to measure the speed of fluid flow. (b) A Venturi tube. Figure 15.21 v1 A2 A1 (a) v2 (b) 15.7 Bernoulli's Equation From the equation of continuity, A1v 1 (2) v1 A2 v A1 2 A2v 2 , we find that 473 Substituting this expression into equation (1) gives P1 1 2 A2 A1 2 v 22 P2 1 2 v 22 2(P1 (A12 P2) A22) v2 A1 ! We can use this result and the continuity equation to obtain an expression for v1 . Because A2 A1 , Equation (2) shows us that v 2 v 1 . This result, together with equation (1), indicates that P1 P2 . In other words, the pressure is reduced in the constricted part of the pipe. This result is somewhat analogous to the following situation: Consider a very crowded room in which people are squeezed together. As soon as a door is opened and people begin to exit, the squeezing (pressure) is least near the door, where the motion (flow) is greatest. EXAMPLE 15.9 A Good Trick mass of a dime is m 2.24 g, and its surface area is A 2.50 10 4 m2. How hard are you blowing when the dime rises and travels into the tumbler? It is possible to blow a dime off a table and into a tumbler. Place the dime about 2 cm from the edge of the table. Place the tumbler on the table horizontally with its open edge about 2 cm from the dime, as shown in Figure 15.22a. If you blow forcefully across the top of the dime, it will rise, be caught in the airstream, and end up in the tumbler. The Solution Figure 15.22b indicates we must calculate the upward force acting on the dime. First, note that a thin stationary layer of air is present between the dime and the table. When you blow across the dime, it deflects most of the moving air from your breath across its top, so that the air above the dime has a greater speed than the air beneath it. This fact, together with Bernoulli's equation, demonstrates that the air moving across the top of the dime is at a lower pressure than the air beneath the dime. If we neglect the small thickness of the dime, we can apply Equation 15.8 to obtain Pabove 1 2 v2 above Pbeneath 1 2 v2 beneath 2 cm 2 cm (a) Because the air beneath the dime is almost stationary, we can neglect the last term in this expression and write the difference as Pbeneath Pabove 1 v 2 above . If we multiply this pres2 sure difference by the surface area of the dime, we obtain the upward force acting on the dime. At the very least, this upward force must balance the gravitational force acting on the dime, and so, taking the density of air from Table 15.1, we can state that Fg FBernoulli Fg mg (Pbeneath 2mg A Pabove)A (1 v 2 )A above 2 v above v above ! ! 2(2.24 10 3 kg)(9.80 m/s2) (1.29 kg/m3)(2.50 10 4 m2) 11.7 m/s (b) Figure 15.22 The air you blow must be moving faster than this if the upward force is to exceed the weight of the dime. Practice this trick a few times and then impress all your friends! 474 CHAPTER 15 Fluid Mechanics EXAMPLE 15.10 Torricelli's Law which the liquid leaves the hole when the liquid's level is a distance h above the hole. An enclosed tank containing a liquid of density has a hole in its side at a distance y1 from the tank's bottom (Fig. 15.23). The hole is open to the atmosphere, and its diameter is much smaller than the diameter of the tank. The air above the liquid is maintained at a pressure P. Determine the speed at Solution Because A2 W A1 , the liquid is approximately at rest at the top of the tank, where the pressure is P. Applying Bernoulli's equation to points 1 and 2 and noting that at the hole P1 is equal to atmospheric pressure P0 , we find that P0 But y 2 y1 1 2 2 P A2 v 12 gy 1 P gy 2 h ; thus, this expression reduces to v1 h y2 y1 1 A1 P0 v1 ! 2(P P0) 2gh Figure 15.23 When P is much larger than atmospheric pressure P0 , the liquid speed as the liquid passes through the hole in the side of the container is given approximately by v 1 !2(P P0)/ . When P is much greater than P0 (so that the term 2gh can be neglected), the exit speed of the water is mainly a function of P. If the tank is open to the atmosphere, then P P0 and v 1 !2gh. In other words, for an open tank, the speed of liquid coming out through a hole a distance h below the surface is equal to that acquired by an object falling freely through a vertical distance h. This phenomenon is known as Torricelli's law. Optional Section 15.8 F OTHER APPLICATIONS OF BERNOULLI'S EQUATION Figure 15.24 Streamline flow around an airplane wing. The pressure above the wing is less than the pressure below, and a dynamic lift upward results. The lift on an aircraft wing can be explained, in part, by the Bernoulli effect. Airplane wings are designed so that the air speed above the wing is greater than that below the wing. As a result, the air pressure above the wing is less than the pressure below, and a net upward force on the wing, called lift, results. Another factor influencing the lift on a wing is shown in Figure 15.24. The wing has a slight upward tilt that causes air molecules striking its bottom to be deflected downward. This deflection means that the wing is exerting a downward force on the air. According to Newton's third law, the air must exert an equal and opposite force on the wing. Finally, turbulence also has an effect. If the wing is tilted too much, the flow of air across the upper surface becomes turbulent, and the pressure difference across the wing is not as great as that predicted by Bernoulli's equation. In an extreme case, this turbulence may cause the aircraft to stall. In general, an object moving through a fluid experiences lift as the result of any effect that causes the fluid to change its direction as it flows past the object. Some factors that influence lift are the shape of the object, its orientation with respect to the fluid flow, any spinning motion it might have, and the texture of its surface. For example, a golf ball struck with a club is given a rapid backspin, as shown in Figure 15.25a. The dimples on the ball help "entrain" the air to follow the curvature of the ball's surface. This effect is most pronounced on the top half of the ball, where the ball's surface is moving in the same direction as the air flow. Figure 15.25b shows a thin layer of air wrapping part way around the ball and being deflected downward as a result. Because the ball pushes the air down, the air must push up on the ball. Without the dimples, the air is not as well entrained, 15.8 Other Applications of Bernoulli's Equation 475 QuickLab You can easily demonstrate the effect of changing fluid direction by lightly holding the back of a spoon against a stream of water coming from a faucet. You will see the stream "attach" itself to the curvature of the spoon and be deflected sideways. You will also feel the third-law force exerted by the water on the spoon. (a) (b) Figure 15.25 (a) A golf ball is made to spin when struck by the club. (b) The spinning ball experiences a lifting force that allows it to travel much farther than it would if it were not spinning. and the golf ball does not travel as far. For the same reason, a tennis ball's fuzz helps the spinning ball "grab" the air rushing by and helps deflect it. A number of devices operate by means of the pressure differentials that result from differences in a fluid's speed. For example, a stream of air passing over one end of an open tube, the other end of which is immersed in a liquid, reduces the pressure above the tube, as illustrated in Figure 15.26. This reduction in pressure causes the liquid to rise into the air stream. The liquid is then dispersed into a fine spray of droplets. You might recognize that this so-called atomizer is used in perfume bottles and paint sprayers. The same principle is used in the carburetor of a gasoline engine. In this case, the low-pressure region in the carburetor is produced by air drawn in by the piston through the air filter. The gasoline vaporizes in that region, mixes with the air, and enters the cylinder of the engine, where combustio a solid iron sphere that has a diameter of 3.00 cm. 2. Find the order of magnitude of the density of the nucleus of an atom. What does this result suggest concerning the structure of matter? (Visualize a nucleus as protons and neutrons closely packed together. Each has mass 1.67 10 27 kg and radius on the order of 10 15 m.) 3. A 50.0-kg woman balances on one heel of a pair of highheeled shoes. If the heel is circular and has a radius of 0.500 cm, what pressure does she exert on the floor? 4. The four tires of an automobile are inflated to a gauge pressure of 200 kPa. Each tire has an area of 0.024 0 m2 in contact with the ground. Determine the weight of the automobile. 5. What is the total mass of the Earth's atmosphere? (The radius of the Earth is 6.37 106 m, and atmospheric pressure at the Earth's surface is 1.013 105 N/m2.) Section 15.2 Variation of Pressure with Depth 6. (a) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is 1 024 kg/m3 and that the air above exerts a pressure of 101.3 kPa. (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of 30.0 cm exert to counterbalance the force exerted by the water? 7. The spring of the pressure gauge shown in Figure 15.2 has a force constant of 1 000 N/m, and the piston has a diameter of 2.00 cm. When the gauge is lowered into water, at what depth does the piston move in by 0.500 cm? 8. The small piston of a hydraulic lift has a cross-sectional area of 3.00 cm2, and its large piston has a cross-sectional area of 200 cm2 (see Fig. 15.5a). What force must be applied to the small piston for it to raise a load of 15.0 kN? (In service stations, this force is usually generated with the use of compressed air.) Problems 479 (a) (b) Figure P15.10 WEB 9. What must be the contact area between a suction cup (completely exhausted) and a ceiling if the cup is to support the weight of an 80.0-kg student? 10. (a) A very powerful vacuum cleaner has a hose 2.86 cm in diameter. With no nozzle on the hose, what is the weight of the heaviest brick that the cleaner can lift (Fig. P15.10)? (b) A very powerful octopus uses one sucker of diameter 2.86 cm on each of the two shells of a clam in an attempt to pull the shells apart. Find the greatest force that the octopus can exert in salt water 32.3 m in depth. (Caution: Experimental verification can be interesting, but do not drop a brick on your foot. Do not overheat the motor of a vacuum cleaner. Do not get an octopus mad at you.) 11. For the cellar of a new house, a hole with vertical sides descending 2.40 m is dug in the ground. A concrete foundation wall is built all the way across the 9.60-m width of the excavation. This foundation wall is 0.183 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by 2.40 m 9.60 m 9.80 m/s2 0.183 m 41.3 kN 1 000 kg/m3 a Figure P15.13 14. The tank shown in Figure P15.14 is filled with water to a depth of 2.00 m. At the bottom of one of the side walls is a rectangular hatch 1.00 m high and 2.00 m wide. The hatch is hinged at its top. (a) Determine the force that the water exerts on the hatch. (b) Find the torque exerted about the hinges. 2.00 m 1.00 m 2.00 m 12. A swimming pool has dimensions 30.0 m 10.0 m and a flat bottom. When the pool is filled to a depth of 2.00 m with fresh water, what is the force caused by the water on the bottom? On each end? On each side? 13. A sealed spherical shell of diameter d is rigidly attached to a cart that is moving horizontally with an acceleration a, as shown in Figure P15.13. The sphere is nearly filled with a fluid having density and also contains one small bubble of air at atmospheric pressure. Find an expression for the pressure P at the center of the sphere. Figure P15.14 15. Review Problem. A sormine the densities of the glycerin and the sphere. 28. A frog in a hemispherical pod finds that he just floats without sinking into a sea of blue-green ooze having a density of 1.35 g/cm3 (Fig. P15.28). If the pod has a radius of 6.00 cm and a negligible mass, what is the mass of the frog? maintains a constant volume and that the density of air decreases with the altitude z according to the expresz/8 000, where z is in meters and sion air 0e 1.25 kg/m3 is the density of air at sea level. 0 30. Review Problem. A long cylindrical tube of radius r is weighted on one end so that it floats upright in a fluid having a density . It is pushed downward a distance x from its equilibrium position and then released. Show that the tube will execute simple harmonic motion if the resistive effects of the water are neglected, and determine the period of the oscillations. 31. A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 104 kg. To dive, this submarine takes on mass in the form of seawater. Determine the amount of mass that the submarine must take on if it is to descend at a constant speed of 1.20 m/s, when the resistive force on it is 1 100 N in the upward direction. Take 1.03 103 kg/m3 as the density of seawater. 32. The United States possesses the eight largest warships in the world -- aircraft carriers of the Nimitz class -- and it is building one more. Suppose that one of the ships bobs up to float 11.0 cm higher in the water when 50 fighters take off from it at a location where g 9.78 m/s2. The planes have an average mass of 29 000 kg. Find the horizontal area enclosed by the waterline of the ship. (By comparison, its flight deck has an area of 18 000 m2.) Section 15.5 Fluid Dynamics Section 15.6 Streamlines and the Equation of Continuity Section 15.7 Bernoulli's Equation 33. (a) A water hose 2.00 cm in diameter is used to fill a 20.0-L bucket. If it takes 1.00 min to fill the bucket, what is the speed v at which water moves through the hose? (Note: 1 L 1 000 cm3.) (b) If the hose has a nozzle 1.00 cm in diameter, find the speed of the water at the nozzle. 34. A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If the pressure of the water in the larger pipe is 8.00 104 Pa and the pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes? 35. A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0 m below the water level. If the rate of flow from the leak is 2.50 10 3 m3/min, determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole. 36. Through a pipe of diameter 15.0 cm, water is pumped from the Colorado River up to Grand Canyon Village, located on the rim of the canyon. The river is at an elevation of 564 m, and the village is at an elevation of 2 096 m. (a) What is the minimum pressure at which the water must be pumped if it is to arrive at the village? Figure P15.28 29. How many cubic meters of helium are required to lift a balloon with a 400-kg payload to a height of 8 000 m? (Take He 0.180 kg/m3.) Assume that the balloon 482 CHAPTER 15 Fluid Mechanics (b) If 4 500 m3 are pumped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow? (Note: You may assume that the acceleration due to gravity and the density of air are constant over this range of elevations.) 37. Water flows through a fire hose of diameter 6.35 cm at a rate of 0.012 0 m3/s. The fire hose ends in a nozzle with an inner diameter of 2.20 cm. What is the speed at which the water exits the nozzle? 38. Old Faithful Geyser in Yellowstone National Park erupts at approximately 1-h intervals, and the height of the water column reaches 40.0 m (Fig. P15.38). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of these drops to determine the speed at which the water leaves the ground. (b) Treating the rising stream as an ideal fluid in streamline flow, use Berand determine the value of that acceleration. (b) How long does it take for the top of the shell to reach the water's surface? 61. An incompressible, nonviscous fluid initially rests in the vertical portion of the pipe shown in Figure P15.61a, where L 2.00 m. When the valve is opened, the fluid flows into the horizontal section of the pipe. What is the speed of the fluid when all of it is in the horizontal section, as in Figure P15.61b? Assume that the cross-sectional area of the entire pipe is constant. 485 Figure P15.63 L Valve closed Valve opened v L (a) (b) Figure P15.61 62. Review Problem. A uniform disk with a mass of 10.0 kg and a radius of 0.250 m spins at 300 rev/min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at an average distance of 0.220 m from the axis. The coefficient of friction between the pad and the disk is 0.500. A piston in a cylinder with a diameter of 5.00 cm presses the brake pad against the disk. Find the pressure that the brake fluid in the cylinder must have. 63. Figure P15.63 shows Superman attempting to drink water through a very long straw. With his great strength, he achieves maximum possible suction. The walls of the tubular straw do not collapse. (a) Find the maximum height through which he can lift the water. (b) Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw. 64. Show that the variation of atmospheric pressure with altitude is given by P P0e h, where 0g /P0 , P0 is atmospheric pressure at some reference level y 0, and 0 is the atmospheric density at this level. Assume that the decrease in atmospheric pressure with increasing altitude is given by Equation 15.4, so that dP/dy g, and assume that the density of air is proportional to the pressure. 65. A cube of ice whose edge measures 20.0 mm is floating in a glass of ice-cold water with one of its faces parallel to the water's surface. (a) How far below the water surface is the bottom face of the block? (b) Ice-cold ethyl alcohol is gently poured onto the water's surface to form a layer 5.00 mm thick above the water. The alcohol does not mix with the water. When the ice cube again attains hydrostatic equilibrium, what is the distance from the top of the water to the bottom face of the block? (c) Additional cold ethyl alcohol is poured onto the water's surface until the top surface of the alcohol coincides with the top surface of the ice cube (in 486 CHAPTER 15 Fluid Mechanics hydrostatic equilibrium). How thick is the required layer of ethyl alcohol? 66. Review Problem. A light balloon filled with helium with a density of 0.180 kg/m3 is tied to a light string of length L 3.00 m. The string is tied to the ground, forming an "inverted" simple pendulum, as shown in Figure P15.66a. If the balloon is displaced slightly from its equilibrium position as shown in Figure P15.66b, (a) show that the ensuing motion is simple harmonic and (b) determine the period of the motion. Take the density of air to be 1.29 kg/m3 and ignore any energy loss due to air friction. He Air Air He L g g L (a) (b) Figure P15.66 67. The water supply of a building is fed through a main 6.00-cm-diameter pipe. A 2.00-cm-diameter faucet tap located 2.00 m above the main pipe is observed to fill a 25.0-L container in 30.0 s. (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the 6-cm main pipe? (Assume that the faucet is the only "leak" in the building.) 68. The spirit-in-glass thermometer, invented in Florence, Italy, around 1654, consists of a tube of liquid (the spirit) containing a number of submerged glass spheres with slightly different masses (Fig. P15.68). At sufficiently low temperatures, all the spheres float, but as the temperature rises, the spheres sink one after the other. The device is a crude but interesting tool for measuring temperature. Suppose that the tube is filled with ethyl alcohol, whose density is 0.789 45 g/cm that of water, the volume of water that weighs the same as the anchor is greater than the volume of the anchor. 15.9 As the water falls, its speed increases. Because the flow rate Av must remain constant at all cross sections (see Eq. 15.7), the stream must become narrower as the speed increases. 15.10 The rapidly moving air characteristic of a tornado is at a pressure below atmospheric pressure. The stationary air inside the building remains at atmospheric pressure. The pressure difference results in an outward force on the roof and walls, and this force can be great enough to lift the roof off the building. Opening the windows helps to equalize the inside and outside pressures. P U Z Z L E R A simple seismograph can be constructed with a spring-suspended pen that draws a line on a slowly unrolling strip of paper. The paper is mounted on a structure attached to the ground. During an earthquake, the pen remains nearly stationary while the paper shakes beneath it. How can a few jagged lines on a piece of paper allow scientists at a seismograph station to determine the distance to the origin of an earthquake? (Ken M. Johns/Photo Researchers, Inc.) c h a p t e r Wave Motion Chapter Outline 16.1 Basic Variables of Wave Motion 16.2 Direction of Particle Displacement 16.6 Reflection and Transmission 16.7 Sinusoidal Waves 16.8 Rate of Energy Transfer by Sinusoidal Waves on Strings 16.3 One-Dimensional Traveling Waves 16.9 (Optional) The Linear Wave Equation 16.4 Superposition and Interference 16.5 The Speed of Waves on Strings 490 16.1 Wave Motion 491 ost of us experienced waves as children when we dropped a pebble into a pond. At the point where the pebble hits the water's surface, waves are created. These waves move outward from the creation point in expanding circles until they reach the shore. If you were to examine carefully the motion of a leaf floating on the disturbed water, you would see that the leaf moves up, down, and sideways about its original position but does not undergo any net displacement away from or toward the point where the pebble hit the water. The water molecules just beneath the leaf, as well as all the other water molecules on the pond's surface, behave in the same way. That is, the water wave moves from the point of origin to the shore, but the water is not carried with it. An excerpt from a book by Einstein and Infeld gives the following remarks concerning wave phenomena:1 A bit of gossip starting in Washington reaches New York [by word of mouth] very quickly, even though not a single individual who takes part in spreading it travels between these two cities. There are two quite different motions involved, that of the rumor, Washington to New York, and that of the persons who spread the rumor. The wind, passing over a field of grain, sets up a wave which spreads out across the whole field. Here again we must distinguish between the motion of the wave and the motion of the separate plants, which undergo only small oscillations... The particles constituting the medium perform only small vibrations, but the whole motion is that of a progressive wave. The essentially new thing here is that for the first time we consider the motion of something which is not matter, but energy propagated through matter. The world is full of waves, the two main types being mechanical waves and electromagnetic waves. We have already mentioned examples of mechanical waves: sound waves, water waves, and "grain waves." In each case, some physical medium is being disturbed -- in our three particular examples, air molecules, water molecules, and stalks of grain. Electromagnetic waves do not require a medium to propagate; some examples of electromagnetic waves are visible light, radio waves, television signals, and x-rays. Here, in Part 2 of this book, we study only mechanical waves. The wave concept is abstract. When we observe what we call a water wave, what we see is a rearrangement of the water's surface. Without the water, there would be no wave. A wave traveling on a string would not exist without the string. Sound waves could not travel through air if there were no air molecules. With mechanical waves, what we interpret as a wave corresponds to the propagation of a disturbance through a medium. M Interference patterns produced by outwardspreading waves from many drops of liquid falling into a body of water. 1 A. Einstein and L. Infeld, The Evolution of Physics, New York, Simon & Schuster, 1961. Excerpt from "What Is a Wave?" 492 CHAPTER 16 Wave Motion The mechanical waves discussed in this chapter require (1) some source of disturbance, (2) a medium that can be disturbed, and (3) some physical connection through which adjacent portions of the medium can influence each other. We shall find that all waves carry energy. The amount of energy transmitted through a medium and the mechanism responsible for that transport of energy differ from case to case. For instance, the power of ocean waves during a storm is much greater than the power of sound waves generated by a single human voice. 16.1 y BASIC VARIABLES OF WAVE MOTION x Figure 16.1 The wavelength of a wave is the distance between adjacent crests, adjacent troughs, or any other comparable adjacent identical points. Imagine you are floating on a raft in a large lake. You slowly bob up and down as waves move past you. As you look out over the lake, you may be able to see the individual waves approaching. The point at which the displacement of the water from its normal level is highest is called the crest of the wave. The distance from one crest to the next is called the wavelength (Greek letter lambda). More generally, the wavelength is the minimum distance between any two identical points (such as the crests) on adjacent waves, as shown in Figure 16.1. If you count the number of seconds between the arrivals of two adjacent waves, you are measuring the period T of the waves. In general, the period is the time required for two identical points (such as the crests) of adjacent waves to pass by a point. The same information is more often given by the inverse of the period, which is called the frequency f. In general, the frequency of a periodic wave is the number of crests (or troughs, or any other point on the wave) that pass a given point in a unit time interval. The maximum displacement of a particle of the medium is called the amplitude A of the wave. For our water wave, this represents the highest distance of a water molecule above the undisturbed surface of the water as the wave passes by. Waves travel with a specific speed, and this speed depends on the properties of the medium being disturbed. For instance, sound waves travel through roomtemperature air with a speed of about 343 m/s (781 mi/h), whereas they travel through most solids with a speed greater than 343 m/s. 16.2 DIRECTION OF PARTICLE DISPLACEMENT One way to demonstrate wave motion is to flick one end of a long rope that is under tension and has its opposite end fixed, as shown in Figure 16.2. In this manner, a single wave bump (called a wave pulse) is formed and travels along the rope with a definite speed. This type of disturbance is called a traveling wave, and Figure 16.2 represents four consecutive "snapshots" of the creation and propagation of the traveling wave. The rope is the medium through which the wave travels. Such a single pulse, in contrast to a train of pulses, has no frequency, no period, and no wavelength. However, the pulse does have definite amplitude and definite speed. As we shall see later, the properties of this particular medium that determine the speed of the wave are the tension in the rope and its mass per unit length. The shape of the wave pulse changes very little as it travels along the rope.2 As the wave pulse travels, each small segment of the rope, as it is disturbed, moves in a direction perpendicular to the wave motion. Figure 16.3 illustrates this 2 Strictly speaking, the pulse changes shape and gradually spreads out during the motion. This effect is called dispersion and is common to many mechanical waves, as well as to electromagnetic waves. We do not consider dispersion in this chapter. 16.2 Direction of Particle Displacement 493 P P P P Figure 16.2 A wave pulse traveling down a stretched rope. The shape of the pulse is approximately unchanged as it travels along the rope. Figure 16.3 A pulse traveling on a stretched rope is a transverse wave. The direction of motion of any element P of the rope (blue arrows) is perpendicular to the direction of wave motion (red arrows). point for one particular segment, labeled P. Note that no part of the rope ever moves in the direction of the wave. A traveling wave that causes the particles of the disturbed medium to move perpendicular to the wave motion is called a transverse wave. Transverse wave Compare this with another type of wave -- one moving down a long, stretched spring, as shown in Figure 16.4. The left end of the spring is pushed briefly to the right and then pulled briefly to the left. This movement creates a sudden compression of a region of the coils. The compressed region travels along the spring (to the right in Figure 16.4). The compressed region is followed by a region where the coils are extended. Notice that the direction of the displacement of the coils is parallel to the direction of propagation of the compressed region. A traveling wave that causes the particles of the medium to move parallel to the direction of wave motion is called a longitudinal wave. Sound waves, which we shall discuss in Chapter 17, are another example of longitudinal waves. The disturbance in a sound wave is a series of high-pressure and low-pressure regions that travel through air or any other material medium. Compressed Compressed Longitudinal wave Stretched Stretched Figure 16.4 A longitudinal wave along a stretched spring. The displacement of the coils is in the direction of the wave motion. Each compressed region is followed by a stretched region. 494 CHAPTER 16 Wave Motion Wave motion Crest Trough Figure 16.5 The motion of water molecules on the surface of deep water in which a wave is propagating is a combination of transverse and longitudinal displacements, with the result that molecules at the surface move in nearly circular paths. Each molecule is displaced both horizontally and vertically from its equilibrium position. QuickLab Make a "telephone" by poking a small hole in the bottom of two paper cups, threading a string through the holes, and tying knots in the ends of the string. If you speak into one cup while pulling the string taut, a friend can hear your voice in the other cup. What kind of wave is present in the string? Some waves in nature exhibit a combination of transverse and longitudinal displacements. Surface water waves are a good example. When a water wave travels on the surface of deep water, water molecules at the surface move in nearly circular paths, as shown in Figure 16.5. Note that the disturbance has both transverse and longitudinal components. The transverse displacement is seen in Figure 16.5 as the variations in vertical position of the water molecules. The longitudinal displacement can be explained as follows: As the wave passes over the water's surface, water molecules at the crests move in the direction of propagation of the wave, whereas molecules at the troughs move in the direction opposite the propagation. Because the molecule at the labeled crest in Figure 16.5 will be at a trough after half a period, its movement in the direction of the propagation of the wave will be canceled by its movement in the opposite direction. This holds for every other water molecule disturbed by the wave. Thus, there is no net displacement of any water molecule during one complete cycle. Although the molecules experience no net displacement, the wave propagates along the surface of the water. The three-dimensional waves that travel out from the point under the Earth's surface at which an earthquake occurs are of both types -- transverse and longitudinal. The longitudinal waves are the faster of the two, traveling at speeds in the range of 7 to 8 km/s near the surface. These are called P waves, with "P" standing for primary because they travel faster than the transverse waves and arrive at a seismograph first. The slower transverse waves, called S waves (with "S" standing for secondary), travel through the Earth at 4 to 5 km/s near the surface. By recording the time interval between the arrival of these two sets of waves at a seismograph, the distance from the seismograph to the point of origin of the waves can be determined. A single such measurement establishes an imaginary sphere centered on the seismograph, with the radius of the sphere determined by the difference in arrival times of the P and S waves. The origin of the waves is located somewhere on that sphere. The imaginary spheres from three or more monitoring stations located far apart from each other intersect at one region of the Earth, and this region is where the earthquake occurred. Quick Quiz 16.1 (a) In a long line of people waiting to buy tickets, the first person leaves and a pulse of motion occurs as people step forward to fill the gap. As each person steps forward, the gap moves through the line. Is the propagation of this gap transverse or longitudinal? (b) Consider the "wave" at a baseball game: people stand up and shout as the wave arrives at their location, and the resultant pulse moves around the stadium. Is this wave transverse or longitudinal? 16.3 One-Dimensional Traveling Waves 495 16.3 ONE-DIMENSIONAL TRAVELING WAVES Consider a wave pulse traveling to the right with constant speed v on a long, taut string, as shown in Figure 16.6. The pulse moves along the x axis (the axis of the string), and the transverse (vertical) displacement of the string (the medium) is measured along the y axis. Figure 16.6a represents the shape and position of the pulse at time t 0. At this time, the shape of the pulse, whatever it may be, can be represented as y f(x). That is, y, which is the vertical position of any point on the string, is some definite function of x. The displacement y, sometimes called the wave function, depends on both x and t. For this reason, it is often written y(x, t), which is read "y as a function of x and t." Consider a particular point P on the string, identified by a specific value of its x coordinate. Before the pulse arrives at P, the y coordinate of this point is zero. As the wave passes P, the y coordinate of this point increases, reaches a maximum, and then decreases to zero. Therefore, the wave function y represents the y coordinate of any point P of the medium at any time t. Because its speed is v, the wave pulse travels to the right a distance vt in a time t (see Fig. 16.6b). If the shape of the pulse does not change with time, we can represent the wave function y for all times after t 0. Measured in a stationary reference frame having its origin at O, the wave function is y f(x vt) (16.1) Wave traveling to the right If the wave pulse travels to the left, the string displacement is y f(x vt) (16.2) Wave traveling to the left For any given time t, the wave function y as a function of x defines a curve representing the shape of the pulse at this time. This curve is equivalent to a "snapshot" of the wave at this time. For a pulse that moves without changing shape, the speed of the pulse is the same as that of any feature along the pulse, such as the crest shown in Figure 16.6. To find the speed of the pulse, we can calculate how far the crest moves in a short time and then divide this distance by the time interval. To follow the motion of the crest, we must substitute some particular value, say x 0 , in Equation 16.1 for x vt. Regardless of how x and t change individually, we must require that x vt x 0 in order to stay with the crest. This expression therefore represents the equation of motion of the crest. At t 0, the crest is at x x 0 ; at a y y vt v P v A P O (a) Pulse at t = 0 x O (b) Pulse at time t x Figure 16.6 A one-dimensional wave pulse traveling to the right with a speed v. (a) At t 0, the shape of the pulse is given by y f (x). (b) At some later time t, the shape remains unchanged and thit is convenient to describe the motion of a small segment of the string in a moving frame of reference. (b) In the moving frame of reference, the small segment of length s moves to the left with speed v. The net force on the segment is in the radial direction because the horizontal components of the tension force cancel. First, let us verify that this expression is dimensionally correct. The dimensions of T are ML/T 2, and the dimensions of are M/L. Therefore, the dimensions of T/ are L2/T 2; hence, the dimensions of T/ are L/T -- indeed, the dimensions of speed. No other combination of T and is dimensionally correct if we assume that they are the only variables relevant to the situation. Now let us use a mechanical analysis to derive Equation 16.4. On our string under tension, consider a pulse moving to the right with a uniform speed v measured relative to a stationary frame of reference. Instead of staying in this reference frame, it is more convenient to choose as our reference frame one that moves along with the pulse with the same speed as the pulse, so that the pulse is at rest within the frame. This change of reference frame is permitted because Newton's laws are valid in either a stationary frame or one that moves with constant velocity. In our new reference frame, a given segment of the string initially to the right of the pulse moves to the left, rises up and follows the shape of the pulse, and then continues to move to the left. Figure 16.11a shows such a segment at the instant it is located at the top of the pulse. The small segment of the string of length s shown in Figure 16.11a, and magnified in Figure 16.11b, forms an approximate arc of a circle of radius R. In our moving frame of reference (which is moving to the right at a speed v along with the pulse), the shaded segment is moving to the left with a speed v. This segment has a centripetal acceleration equal to v 2/R, which is supplied by components of the tension T in the string. The force T acts on either side of the segment and tangent to the arc, as shown in Figure 16.11b. The horizontal components of T cancel, and each vertical component T sin acts radially toward the center of the arc. Hence, the total radial force is 2T sin . Because the segment is small, is small, and we can use the small-angle approximation sin . Therefore, the total radial force is Fr 2T sin 2T s. Because the segment forms part of a circle The segment has a mass m at the center, s R(2 ), and hence and subtends an angle 2 m s 2 R 16.5 The Speed of Waves on Strings 501 If we apply Newton's second law to this segment, the radial component of motion gives Fr 2T ma mv 2 R 2 R v2 R Solving for v gives Equation 16.4. Notice that this derivation is based on the assumption that the pulse height is small relative to the length of the string. Using this assumption, we were able to . Furthermore, the model assumes that the tenuse the approximation sin sion T is not affected by the presence of the pulse; thus, T is the same at all points on the string. Finally, this proof does not assume any particular shape for the pulse. Therefore, we conclude that a pulse of any shape travels along the string with speed v T/ without any change in pulse shape. EXAMPLE 16.2 The Speed of a Pulse on a Cord T mg (2.00 kg)(9.80 m/s2) 19.6 N A uniform cord has a mass of 0.300 kg and a length of 6.00 m (Fig. 16.12). The cord passes over a pulley and supports a 2.00kg object. Find the speed of a pulse traveling along this cord. Solution The tension T in the cord is equal to the weight of the suspended 2.00-kg mass: 5.00 m (This calculation of the tension neglects the small mass of the cord. Strictly speaking, the cord can never be exactly horizontal, and therefore the tension is not uniform.) The mass per unit length of the cord is m 0.300 kg 6.00 m 0.050 0 kg/m Therefore, the wave speed is 1.00 m v T 0.253 s. 19.6 N 0.050 0 kg/m 19.8 m/s 2.00 kg Exercise Answer Figure 16.12 The tension T in the cord is maintained by the suspended object. The speed of any wave traveling along the cord is given by v T/ . Find the time it takes the pulse to travel from the wall to the pulley. Quick Quiz 16.3 Suppose you create a pulse by moving the free end of a taut string up and down once with your hand. The string is attached at its other end to a distant wall. The pulse reaches the wall in a time t. Which of the following actions, taken by itself, decreases the time it takes the pulse to reach the wall? More than one choice may be correct. (a) Moving your hand more quickly, but still only up and down once by the same amount. (b) Moving your hand more slowly, but still only up and down once by the same amount. (c) Moving your hand a greater distance up and down in the same amount of time. (d) Moving your hand a lesser distance up and down in the same amount of time. (e) Using a heavier string of the same length and under the same tension. (f) Using a lighter string of the same length and under the same tension. (g) Using a string of the same linear mass density but under decreased tension. (h) Using a string of the same linear mass density but under increased tension. 502 CHAPTER 16 Wave Motion 16.6 REFLECTION AND TRANSMISSION Incident pulse (a) (b) (c) (d) (e) Reflected pulse Figure 16.13 The reflection of a traveling wave pulse at the fixed end of a stretched string. The reflected pulse is inverted, but its shape is unchanged. Incident pulse (a) We have discussed traveling waves moving through a uniform medium. We now consider how a traveling wave is affected when it encounters a change in the medium. For example, consider a pulse traveling on a string that is rigidly attached to a support at one end (Fig. 16.13). When the pulse reaches the support, a severe change in the medium occurs -- the string ends. The result of this change is that the wave undergoes reflection -- that is, the pulse moves back along the string in the opposite direction. Note that the reflected pulse is inverted. This inversion can be explained as follows: When the pulse reaches the fixed end of the string, the string produces an upward force on the support. By Newton's third law, the support must exert an equal and opposite (downward) reaction force on the string. This downward force causes the pulse to invert upon reflection. Now consider another case: this time, the pulse arrives at the end of a string that is free to move vertically, as shown in Figure 16.14. The tension at the free end is maintained because the string is tied to a ring of negligible mass that is free to slide vertically on a smooth post. Again, the pulse is reflected, but this time it is not inverted. When it reaches the post, the pulse exerts a force on the free end of the string, causing the ring to accelerate upward. The ring overshoots the height of the incoming pulse, and then the downward component of the tension force pulls the ring back down. This movement of the ring produces a reflected pulse that is not inverted and that has the same amplitude as the incoming pulse. Finally, we may have a situation in which the boundary is intermediate between these two extremes. In this case, part of the incident pulse is reflected and part undergoes transmission -- that is, some of the pulse passes through the boundary. For instance, suppose a light string is attached to a heavier string, as shown in Figure 16.15. When a pulse traveling on the light string reaches the boundary between the two, part of the pulse is reflected and inverted and part is transmitted to the heavier string. The reflected pulse is inverted for the same reasons described earlier in the case of the string rigidly attached to a support. Note that the reflected pulse has a smaller amplitude than the incident pulse. In Section 16.8, we shall learn that the energy carried by a wave is related to its amplitude. Thus, according to the principle of the conservation of energy, when the pulse breaks up into a reflected pulse and a transmitted pulse at the boundary, the sum of the energies of these two pulses must equal the energy of the incident pulse. Because the reflected pulse contains only part of the energy of the incident pulse, its amplitude must be smaller. (b) Incident pulse (c) Reflected pulse (d) (a) Transmitted pulse Figure 16.15 Figure 16.14 The reflection of a traveling wave pulse at the free end of a stretched string. The reflected pulse is not inverted. Reflected pulse (b) (a) A pulse traveling to the right on a light string attached to a heavier string. (b) Part of the incident pulse is reflected (and inverted), and part is transmitted to the heavier string. 16.7 Sinusoidal Waves Incident pulse 503 (a) Figure 16.16 Reflected pulse Transmitted pulse (b) (a) A pulse traveling to the right on a heavy string attached to a lighter string. (b) The incident pulse is partially reflected and partially transmitted, and the reflected pulse is not inverted. When a pulse traveling on a heavy string strikes the boundary between the heavy string and a lighter one, as shown in Figure 16.16, again part is reflected and part is transmitted. In this case, the reflected pulse is not inverted. In either case, the relative heights of the reflected and transmitted pulses depend on the relative densities of the two strings. If the strings are identical, there is no discontinuity at the boundary and no reflection takes place. According to Equation 16.4, the speed of a wave on a string increases as the mass per unit length of the string decreases. In other words, a pulse travels more slowly on a heavy string than on a light string if both are under the same tension. The following general rules apply to reflected waves: When a wave pulse travels from medium A to medium B and vA Q vB (that is, when B is denser than A), the pulse is inverted upon reflection. When a wave pulse travels from medium A to medium B and vA P vB (that is, when A is denser than B), the pulse is not inverted upon reflection. 16.7 SINUSOIDAL WAVES In this section, we introduce an important wave function whose shape is shown in Figure 16.17. The wave represented by this curve is called a sinusoidal wave because the curve is the same as that of the function sin plotted against . The sinusoidal wave is the simplest example of a periodic continuous wave and can be used to build more complex waves, as we shall see in Section 18.8. The red curve represents a snapshot of a traveling sinusoidal wave at t 0, and the blue curve represents a snapshot of the wave at some later time t. At t 0, the function describing the positions of the particles of the medium through which the sinusoidal wave is traveling can be written y A sin 2 x (16.5) y vt v x where the constant A represents the wave amplitude and the constant is the wavelength. Thus, we see that the position of a particle of the medium is the same whenever x is increased by an integral multiple of . If the wave moves to the right with a speed v, then the wave function at some later time t is y A sin 2 (x vt) (16.6) t=0 t Figure 16.17 That is, the traveling sinusoidal wave moves to the right a distance vt in the time t, as shown in Figure 16.17. Note that the wave function has the form f(x vt) and A one-dimensional sinusoidal wave traveling to the right with a speed v. The red curve represents a snapshot of the wave at t 0, and the blue curve represents a snapshot at some later time t. 504 CHAPTER 16 Wave Motion so represents a wave traveling to the right. If the wave were traveling to the left, the quantity x vt would be replaced by x vt, as we learned when we developed Equations 16.1 and 16.2. By definition, the wave travels a distance of one wavelength in one period T. Therefore, the wave speed, wavelength, and period are related by the expression v (16.7) T Substituting this expression for v into Equation 16.6, we find that y A sin 2 x t T (16.8) This form of the wave function clearly shows the periodic nature of y. At any given , time t (a snapshot of the wave), y has the same value at the positions x, x x 2 , and so on. Furthermore, at any given position x, the value of y is the same at times t, t T, t 2T, and so on. We can express the wave function in a convenient form) (d) Figure 16.19 One method for producing a train of sinusoidal wave pulses on a string. The left end of the string is connected to a blade that is set into oscillation. Every segment of the string, such as the point P, oscillates with simple harmonic motion in the vertical direction. If the wave at t can be written as 0 is as described in Figure 16.19b, then the wave function y A sin(kx t) We can use this expression to describe the motion of any point on the string. The point P (or any other point on the string) moves only vertically, and so its x coordinate remains constant. Therefore, the transverse speed vy (not to be confused with the wave speed v) and the transverse acceleration ay are vy ay dy dt dv y dt y t vy t A cos(kx 2A t) t) (16.16) (16.17) x constant sin(kx x constant In these expressions, we must use partial derivatives (see Section 8.6) because y depends on both x and t. In the operation y/ t, for example, we take a derivative with respect to t while holding x constant. The maximum values of the transverse speed and transverse acceleration are simply the absolute values of the coefficients of the cosine and sine functions: v y, max a y, max A 2A (16.18) (16.19) The transverse speed and transverse acceleration do not reach their maximum values simultaneously. The transverse speed reaches its maximum value ( A) when y 0, whereas the transverse acceleration reaches its maximum value ( 2A) when y A. Finally, Equations 16.18 and 16.19 are identical in mathematical form to the corresponding equations for simple harmonic motion, Equations 13.10 and 13.11. 16.8 Rate of Energy Transfer by Sinusoidal Waves on Strings 507 Quick Quiz 16.4 A sinusoidal wave is moving on a string. If you increase the frequency f of the wave, how do the transverse speed, wave speed, and wavelength change? EXAMPLE 16.4 A Sinusoidally Driven String Because A y 12.0 cm t) 0.120 m, we have (0.120 m) sin(1.57x 31.4t ) The string shown in Figure 16.19 is driven at a frequency of 5.00 Hz. The amplitude of the motion is 12.0 cm, and the wave speed is 20.0 m/s. Determine the angular frequency and angular wave number k for this wave, and write an expression for the wave function. A sin(kx Exercise Solution that Using Equations 16.10, 16.12, and 16.13, we find 2 T 2 f 2 (5.00 Hz) 31.4 rad/s Calculate the maximum values for the transverse speed and transverse acceleration of any point on the string. 3.77 m/s; 118 m/s2. Answer k v 31.4 rad/s 20.0 m/s 1.57 rad/m 16.8 RATE OF ENERGY TRANSFER BY SINUSOIDAL WAVES ON STRINGS As waves propagate through a medium, they transport energy. We can easily demonstrate this by hanging an object on a stretched string and then sending a pulse down the string, as shown in Figure 16.20. When the pulse meets the suspended object, the object is momentarily displaced, as illustrated in Figure 16.20b. In the process, energy is transferred to the object because work must be done for it to move upward. This section examines the rate at which energy is transported along a string. We shall assume a one-dimensional sinusoidal wave in the calculation of the energy transferred. Consider a sinusoidal wave traveling on a string (Fig. 16.21). The source of the energy being transported by the wave is some external agent at the left end of the string; this agent does work in producing the oscillations. As the external agent performs work on the string, moving it up and down, energy enters the system of the string and propagates along its length. Let us focus our attention on a segment of the string of length x and mass m. Each such segment moves vertically with simple harmonic motion. Furthermore, all segments have the same angular frequency and the same amplitude A. As we found in Chapter 13, the elastic potential energy U associated with a particle in simple harmonic motion is U 1ky 2, 2 where the simple harmonic motion is in the y direction. Using the relationship 2 k/m developed in Equations 13.16 and 13.17, we can write this as m (a) m (b) Figure 16.20 (a) A pulse traveling to the right on a stretched string on which an object has been suspended. (b) Energy is transmitted to the suspended object when the pulse arrives. m Figure 16.21 A sinusoidal wave traveling along the x axis on a stretched string. Every segment moves vertically, and every segment has the same total energy. 508 CHAPTER 16 Wave Motion U 1m 2y 2. If we apply this equation to the segment of mass m, we see that the 2 potential energy of this segment is U 1 2( m) 2y 2 Because the mass per unit length of the string is potential energy of the segment as U 1 2( m/ x, we can express the x) 2y 2 As the length of the segment shrinks to zero, comes a differential relationship: dU 1 2( x : dx, and this expression be- dx) 2y 2 We replace the general displacement y of the segment with the wave function for a sinusoidal wave: dU 1 2 2[A sin(kx t)]2 dx 1 2 2A2 sin2(kx t) dx If we take a snapshot of the wave at time t segment is dU 1 2 2A2 0, then the potential energy in a given sin2 kx dx To obtain the total potential energy in one wavelength, we integrate this expression over all the string segments in one wavelength: U 1 2 dU 0 2A2 1 x 2 1 2 2A2 1 4k sin2 kx dx 0 1 2 1 2 2A2 0 2A2(1 2 sin2 kx dx ) 1 4 2A2 sin 2 kx Because it is in motion, each segment of the string also has kinetic energy. When we use this procedure to analyze the total kinetic energy in one wavelength of the string, we obtain the same result: K dK 1 4 2A2 The total energy in one wavelength of the wave is the sum of the potential and kinetic energies: E U K 1 2 2A2 (16.20) As the wave moves along the string, this amount of energy passes by a given point on the string during one period of the oscillation. Thus, the power, or rate of energy transfer, associated with the wave is E Power of a wave 1 2 1 2 2A2 t 2 T A2v 1 2 2A2 T (16.21) This shows that the rate of energy transfer by a sinusoidal wave on a string is proportional to (a) the wave speed, (b) the square of the frequency, and (c) the square of the amplitude. In fact: the rate of energy transfer in any sinusoidal wave is proportional to the square of the angular frequency and to the square of the amplitude. 16.9 The Linear Wave Equation 509 EXAMPLE 16.5 Power Supplied to a Vibrating String oidal waves on the string has the value 2 f 2 (60.0 Hz) 377 s 1 A taut string for which 5.00 10 2 kg/m is under a tension of 80.0 N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm? Using these values in Equation 16.21 for the power, with A 6.00 10 2 m, we obtain 1 2A2v 2 1 2 (5.00 Solution 16.4, v Because f The wave speed on the string is, from Equation T 10 10 2 kg/m)(377 s 1)2 2 5.00 80.0 N 10 2 kg/m 40.0 m/s of the sinus- (6.00 512 W m)2(40.0 m/s) 60.0 Hz, the angular frequency Optional Section 16.9 THE LINEAR WAVE EQUATION In Section 16.3 we introduced the concept of the wave function to represent waves traveling on a string. All wave functions y(x, t) represent solutions of an equation called the linear wave equation. This equation gives a complete description of the wave motion, and from it one can derive an expression for the wave speed. Furthermore, the linear wave equation is basic to many forms of wave motion. In this section, we derive this equation as applied to waves on strings. Suppose a traveling wave is propagating along a string that is under a tension T. Let us consider one small string segment of length x (Fig. 16.22). The ends of the segment make small angles A and B with the x axis. The net force acting on the segment in the vertical direction is Fy T sin B T x A T B B T sin A T(sin B sin A) A Because the angles are small, we can use the small-angle approximation sin tan to express the net force as Fy T(tan B Figure 16.22 tan A) However, the tangents of the angles at A and B are defined as the slopes of the string segment at these points. Because the slope of a curve is given by y/ x, we have Fy T y x y x (16.22) A A segment of a string under tension T. The slopes at points A and B are given by tan A and tan B , respectively. B We now apply Newton's second law to the segment, with the mass of the segment given by m x: 2y F y ma y x (16.23) t2 Combining Equation 16.22 with Equation 16.23, we obtain x 2y t2 2y T y x B y x x A ( y/ x)B ( y/ x)A T t2 (16.24) 510 CHAPTER 16 Wave Motion The right side of this equation can be expressed in a different form if we note that the partial derivative of any function is defined as f x lim f(x x:0 x) x f(x) If we associate f(x x) with ( y/ x)B and f(x) with ( y/ x)A , we see that, in the limit x : 0, Equation 16.24 becomes Linear wave equation 2y 2y T t2 x2 (16.25) This is the linear wave equation as it applies to waves on a string. We now show that the sinusoidal wave function (Eq. 16.11) represents a solution of the linear wave equation. If we take the sinusoidal wave function to be of the form y(x, t) A sin(kx t), then the appropriate derivatives are 2y t2 2y 2A sin(kx t) t) x2 2 k 2A sin(kx Substituting these expressions into Equation 16.25, we obtain T sin(kx t) k 2 sin(kx t) This equation must be true for all values of the variables x and t in order for the sinusoidal wave function to be a solution of the wave equation. Both sides of the equation depend on x and t through the same function sin(kx t). Because this function divides out, we do indeed have an identity, provided that 2 k2 Using the relationship v T 2 /k (Eq. 16.13) in this expression, we see that v2 v T T k2 which is Equation 16.4. This derivation represents another proof of the expression for the wave speed on a taut string. The linear wave equation (Eq. 16.25) is often written in the form Linear wave equation in general 2y x2 1 v2 2y t2 (16.26) This expression applies in general to various types of traveling waves. For waves on strings, y represents the vertical displacement of the string. For sound waves, y corresponds to displacement of air molecules from equilibrium or variations in either the pressure or the density of the gas through which the sound waves are propagating. In the case of electromagnetic waves, y corresponds to electric or magnetic field components. We have shown that the sinusoidal wave function (Eq. 16.11) is one solution of the linear wave equation (Eq. 16.26). Although we do not prove it here, the linear Summary 511 wave equation is satisfied by any wave function having the form y f(x vt). Furthermore, we have seen that the linear wave equation is a direct consequence of Newton's second law applied to any segment of the string. SUMMARY A transverse wave is one in which the particles of the medium move in a direction perpendicular to the direction of the wave velocity. An example is a wave on a taut string. A longitudinal wave is one in which the particles of the medium move in a direction parallel to the direction of the wave velocity. Sound waves in fluids are longitudinal. You should be able to identify examples of both types of waves. Any one-dimensional wave traveling with a speed v in the x direction can be represented by a wave function of the form y f(x vt) (16.1, 16.2) where the positive sign applies to a wave traveling in the negative x direction and the negative sign applies to a wave traveling in the positive x direction. The shape of the wave at any instant in time (a snapshot of the wave) is obtained by holding t constant. The superposition principle specifies that when two or more waves move through a medium, the resultant wave function equals the algebraic sum of the individual wave functions. When two waves combine in space, they interfere to produce a resultant wave. The interference may be constructive (when the individual displacements are in the same direction) or destructive (when the displacements are in opposite directions). The speed of a wave traveling on a taut string of mass per unit length and tension T is T (16.4) v A wave is totally or partially reflected when it reaches the end of the medium in which it propagates or when it reaches a boundary where its speed cif the end at which reflection occurs is free to slide up and down? WEB Section 16.7 Sinusoidal Waves 23. (a) Plot y versus t at x 0 for a sinusoidal wave of the form y (15.0 cm) cos(0.157x 50.3t) , where x and y are in centimeters and t is in seconds. (b) Determine (a) Determine the transverse speed and acceleration of the string at t 0.200 s for the point on the string located at x 1.60 m. (b) What are the wavelength, period, and speed of propagation of this wave? 31. (a) Write the expression for y as a function of x and t for a sinusoidal wave traveling along a rope in the negative x direction with the following characteristics: A 8.00 cm, 80.0 cm, f 3.00 Hz, and y(0, t ) 0 at t 0. (b) Write the expression for y as a function of x and t for the wave in part (a), assuming that y(x, 0) 0 at the point x 10.0 cm. 32. A transverse sinusoidal wave on a string has a period T 25.0 ms and travels in the negative x direction with a speed of 30.0 m/s. At t 0, a particle on the string at Problems x 0 has a displacement of 2.00 cm and travels downward with a speed of 2.00 m/s. (a) What is the amplitude of the wave? (b) What is the initial phase angle? (c) What is the maximum transverse speed of the string? (d) Write the wave function for the wave. 33. A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. Initially, the left end of the string is at the origin. Find (a) the frequency and angular frequency, (b) the angular wave number, and (c) the wave function for this wave. Determine the equation of motion for (d) the left end of the string and (e) the point on the string at x 1.50 m to the right of the left end. (f) What is the maximum speed of any point on the string? 34. A sinusoidal wave on a string is described by the equation y (0.51 cm) sin(kx t) 515 WEB where k 3.10 rad/cm and 9.30 rad/s. How far does a wave crest move in 10.0 s? Does it move in the positive or negative x direction? 35. A wave is described by y (2.00 cm) sin(kx t ), where k 2.11 rad/m, 3.62 rad/s, x is in meters, and t is in seconds. Determine the amplitude, wavelength, frequency, and speed of the wave. 36. A transverse traveling wave on a taut wire has an amplitude of 0.200 mm and a frequency of 500 Hz. It travels with a speed of 196 m/s. (a) Write an equation in SI units of the form y A sin(kx t ) for this wave. (b) The mass per unit length of this wire is 4.10 g/m. Find the tension in the wire. 37. A wave on a string is described by the wave function y (0.100 m) sin(0.50x 20t) quency is halved, (c) both the wavelength and the amplitude are doubled, and (d) both the length of the rope and the wavelength are halved? 41. Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string that has a linear mass density of 4.00 10 2 kg/m. If the source can deliver a maximum power of 300 W and the string is under a tension of 100 N, what is the highest vibrational frequency at which the source can operate? 42. It is found that a 6.00-m segment of a long string contains four complete waves and has a mass of 180 g. The string is vibrating sinusoidally with a frequency of 50.0 Hz and a peak-to-valley displacement of 15.0 cm. (The "peak-to-valley" distance is the vertical distance from the farthest positive displacement to the farthest negative displacement.) (a) Write the function that describes this wave traveling in the positive x direction. (b) Determine the power being supplied to the string. 43. A sinusoidal wave on a string is described by the equation y (0.15 m) sin(0.80x 50t ) where x and y are in meters and t is in seconds. If the mass per unit length of this string is 12.0 g/m, determine (a) the speed of the wave, (b) the wavelength, (c) the frequency, and (d) the power transmitted to the wave. 44. A horizontal string can transmit a maximum power of (without breaking) if a wave with amplitude A and angular frequency is traveling along it. To increase this maximum power, a student folds the string and uses the "double string" as a transmitter. Determine the maximum power that can be transmitted along the "double string," supposing that the tension is constant. (Optional) (a) Show that a particle in the string at x 2.00 m executes simple harmonic motion. (b) Determine the frequency of oscillation of this particular point. Section 16.9 The Linear Wave Equation 45. (a) Evaluate A in the scalar equality (7 3)4 A. (b) Evaluate A, B, and C in the vector equality 7.00 i 3.00 k A i B j C k. Explain how you arrive at your answers. (c) The functional equality or identity A B cos(Cx Dt E) (7.00 mm) cos(3x 4t 2) Section 16.8 Rate of Energy Transfer by Sinusoidal Waves on Strings 38. A taut rope has a mass of 0.180 kg and a length of 3.60 m. What power must be supplied to the rope to generate sinusoidal waves having an amplitude of 0.100 m and a wavelength of 0.500 m and traveling with a speed of 30.0 m/s? 39. A two-dimensional water wave spreads in circular wave fronts. Show that the amplitude A at a distance r from the initial disturbance is proportional to 1/r. (Hint: Consider the energy carried by one outward-moving ripple.) 40. Transverse waves are being generated on a rope under constant tension. By what factor is the required power increased or decreased if (a) the length of the rope is doubled and the angular frequency remains constant, (b) the amplitude is doubled and the angular fre- is true for all values of the variables x and t, which are measured in meters and in seconds, respectively. Evaluate the constants A, B, C, D, and E. Explain how you arrive at your answers. 46. Show that the wave function y e b(x vt ) is a solution of the wave equation (Eq. 16.26), where b is a constant. 47. Show that the wave function y ln[b(x vt )] is a solution to Equation 16.26, where b is a constant. 48. (a) Show that the function y(x, t ) x 2 v 2t 2 is a solution to the wave equation. (b) Show that the function above can be written as f (x vt ) g(x vt ), and determine the functional forms for f and g. (c) Repeat parts (a) and (b) for the function y(x, t ) sin(x) cos(vt ). 516 CHAPTER 16 Wave Motion (a) What are the speed and direction of travel of the wave? (b) What is the vertical displacement of the string at t 0, x 0.100 m? (c) What are the wavelength and frequency of the wave? (d) What is the maximum magnitude of the transverse speed of the string? 52. Motion picture film is projected at 24.0 frames per second. Each frame is a photograph 19.0 mm in height. At what constant speed does the film pass into the projector? 53. Review Problem. A block of mass M, supported by a string, rests on an incline making an angle with the horizontal (Fig. P16.53). The string's length is L, and its mass is m V M. Derive an expression for the time it takes a transverse wave to travel from one end of the string to the other. ADDITIONAL PROBLEMS 49. The "wave" is a particular type of wave pulse that can sometimes be seen propagating through a large crowd gathered at a sporting arena to watch a soccer or American football match (Fig. P16.49). The particles of the medium are the spectators, with zero displacement corresponding to their being in the seated position and maximum displacement corresponding to their being in the standing position and raising their arms. When a large fraction of the spectators participate in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people's reaction time, which is typically on the order of 0.1 s. Estimate the order of magnitude, in minutes, of the time required for such a wave pulse to make one circuit around a large sports stadium. State the quantities you measure or estimate and their values. m, L M Figure P16.53 54. (a) Determine the speed of transverse waves on a string under a tension of 80.0 N if the string has a length of 2.00 m and a mass of 5.00 g. (b) Calculate the power required to generate these waves if they have a wavelength of 16.0 cm and an amplitude of 4.00 cm. 55. Review Problem. A 2.00-kg block hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is 0.500 m, and its mass is 5.00 g. The "spring constant" for the cord is 100 N/m. The block is released and stops at the lowest point. (a) Determine the tension in the cord when the block is at this lowest point. (b) What is the length of the cord in this "stretched" position? (c) Find the speed of a transverse wave in the cord if the block is held in this lowest position. 56. Review Problem. A block of mass M hangs from a rubber cord. The block is supported so that the cord is not stretched. The unstretched length of the cord is L 0 , and its mass is m, much less than M. The "spring constant" for the cord is k. The block is released and stops at the lowest point. (a) Determine the tension in the cord when the block is at this lowest point. (b) What is the length of the cord in this "stretched" position? (c) Find the speed of a transverse wave in the cord if the block is held in this lowest position. Figure P16.49 50. A traveling wave propagates according to the expression y (4.0 cm) sin(2.0x 3.0t ), where x is in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the period, and (e) the direction of travel of the wave. 51. The wave function for a traveling wave on a taut string is (in SI units) y(x, t ) (0.350 m) sin(10 t 3 x /4) WEB Problems 57. A sinusoidal wave in a rope is described by the wave function y (0.20 m) sin(0.75 x 18 t ) 517 where x and y are in meters and t is in seconds. The rope has a linear mass density of 0.250 kg/m. If the tension in the rope is provided by an arrangement like the one illustrated in Figure 16.12, what is the value of the suspended mass? 58. A wire of density is tapered so that its cross-sectional area varies with x, according to the equation A (1.0 10 3x 0.010) cm2 (a) If the wire is subject to a tension T, derive a relationship for the speed of a wave as a function of position. (b) If the wire is aluminum and is subject to a tension of 24.0 N, determine the speed at the origin and at x 10.0 m. 59. A rope of total mass m and length L is suspended vertically. Show that a transverse wave pulse travels the length of the rope in a time t 2L /g. (Hint: First find an expression for the wave speed at any point a distance x from the lower end by considering the tension in the rope as resulting from the weight of the segment below that point.) 60. If mass M is suspended from the bottom of the rope in Problem 59, (a) show that the time for a transverse wave to travel the length of the rope is t 2 t L (M mg m) M 63. Review Problem. An aluminum wire under zero tension at room temperature is clamped at each end. The tension in the wire is increased by reducing the temperature, which results in a decrease in the wire's equilibrium length. What strain ( L/L) results in a transverse wave speed of 100 m/s? Take the cross-sectional area of the wire to be 5.00 10 6 m2, the density of the material to be 2.70 103 kg/m3, and Young's modulus to be 7.00 1010 N/m2 . 64. (a) Show that the speed of longitudinal waves along a spring of force constant k is v kL/ , where L is the unstretched length of the spring and is the mass per unit length. (b) A spring with a mass of 0.400 kg has an unstretched length of 2.00 m and a force constant of 100 N/m. Using the result you obtained in (a), determine the speed of longitudinal waves along this spring. 65. A string of length L consists of two sections: The left half has mass per unit length 0/2, whereas the right half has a mass per unit length 3 3 0/2. Tension in the string is T0 . Notice from the data given that this string has the same total mass as a uniform string of length L and of mass per unit length 0 . (a) Find the speeds v and v at which transverse wave pulses travel in the two sections. Express the speeds in terms of T0 and 0 , and also as multiples of the speed v 0 (T0 / 0)1/2. (b) Find the time required for a pulse to travel from one end of the string to the other. Give your result as a multiple of t 0 L/v 0 . 66. A wave pulse traveling along a string of linear mass density is described by the rel energy is propagated away from the piston by the sound wave.1 To evaluate the rate of energy transfer for the sound wave, we shall evaluate the kinetic energy of this volume of air, which is undergoing simple harmonic motion. We shall follow a procedure similar to that in Section 16.8, in which we evaluated the rate of energy transfer for a wave on a string. As the sound wave propagates away from the piston, the displacement of any volume of air in front of the piston is given by Equation 17.2. To evaluate the kinetic energy of this volume of air, we need to know its speed. We find the speed by taking the time derivative of Equation 17.2: v(x, t) t s(x, t) t [s max cos(kx t)] s max sin(kx t) Imagine that we take a "snapshot" of the wave at t given volume of air at this time is K 1 2 1 2 0. The kinetic energy of a mv 2 1 2 m( s max sin kx)2 1 2 A x( s max sin kx)2 A x( s max)2 sin2 kx where A is the cross-sectional area of the moving air and A x is its volume. Now, as in Section 16.8, we integrate this expression over a full wavelength to find the total kinetic energy in one wavelength. Letting the volume of air shrink to infinitesimal thickness, so that x : dx, we have K dK 0 1 2 A( s max)2 sin2 kx dx 1 2 1 2 1 4 A( s max)2 sin2 kx dx 0 A( s max)2 1 2 A( s max)2 As in the case of the string wave in Section 16.8, the total potential energy for one wavelength has the same value as the total kinetic energy; thus, the total mechaniArea = A v m x Figure 17.4 An oscillating piston transfers energy to the air in the tube, initially causing the volume of air of width x and mass m to oscillate with an amplitude s max . 1 Although it is not proved here, the work done by the piston equals the energy carried away by the wave. For a detailed mathematical treatment of this concept, see Chapter 4 in Frank S. Crawford, Jr., Waves, Berkeley Physics Course, vol. 3, New York, McGraw-Hill Book Company, 1968. 526 CHAPTER 17 Sound Waves cal energy is E K U 1 2 A( s max)2 As the sound wave moves through the air, this amount of energy passes by a given point during one period of oscillation. Hence, the rate of energy transfer is E t 1 2 A( s max)2 T 1 2 A( s max)2 T 1 2 Av( s max)2 where v is the speed of sound in air. We define the intensity I of a wave, or the power per unit area, to be the rate at which the energy being transported by the wave flows through a unit area A perpendicular to the direction of travel of the wave. In the present case, therefore, the intensity is Intensity of a sound wave I A 1 2 v( s max)2 (17.5) Thus, we see that the intensity of a periodic sound wave is proportional to the square of the displacement amplitude and to the square of the angular frequency (as in the case of a periodic string wave). This can also be written in terms of the pressure amplitude Pmax ; in this case, we use Equation 17.4 to obtain I P2 max 2 v (17.6) EXAMPLE 17.3 Hearing Limits tells us that the ear can discern pressure fluctuations as small as 3 parts in 1010 ! We can calculate the corresponding displacement amplitude by using Equation 17.4, recalling that 2 f (see Eqs. 16.10 and 16.12): s max Pmax v 1.11 2.87 10 5 N/m2 (1.20 kg/m3)(343 m/s)(2 1 000 Hz) 10 11 The faintest sounds the human ear can detect at a frequency of 1 000 Hz correspond to an intensity of about 1.00 10 12 W/m2 -- the so-called threshold of hearing. The loudest sounds the ear can tolerate at this frequency correspond to an intensity of about 1.00 W/m2 -- the threshold of pain. Determine the pressure amplitude and displacement amplitude associated with these two limits. Solution First, consider the faintest sounds. Using Equation 17.6 and taking v 343 m/s as the speed of sound waves in air and 1.20 kg/m3 as the density of air, we obtain Pmax !2 vI m !2(1.20 kg/m3)(343 m/s)(1.00 2.87 10 5 10 12 W/m2) N/m2 Because atmospheric pressure is about 105 N/m2, this result This is a remarkably small number! If we compare this result for s max with the diameter of a molecule (about 10 10 m), we see that the ear is an extremely sight side max Wave front r1 r2 Source Ray Figure 17.5 A spherical sound wave propagating radially outward from an oscillating spherical body. The intensity of the spherical wave varies as 1/r 2. Figure 17.6 Spherical waves emitted by a point source. The circular arcs represent the spherical wave fronts that are concentric with the source. The rays are radial lines pointing outward from the source, perpendicular to the wave fronts. 17.4 Spherical and Plane Waves 529 Rays Wave fronts Figure 17.7 Far away from a point source, the wave fronts are nearly parallel planes, and the rays are nearly parallel lines perpendicular to the planes. Hence, a small segment of a spherical wave front is approximately a plane wave. Plane wave front y of Equation 17.8, we conclude that the displacement amplitude s max of a spherical wave must vary as 1/r. Therefore, we can write the wave function (Greek letter psi) for an outgoing spherical wave in the form (r, t) s0 sin(kr r t) (17.9) where s 0 , the displacement amplitude at unit distance from the source, is a constant parameter characterizing the whole wave. It is useful to represent spherical waves with a series of circular arcs concentric with the source, as shown in Figure 17.6. Each arc represents a surface over which the phase of the wave is constant. We call such a surface of constant phase a wave front. The distance between adjacent wave fronts equals the wavelength . The radial lines pointing outward from the source are called rays. Now consider a small portion of a wave front far from the source, as shown in Figure 17.7. In this case, the rays passing through the wave front are nearly parallel to one another, and the wave front is very close to being planar. Therefore, at distances from the source that are great compared with the wavelength, we can approximate a wave front with a plane. Any small portion of a spherical wave far from its source can be considered a plane wave. Figure 17.8 illustrates a plane wave propagating along the x axis, which means that the wave fronts are parallel to the yz plane. In this case, the wave function depends only on x and t and has the form (x, t) A sin(kx t) (17.10) x v z Figure 17.8 A representation of a plane wave moving in the positive x direction with a speed v. The wave fronts are planes parallel to the yz plane. Representation of a plane wave That is, the wave function for a plane wave is identical in form to that for a onedimensional traveling wave. The intensity is the same at all points on a given wave front of a plane wave. EXAMPLE 17.5 Intensity Variations of a Point Source I av A point source emits sound waves with an average power output of 80.0 W. (a) Find the intensity 3.00 m from the source. A point source emits energy in the form of spherical waves (see Fig. 17.5). At a distance r from the source, the power is distributed over the surface area of a sphere, 4 r 2. Therefore, the intensity at the distance r is given by Equation 17.8: 4 r2 80.0 W 4 (3.00 m)2 0.707 W/m2 Solution an intensity that is close to the threshold of pain. (b) Find the distance at which the sound level is 40 dB. Solution We can find the intensity at the 40-dB sound level by using Equation 17.7 with I 0 1.00 10 12 W/m2: 530 I I0 log I 0 log I log I I CHAPTER 17 Sound Waves Using this value for I in Equation 17.8 and solving for r, we obtain 10 log log I 40 dB 40 10 4 8 1.00 10 8 4 log 10 12 r ! av 4 I ! 4 80.0 W 1.00 10 8 W/m2 2.52 10 4 m which equals about 16 miles! W/m2 17.5 QuickLab (Before attempting to do this QuickLab, you should check to see whether it is legal to sound a horn in your area.) Sound your car horn while driving toward and away from a friend in a campus parking lot or on a country road. Try this at different speeds while driving toward and past the friend (not at the friend). Do the frequencies of the sounds your friend hears agree with what is described in the text? THE DOPPLER EFFECT Perhaps you have noticed how the sound of a vehicle's horn changes as the vehicle moves past you. The fr f v (observer moving away from source) (17.12) In general, whenever an observer moves with a speed vO relative to a stationary source, the frequency heard by the observer is Frequency heard with an observer in motion f 1 vO f v (17.13) where the positive sign is used when the observer moves toward the source and the negative sign is used when the observer moves away from the source. Now consider the situation in which the source is in motion and the observer is at rest. If the source moves directly toward observer A in Figure 17.11a, the wave fronts heard by the observer are closer together than they would be if the source were not moving. As a result, the wavelength measured by observer A is shorter than the wavelength of the source. During each vibration, which lasts for a time T (the period), the source moves a distance v ST v S /f and the wavelength is (b) Observer B S vS Observer A (a) Figure 17.11 (a) A source S moving with a speed vS toward a stationary observer A and away from a stationary observer B. Observer A hears an increased frequency, and observer B hears a decreased frequency. (b) The Doppler effect in water, observed in a ripple tank. A point source is moving to the right with speed vS . 17.5 The Doppler Effect 533 shortened by this amount. Therefore, the observed wavelength vS f Because v/f, the frequency heard by observer A is f v v vS f 1 1 vS v f v f v vS f is f (17.14) "I love hearing that lonesome wail of the train whistle as the magnitude of the frequency of the wave changes due to the Doppler effect." That is, the observed frequency is increased whenever the source is moving toward the observer. When the source moves away from a stationary observer, as is the case for observer B in Figure 17.11a, the observer measures a wavelength that is greater than and hears a decreased frequency: f 1 1 vS v f (17.15) Combining Equations 17.14 and 17.15, we can express the general relationship for the observed frequency when a source is moving and an observer is at rest as f 1 1 vS v f (17.16) Frequency heard with source in motion Finally, if both source and observer are in motion, we find the following general relationship for the observed frequency: f v v vO f vS (17.17) Frequency heard with observer and source in motion In this expression, the upper signs ( vO and vS ) refer to motion of one toward the other, and the lower signs ( vO and vS ) refer to motion of one away from the other. A convenient rule concerning signs for you to remember when working with all Doppler-effect problems is as follows: The word toward is associated with an increase in observed frequency. The words away from are associated with a decrease in observed frequency. Although the Doppler effect is most typically experienced with sound waves, it is a phenomenon that is common to all waves. For example, the relative motion of source and observer produces a frequency shift in light waves. The Doppler effect is used in police radar systems to measure the speeds of motor vehicles. Likewise, astronomers use the effect to determine the speeds of stars, galaxies, and other celestial objects relative to the Earth. 534 CHAPTER 17 Sound Waves EXAMPLE 17.6 The Noisy Siren f v v vO f vS 343 m/s 343 m/s 24.6 m/s (400 Hz) 33.5 m/s As an ambulance travels east down a highway at a speed of 33.5 m/s (75 mi/h), its siren emits sound at a frequency of 400 Hz. What frequency is heard by a person in a car traveling west at 24.6 m/s (55 mi/h) (a) as the car approaches the ambulance and (b) as the car moves away from the ambulance? (a) We can use Equation 17.17 in both cases, taking the speed of sound in air to be v 343 m/s. As the ambulance and car approach each other, the person in the car hears the frequency f v v vO f vS 343 m/s 343 m/s 24.6 m/s (400 Hz) 33.5 m/s 338 Hz The change in frequency detected by the person in the car is 475 338 137 Hz, which is more than 30% of the true frequency. Solution Exercise Suppose the car is parked on the side of the highway as the ambulance speeds by. What frequency does the persith the displacement wave. The relationship between s max and Pmax is given by Pmax v s max P2 max 2 v (17.4) The intensity of a periodic sound wave, which is the power per unit area, is I 1 2 v( s max)2 (17.5, 17.6) The sound level of a sound wave, in decibels, is given by 10 log I I0 (17.7) The constant I 0 is a reference intensity, usually taken to be at the threshold of hearing (1.00 10 12 W/m2 ), and I is the intensity of the sound wave in watts per square meter. The intensity of a spherical wave produced by a point source is proportional to the average power emitted and inversely proportional to the square of the distance from the source: I av 4 r2 (17.8) The change in frequency heard by an observer whenever there is relative motion between a source of sound waves and the observer is called the Doppler effect. The observed frequency is f v v vO f vS (17.17) The upper signs ( vO and vS ) are used with motion of one toward the other, and the lower signs ( vO and vS ) are used with motion of one away from the other. You can also use this formula when vO or vS is zero. Problems 537 QUESTIONS 1. Why are sound waves characterized as longitudinal? 2. If an alarm clock is placed in a good vacuum and then activated, no sound is heard. Explain. 3. A sonic ranger is a device that determines the position of an object by sending out an ultrasonic sound pulse and measuring how long it takes for the sound wave to return after it reflects from the object. Typically, these devices cannot reliably detect an object that is less than half a meter from the sensor. Why is that? 4. In Example 17.5, we found that a point source with a power output of 80 W reduces to a sound level of 40 dB at a distance of about 16 miles. Why do you suppose you cannot normally hear a rock concert that is going on 16 miles away? (See Table 17.2.) 5. If the distance from a point source is tripled, by what factor does the intensity decrease? 6. Explain how the Doppler effect is used with microwaves to determine the speed of an automobile. 7. Explain what happens to the frequency of your echo as you move in a vehicle toward a canyon wall. What happens to the frequency as you move away from the wall? 8. Of the following sounds, which is most likely to have a sound level of 60 dB -- a rock concert, the turning of a page in this text, normal conversation, or a cheering crowd at a football game? 9. Estimate the decibel level of each of the sounds in the previous question. 10. A binary star system consists of two stars revolving about their common center of mass. If we observe the light reaching us from one of these stars as it makes one complete revolution, what does the Doppler effect predict will happen to this light? 11. How can an object move with respect to an observer so that the sound from it is not shifted in frequency? 12. Why is it not possible to use sonar (sound waves) to determine the speed of an object traveling faster than the speed of sound in a given medium? 13. Why is it so quiet after a snowfall? 14. Why is the intensity of an echo less than that of the original sound? 15. If the wavelength of a sound source is reduced by a factor of 2, what happens to its frequency? Its speed? 16. In a recent discovery, a nearby star was found to have a large planet orbiting about it, although the planet could not be seen. In terms of the concept of a system rotating about its center of mass and the Doppler shift for light (which is in many ways similar to that for sound), explain how an astronomer could determine the presence of the invisible planet. 17. A friend sitting in her car far down the road waves to you and beeps her horn at the same time. How far away must her car be for you to measure the speed of sound to two significant figures by measuring the time it takes for the sound to reach you? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive sound wave is described by the displacement s(x, t ) (2.00 m) cos[(15.7 m 1)x (858 s 1)t ] 10. 11. 12. 13. 14. WEB 15. (a) Find the amplitude, wavelength, and speed of this wave. (b) Determine the instantaneous displacement of the molecules at the position x 0.050 0 m at t 3.00 ms. (c) Determine the maximum speed of a molecule's oscillatory motion. As a sound wave travels through the air, it produces pressure variations (above and below atmospheric pressure) that are given by P 1.27 sin( x 340 t) in SI units. Find (a) the amplitude of the pressure variations, (b) the frequency of the sound wave, (c) its wavelength in air, and (d) its speed. Write an expression that describes the pressure variation as a function of position and time for a sinusoidal sound wave in air, if 0.100 m and Pmax 0.200 Pa. Write the function that describes the displacement wave corresponding to the pressure wave in Problem 11. The tensile stress in a thick copper bar is 99.5% of its elastic breaking point of 13.0 1010 N/m2. A 500-Hz sound wave is transmitted through the material. (a) What displacement amplitude will cause the bar to break? (b) What is the maximum speed of the particles at this moment? Calculate the pressure amplitude of a 2.00-kHz sound wave in air if the displacement amplitude is equal to 2.00 10 8 m. An experimenter wishes to generate in air a sound wave that has a displacement amplitude of 5.50 10 6 m. The pressure amplitude is to be limited to 8.40 10 1 Pa. What is the minimum wavelength the sound wave can have? 21. A family ice show is held in an enclosed arena. The skaters perform to music with a sound level of 80.0 dB. This is too loud for your baby, who consequently yells at a level of 75.0 dB. (a) What total sound intensity engulfs you? (b) What is the combined sound level? Section 17.4 Spherical and Plane Waves 22. For sound radiating from a point source, show that the difference in sound levels, 1 and 2 , at two receivers is related to the ratio of the distances r 1 and r 2 from the source to the receivers by the expression 2 1 20 log r1 r2 23. A fireworks charge is detonated many meters above the ground. At a distance of 400 m from the explosion, the acoustic pressure reaches a maximum of 10.0 N/m2. Assume that the speed of sound is constant at 343 m/s throughout the atmosphere over the region considered, that the ground absorbs all the sound falling on it, and that the air absorbs sound energy as described by the rate 7.00 dB/km. What is the sound level (in decibels) at 4.00 km from the explosion? 24. A loudspeaker is placed between two observers who are 110 m apart, along the line connecting them. If one observer records a sound level of 60.0 dB and the other records a sound level of 80.0 dB, how far is the speaker from each observer? Problems 25. Two small speakers emit spherical sound waves of different frequencies. Speaker A has an output of 1.00 mW, and speaker B has an output of 1.50 mW. Determine the sound level (in decibels) at point C (Fig. P17.25) if (a) only speaker A emits sound, (b) only speaker B emits sound, (c) both speakers emit sound. A 539 radiates uniformly in all horizontal and upward directions. Find the sound level 1.00 km away. 32. A spherical wave is radiating from a point source and is described by the wave function P (r, t ) 25.0 r sin(1.25r 1 870t ) 3.00 m where P is in pascals, r in meters, and t in seconds. (a) What is the pressure amplitude 4.00 m from the source? (b) Determine the speed of the wave and hence the material the wave might be traveling through. (c) Find the sound level of the wave, in decibels, 4.00 m from the source. (d) Find the instantaneous pressure 5.00 m from the source at 0.080 0 s. Section 17.5 The Doppler Effect 4.00 m C 2.00 m B Figure P17.25 26. An experiment requires a sound intensity of 1.20 W/m2 at a distance of 4.00 m from a speaker. What power output is required? Assume that the speaker radiates sound equally in all directions. 27. A source of sound (1 000 Hz) emits uniformly in all directions. An observer 3.00 m from the source measurt him now? The ocean floor is underlain by a layer of basalt that constitutes the crust, or uppermost layer, of the Earth in that region. Below the crust is found denser peridotite rock, which forms the Earth's mantle. The boundary between these two layers is called the Mohorovicic discontinuity ("Moho" for short). If an explosive charge is set off at the surface of the basalt, it generates a seismic wave that is reflected back out at the Moho. If the speed of the wave in basalt is 6.50 km/s and the two-way travel time is 1.85 s, what is the thickness of this oceanic crust? A worker strikes a steel pipeline with a hammer, generating both longitudinal and transverse waves. Reflected waves return 2.40 s apart. How far away is the reflection point? (For steel, vlong 6.20 km/s and vtrans 3.20 km/s.) For a certain type of steel, stress is proportional to strain with Young's modulus as given in Table 12.1. The steel has the density listed for iron in Table 15.1. It bends permanently if subjected to compressive stress greater than its elastic limit, 400 MPa, also called its yield strength. A rod 80.0 cm long, made of this steel, is projected at 12.0 m/s straight at a hard wall. (a) Find the speed of compressional waves moving along the rod. (b) After the front end of the rod hits the wall and stops, the back end of the rod keeps moving, as described by Newton's first law, until it is stopped by the excess pressure in a sound wave moving back through the rod. How much time elapses before the back end of the rod gets the message? (c) How far has the back end of the rod moved in this time? (d) Find the strain in the rod and (e) the stress. (f) If it is not to fail, show that the maximum impact speed a rod can have is given by the expression /! Y . To determine her own speed, a sky diver carries a buzzer that emits a steady tone at 1 800 Hz. A friend at the landing site on the ground directly below the sky diver listens to the amplified sound he receives from the buzzer. Assume that the air is calm and that the speed 541 48. 49. of sound is 343 m/s, independent of altitude. While the sky diver is falling at terminal speed, her friend on the ground receives waves with a frequency of 2 150 Hz. (a) What is the sky diver's speed of descent? (b) Suppose the sky diver is also carrying sound-receiving equipment that is sensitive enough to detect waves reflected from the ground. What frequency does she receive? 54. A train whistle ( f 400 Hz) sounds higher or lower in pitch depending on whether it is approaching or receding. (a) Prove that the difference in frequency between the approaching and receding train whistle is f 1 2(u/v) f (u 2/v 2) 50. where u is the speed of the train and v is the speed of sound. (b) Calculate this difference for a train moving at a speed of 130 km/h. Take the speed of sound in air to be 340 m/s. 55. A bat, moving at 5.00 m/s, is chasing a flying insect. If the bat emits a 40.0-kHz chirp and receives back an echo at 40.4 kHz, at what relative speed is the bat moving toward or away from the insect? (Take the speed of sound in air to be v 340 m/s.) 51. 52. Figure P17.55 56. A supersonic aircraft is flying parallel to the ground. When the aircraft is directly overhead, an observer on the ground sees a rocket fired from the aircraft. Ten seconds later the observer hears the sonic boom, which is followed 2.80 s later by the sound of the rocket engine. What is the Mach number of the aircraft? 57. A police car is traveling east at 40.0 m/s along a straight road, overtaking a car that is moving east at 30.0 m/s. The police car has a malfunctioning siren that is stuck at 1 000 Hz. (a) Sketch the appearance of the wave fronts of the sound produced by the siren. Show the 53. 542 CHAPTER 17 Sound Waves 63. A meteoroid the size of a truck enters the Earth's atmosphere at a speed of 20.0 km/s and is not significantly slowed before entering the ocean. (a) What is the Mach angle of the shock wave from the meteoroid in the atmosphere? (Use 331 m/s as the sound speed.) (b) Assuming that the meteoroid survives the imcm 40.0 cm 50.0 cm. f Answers to Quick Quizzes 25.0 m/s vS fS 543 S O vO fO (a) (b) sity values and Young's moduli for the three materials are 1 2.70 10 3 kg/m3, Y1 7.00 10 10 N/m2; 11.3 10 3 kg/m3, Y2 1.60 10 10 N/m2; 2 8.80 10 3 kg/m3, Y3 11.0 10 10 N/m2. 3 (a) If L 3 1.50 m, what must the ratio L 1 /L 2 be if a sound wave is to travel the combined length of rods 1 and 2 in the same time it takes to travel the length of rod 3? (b) If the frequency of the source is 4.00 kHz, determine the phase difference between the wave traveling along rods 1 and 2 and the one traveling along rod 3. Figure P17.69 L1 L2 2 3 L3 where O and S are defined in Figure P17.69a. (a) Show that if the observer and source are moving away from each other, the preceding equation reduces to Equation 17.17 with lower signs. (b) Use the preceding equation to solve the following problem. A train moves at a constant speed of 25.0 m/s toward the intersection shown in Figure P17.69b. A car is stopped near the intersection, 30.0 m from the tracks. If the train's horn emits a frequency of 500 Hz, what frequency is heard by the passengers in the car when the train is 40.0 m from the intersection? Take the speed of sound to be 343 m/s. 70. Figure 17.5 illustrates that at distance r from a point source with power av , the wave intensity is I 2 av /4 r . Study Figure 17.11a and prove that at distance r straight in front of a point source with power av , moving with constant speed vS , the wave intensity is I av 1 Figure P17.71 72. The volume knob on a radio has what is known as a "logarithmic taper." The electrical device connected to the knob (called a potentiometer) has a resistance R whose logarithm is proportional to the angular position of the knob: that is, log R . If the intensity of the sound I (in watts per square meter) produced by the speaker is proportional to the resistance R, show that the sound level (in decibels) is a linear function of . 73. The smallest wavelength possible for a sound wave in air is on the order of the separation distance between air molecules. Find the order of magnitude of the highestfrequency sound wave possible in air, assuming a wave speed of 343 m/s, a density of 1.20 kg/m3, and an average molecular mass of 4.82 10 26 kg. v v vS 4 r2 71. Three metal rods are located relative to each other as shown in Figure P17.71, where L 1 L 2 L 3 . The den- ANSWERS TO QUICK QUIZZES 17.1 The only correct answer is (c). Although the speed of a wave is given by the product of its wavelength and frequency, it is not affected by changes in either one. For example, if the sound from a musical instrument increases in frequency, the wavelength decreases, and thus v f remains constant. The amplitude of a sound wave determines the size of the oscillations of air molecules but does not affect the speed of the wave through the air. 17.2 The ground tremor represents a sound wave moving through the Earth. Sound waves move faster through the Earth than through air because rock and other ground materials are much stiffer against compression. Therefore -- the vibration through the ground and the sound in the air having started together -- the vibration through the ground reaches the observer first. 17.3 Because the bottom of the bottle does not allow molecular motion, the displacement in this region is at its minimum value. Because the pressure variation is a maximum when the displacement is a minimum, the pressure variation at the bottom is a maximum. 17.4 (a) 10 dB. If we call the intensity of each violin I, the total intensity when all the violins are playing is I 9I 10I. Therefore, the addition of the nine violins increases the intensity of the sound over that of one violin by a factor of 10. From Equation 17.7 we see that an increase in intensity by a factor of 10 increases the sound level by 10 dB. (b) 13 dB. The intensity is now increased by a factor of 20 over that of a single violin. 17.5 The Mach number is the ratio of the plane's speed (which does not change) to the speed of sound, which is greater in the warm air than in the cold, as we learned 544 CHAPTER 17 Sound Waves the denominator: f v sound v sound v wind f v wind in Section 17.1 (see Quick Quiz 17.1). The denominator of this fraction increases while the numerator stays constant. Therefore, the fraction as a whole -- the Mach number -- decreases. 17.6 (a) In the reference frame of the air, the observer is moving toward the source at the wind speed through stationary air, and the source is moving away from the observer with the same speed. In Equation 17.17, therefore, a plus sign is needed in both the numerator and meaning the observed frequency is the same as if no wind were blowing. (b) The observer "sees" the sound waves coming toward him at a higher speed (v sound v wind). (c) At this higher speed, he attributes a greater wavelength (v sound v wind)/f to the wave. 2.2 This is the Nearest One Head 545 P U Z Z L E R A speaker for a stereo system operates even if the wires connecting it to the amplifier are reversed, that is, for and for (or red for black and black for red). Nonetheless, the owner's manual says that for best performance you should be careful to connect the two speakers properly, so that they are "in phase." Why is this such an important consideration for the quality of the sound you hear? (George Semple) c h a p t e r Superposition and Standing Waves Chapter Outline 18.1 Superposition and Interference of Sinusoidal Waves 18.6 (Optional) Standing Waves in Rods and Plates 18.2 Standing Waves 18.3 Standing Waves in a String Fixed at Both Ends 18.7 Beats: Interference in Time 18.8 (Optional) Non-Sinusoidal Wave Patterns 18.4 Resonance 18.5 Standing Waves in Air Columns 545 546 CHAPTER 18 Superposition and Standing Waves I mportant in the study of waves is the combined effect of two or more waves traveling in the same medium. For instance, what happens to a string when a wave traveling along it hits a fixed end and is reflected back on itself ? What is the air pressure variation at a particular seat in a theater when the instruments of an orchestra sound together? When analyzing a linear medium -- that is, one in which the restoring force acting on the particles of the medium is proportional to the displacement of the particles -- we can apply the principle of superposition to determine the resultant disturbance. In Chapter 16 we discussed this principle as it applies to wave pulses. In this chapter we study the superposition principle as it applies to sinusoidal waves. If the sinusoidal waves that combine in a linear medium have the same frequency and wavelength, a stationary pattern -- called a standing wave -- can be produced at certain frequencies under certain circumstances. For example, a taut string fixed at both ends has a discrete set of oscillation patterns, called modes of vibration, that are related to the tension and linear mass density of the string. These modes of vibration are found in stringed musical instruments. Other musical instruments, such as the organ and the flute, make use of the natural frequencies of sound waves in hollow pipes. Such frequencies are related to the length and shape of the pipe and depend on whether the pipe is open at both ends or open at one end and closed at the other. We also consider the superposition and interference of waves having different frequencies and wavelengths. When two sound waves having nearly the same frequency interfere, we hear variations in the loudness called beats. The beat frequency corresponds to the rate of alternation between constructive and destructive interference. Finally, we discuss how any non-sinusoidal periodic wave can be described as a sum of sine and cosine functions. 18.1 SUPERPOSITION AND INTERFERENCE OF SINUSOIDAL WAVES 9.6 & 9.7 Imagine that you are standing in a swimming pool and that a beach ball is floating a couple of meters away. You use your right hand to send a series of waves toward the beach ball, causing it to repeatedly move upward by 5 cm, return to its original position, and then move downward by 5 cm. After the water becomes still, you use your left hand to send an identical set of waves toward the beach ball and observe the same behavior. What happens if you use both hands at the same time to send two waves toward the beach ball? How the beach ball responds to the waves depends on whether the waves work together (that is, both waves make the beach ball go up at the same time and then down at the same time) or work against each other (that is, one wave tries to make the beach ball go up, while the other wave tries to make it go down). Because it is possible to have two or more waves in the same location at the same time, we have to consider how waves interact with each other and with their surroundings. The superposition principle states that when two or more waves move in the same linear medium, the net displacement of the medium (that is, the resultant wave) at any point equals the algebraic sum of all the displacements caused by the individual waves. Let us apply this principle to two sinusoidal waves traveling in the same direction in a linear medium. If the two waves are traveling to the right and have the same frequency, wavelength, and amplitude but differ in phase, we can 18.1 Superposition and Interference of Sinusoidal Waves 547 express their individual wave functions as y1 A sin(kx t) y2 A sin(kx t ) where, as usual, k 2 / , 2 f, and is the phase constant, which we introduced in the context of simple harmonic motion in Chapter 13. Hence, the resultant wave function y is y y1 y2 A[sin(kx t) sin(kx t )] To simplify this expression, we use the trigonometric identity sin a If we let a kx tion y reduces to t and b y sin b kx 2A cos 2 cos t a 2 b sin a 2 b , we find that the resultant wave funcsin kx t Resultant of two traveling sinusoidal waves 2 2 This result has several important features. The resultant wave function y also is sinusoidal and has the same frequency and wavelength as the individual waves, since the sine function incorporates the same values of k and that appear in the original wave functions. The amplitude of the resultant wave is 2A cos( /2), and its phase is /2. If the phase constant equals 0, then cos( /2) cos 0 1, and the amplitude of the resultant wave is 2A -- twice the amplitude of either individual wave. In this case, in which 0, the waves are said to be everywhere in phase and thus interfere constructively. That is, the crests and troughs of the individual waves y 1 and y 2 occur at the same positions and combine to form the red curve y of amplitude 2A shown in Figure 18.1a. Because the individual waves are in phase, they are indistinguishable in Figure 18.1a, in which they appear as a single blue curve. In general, constructive interference occurs when cos( /2) 1. This is true, for example, when 0, 2 , 4 , . . . rad -- that is, when is an even multiple of . When is equal to rad or to any odd multiple of , then cos( /2) cos( /2) 0, and the crests of one wave occur at the same positions as the troughs of the second wave (Fig. 18.1b). Thus, the resultant wave has zero amplitude everywhere, as a consequence of destructive interference. Finally, when the phase constant has an arbitrary value other than 0 or other than an integer multiple of rad (Fig. 18.1c), the resultant wave has an amplitude whose value is somewhere between 0 and 2A. Constructive interference Destructive interference Interference of Sound Waves One simple device for demonstrating interference of sound waves is illustrated in Figure 18.2. Sound from a loudspeaker S is sent into a tube at point P, where there is a T-shaped junction. Half of the sound power travels in one direction, and half travels in the opposite direction. Thus, the sound waves that reach the receiver R can travel along either of the two paths. The distance along any path from speaker to receiver is called the path length r. The lower path length r 1 is fixed, but the upper path length r 2 can be varied by sliding the U-shaped tube, which is similar to that on a slide trombone. When the difference in the path lengths r r2 r1 is either zero or some integer multiple of the wavelength (sound in air, 343 m/s: f v 343 m/s 0.26 m 1.3 kHz Exercise If the oscillator frequency is adjusted such that the first location at which a listener hears no sound is at a distance of 0.75 m from O, what is the new frequency? 0.63 kHz. Answer P O r2 0.350 m 3.00 m 1.85 m 8.00 m Figure 18.3 550 CHAPTER 18 Superposition and Standing Waves You can now understand why the speaker wires in a stereo system should be connected properly. When connected the wrong way -- that is, when the positive (or red) wire is connected to the negative (or black) terminal -- the speakers are said to be "out of phase" because the sound wave coming from one speaker destructively interferes with the wave coming from the other. In this situation, one speaker cone moves outward while the other moves inward. Along a line midway between the two, a rarefaction region from one speaker is superposed on a condensation region from the other speaker. Although the two sounds probably do not completely cancel each other (because the left and right stereo signals are usually not identical), a substantial loss of sound quality still occurs at points along this line. 18.2 STANDING WAVES The sound waves from the speakers in Example 18.1 left the speakers in the forward direction, and we considered interference at a point in space in front of the speakers. Suppose that we turn the speakers so that they face each other and then have them emit sound of the same frequency and amplitude. We now have a situation in which two identical waves travel in opposite directions in the same medium. These waves combine in accordance with the superposition principle. We can analyze such a situation by considering wave functions for two transverse sinusoidal waves having the same amplitude, frequency, and wavelength but traveling in opposite directions in the same medium: y1 A sin(kx t) y2 A sin(kx t) where y 1 represents a wave traveling to the right and y 2 represents one traveling to the left. Adding these two functions gives the resultant wave function y: y y1 y2 A sin(kx t) A sin(kx b) t) cos a sin b, this When we use the trigonometric identity sin(a expression reduces to Wave function for a standing wave sin a cos b y (2A sin kx) cos t (18.3) which is the wave function of a standing wave. A standing wave, such as the one shown in Figure 18.4, is an oscillation pattern with a stationary outline that results from the superposition of two identical waves traveling in opposite directions. Notice that Equation 18.3 does not contain a function of kx t. Thus, it is not an expression for a traveling wave. If we observe a standing wave, we have no sense of motion in the direction of propagation of either of the original waves. If we compare this equation with Equation 13.3, we see that Equation 18.3 describes a special kind of simple harmonic motion. Every particle of the medium oscillates in simple harmonic motion with the same frequency (according to the cos t factor in the equation). However, the amplitude of the simple harmonic motion of a given particle (given by the factor 2A sin kx, the coefficient of the cosine function) depends on the location x of the particle in the medium. We need to distinguish carefully between the amplitude A of the individual waves and the amplitude 2A sin kx of the simple harmonic motion of the particles of the medium. A given particle in a standing wave vibrates within the constraints of the envelope function 2A sin kx, where x is the particle's position in the medium. This is in contrast to the situation in a traveling sinusoidal wave, in which all particles oscillate with the 18.2 Standing Waves 551 Antinode Node Antinode Node 2A sin kx Figure 18.4 Multiflash photograph of a standing wave on a string. The time behavior of the vertical displacement from equilibrium of an individual particle of the string is given by cos t. That is, each particle vibrates at an angular frequency . The amplitude of the vertical oscillation of any particle on the string depends on the horizontal position of the particle. Each particle vibrates within the confines of the envelope function 2A sin kx. same amplitude and the same frequency and in which the amplitude of the wave is the same as the amplitude of the simple harmonic motion of the particles. The maximum displacement of a particle of the medium has a minimum value of zero when x satisfies the condition sin kx 0, that is, when kx Because k ,2 ,3 , . . . 2 / , these values for kx give x 2 , , 3 , . . . 2 n 2 n 0, 1, 2, 3, . . . (18.4) Position of nodes These points of zero displacement are called nodes. The particle with the greatest possible displacement from equilibrium has an amplitude of 2A, and we define this as the amplitude of the standing wave. The positions in the medium at which this maximum displacement occurs are called antinodes. The antinodes are located at positions for which the coordinate x satisfies the condition sin kx 1, that is, when kx 2 , 3 5 , , . . . 2 2 Thus, the positions of the antinodes are given by x 4 , 3 5 , , . . . 4 4 n 4 n 1, 3, 5, . . . (18.5) Position of antinodes In examining Equations 18.4 and 18.5, we note the following important features of the locations of nodes and antinodes: The distance between adjacent antinodes is equal to /2. The distance between adjacent nodes is equal to /2. The distance between a node and an adjacent antinode is /4. Displacement patterns of the particles of the medium produced at various times by two waves traveling in opposite directions are shown in Figure 18.5. The blue and green curves are the individual traveling waves, and the red curves are 552 CHAPTER 18 Superposition and Standing Waves y1 y1 y1 y2 A y A y2 y2 A A N N N N N y y N A N N A (c) t = T/2 N N N (a) t=0 N N (a) t = 0 (b) t = T/4 Figure 18.5 Standing-wave patterns produced at various times by two waves of equal amplitude traveling in opposite directions. For the resultant wave y, the nodes (N) are points of zero displacement, and the antinodes (A) are points of maximum displacement. the displacement patterns. At t 0 (Fig. 18.5a), the two traveling waves are in phase, giving a displacement pattern in which each particle of the medium is experiencing its maximum displacement from equilibrium. One quarter of a period later, at t T/4 (Fig. 18.5b), the traveling waves have moved one quarter of a wavelength (one to the right and the other to the left). At this time, the traveling waves are out of phase, and each particle of the medium is passing through the equilibrium position in its simple harmonic motion. The result is zero displacement for particles at all values of x -- that is, the displacement pattern is a straight line. At t T/2 (Fig. 18.5c), the traveling waves are again in phase, producing a displacement pattern that is inverted relative to the t 0 pattern. In the standing wave, the particles of the medium alternate in time between the extremes shown in Figure 18.5a and c. (b) t = T/ 8 (c) t = T/4 (d) t = 3T/ 8 (e) t = T/ 2 Energy in a Standing Wave It is instructive to describe the energy associated with the particles of a medium in which a standing wave exists. Consider a standing wave formed on a taut string fixed at both ends, as shown in Figure 18.6. Except for the nodes, which are always stationary, all points on the string oscillate vertically with the same frequency but with different amplitudes of simple harmonic motion. Figure 18.6 represents snapshots of the standing wave at various times over one half of a period. In a traveling wave, energy is transferred along with the wave, as we discussed in Chapter 16. We can imagine this transfer to be due to work done by one segment of the string on the next segment. As one segment moves upward, it exerts a force on the next segment, moving it through a displacement -- that is, work is done. A particle of the string at a node, however, experiences no displacement. Thus, it cannot do work on the neighboring segment. As a result, no energy is transmitted along the string across a node, and energy does not propagate in a standing wave. For this reason, standing waves are ofteposition of several normal modes. Exactly which normal modes are present depends on how the oscillation is started. For example, when a guitar string is plucked near its middle, the modes shown in Figure 18.7b and d, as well as other modes not shown, are excited. In general, we can describe the normal modes of oscillation for the string by imposing the requirements that the ends be nodes and that the nodes and antinodes be separated by one fourth of a wavelength. The first normal mode, shown in Figure 18.7b, has nodes at its ends and one antinode in the middle. This is the longestwavelength mode, and this is consistent with our requirements. This first normal mode occurs when the wavelength 1 is twice the length of the string, that is, 2L. The next normal mode, of wavelength 2 (see Fig. 18.7c), occurs when the 1 wavelength equals the length of the string, that is, 2 L. The third normal mode (see Fig. 18.7d) corresponds to the case in which 3 2L/3. In general, the wavelengths of the various normal modes for a string of length L fixed at both ends are Wavelengths of normal modes n 2L n n 1, 2, 3, . . . (18.6) where the index n refers to the nth normal mode of oscillation. These are the possible modes of oscillation for the string. The actual modes that are excited by a given pluck of the string are discussed below. The natural frequencies associated with these modes are obtained from the relationship f v/ , where the wave speed v is the same for all frequencies. Using Equation 18.6, we find that the natural frequencies fn of the normal modes are Frequencies of normal modes as functions of wave speed and length of string fn v n n v 2L n 1, 2, 3, . . . (18.7) Because v T/ (see Eq. 16.4), where T is the tension in the string and is its linear mass density, we can also express the natural frequencies of a taut string as Frequencies of normal modes as functions of string tension and linear mass density fn n 2L T n 1, 2, 3, . . . (18.8) 18.3 Standing Waves in a String Fixed at Both Ends 555 Multiflash photographs of standing-wave patterns in a cord driven by a vibrator at its left end. 1). The double-loop pattern repThe single-loop pattern represents the first normal mode (n 2), and the triple-loop pattern represents the third norresents the second normal mode (n 3). mal mode (n The lowest frequency f 1 , which corresponds to n 1, is called either the fundamental or the fundamental frequency and is given by f1 1 2L T (18.9) Fundamental frequency of a taut string The frequencies of the remaining normal modes are integer multiples of the fundamental frequency. Frequencies of normal modes that exhibit an integermultiple relationship such as this form a harmonic series, and the normal modes are called harmonics. The fundamental frequency f 1 is the frequency of the first harmonic; the frequency f 2 2f 1 is the frequency of the second harmonic; and the frequency f n nf 1 is the frequency of the nth harmonic. Other oscillating systems, such as a drumhead, exhibit normal modes, but the frequencies are not related as integer multiples of a fundamental. Thus, we do not use the term harmonic in association with these types of systems. In obtaining Equation 18.6, we used a technique based on the separation distance between nodes and antinodes. We can obtain this equation in an alternative manner. Because we require that the string be fixed at x 0 and x L, the wave function y(x, t) given by Equation 18.3 must be zero at these points for all times. That is, the boundary conditions require that y(0, t) 0 and that y(L, t) 0 for all values of t. Because the standing wave is described by y (2A sin kx) cos t, the first boundary condition, y(0, t) 0, is automatically satisfied because sin kx 0 at x 0. To meet the second boundary condition, y(L, t) 0, we require that sin kL 0. This condition is satisfied when the angle kL equals an integer multiple of rad. Therefore, the allowed values of k are given by 1 k nL Because k n 2 / n, n n 1, 2, 3, . . . (18.10) we find that 2 n L n or n 2L n QuickLab Compare the sounds of a guitar string plucked first near its center and then near one of its ends. More of the higher harmonics are present in the second situation. Can you hear the difference? which is identical to Equation 18.6. Let us now examine how these various harmonics are created in a string. If we wish to excite just a single harmonic, we need to distort the string in such a way that its distorted shape corresponded to that of the desired harmonic. After being released, the string vibrates at the frequency of that harmonic. This maneuver is difficult to perform, however, and it is not how we excite a string of a musical in1 We exclude n 0 because this value corresponds to the trivial case in which no wave exists (k 0). 556 CHAPTER 18 Superposition and Standing Waves strument. If the string is distorted such that its distorted shape is not that of just one harmonic, the resulting vibration includes various harmonics. Such a distortion occurs in musical instruments when the string is plucked (as in a guitar), bowed (as in a cello), or struck (as in a piano). When the string is distorted into a non-sinusoidal shape, only waves that satisfy the boundary conditions can persist on the string. These are the harmonics. The frequency of a stringed instrument can be varied by changing either the tension or the string's length. For example, the tension in guitar and violin strings is varied by a screw adjustment mechanism or by tuning pegs located on the neck of the instrument. As the tension is increased, the frequency of the normal modes increases in accordance with Equation 18.8. Once the instrument is "tuned," players vary the frequency by moving their fingers along the neck, thereby changing the length of the oscillating portion of the string. As the length is shortened, the frequency increases because, as Equation 18.8 specifies, the normal-mode frequencies are inversely proportional to string length. EXAMPLE 18.3 Give Me a C Note! Setting up the ratio of these frequencies, we find that f 1A f 1C TA TC Middle C on a piano has a fundamental frequency of 262 Hz, and the first A above middle C has a fundamental frequency of 440 Hz. (a) Calculate the frequencies of the next two harmonics of the C string. Knowing that the frequencies of higher harmonics are integer multiples of the fundamental frequency f 1 262 Hz, we find that f2 f3 2f 1 3f 1 524 Hz 786 Hz TA TC 2 Solution f 1A f 1C 440 262 2 2.82 (c) With respect to a real piano, the assumption we made in (b) is only partially true. The string densities are equal, but the length of the A string is only 64 percent of the length of the C string. What is the ratio of their tensions? Solution frequencies: Using Equation 18.8 again, we set up the ratio of f 1A f 1C TA TC LC LA (b) If the A and C strings have the same linear mass density and length L, determine the ratio of tensions in the two strings. Using Equation 18.8 for the two strings vibrating at their fundamental frequencies gives f 1A 1 2L TA TC 440 262 2 100 64 1.16 TA TC Solution (0.64)2 TA and f 1C 1 2L TC EXAMPLE 18.4 Guitar Basics speed of the wave on the string, v 2L f n n 2(0.640 m) (330 Hz) 1 422 m/s The high E string on a guitar measures 64.0 cm in length and has a fundamental frequency of 330 Hz. By pressing down on it at the first fret (Fig. 18.8), the string is shortened so that it plays an F note that has a frequency of 350 Hz. How far is the fret from the neck end of the string? Equation 18.7 relates the string's length to the fundamental frequency. With n 1, we can solve for the Solution Because we have not adjusted the tuning peg, the tension in the string, and hence the wave speed, remain constant. We can again use Equation 18.7, this time solving for L and sub- 18.4 Resonance 557 stituting the new frequency to find the shortened string length: L n v 2f n (1) 422 m/s 2(350 Hz) 0.603 m The difference between this length and the measured length of 64.0 cm is the distance from the fret to the neck end of the string, or 3.70 cm. Figure 18.8 Playing an F note oen end of an air column is not exactly a displacement antinode. A condensation reaching an open end does not reflect until it passes beyond the end. For a thin-walled tube of circular cross section, this end correction is approximately 0.6R, where R is the tube's radius. Hence, the effective length of the tube is longer than the true length L. We ignore this end correction in this discussion. 2 560 CHAPTER 18 Superposition and Standing Waves You may wonder how a sound wave can reflect from an open end, since there may not appear to be a change in the medium at this point. It is indeed true that the medium through which the sound wave moves is air both inside and outside the pipe. Remember that sound is a pressure wave, however, and a compression region of the sound wave is constrained by the sides of the pipe as long as the region is inside the pipe. As the compression region exits at the open end of the pipe, the constraint is removed and the compressed air is free to expand into the atmosphere. Thus, there is a change in the character of the medium between the inside of the pipe and the outside even though there is no change in the material of the medium. This change in character is sufficient to allow some reflection. The first three normal modes of oscillation of a pipe open at both ends are shown in Figure 18.14a. When air is directed against an edge at the left, longitudinal standing waves are formed, and the pipe resonates at its natural frequencies. All normal modes are excited simultaneously (although not with the same amplitude). Note that both ends are displacement antinodes (approximately). In the first normal mode, the standing wave extends between two adjacent antinodes, L A N A 1 = 2L v v f1 = -- = -- 1 2L 2 = L v f2 = -- = 2f1 L 2 3 = -- L 3 3v f3 = -- = 3f1 2L (a) Open at both ends First harmonic A N A N A Second harmonic A N A N A NA Third harmonic A N 1 = 4L v v f1 = -- = -- 1 4L 4 3 = -- L 3 3v f3 = -- = 3f1 4L 4 5 = -- L 5 5v f5 = -- = 5f1 4L (b) Closed at one end, open at the other First harmonic A N A N Third harmonic A N A N A N Fifth harmonic Figure 18.14 Motion of air molecules in standing longitudinal waves in a pipe, along with schematic representations of the waves. The graphs represent the displacement amplitudes, not the pressure amplitudes. (a) In a pipe open at both ends, the harmonic series created consists of all integer multiples of the fundamental frequency: f 1 , 2f 1 , 3f 1 , . . . . (b) In a pipe closed at one end and open at the other, the harmonic series created consists of only odd-integer multiples of the fundamental frequency: f 1 , 3f 1 , 5f 1 , . . . . 18.5 Standing Waves in Air Columns 561 which is a distance of half a wavelength. Thus, the wavelength is twice the length of the pipe, and the fundamental frequency is f 1 v/2L. As Figure 18.14a shows, the frequencies of the higher harmonics are 2f 1 , 3f 1 , . . . . Thus, we can say that in a pipe open at both ends, the natural frequencies of oscillation form a harmonic series that includes all integral multiples of the fundamental frequency. Because all harmonics are present, and because the fundamental frequency is given by the same expression as that for a string (see Eq. 18.7), we can express the natural frequencies of oscillation as fn n v 2L n 1, 2, 3 . . . (18.11) Natural frequencies of a pipe open at both ends Despite the similarity between Equations 18.7 and 18.11, we must remember that v in Equation 18.7 is the speed of waves on the string, whereas v in Equation 18.11 is the speed of sound in air. If a pipe is closed at one end and open at the other, the closed end is a displacement node (see Fig. 18.14b). In this case, the standing wave for the fundamental mode extends from an antinode to the adjacent node, which is one fourth of a wavelength. Hence, the wavelength for the first normal mode is 4L, and the fundamental frequency is f 1 v/4L. As Figure 18.14b shows, the higher-frequency waves that satisfy our conditions are those that have a node at the closed end and an antinode at the open end; this means that the higher harmonics have frequencies 3f 1 , 5f 1 , . . . : In a pipe closed at one end and open at the other, the natural frequencies of oscillation form a harmonic series that includes only odd integer multiples of the fundamental frequency. We express this result mathematically as fn n v 4L n 1, 3, 5, . . . (18.12) QuickLab Blow across the top of an empty sodapop bottle. From a measurement of the height of the bottle, estimate the frequency of the sound you hear. Note that the cross-sectional area of the bottle is not constant; thus, this is not a perfect model of a cylindrical air column. Natural frequencies of a pipe closed at one end and open at the other It is interesting to investigate what happens to the frequencies of instruments based on air columns and strings during a concert as the temperature rises. The sound emitted by a flute, for example, becomes sharp (increases in frequency) as it warms up because the speed of sound increases in the increasingly warmer air inside the flute (consider Eq. 18.11). The sound produced by a violin becomes flat (decreases in frequency) as the strings expand thermally because the expansion causes their tension to decrease (see Eq. 18.8). Quick Quiz 18.4 A pipe open at both ends resonates at a fundamental frequency f open . When one end is covered and the pipe is again made to resonate, the fundamental frequency is fclosed . Which of the following expressions describes how these two resonant frequencies compare? (a) f closed f open (b) f closed 1 f open (c) f closed 2f open (d) f closed 3 f open 2 2 562 CHAPTER 18 Superposition and Standing Waves EXAMPLE 18.5 Wind in a Culvert In this case, only odd harmonics are present; hence, the next two harmonics have frequencies f 3 f5 5f 1 349 Hz. 3f 1 209 Hz and A section of drainage culvert 1.23 m in length makes a howling noise when the wind blows. (a) Determine the frequencies of the first three harmonics of the culvert if it is open at both ends. Take v 343 m/s as the speed of sound in air. Solution The frequency of the first harmonic of a pipe open at both ends is f1 v 2L 343 m/s 2(1.23 m) 139 Hz (c) For the culvert open at both ends, how many of the harmonics present fall within the normal human hearing range (20 to 17 000 Hz)? Because both ends are open, all harmonics are present; thus, f2 2f 1 278 Hz and f 3 3f 1 417 Hz. (b) What are the three lowest natural frequencies of the culvert if it is blocked at one end? Solution Because all harmonics are present, we can express the frequency of the highest harmonic heard as f n nf 1 , where n is the number of harmonics that we can hear. For f n 17 000 Hz, we find that the number of harmonics present in the audible range is n 17 000 Hz 139 Hz 122 Solution one end is The fundamental frequency of a pipe closed at f1 v 4L 343 m/s 4(1.23 m) 69.7 Hz Only the first few harmonics are of sufficient amplitude to be heard. EXAMPLE 18.6 Measuring the Frequency of a Tuning Fork of the tuning fork is constant, the next two normal modes (see Fig. 18.15b) correspond to lengths of L 3 /4 0.270 m and L 5 /4 0.450 m. A simple apparatus for demonstrating resonance in an air column is depicted in Figure 18.15. A vertical pipe open at both ends is partially submerged in water, and a tuning fork vibrating at an unknown frequency is placed near the top of the pipe. The length L of the air column can be adjusted by moving the pipe vertically. The sound waves generated by the fork are reinforced when L corresponds to one of the resonance frequencies of the pipe. For a certain tube, the smallest value of L for which a peak occurs in the sound intensity is 9.00 cm. What are (a) the frequency of the tuning fork and (b) the value of L for the next two resonance frequencies? f=? /4 3/4 5/4 Solution (a) Although the pipe is open at its lower end to allow the water to enter, the water's surface acts like a wall at one end. Therefore, this setup represents a pipe closed at one end, and so the fundamental frequency is f 1 v/4L. Taking v 343 m/s for the speed of sanding waves in strings and pipes are common examples of spatial interference. We now consider another type of interference, one that results from the superposition of two waves having slightly different frequencies. In this case, when the two waves are observed at the point of superposition, they are periodically in and out of phase. That is, there is a temporal (time) alternation between constructive and destructive interference. Thus, we refer to this phenomenon as interference in time or temporal interference. For example, if two tuning forks of slightly different frequencies are struck, one hears a sound of periodically varying intensity. This phenomenon is called beating: Definition of beating Beating is the periodic variation in intensity at a given point due to the superposition of two waves having slightly different frequencies. 18.7 Beats: Interference in Time 565 The number of intensity maxima one hears per second, or the beat frequency, equals the difference in frequency between the two sources, as we shall show below. The maximum beat frequency that the human ear can detect is about 20 beats/s. When the beat frequency exceeds this value, the beats blend indistinguishably with the compound sounds producing them. A piano tuner can use beats to tune a stringed instrument by "beating" a note against a reference tone of known frequency. The tuner can then adjust the string tension until the frequency of the sound it emits equals the frequency of the reference tone. The tuner does this by tightening or loosening the string until the beats produced by it and the reference source become too infrequent to notice. Consider two sound waves of equal amplitude traveling through a medium with slightly different frequencies f 1 and f 2 . We use equations similar to Equation 16.11 to represent the wave functions for these two waves at a point that we choose as x 0: y1 y2 A cos A cos 1t 2t A cos 2 f 1t A cos 2 f 2t Using the superposition principle, we find that the resultant wave function this at point is y y1 y2 A(cos 2 f 1t cos 2 f 2t) The trigonometric identity cos a cos b 2 cos a 2 b cos a 2 b allows us to write this expression in the form y 2 A cos 2 f1 2 f2 t cos 2 f1 2 f2 t (18.13) Resultant of two waves of different frequencies but equal amplitude Graphs of the individual waves and the resultant wave are shown in Figure 18.18. From the factors in Equation 18.13, we see that the resultant sound for a listener standing at any given point has an effective frequency equal to the average frequency ( f 1 f 2)/2 and an amplitude given by the expression in the square y (a) t y (b) t Figure 18.18 Beats are formed by the combination of two waves of slightly different frequencies. (a) The individual waves. (b) The combined wave has an amplitude (broken line) that oscillates in time. 566 CHAPTER 18 Superposition and Standing Waves brackets: Aresultant 2A cos 2 f1 2 f2 t (18.14) That is, the amplitude and therefore the intensity of the resultant sound vary in time. The broken blue line in Figure 18.18b is a graphical representation of Equation 18.14 and is a sine wave varying with frequency ( f 1 f 2)/2. Note that a maximum in the amplitude of the resultant sound wave is detected whenever cos 2 f1 2 f2 t 1 This means there are two maxima in each period of the resultant wave. Because the amplitude varies with frequency as ( f 1 f 2)/2, the number of beats per second, or the beat frequency fb , is twice this value. That is, Beat frequency fb f1 f2 (18.15) For instance, if one tuning fork vibrates at 438 Hz and a second one vibrates at 442 Hz, the resultant sound wave of the combination has a frequency of 440 Hz (the musical note A) and a beat frequency of 4 Hz. A listener would hear a 440-Hz sound wave go through an intensity maximum four times every second. Optional Section 18.8 9.6 NON-SINUSOIDAL WAVE PATTERNS (a) Tuning fork t (b) Flute t (c) Clarinet t Figure 18.19 Sound wave patterns produced by (a) a tuning fork, (b) a flute, and (c) a clarinet, each at approximately the same frequency. The sound-wave patterns produced by the majority of musical instruments are non-sinusoidal. Characteristic patterns produced by a tuning fork, a flute, and a clarinet, each playing the same note, are shown in Figure 18.19. Each instrument has its own characteristic pattern. Note, however, that despite the differences in the patterns, each pattern is periodic. This point is important for our analysis of these waves, which we now discuss. We can distinguish the sounds coming from a trumpet and a saxophone even when they are both playing the same note. On the other hand, we may have difficulty distinguishing a note played on a clarinet from the same note played on an oboe. We can use the pattern of the sound waves from various sources to explain these effects. The wave patterns produced by a musical instrument are the result of the superposition of various harmonics. This superposition results in the corresponding richness of musical tones. The human perceptive response associated with various mixtures of harmonics is the quality or timbre of the sound. For instance, the sound of the trumpet is perceived to have a "brassy" quality (that is, we have learned to associate the adjective brassy with that sound); this quality enables us to distinguish the sound of the trumpet from that of the saxophone, whose quality is perceived as "reedy." The clarinet and oboe, however, are both straight air columns excited by reeds; because of this similarity, it is more difficult for the ear to distinguish them on the basis of their sound quality. The problem of analyzing non-sinusoidal wave patterns appears at first sight to be a formidable task. However, if the wave pattern is periodic, it can be represented as closely as desired by the combination of a sufficiently large number of si- 18.8 Non-Sinusoidal Wave Patterns 567 Relative intensity Relative intensity 1 2 3 4 5 6 Harmonics (a) 1 2 3 4 5 6 7 Harmonics (b) Relative intensity Tuning fork Clarinet Flute 1 2 3 4 5 6 7 8 9 Harmonics (c) Figure 18.20 Harmonics of the wave patterns shown in Figure 18.19. Note the variations in intensity of the various harmonics. nusoidal waves that form a harmonic series. In fact, we can represent any periodic function as a series of sine and cosine terms by using a mathematical technique based on Fourier's theorem.3 The corresponding sum of terms that represents the periodic wave pattern is called a Fourier series. Let y(t) be any function that is periodic in time with period T, such that y(t T ) y(t). Fourier's theorem states that this function can be written as y(t) n (An sin 2 f nt Bn cos 2 f nt) (18.16) Fourier's theorem where the lowest frequency is f 1 1/T. The higher frequencies are integer multiples of the fundamental, f n nf 1 , and the coefficients An and Bn represent the amplitudes of the various waves. Figure 18.20 represents a harmonic analysis of the wave patterns shown in Figure 18.19. Note that a struck tuning fork produces only one harmonic (the first), whereas the flute and clarinet produce the first and many higher ones. Note the variation in relative intensity of the various harmonics for the flute and the clarinet. In general, any musical sound consists of a fundamental frequency f plus other frequencies that are integer multiples of f, all having different intensities. We have discussed the analysis of a wave pattern using Fourier's theorem. The analysis involves determining the coefficients of the harmonics in Equation 18.16 from a knowledge of the wave pattern. The reverse process, called Fourier synthesis, can also be performed. In this process, the various harmonics are added together to form a resultant wave pattern. As an example of Fourier synthesis, consider the building of a square wave, as shown in Figure 18.21. The symmetry of the square wave results in only odd multiples of the fundamental frequency combining in its synthesis. In Figure 18.21a, the orange curve shows the combination of f and 3f. In Figure 18.21b, we have added 5f to the combination and obtained the green curve. Notice how the genernon of wave interference apply only to sinusoidal waves? 3. When two waves interfere constructively or destructively, is there any gain or loss in energy? Explain. 4. A standing wave is set up on a string, as shown in Figure 18.6. Explain why no energy is transmitted along the string. 5. What is common to all points (other than the nodes) on a string supporting a standing wave? 6. What limits the amplitude of motion of a real vibrating system that is driven at one of its resonant frequencies? 7. In Balboa Park in San Diego, CA, there is a huge outdoor organ. Does the fundamental frequency of a particular 570 CHAPTER 18 Superposition and Standing Waves chalkboard sets a larger number of air molecules into vibration. Thus, the chalkboard is a better radiator of sound than the tuning fork. How does this affect the length of time during which the fork vibrates? Does this agree with the principle of conservation of energy? To keep animals away from their cars, some people mount short thin pipes on the front bumpers. The pipes produce a high-frequency wail when the cars are moving. How do they create this sound? Guitarists sometimes play a "harmonic" by lightly touching a string at the exact center and plucking the string. The result is a clear note one octave higher than the fundamental frequency of the string, even though the string is not pressed to the fingerboard. Why does this happen? If you wet your fingers and lightly run them around the rim of a fine wine glass, a high-frequency sound is heard. Why? How could you produce various musical notes with a set of wine glasses, each of which contains a different amount of water? Despite a reasonably steady hand, one often spills coffee when carrying a cup of it from one place to another. Discuss resonance as a possible cause of this difficulty, and devise a means for solving the problem. 8. 9. 10. 11. 12. 13. 14. pipe of this organ change on hot and cold days? How about on days with high and low atmospheric pressure? Explain why your voice seems to sound better than usual when you sing in the shower. What is the purpose of the slide on a trombone or of the valves on a trumpet? Explain why all harmonics are present in an organ pipe open at both ends, but only the odd harmonics are present in a pipe closed at one end. Explain how a musical instrument such as a piano may be tuned by using the phenomenon of beats. An airplane mechanic notices that the sound from a twinengine aircraft rapidly varies in loudness when both engines are running. What could be causing this variation from loudness to softness? Why does a vibrating guitar string sound louder when placed on the instrument than it would if it were allowed to vibrate in the air while off the instrument? When the base of a vibrating tuning fork is placed against a chalkboard, the sound that it emits becomes louder. This is due to the fact that the vibrations of the tuning fork are transmitted to the chalkboard. Because it has a larger area than that of the tuning fork, the vibrating 15. 16. 17. 18. PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 18.1 Superposition and Interference of Sinusoidal Waves WEB 1. Two sinusoidal waves are described by the equations y1 and y2 (5.00 m) sin[ (4.00x 1 200t 0.250)] where x, y 1 , and y 2 are in meters and t is in seconds. (a) What is the amplitude of the resultant wave? (b) What is the frequency of the resultant wave? 2. A sinusoidal wave is described by the equation y1 (0.080 0 m) sin[2 (0.100x 80.0t )] (5.00 m) sin[ (4.00x 1 200t )] where y 1 and x are in meters and t is in seconds. Write an expression for a wave that has the same frequency, amplitude, and wavelength as y 1 but which, when added to y 1 , gives a resultant with an amplitude of 83 cm. 3. Two waves are traveling in the same direction along a stretched ld's swing are 2.00 m long. At what frequency should a big brother push to make the child swing with greatest amplitude? 28. Standing-wave vibrations are set up in a crystal goblet with four nodes and four antinodes equally spaced Problems around the 20.0-cm circumference of its rim. If transverse waves move around the glass at 900 m/s, an opera singer would have to produce a high harmonic with what frequency to shatter the glass with a resonant vibration? 29. An earthquake can produce a seiche (pronounced "saysh") in a lake, in which the water sloshes back and forth from end to end with a remarkably large amplitude and long period. Consider a seiche produced in a rectangular farm pond, as diagrammed in the cross-sectional view of Figure P18.29 (figure not drawn to scale). Suppose that the pond is 9.15 m long and of uniform depth. You measure that a wave pulse produced at one end reaches the other end in 2.50 s. (a) What is the wave speed? (b) To produce the seiche, you suggest that several people stand on the bank at one end and paddle together with snow shovels, moving them in simple harmonic motion. What must be the frequency of this motion? 573 Figure P18.29 30. The Bay of Fundy, Nova Scotia, has the highest tides in the world. Assume that in mid-ocean and at the mouth of the bay, the Moon's gravity gradient and the Earth's rotation make the water surface oscillate with an amplitude of a few centimeters and a period of 12 h 24 min. At the head of the bay, the amplitude is several meters. Argue for or against the proposition that the tide is amplified by standing-wave resonance. Suppose that the bay has a length of 210 km and a depth everywhere of 36.1 m. The speed of long-wavelength water waves is given by gd, where d is the water's depth. WEB 31. Calculate the length of a pipe that has a fundamental frequency of 240 Hz if the pipe is (a) closed at one end and (b) open at both ends. 32. A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f 0.680 kHz. For what values of L will the tube resonate with the speaker? 33. The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as a pipe open at both ends. (a) Find the frequency of the lowest note that a piccolo can play, assuming that the speed of sound in air is 340 m/s. (b) Opening holes in the side effectively shortens the length of the resonant column. If the highest note that a piccolo can sound is 4 000 Hz, find the distance between adjacent antinodes for this mode of vibration. 34. The fundamental frequency of an open organ pipe corresponds to middle C (261.6 Hz on the chromatic musical scale). The third resonance of a closed organ pipe has the same frequency. What are the lengths of the two pipes? 35. Estimate the length of your ear canal, from its opening at the external ear to the eardrum. (Do not stick anything into your ear!) If you regard the canal as a tube that is open at one end and closed at the other, at approximately what fundamental frequency would you expect your hearing to be most sensitive? Explain why you can hear especially soft sounds just around this frequency. 36. An open pipe 0.400 m in length is placed vertically in a cylindrical bucket and nearly touches the bottom of the bucket, which has an area of 0.100 m2. Water is slowly poured into the bucket until a sounding tuning fork of frequency 440 Hz, held over the pipe, produces resonance. Find the mass of water in the bucket at this moment. 37. A shower stall measures 86.0 cm 86.0 cm 210 cm. If you were singing in this shower, which frequencies would sound the richest (because of resonance)? Assume that the stall acts as a pipe closed at both ends, with nodes at opposite sides. Assume that the voices of various singers range from 130 Hz to 2 000 Hz. Let the speed of sound in the hot shower stall be 355 m/s. 38. When a metal pipe is cut into two pieces, the lowest resonance frequency in one piece is 256 Hz and that for the other is 440 Hz. (a) What resonant frequency would have been produced by the original length of pipe? (b)ork oscillating at 256 Hz. He walks toward a wall at a constant speed of 1.33 m/s. (a) What beat frequency does he observe between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of 5.00 Hz? (Optional) Section 18.8 Non-Sinusoidal Wave Patterns 52. Suppose that a flutist plays a 523-Hz C note with first harmonic displacement amplitude A1 100 nm. From Figure 18.20b, read, by proportion, the displacement amplitudes of harmonics 2 through 7. Take these as the values A2 through A7 in the Fourier analysis of the sound, and assume that B 1 B 2 . . . B 7 0. Construct a graph of the waveform of the sound. Your waveform will not look exactly like the flute waveform in Figure 18.19b because you simplify by ignoring cosine terms; nevertheless, it produces the same sensation to human hearing. Problems 53. An A-major chord consists of the notes called A, C , and E. It can be played on a piano by simultaneously striking strings that have fundamental frequencies of 440.00 Hz, 554.37 Hz, and 659.26 Hz. The rich consonance of the chord is associated with the near equality of the frequencies of some of the higher harmonics of the three tones. Consider the first five harmonics of each string and determine which harmonics show near equality. 575 56. On a marimba (Fig. P18.56), the wooden bar that sounds a tone when it is struck vibrates in a transverse standing wave having three antinodes and two nodes. The lowest-frequency note is 87.0 Hz; this note is produced by a bar 40.0 cm long. (a) Find the speed of transverse waves on the bar. (b) The loudness of the emitted sound is enhanced by a resonant pipe suspended vertically below the center of the bar. If the pipe is open at the top end only and the speed of sound in air is 340 m/s, what is the length of the pipe required to resonate with the bar in part (a)? ADDITIONAL PROBLEMS 54. Review Problem. For the arrangement shown in Figure P18.54, 30.0 , the inclined plane and the small pulley are frictionless, the string supports the mass M at the bottom of the plane, and the string has a mass m that is small compared with M. The system is in equilibrium, and the vertical part of the string has a length h. Standing waves are set up in the vertical section of the string. Find (a) the tension in the string, (b) the whole length of the string (ignoring the radius of curvature of the pulley), (c) the mass per unit length of the string, (d) the speed of waves on the string, (e) the lowest-frequency standing wave, (f) the period of the standing wave having three nodes, (g) the wavelength of the standing wave having three nodes, and (h) the frequency of the beats resulting from the interference of the sound wave of lowest frequency generated by the string with another sound wave having a frequency that is 2.00% greater. Figure P18.56 Marimba players in Mexico City. (Murray Greenberg) h M Figure P18.54 55. Two loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speakers. The speakers are being driven by a single oscillator at a frequency of 300 Hz. (a) What is the phase difference between the two waves when they reach the observer? (b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound? 57. Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and is sounding its whistle, a beat frequency of 2.00 Hz is heard from a train moving nearby. What are the two possible speeds and directions that the moving train can have? 58. A speaker at the front of a room and an identical speaker at the rear of the room are being driven by the same oscillator at 456 Hz. A student walks at a uniform rate of 1.50 m/s along the length of the room. How many beats does the student hear per second? 59. While Jane waits on a railroad platform, she observes two trains approaching from the same direction at equal speeds of 8.00 m/s. Both trains are blowing their whistles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane, but before the second train passes her, she hears beats having a frequency of 4.00 Hz. What is the frequency of the trains' whistles? 60. A string fixed at both ends and having a mass of 4.80 g, a length of 2.00 m, and a tension of 48.0 N vibrates in its second (n 2) natural mode. What is the wavelength in air of the sound emitted by this vibrating string? 576 CHAPTER 18 Superposition and Standing Waves the fundamental frequency and length of the pipe. (Use v 340 m/s.) 68. Two waves are described by the equations y 1(x, t ) and y 2(x, t ) 10 cos(2.0x 10t ) 5.0 sin(2.0x 10t ) WEB 61. A string 0.400 m in length has a mass per unit length of 9.00 10 3 kg/m. What must be the tension in the string if its second harmonic is to have the same frequency as the second resonance mode of a 1.75-m-long pipe open at one end? 62. In a major chord on the physical pitch musical scale, the frequencies are in the ratios 4: 5: 6: 8. A set of pipes, closed at one end, must be cut so that, when they are sounded in their first normal mode, they produce a major chord. (a) What is the ratio of the lengths of the pipes? (b) What are the lengths of the pipes needed if the lowest frequency of the chord is 256 Hz? (c) What are the frequencies of this chord? 63. Two wires are welded together. The wires are made of the same material, but the diameter of one wire is twice that of the other. They are subjected to a tension of 4.60 N. The thin wire has a length of 40.0 cm and a linear mass density of 2.00 g/m. The combination is fixed at both ends and vibrated in such a way that two antinodes are present, with the node between them being right at the weld. (a) What is the frequency of vibration? (b) How long is the thick wire? 64. Two identical strings, each fixed at both ends, are arranged near each other. If string A starts oscillating in its first normal mode, string B begins vibrating in its third (n 3) natural mode. Determine the ratio of the tension of string B to the tension of string A. 65. A standing wave is set up in a string of variable length and tension by a vibrator of variable frequency. When the vibrator has a frequency f, in a string of length L and under a tension T, n antinodes are set up in the string. (a) If the length of the string is doubled, by what factor should the frequency be changed so that the same number of antinodes is produced? (b) If the frequency and length are held constant, what tension produces n 1 antinodes? (c) If the frequency is tripled and the length of the string is halved, by what factor should the tension be changed so that twice as many antinodes are produced? 66. A 0.010 0-kg, 2.00-m-long wire is fixed at both ends and vibrates in its simplest mode under a tension of 200 N. When a tuning fork is placed near the wire, a beat frequency of 5.00 Hz is heard. (a) What could the frequency of the tuning fork be? (b) What should the tension in the wire be if the beats are to disappear? 67. If two adjacent natural frequencies of an organ pipe are determined to be 0.550 kHz and 0.650 kHz, calculate where x is in meters and t is in seconds. Show that the resulting wave is sinusoidal, and determine the amplitude and phase of this sinusoidal wave. 69. The wave function for a standing wave is given in Equation 18.3 as y (2A sin kx) cos t. (a) Rewrite this wave function in terms of the wavelength and the wave speed v of the wave. (b) Write the wave function of the simplest standing-wave vibration of a stretched string of length L. (c) Write the wave function for the second harmonic. (d) Generalize these results, and write the wave function for the nth resonance vibration. 70. Review Problem. A 12.0-kg mass hangs in equilibrium from a string with a total length of L 5.00 m and a linear mass density of 0.001 00 kg/m. The string is wrapped around two light, frictionless pulleys that are separated by a distance of d 2.00 m (Fig. P18.70a). (a) Determine the tension in the string. (b) At what frequency must the string between the pulleys vibrate to form the standing-wave pattern shown in Figure P18.70b? d d g m (a) m (b) Figure P18.70 ANSWERS TO QUICK QUIZZES 18.1 At the antinodes. All particles have the same period T 2 / , but a particle at an antinode must travel through the greatest vertical distance in this amount of time and therefore must travel fastest. 18.2 For each natural frequency of the glass, the standing wave must "fit" exactly around the rim. In Figure 18.12a we see three antinodes on the near side of the glass, and thus there must be another three on the far side. This Answers to Quick Quizzes corresponds to three complete waves. In a top view, the wave pattern looks like this (although we have greatly exaggerated the amplitude): 577 18.3 At highway speeds, a car crosses the ridges on the rumble strip at a rate that matches one of the car's natural frequencies of oscillation. This causes the car to oscillate substantially more than when it is traveling over the randomly spaced bumps of regular pavement. This sudden resonance oscillation alerts the driver that he or she must pay attention. 18.4 (b). With both ends open, the pipe has a fundamental frequency given by Equation 18.11: f open v/2L. With one end closed, the pipe has a fundamental frequency given by Equation 18.12: 1 v 1 v f closed f 4L 2 2L 2 open P U Z Z L E R After this bottle of champagne was shaken, the cork was popped off and champagne spewed everywhere. Contrary to common belief, shaking a champagne bottle before opening it does not increase the pressure of the carbon dioxide (CO2 ) inside. In fact, if you know the trick, you can open a thoroughly shaken bottle without spraying a drop. What's the secret? And why isn't the pressure inside the bottle greater after the bottle is shaken? (Steve Niedorf/The Image Bank) c h a p t e r Temperature Chapter Outline 19.1 Temperature and the Zeroth Law of Thermodynamics 19.4 Thermal Expansion of Solids and Liquids 19.2 Thermometers and the Celsius Temperature Scale 19.5 Macroscopic Description of an Ideal Gas 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 580 19.1 Temperature and the Zeroth Law of Thermodynamics 581 I n our study of mechanics, we carefully defined such concepts as mass, force, and kinetic energy to facilitate our quantitative approach. Likewise, a quantitative description of thermal phenomena requires a careful definition of such important terms as temperature, heat, and internal energy. This chapter begins with a look at these three entities and with a description of one of the laws of thermodynamics (the poetically named "zeroth law"). We then discuss the three most common temperature scales -- Celsius, Fahrenheit, and Kelvin. Next, we consider why the composition of a body is an important factor when we are dealing with thermal phenomena. For example, gases expand appreciably when heated, whereas liquids and solids expand only slightly. If a gas is not free to expand as it is heated, its pressure increases. Certain substances may melt, boil, burn, or explode when they are heated, depending on their composition and structure. This chapter concludes with a study of ideal gases on the macroscopic scale. Here, we are concerned with the relationships among such quantities as pressure, volume, and temperature. Later on, in Chapter 21, we shall examine gases on a microscopic scale, using a model that represents the components of a gas as small particles. 19.1 TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS 10.3 & 10.4 We often associate the concept of temperature with how hot or cold an object feels when we touch it. Thus, our senses provide us with a qualitative indication of temperature. However, our senses are unreliable and often mislead us. For example, if we remove a metal ice tray and a cardboard box of frozen vegetables from the freezer, the ice tray feels colder than the box even though both are at the same temperature. The two objects feel different because metal is a better thermal conductor than cardboard is. What we need, therefore, is a reliable and reproducible method for establishing the relative hotness or coldness of bodies. Scientists have developed a variety of thermometers for making such quantitative measurements. We are all familiar with the fact that two objects at different initial temperatures eventually reach some intermediate temperature when placed in contact with each other. For example, when a scoop of ice cream is placed in a roomtemperature glass bowl, the ice cream melts and the temperature of the bowl decreases. Likewise, when an ice cube is dropped into a cup of hot coffee, it melts and the coffee's temperature decreases. To understand the concept of temperature, it is useful to define two oftenused phrases: thermal contact and thermal equilibrium. To grasp the meaning of thermal contact, let us imagine that two objects are placed in an insulated container such that they interact with each other but not with the rest of the world. If the objects are at different temperatures, energy is exchanged between them, even if they are initially not in physical contact with each other. Heat is the transfer of energy from one object to another object as a result of a difference in temperature between the two. We shall examine the concept of heat in greater detail in Chapter 20. For purposes of the current discussion, we assume that two objects are in thermal contact with each other if energy can be exchanged between them. Thermal equilibrium is a situation in which two objects in thermal contact with each other cease to exchange energy by the process of heat. Let us consider two objects A and B, which are not in thermal contact, and a third object C, which is our thermometer. We wish to determine whether A and B Molten lava flowing down a mountain in Kilauea, Hawaii. The temperature of the hot lava flowing from a central crater decreases until the lava is in thermal equilibrium with its surroundings. At that equilibrium temperature, the lava has solidified and formed the mountains. QuickLab Fill three cups with tap water: one hot, one cold, and one lukewarm. Dip your left index finger into the hot water and your right index finger into the cold water. Slowly count to 20, then quickly dip both fingers into the lukewarm water. What do you feel? 582 CHAPTER 19 Temperature are in thermal equilibrium with each other. The thermometer (object C) is first placed in thermal contact with object A until thermal equilibrium is reached. From that moment on, the thermometer's reading remains constant, and we record this reading. The thermometer is then removed from object A and placed in thermal contact with object B. The reading is again recorded after thermal equilibrium is reached. If the two readings are the same, then object A and object B are in thermal equilibrium with each other. We can summarize these results in a statement known as the zeroth law of thermodynamics (the law of equilibrium): Zeroth law of thermodynamics If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other. This statement can easily be proved experimentally and is very important because it enables us to define temperature. We can think of temperature as the property that determines whether an object is in thermal equilibrium with other objects. Two objects in thermal equilibrium with each other are at the same temperature. Conversely, if two objects have different temperatures, then they are not in thermal equilibrium with each other. 19.2 THERMOMETERS AND THE CELSIUS TEMPERATURE SCALE Thermometers are devices that are used to define and measure temperatures. All thermometers are based on the principle that some physical property of a system changes as the system's temperature changes. Some physical properties that change with temperature are (1) the volume of a liquid, (2) the length of a solid, (3) the pressure of a gas at constant volume, (4) the volume of a gas at constant pressure, (5) the electric resistance of a conductor, and (6) the color of an object. For a given subone shown in Figure 20.4. The value of the integral in Equation 20.8 is the area bounded by such a curve. Thus, we can say that the work done by a gas in the expansion from an initial state to a final state is the area under the curve connecting the states in a PV diagram. As Figure 20.4 shows, the work done in the expansion from the initial state i to the final state f depends on the path taken between these two states, where the path on a PV diagram is a description of the thermodynamic process through which the system is taken. To illustrate this important point, consider several paths connecting i and f (Fig. 20.5). In the process depicted in Figure 20.5a, the pressure of the gas is first reduced from Pi to Pf by cooling at constant volume Vi . The gas then expands from Vi to Vf at constant pressure Pf . The value of the work done along this path is equal to the area of the shaded rectangle, which is equal to Work equals area under the curve in a PV diagram. 616 CHAPTER 20 P Pi i Heat and the First Law of Thermodynamics P Pi i Pi P i Pf Vi (a) Vf f V Pf Vi (b) Vf f V Pf Vi (c) Vf f V Figure 20.5 The work done by a gas as it is taken from an initial state to a final state depends on the path between these states. Work done depends on the path between the initial and final states. Pf (Vf Vi ). In Figure 20.5b, the gas first expands from Vi to Vf at constant pressure Pi . Then, its pressure is reduced to Pf at constant volume Vf . The value of the work done along this path is Pi(Vf Vi ), which is greater than that for the process described in Figure 20.5a. Finally, for the process described in Figure 20.5c, where both P and V change continuously, the work done has some value intermediate between the values obtained in the first two processes. Therefore, we see that the work done by a system depends on the initial and final states and on the path followed by the system between these states. The energy transfer by heat Q into or out of a system also depends on the process. Consider the situations depicted in Figure 20.6. In each case, the gas has the same initial volume, temperature, and pressure and is assumed to be ideal. In Figure 20.6a, the gas is thermally insulated from its surroundings except at the bottom of the gas-filled region, where it is in thermal contact with an energy reservoir. An energy reservoir is a source of energy that is considered to be so great that a finite transfer of energy from the reservoir does not change its temperature. The piston is held at its initial position by an external agent -- a hand, for instance. When the force with which the piston is held is reduced slightly, the piston rises very slowly to its final position. Because the piston is moving upward, the gas is doing work on Insulating wall Final position Insulating wall Vacuum Membrane Gas at Ti Initial position Gas at Ti Energy reservoir at Ti (a) (b) Figure 20.6 (a) A gas at temperature Ti expands slowly while absorbing energy from a reservoir in order to maintain a constant temperature. (b) A gas expands rapidly into an evacuated region after a membrane is broken. 20.5 The First Law of Thermodynamics 617 the piston. During this expansion to the final volume Vf , just enough energy is transferred by heat from the reservoir to the gas to maintain a constant temperature Ti . Now consider the completely thermally insulated system shown in Figure 20.6b. When the membrane is broken, the gas expands rapidly into the vacuum until it occupies a volume Vf and is at a pressure Pf . In this case, the gas does no work because there is no movable piston on which the gas applies a force. Furthermore, no energy is transferred by heat through the insulating wall. The initial and final states of the ideal gas in Figure 20.6a are identical to the initial and final states in Figure 20.6b, but the paths are different. In the first case, the gas does work on the piston, and energy is transferred slowly to the gas. In the second case, no energy is transferred, and the value of the work done is zero. Therefore, we conclude that energy transfer by heat, like work done, depends on the initial, final, and intermediate states of the system. In other words, because heat and work depend on the path, neither quantity is determined solely by the end points of a thermodynamic process. This device, called Hero's engine, was invented around 150 B.C. by Hero in Alexandria. When water is boiled in the flask, which is suspended by a cord, steam exits through two tubes at the sides (in opposite directions), creating a torque that rotates the flask. 20.5 10.6 THE FIRST LAW OF THERMODYNAMICS When we introduced the law of conservation of mechanical energy in Chapter 8, we stated that the mechanical energy of a system is constant in the absence of nonconservative forces such as friction. That is, we did not include changes in the internal energy of the system in this mechanical model. The first law of thermodynamics is a generalization of the law of conservation of energy that encompasses changes in internal energy. It is a universally valid law that can be applied to many processes and provides a connection between the microscopic and macroscopic worlds. We have discussed two ways in which energy can be transferred between a system and its surroundings. One is work done by the system, which requires that there be a macroscopic displacement of the point of application of a force (or pressure). The other is heat, which occurs through random collisions between the molecules of the system. Both mechanisms result in a change in the internal energy of the system and therefore usually result in measurable changes in the macroscopic variables of the system, such as the pressure, temperature, and volume of a gas. To better understand these ideas on a quantitative basis, suppose that a system undergoes a change from an initial state to a final state. During this change, energy transfer by heat Q to the system occurs, and work W is done by the system. As an example, suppose that the system is a gas in which the pressure and volume change from Pi and Vi to Pf and Vf . If the quantity Q W is measured for various paths connecting the initial and final equilibrium states, we find that it is the same for all paths connecting the two states. We conclude that the quantity Q W is determined completely by the initial and final states of the system, and we call this quantity the change in the internal energy of the system. Although Q and W both depend on the path, the quantity Q W is independent of the path. If we use the sumbol E int to represent the internal energy, then the change in internal energy E int can be expressed as5 E int Q W (20.9) Q W is the change in internal energy First-law equation 5 It is an unfortunate accident of history that the traditional symbol for internal energy is U, which is also the traditional symbol for potential energy, as introduced in Chapter 8. To avoid confusion between potential energy and internal energy, we use the symbol E int for internal energy in this book. If you take an advanced course in thermodynamics, however, be prepared to see U used as the symbol for internal energy. 618 CHAPTER 20 Heat and the First Law of Thermodynamics where all quantities must have the same units of measure for energy.6 Equation 20.9 is known as the first-law equation and is a key concept in many applications. As a reminder, we use the convention that Q is positive when energy enters the system and negative when energy leaves the system, and that W is positive when the system does work on the surroundings and negative when work is done on the system. When a system undergoes an infinitesimal change in state in which a small amount of energy dQ is transferred by heat and a small amount of work dW is done, the internal energy changes by a small amount dE int . Thus, for infinitesimal processes we can express the first-law equation as7 First-law equation for infinitesimal changes dE int dQ dW Isolated system The first-law equation is an energy conservation equation specifying that the only type of energy that changes in the system is the internal energy E int . Let us look at some special cases in which this condition exists. First, let us consider an isolated system -- that is, one that does not interact with its surroundings. In this case, no energy transfer by heat takes place and the value of the work done by the system is zero; hence, the internal energy remains constant. That is, because Q W 0, it follows that E int 0, and thus E int , i E int , f . We conclude that the internal energy Eint of an isolated system remains constant. Next, we consider the case of a system (one not isolated from its surroundings) that is taken through a cyclic process -- that is, a process that starts and ends at the same state. In this case, the change in the internal energy must again be zero, and therefore the energy Q added to the system must equal the work W done by the system during the cycle. That is, in a cyclic process, E int 0 and Q W Cyclic process On a PV diagram, a cyclic process appears as a closed curve. (The processes described in Figure 20.5 are represented by open curves because the initial and final states differ.) It can be shown that in a cyclic process, the net work done by the system per cycle equals the area enclosed by the path representing the process on a PV diagram. If the value of the work done by the system during some process is zero, then the change in internal energy E int equals the energy transfer Q into or out of the system: E int Q If energy enters the system, then Q is positive and the internal energy increases. For a gas, we can associate this increase in internal energy with an increase in the kinetic energy of the molecules. Conversely, if no energy transfer occurs during some process but work is done by the system, then the change in internal energy equals the negative value of the work done by the system: E int 6 W For the definition of work from our mechanics studies, the first law would be written as E int Q W because energy transfer into the system by either work or heat would increase the internal energy of the system. Because of the reversal of the definition of positive work discussed in Section 20.4, the first law appears as in Equation 20.9, with a minus sign. 7 Note that dQ and dW are not true differential quantities; however, dE int is. Because dQ and dW are inexact differentials, they are often represented by the symbols dQ and d W . For further details on this point, see an advanced text on thermodynamics, such as R. P. Bauman, Modern Thermodynamics and Statistical Mechanics, New York, Macmillan Publishing Co., 1992. 20.6 Some Applications of the First Law of Thermodynamics 619 For example, if a gas is compressed by a moving piston in an insulated cylinder, no energy is transferred by heat and the work done by the gas is negative; thus, the internal energy increases because kinetic energy is transferred from the moving piston to the gas molecules. On a microscopic scale, no distinction exists between the result of heat and that of work. Both heat and work can produce a change in the internal energy of a system. Although the macroscopic quantities Q and W are not properties of a system, they are related to the change of the internal energy of a system through the first-law equation. Once we define a process, or path, we can either calculate or measure Q and W, and we can find the change in the system's internal energy using the first-law equation. One of the important consequences of the first law of thermodynamics is that there exists a quantity known as internal energy whose value is determined by the state of the system. The internal energy function is therefore called a state function. 20.6 SOME APPLICATIONS OF THE FIRST LAW OF THERMODYNAMICS In an adiabatic process, Q 0. Before we apply the first law of thermodynamics to specific systems, it is useful for us to first define some common thermodynamic processes. An adiabatic process is one during which no energy enters or leaves the system by heat -- that is, Q 0. An adiabatic process can be achieved either by thermally insulating the system from its surroundings (as shown in Fig. 20.6b) or by performing the process rapidly, so that there is little time for energy to transfer by heat. Applying the first law of thermodynamics to an adiabatic process, we see that E int W (adiabatic process) (20.10) First-law equation for an adiabatic process From this result, we see that if a gas expands adiabatically such that W is positive, then E int is negative and the temperature of the gas decreases. Conversely, the temperature of a gas increases when the gas is compressed adiabatically. Adiabatic processes are very important in engineering practice. Some common examples are the expansion of hot gases in an internal combustion engine, the liquefaction of gases in a cooling system, and the compression stroke in a diesel engine. The process described in Figure 20.6b, called an adiabatic free expansion, is unique. The process is adiabatic because it takes place in an insulated container. Because the gas expands into a vacuum, it does not apply a force on a piston as was depicted in Figure 20.6a, so no work is done on or by the gas. Thus, in this adiabatic process, both Q 0 and W 0. As a result, E int 0 for this process, as we can see from the first law. That is, the initial and final internal energies of a gas are equal in an adiabatic free expansion. As we shall see in the next chapter, the internal energy of an ideal gas depends only on its temperature. Thus, we expect no change in temperature during an adiabatic free expansion. This prediction is in accord with the results of experiments performed at low pressures. (Experiments performed at high pressures for real gases show a slight decrease or increase in temperature after the expansion. This change is due to intermolecular interactions, which represent a deviation from the model of an ideal gas.) A process that occurs at constant pressure is called an isobaric process. In such a process, the values of the heat and the work are both usually nonzero. The In an adiabatic free expansion, E int 0. In an isobaric process, P remains constant. 620 CHAPTER 20 Heat and the First Law of Thermodynamics work done by the gas is simply W P(Vf Vi ) (isobaric process) (20.11) where P is the constant pressure. A process that takes place at constant volume is called an isovolumetric process. In such a process, the value of the work done is clearly zero because the volume does not change. Hence, from the first law we see that in an isovolumetric process, because W 0, First-law equation for a constantvolume process E int Q (isovolumetric process) (20.12) In an isothermal process, T remains constant. This expression specifies that if energy is added by heat to a system kept at constant volume, then all of the transferred energy remains in the system as an increase of the internal energy of the system. For example, when a can of spray paint is thrown into a fire, energy enters the system (the gas in the can) by heat through the metal walls of the can. Consequently, the temperature, and thus the pressure, in the can increases until the can possibly explodes. A process that occurs at constant temperature is called an isothermal process. A plot of P versus V at constant temperature for an ideal gas yields a hyperbolic curve called an isotherm. The internal energy of an ideal gas is a function of temperature only. Hence, in an isothermal process involving an ideal gas, E int 0. For an isothermal process, then, we conclude from the first law that the energy transfer Q must be equal to the work done by the gas -- that is, Q W. Any energy that enters the system by heat is transferred out of the system by work; as a result, no change of the internal energy of the system occurs. Quick Quiz 20.4 In the last three columns of the following table, fill in the boxes with situation, the system to be considered is identified. Situation System Q W , , or 0. For each E int P Isotherm Pi i PV = constant (a) Rapidly pumping up a bicycle tire (b) Pan of room-temperature water sitting on a hot stove (c) Air quickly leaking out of a balloon Air in the p section, we found that the temperature of a gas is a measure of the average translational kinetic energy of the gas molecules. This kinetic energy is associated with the motion of the center of mass of each molecule. It does not include the energy associated with the internal motion of the molecule -- namely, vibrations and rotations about the center of mass. This should not be surprising because the simple kinetic theory model assumes a structureless molecule. In view of this, let us first consider the simplest case of an ideal monatomic gas, that is, a gas containing one atom per molecule, such as helium, neon, or argon. When energy is added to a monatomic gas in a container of fixed volume (by heating, for example), all of the added energy goes into increasing the translational kinetic energy of the atoms. There is no other way to store the energy in a monatomic gas. Therefore, from Equation 21.6, we see that the total internal energy E int of N molecules (or n mol) of an ideal monatomic gas is E int 3 2 Nk BT 3 2 nRT (21.10) Note that for a monatomic ideal gas, E int is a function of T only, and the functional relationship is given by Equation 21.10. In general, the internal energy of an ideal gas is a function of T only, and the exact relationship depends on the type of gas, as we shall soon explore. Quick Quiz 21.2 How does the internal energy of a gas change as its pressure is decreased while its volume is increased in such a way that the process follows the isotherm labeled T in Figure 21.4? (a) E int increases. (b) E int decreases. (c) Eint stays the same. (d) There is not enough information to determine E int . If energy is transferred by heat to a system at constant volume, then no work is done by the system. That is, W P dV 0 for a constant-volume process. Hence, from the first law of thermodynamics, we see that Q E int (21.11) In other words, all of the energy transferred by heat goes into increasing the internal energy (and temperature) of the system. A constant-volume process from i to f is described in Figure 21.4, where T is the temperature difference between the two isotherms. Substituting the expression for Q given by Equation 21.8 into 21.2 Molar Specific Heat of an Ideal Gas 647 Equation 21.11, we obtain E int nCV T (21.12) P Isotherms If the molar specific heat is constant, we can express the internal energy of a gas as E int nCVT i f f T + T T V This equation applies to all ideal gases -- to gases having more than one atom per molecule, as well as to monatomic ideal gases. In the limit of infinitesimal changes, we can use Equation 21.12 to express the molar specific heat at constant volume as CV 1 dE int n dT (21.13) Let us now apply the results of this discussion to the monatomic gas that we have been studying. Substituting the internal energy from Equation 21.10 into Equation 21.13, we find that (21.14) CV 3 R 2 This expression predicts a value of CV 3 R 12.5 J/mol K for all monatomic 2 gases. This is in excellent agreement with measured values of molar specific heats for such gases as helium, neon, argon, and xenon over a wide range of temperatures (Table 21.2). Now suppose that the gas is taken along the constant-pressure path i : f shown in Figure 21.4. Along this path, the temperature again increases by T. The energy that must be transferred by heat to the gas in this process is Q nCP T. Because the volume increases in this process, the work done by the gas is W P V, where P is the constant pressure at which the process occurs. Applying Figure 21.4 Energy is transferred by heat to an ideal gas in two ways. For the constant-volume path i : f, all the energy goes into increasing the internal energy of the gas because no work is done. Along the constant-pressure path i : f , part of the energy transferred in by heat is transferred out by work done by the gas. TABLE 21.2 Molar Specific Heats of Various Gases Molar Specific Heat ( J/mol K)a Gas CP CV 12.5 12.5 12.7 12.3 20.4 20.8 21.1 21.0 25.7 28.5 31.4 27.0 27.1 CP CV CP /C V 1.67 1.67 1.64 1.69 1.41 1.40 1.40 1.40 1.35 1.30 1.29 1.30 1.31 Monatomic Gases He 20.8 Ar 20.8 Ne 20.8 Kr 20.8 Diatomic Gases H2 28.8 N2 29.1 O2 29.4 CO 29.3 Cl2 34.7 Polyatomic Gases CO2 37.0 SO2 40.4 H2O 35.4 CH4 35.5 a All 8.33 8.33 8.12 8.49 8.33 8.33 8.33 8.33 8.96 8.50 9.00 8.37 8.41 values except that for water were obtained at 300 K. 648 CHAPTER 21 The Kinetic Theory of Gases the first law to this process, we have E int Q W nCP T P V (21.15) In this case, the energy added to the gas by heat is channeled as follows: Part of it does external work (that is, it goes into moving a piston), and the remainder increases the internal energy of the gas. But the change in internal energy for the process i : f is equal to that for the process i : f because E int depends only on temperature for an ideal gas and because T is the same for both processes. In addition, because PV nRT, we note that for a constant-pressure process, P V nR T. Substituting this value for P V into Equation 21.15 with E int nCV T (Eq. 21.12) gives nCV T CP CV nCP T R nR T (21.16) Ratio of molar specific heats for a monatomic ideal gas This expression applies to any ideal gas. It predicts that the molar specific heat of an ideal gas at constant pressure is greater than the molar specific heat at constant volume by an amount R, the universal gas constant (which has the value 8.31 J/mol K). This expression is applicable to real gases, as the data in Table 21.2 show. Because CV 3 R for a monatomic ideal gas, Equation 21.16 predicts a value 2 CP 5 R 20.8 J/mol K for the molar specific heat of a monatomic gas at con2 stant pressure. The ratio of these heat capacities is a dimensionless quantity (Greek letter gamma): CP (5/2)R 5 (21.17) 1.67 CV (3/2)R 3 Theoretical values of CP and are in excellent agreement with experimental values obtained for monatomic gases, but they are in serious disagreement with the values for the more complex gases (see Table 21.2). This is not surprising because the value CV 3 R was derived for a monatomic ideal gas, and we expect some ad2 ditional contribution to the molar specific heat from the internal structure of the more complex molecules. In Section 21.4, we describe the effect of molecular structure on the molar specific heat of a gas. We shall find that the internal energy -- and, hence, the molar specific heat -- of a complex gas must include contributions from the rotational and the vibrational motions of the molecule. We have seen that the molar specific heats of gases at constant pressure are greater than the molar specific heats at constant volume. This difference is a consequence of the fact that in a constant-volume process, no work is done and all of the energy transferred by heat goes into increasing the internal energy (and temperature) of the gas, whereas in a constant-pressure process, some of the energy transferred by heat is transferred out as work done by the gas as it expands. In the case of solids and liquids heated at constant pressure, very little work is done because the thermal expansion is small. Consequently, CP and CV are approximately equal for solids and liquids. EXAMPLE 21.2 Heating a Cylinder of Helium A cylinder contains 3.00 mol of helium gas at a temperature of 300 K. (a) If the gas is heated at constant volume, how much energy must be transferred by heat to the gas for its temperature to increase to 500 K ? Solution Because CV For the constant-volume process, we have Q1 nCV T T 200 K, we 12.5 J/mol K for helium and 21.3 Adiabatic Processes for an Ideal Gas obtain Q1 (3.00 mol)(12.5 J/mol K)(200 K) 7.50 10 3 J Q2 nCP T 12.5 (3.00 mol)(20.8 J/mol K)(200 K) 10 3 J 649 (b) How much energy must be transferred by heat to the gas at constant pressure to raise the temperature to 500 K? Exercise process? What is the work done by the gas in this isobaric Solution Making use of Table 21.2, we obtain Answer W Q2 Q1 5.00 10 3 J. 21.3 ADIABATIC PROCESSES FOR AN IDEAL GAS As we noted in Section 20.6, an adiabatic process is one in which no energy is transferred by heat between a system and its surroundings. For example, if 1 !k BT/m (21.27) (21.28) (21.29) rms speed Average speed Most probable speed The details of these calculations are left for the student (see Problems 41 and 62). From these equations, we see that v rms v v mp Figure 21.12 represents speed distribution curves for N 2 . The curves were obtained by using Equation 21.26 to evaluate the distribution function at various speeds and at two temperatures. Note that the peak in the curve shifts to the right 1 For the derivation of this expression, see an advanced textbook on thermodynamics, such as that by R. P. Bauman, Modern Thermodynamics with Statistical Mechanics, New York, Macmillan Publishing Co., 1992. 658 CHAPTER 21 The Kinetic Theory of Gases Nv , number of molecules per unit speed interval (molecules/m/s) 200 T = 300 K 160 120 80 T = 900 K 40 vmp Curves calculated for N = 105 nitrogen molecules vv rms 200 400 600 800 v (m/s) 1000 1200 1400 1600 The speed distribution function for 105 nitrogen molecules at 300 K and 900 K. The total area under either curve is equal to the total number of molecules, which in this case v mp . equals 105. Note that v rms v Figure 21.12 as T increases, indicating that the average speed increases with increasing temperature, as expected. The asymmetric shape of the curves is due to the fact that the lowest speed possible is zero while the upper classical limit of the speed is infinity. Quick Quiz 21.3 Consider the two curves in Figure 21.12. What is represented by the area under each of the curves between the 800-m/s and 1 000-m/s marks on the horizontal axis? QuickLab Fill one glass with very hot tap water and another with very cold water. Put a single drop of food coloring in each glass. Which drop disperses faster? Why? The evaporation process Equation 21.26 shows that the distribution of molecular speeds in a gas depends both on mass and on temperature. At a given temperature, the fraction of molecules with speeds exceeding a fixed value increases as the mass decreases. This explains why lighter molecules, such as H 2 and He, escape more readily from the Earth's atmosphere than do heavier molecules, such as N 2 and O2 . (See the discussion of escape speed in Chapter 14. Gas molecules escape even more readily from the Moon's surface than from the Earth's because the escape speed on the Moon is lower than that on the Earth.) The speed distribution curves for molecules in a liquid are similar to those shown in Figure 21.12. We can understand the phenomenon of evaporation of a liquid from this distribution in speeds, using the fact that some molecules in the liquid are more energetic than others. Some of the faster-moving molecules in the liquid penetrate the surface and leave the liquid even at temperatures well below the boiling point. The molecules that escape the liquid by evaporation are those that have sufficient energy to overcome the attractive forces of the molecules in the liquid phase. Consequently, the molecules left behind in the liquid phase have a lower average kinetic energy; as a result, the temperature of the liquid decreases. Hence, evaporation is a cooling process. For example, an alcoholsoaked cloth often is placed on a feverish head to cool and comfort a patient. 21.7 Mean Free Path 659 EXAMPLE 21.6 A System of Nine Particles v2 (5.00 2 8.00 2 12.0 2 12.0 2 12.0 2 14.0 2 14.0 2 17.0 2 20.0 2 ) m 9 178 m2/s2 Hence, the rms speed is v rms Nine particles have speeds of 5.00, 8.00, 12.0, 12.0, 12.0, 14.0, 14.0, 17.0, and 20.0 m/s. (a) Find the particles' average speed. The average speed is the sum of the speeds divided by the total number of particles: (5.00 v 8.00 12.0 12.0 12.0 14.0 14.0 17.0 20.0) m/s 9 Solution !v 2 !178 m2/s2 13.3 m/s (c) What is the most probable speed of the particles? 12.7 m/s (b) What is the rms speed? Solution Three of the particles have a speed of 12 m/s, two have a speed of 14 m/s, and the remaining have different speeds. Hence, we see that the most probable speed vmp is 12 m/s. Solution The average value of the square of the speed is Optional Section 21.7 MEAN FREE PATH Most of us are familiar with the fact that the strong odor associated with a gas such as ammonia may take a fraction of a minute to diffuse throughout a room. However, because average molecular speeds are typically several hundred meters per second at room temperature, we might expect a diffusion time much less than 1 s. But, as we saw in Quick Quiz 21.1, molecules collide with one other because they are not geometrical points. Therefore, they do not travel from one side of a room to the other in a straight line. Between collisions, the molecules move with constant speed along straight lines. The average distance between collisions is called the mean free path. The path of an individual molecule is random and resembles that shown in Figure 21.13. As we would expect from this description, the mean free path is related to the diameter of the molecules and the density of the gas. We now describe how to estimate the mean free path for a gas molecule. For this calculation, we assume that the molecules are spheres of diameter d. We see from Figure 21.14a that no two molecules collide unless their centers are less than a distance d apart as they approach each other. An equivalent way to describe the Figure 21.13 d Actual collision d (a) Equivalent collision A molecule moving through a gas collides with other molecules in a random fashion. This behavior is sometimes referred to as a random-walk process. The mean free path increases as the number of molecules per unit volume decreases. Note that the motion is not limited to the plane of the paper. 2d (b) Figure 21.14 (a) Two spherical molecules, each of diameter d, collide if their centers are within a distance d of each other. (b) The collision between the two molecules is equivalent to a point molecule's colliding with a molecule having an effective diameter of 2d. 660 CHAPTER 21 The Kinetic Theory of Gases 2d vt Figure 21.15 In a time t, a molecule of effective diameter 2d sweeps out a cylinder of length vt, where v is its average speed. In this time, it collides with every point molecule within this cylinder. collisions is to imagine that one of the molecules has a diameter 2d and that the rest are geometrical points (Fig. 21.14b). Let us choose the large molecule to be one moving with the average speed v. In a time t, this molecule travels a distance vt. In this time interval, the molecule sweeps out a cylinder having a cross-sectional area d 2 and a length vt (Fig. 21.15). Hence, the volume of the cylinder is d 2 vt. If nV is the number of molecules per unit volume, then the number of point-size molecules in the cylinder is ( d 2 vt)n V . The molecule of equivalent diameter 2d collides with every molecule in this cylinder in the time t. Hence, the number of collisions in the time t is equal to the number of molecules in the cylinder, ( d 2 vt)n V . The mean free path equals the average distance vt traveled in a time t divided by the number of collisions that occur in that time: vt 1 2 vt)n ( d d 2n V V Because the number of collisions in a time t is ( d 2 vt)n V , the number of collisions per unit time, or collision frequency f, is f d 2 vn V The inverse of the collision frequency is the average time between collisions, known as the mean free time. Our analysis has assumed that molecules in the cylinder are stationary. When the motion of these molecules is included in the calculation, the correct results are Mean free path 1 !2 d 2n V f !2 d 2 vn V v (21.30) (21.31) Collision frequency EXAMPLE 21.7 Bouncing Around in the Air This value is about 103 times greater than the molecular diameter. (b) On average, how frequently does one molecule collide with another? Approximate the air around you as a collection of nitrogen molecules, each of which has a diameter of 2.00 10 10 m. (a) How far does a typical molecule move before it collides with another molecule? Solution Assuming that the gas is ideal, we can use the equation PV Nk BT to obtain the number of molecules per unit volume under typical room conditions: nV N V 2.50 P k BT 1.01 on the first day of the month, when was the last one likely to have been ladled out of the pot? (c) The broth has been simmering for centuries, through wars, earthquakes, and stove repairs. Suppose that the water that was in the pot long ago has thoroughly mixed into the Earth's hydrosphere, of mass 1.32 1021 kg. How many of the water molecules originally in the pot are likely to be present in it again today? 66. Review Problem. (a) If it has enough kinetic energy, a molecule at the surface of the Earth can escape the Earth's gravitation. Using the principle of conservation of energy, show that the minimum kinetic energy needed for escape is mgR, where m is the mass of the molecule, g is the free-fall acceleration at the surface of the Earth, and R is the radius of the Earth. (b) Calculate the temperature for which the minimum escape kinetic energy is ten times the average kinetic energy of an oxygen molecule. 67. Using multiple laser beams, physicists have been able to cool and trap sodium atoms in a small region. In one experiment, the temperature of the atoms was reduced to 0.240 mK. (a) Determine the rms speed of the sodium atoms at this temperature. The atoms can be trapped for about 1.00 s. The trap has a linear dimension of roughly 1.00 cm. (b) Approximately how long would it take an atom to wander out of the trap region if there were no trapping action? 668 CHAPTER 21 The Kinetic Theory of Gases ANSWERS TO QUICK QUIZZES 21.1 Although a molecule moves very rapidly, it does not travel far before it collides with another molecule. The collision deflects the molecule from its original path. Eventually, a perfume molecule will make its way from one end of the room to the other, but the path it takes is much longer than the straight-line distance from the perfume bottle to your nose. 21.2 (c) E int stays the same. According to Equation 21.10, E int is a function of temperature only. Along an isotherm, T is constant by definition. Therefore, the internal energy of the gas does not change. 21.3 The area under each curve represents the number of molecules in that particular velocity range. The T 900 K curve has many more molecules moving between 800 m/s and 1000 m/s than does the T 300 K curve. 2.2 This is the Nearest One Head 669 P U Z Z L E R The purpose of a refrigerator is to keep its contents cool. Beyond the attendant increase in your electricity bill, there is another good reason you should not try to cool the kitchen on a hot day by leaving the refrigerator door open. What might this reason be? (Charles D. Winters) c h a p t e r Heat Engines, Entropy, and the Second Law of Thermodynamics Chapter Outline 22.1 Heat Engines and the Second Law of Thermodynamics 22.2 Reversible and Irreversible Processes 22.5 Heat Pumps and Refrigerators 22.6 Entropy 22.7 Entropy Changes in Irreversible Processes 22.3 The Carnot Engine 22.4 Gasoline and Diesel Engines 22.8 (Optional) Entropy on a Microscopic Scale 669 670 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics T he first law of thermodynamics, which we studied in Chapter 20, is a statement of conservation of energy, generalized to include internal energy. This law states that a change in internal energy in a system can occur as a result of energy transfer by heat or by work, or by both. As was stated in Chapter 20, the law makes no distinction between the results of heat and the results of work -- either heat or work can cause a change in internal energy. However, an important distinction between the two is not evident from the first law. One manifestation of this distinction is that it is impossible to convert internal energy completely to mechanical energy by taking a substance through a thermodynamic cycle such as in a heat engine, a device we study in this chapter. Although the first law of thermodynamics is very important, it makes no distinction between processes that occur spontaneously and those that do not. However, we find that only certain types of energy-conversion and energy-transfer processes actually take place. Th the base of the cylinder and does work WAB in raising the piston. 2. In process B : C (Fig. 22.9b), the base of the cylinder is replaced by a thermally nonconducting wall, and the gas expands adiabatically -- that is, no energy enters or leaves the system. During the expansion, the temperature of the gas decreases from Th to Tc and the gas does work WBC in raising the piston. 3. In process C : D (Fig. 22.9c), the gas is placed in thermal contact with an energy reservoir at temperature Tc and is compressed isothermally at temperature Tc . During this time, the gas expels energy Q c to the reservoir, and the work done by the piston on the gas is WCD . 4. In the final process D : A (Fig. 22.9d), the base of the cylinder is replaced by a nonconducting wall, and the gas is compressed adiabatically. The temperature of the gas increases to Th , and the work done by the piston on the gas is WDA . The net work done in this reversible, cyclic process is equal to the area enclosed by the path ABCDA in Figure 22.10. As we demonstrated in Section 22.1, because the change in internal energy is zero, the net work W done in one cycle equals the net energy transferred into the system, Q h Q c . The thermal efficiency of the engine is given by Equation 22.2: e W Qh Qh Qh Qc 1 Qc Qh P A Qh B W D Qc C Tc V Th Figure 22.10 PV diagram for the Carnot cycle. The net work done, W, equals the net energy received Q c . Note that in one cycle, Q h E int 0 for the cycle. In Example 22.2, we show that for a Carnot cycle Qc Qh Tc Th (22.3) Ratio of energies for a Carnot cycle Hence, the thermal efficiency of a Carnot engine is eC 1 Tc Th (22.4) Efficiency of a Carnot engine This result indicates that all Carnot engines operating between the same two temperatures have the same efficiency. 678 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics Equation 22.4 can be applied to any working substance operating in a Carnot cycle between two energy reservoirs. According to this equation, the efficiency is zero if Tc Th , as one would expect. The efficiency increases as Tc is lowered and as Th is raised. However, the efficiency can be unity (100%) only if Tc 0 K. Such reservoirs are not available; thus, the maximum efficiency is always less than 100%. In most practical cases, Tc is near room temperature, which is about 300 K. Therefore, one usually strives to increase the efficiency by raising Th . EXAMPLE 22.2 Efficiency of the Carnot Engine pression for P and substituting into (2), we obtain nRT V V which we can write as TV 1 Show that the efficiency of a heat engine operating in a Carnot cycle using an ideal gas is given by Equation 22.4. During the isothermal expansion (process A : B in Figure 22.9), the temperature does not change. Thus, the internal energy remains constant. The work done by a gas during an isothermal expansion is given by Equation 20.13. According to the first law, this work is equal to Q h , the energy absorbed, so that Qh W AB nRTh ln VB VA Solution constant constant where we have absorbed nR into the constant right-hand side. Applying this result to the adiabatic processes B : C and D : A, we obtain ThVB ThVA 1 1 TcVC TcVD 1 1 In a similar manner, the energy transferred to the cold reservoir during the isothermal compression C : D is Qc W CD V nRTc ln C VD Dividing the first equation by the second, we obtain (VB /VA ) (3) VB VA VC VD 1 (VC /VD ) 1 We take the absolute value of the work because we are defining all values of Q for a heat engine as positive, as mentioned earlier. Dividing the second expression by the first, we find that (1) Qc Qh Tc ln(VC /VD ) Th ln(VB /VA ) Substituting (3) into (1), we find that the logarithmic terms cancel, and we obtain the relationship Qc Qh Tc Th We now show that the ratio of the logarithmic quantities is unity by establishing a relationship between the ratio of volumes. For any quasi-static, adiabatic process, the pressure and volume are related by Equation 21.18: (2) PV constant Using this result and Equation 22.2, we see that the thermal efficiencye by an individual cylinder that has an initial volume of VA (2.00 10 3 m3 )/4 0.500 10 3 m3. Because the compression ratio is quite high, we approximate the maximum cylinder volume to be the displacement volume. Using the initial pressure PA 100 kPa and initial temperature TA 300 K, we can calculate the mass of the air in the cylinder using the ideal gas law: We find the temperature at D from the ideal gas law: TD PDVD mR 792 K Now that we have the temperatures at the beginning and the end of each process, we can calculate the net energy transfer by heat and the net work done by each cylinder every two cycles: Qh Qc W net Q in Q out Q in mc P(TC mc V (TD Q out TB ) TA ) 0.601 kJ 0.205 kJ (264 kPa)(0.500 10 3 m3 ) (5.81 10 4 kg)(0.287 kPa m3/kg K) 0.396 kJ The efficiency is e W net /Q in 66%. The net power for the four-cylinder engine operating at 3 000 rpm is net 4 1 2 rev (3 000 rev/min) (1 min/60 s) (0.396 kJ) 53 hp 39.6 kW Of course, modern engine design goes beyond this simple thermodynamic treatment, which uses idealized cycles. 684 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics 22.5 HEAT PUMPS AND REFRIGERATORS In Section 22.1 we introduced a heat pump as a mechanical device that moves energy from a region at lower temperature to a region at higher temperature. Heat pumps have long been used for cooling homes and buildings, and they are now becoming increasingly popular for heating them as well. The heat pump contains two sets of metal coils that can exchange energy by heat with the surroundings: one set on the outside of the building, in contact with the air or buried in the ground; and the other set in the interior of the building. In the heating mode, a circulating fluid flowing through the coils absorbs energy from the outside and releases it to the interior of the building from the interior coils. The fluid is cold and at low pressure when it is in the external coils, where it absorbs energy by heat from either the air or the ground. The resulting warm fluid is then compressed and enters the interior coils as a hot, high-pressure fluid, where it releases its stored energy to the interior air. An air conditioner is simply a heat pump operating in the cooling mode, with its exterior and interior coils interchanged. Energy is absorbed into the circulating fluid in the interior coils; then, after the fluid is compressed, energy leaves the fluid through the external coils. The air conditioner must have a way to release energy to the outside. Otherwise, the work done on the air conditioner would represent energy added to the air inside the house, and the temperature would increase. In the same manner, a refrigerator cannot cool the kitchen if the refrigerator door is left open. The amount of energy leaving the external coils (Fig. 22.14) behind or underneath the refrigerator is greater than the amount of energy removed from the food or from the air in the kitchen if the door is left open. The difference between the energy out and the energy in is the work done by the electricity supplied to the refrigerator. Figure 22.15 is a schematic representation of a heat pump. The cold temperature is Tc , the hot temperature is Th , and the energy absorbed by the circulating fluid is Q c . The heat pump does work W on the fluid, and the energy transferred from the pump to the building in the heating mode is Q h . The effectiveness of a heat pump is described in terms of a number called the coefficient of performance (COP). In the heating mode, the COP is defined as the ratio of the energy transferred to the hot reservoir to the work required to transfer that energy: COP (heating mode) Energy transferred at high temperature Work done by pump Qh W (22.6) Figure 22.14 The coils on the back of a refrigerator transfer energy by heat to the air. The second law of thermodynamics states that this amount of energy must be greater than the amount of energy removed from the contents of the refrigerator (or from the air in the kitchen, if the refrigerator door is left open). Note that the COP is transferred by heat when the system follows a reversible path between the states, then the change in entropy dS is equal to this amount of energy for the reversible process divided by the absolute temperature of the system: Clausius definition of change in entropy dS dQ r T (22.8) We have assumed that the temperature is constant because the process is infinitesimal. Since we have claimed that entropy is a state function, the change in entropy during a process depends only on the end points and therefore is independent of the actual path followed. The subscript r on the quantity dQ r is a reminder that the transferred energy is to be measured along a reversible path, even though the system may actually have followed some irreversible path. When energy is absorbed by the system, dQ r is positive and the entropy of the system increases. When energy is expelled by the system, dQ r is negative and the entropy of the system decreases. Note that Equation 22.8 defines not entropy but rather the change in entropy. Hence, the meaningful quantity in describing a process is the change in entropy. Entropy was originally formulated as a useful concept in thermodynamics; however, its importance grew tremendously as the field of statistical mechanics developed because the analytical techniques of statistical mechanics provide an alternative means of interpreting entropy. In statistical mechanics, the behavior of a substance is described in terms of the statistical behavior of its atoms and molecules. One of the main results of this treatment is that isolated systems tend toward disorder and that entropy is a measure of this disorder. For example, consider the molecules of a gas in the air in your room. If half of the gas molecules had velocity vectors of equal magnitude directed toward the left and the other half had velocity vectors of the same magnitude directed toward the right, the situation would be very ordered. However, such a situation is extremely unlikely. If you could actually view the molecules, you would see that they move haphazardly in all directions, bumping into one another, changing speed upon collision, some going fast and others going slowly. This situation is highly disordered. The cause of the tendency of an isolated system toward disorder is easily explained. To do so, we distinguish between microstates and macrostates of a system. A microstate is a particular description of the properties of the individual molecules of the system. For example, the description we just gave of the velocity vectors of the air molecules in your room being very ordered refers to a particular microstate, and the more likely likely haphazard motion is another microstate -- one that represents disorder. A macrostate is a description of the conditions of the system from a macroscopic point of view and makes use of macroscopic variables such as pressure, density, and temperature. For example, in both of the microstates described for the air molecules in your room, the air molecules are distributed uniformly throughout the volume of the room; this uniform density distribution is a macrostate. We could not distinguish between our two microstates by making a macroscopic measurement -- both microstates would appear to be the same macroscopically, and the two macrostates corresponding to these microstates are equivalent. For any given macrostate of the system, a number of microstates are possible, or accessible. Among these microstates, it is assumed that all are equally probable. However, when all possible microstates are examined, it is found that far more of them are disordered than are ordered. Because all of the microstates are equally 22.6 Entropy 687 probable, it is highly likely that the actual macrostate is one resulting from one of the highly disordered microstates, simply because there are many more of them. Similarly, the probability of a macrostate's forming from disordered microstates is greater than the probability of a macrostate's forming from ordered microstates. All physical processes that take place in a system tend Assuming that CV is constant over the interval in question, and integrating Equation 22.11 from the initial state to the final state, we obtain f S i dQ r T nCV ln Tf Ti nR ln Vf Vi (22.12) This expression demonstrates mathematically what we argued earlier -- that S depends only on the initial and final states and is independent of the path between the states. Also, note in Equation 22.12 that S can be positive or negative, depending on the values of the initial and final volumes and temperatures. Finally, for a cyclic process (Ti Tf and Vi Vf ), we see from Equation 22.12 that S 0. This is evidence that entropy is a state function. EXAMPLE 22.6 Change in Entropy -- Melting Making use of Equations 22.9 and that for the latent heat of fusion Q mL f (Eq. 20.6), we find that S dQ r T 1 Tm dQ Q Tm mL f Tm A solid that has a latent heat of fusion Lf melts at a temperature Tm . (a) Calculate the change in entropy of this substance when a mass m of the substance melts. Let us assume that the melting occurs so slowly that it can be considered a reversible process. In this case the temperature can be regarded as constant and equal to Tm . Solution 22.7 Entropy Changes in Irreversible Processes Note that we are able to remove Tm from the integral because the process is isothermal. Note also that S is positive. This means that when a solid melts, its entropy increases because the molecules are much more disordered in the liquid state than they are in the solid state. The positive value for S also means that the substance in its liquid state does not spontaneously transfer energy from itself to the surroundings and freeze because to do so would involve a spontaneous decrease in entropy. (b) Estimate the value of the change in entropy of an ice cube when it melts. 689 Solution Let us assume an ice tray makes cubes that are about 3 cm on a side. The volume per cube is then (very roughly) 30 cm3. This much liquid water has a mass of 30 g. From Table 20.2 we find that the latent heat of fusion of ice is 3.33 105 J/kg. Substituting these values into our answer for part (a), we find that S mL f Tm (0.03 kg)(3.33 273 K 10 5 J/kg) 4 10 1 J/K We retain only one significant figure, in keeping with the nature of our estimations. 22.7 ENTROPY CHANGES IN IRREVERSIBLE PROCESSES By definition, calculation of the change in entropy requires information about a reversible path connecting the initial and final equilibrium states. To calculate changes in entropy for real (irreversible) processes, we must remember that entropy (like internal energy) depends only on the state of the system. That is, entropy is a state function. Hence, the change in entropy when a system moves between any two equilibrium states depends only on the initial and final states. We can show that if this were not the case, the second law of thermodynamics would be violated. We now calculate the entropy change in some irreversible process between two equilibrium states by devising a reversible process (or series of reversible processes) between the same two states and computing S dQ r /T for the reversible process. In irreversible processes, it is critically important that we distinguish between Q , the actual energy transfer in the process, and Q r , the energy that would have been transferred by heat along a reversible path. Only Q r is the correct value to be used in calculating the entropy change. As we shall see in the following examples, the change in entropy for a system and its surroundings is always positive for an irreversible process. In general, the total entropy -- and therefore the disorder -- always increase in an irreversible process. Keeping these considerations in mind, we can state the second law of thermodynamics as follows: The total entropy of an isolated system that undergoes a change can never decrease. Furthermore, if the process is irreversible, then the total entropy of an isolated system always increases. In a reversible process, the total entropy of an isolated system remains constant. When dealing with a system that is not isolated from its surroundings, remember that the increase in entropy described in the second law is that of the system and its surroundings. When a system and its surroundings interact in an irreversible process, the increase in entropy of one is greater than the decrease in entropy of the other. Hence, we conclude that the change in entropy of the Universe must be greater than zero for an irreversible process and equal to zero for a reversible process. Ultimately, the entropy of the Universe should reach a maximum value. At this value, the Universe will be in a state of uniform temperature and density. All physical, chemical, and biological processes will cease because a state of perfect disorder implies that no energy is available for doing work. This gloomy state of affairs is sometimes referred to as the heat death of the Universe. 690 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics Quick Quiz 22.3 In the presence of sunlight, a tree rearranges an unorganized collection of carbon dioxide and water molecules into the highly ordered collection of molecules we see as leaves and branches. True or false: This reduction of entropy in the tree is a violation of the second law of thermodynamics. Explain your response. Entropy Change in Thermal Conduction Let us now consider a system consisting of a hot reservoir and a cold reservoir in thermal contact with each other and isolated from the rest of the Universe. A process occurs during which energy Q is transferred by heat from the hot reservoir at temperature Th to the cold reservoir at temperature Tc . Because the cold reservoir absorbs energy Q , its entropy increases by Q /Tc . At the same time, the hot reservoir loses energy Q , and so its entropy change is Q /Th . Because Th Tc , the increase in entropy of the cold reservoir is greater than the decrease in entropy of the hot reservoir. Therefore, the change in entropy of the system (and of the Universe) is greater than zero: SU Q Tc Q Th 0 EXAMPLE 22.7 Which Way Does the Energy Flow? that of our two-object system, which is SU Sc Sh 0.007 9 J/K A large, cold object is at 273 K, and a large, hot object is at 373 K. Show that it is impossible for a small amount of energy -- for example, 8.00 J -- to be transferred spontaneously from the cold object to the hot one without a decrease in the entropy of the Universe and therefore a violation of the second law. Solution We assume that, during the energy transfer, the two objects do not undergo a temperature change. This is not a necessary assumption; we make it only to avoid using integral calculus in our calculations. The process as described is irreversible, and so we must find an equivalent reversible process. It is sufficient to assume that the objects are connected by a poor thermal conductor whose temperature spans the range from 273 K to 373 K. This conductor transfers energy slowly, and its state does not change during the process. Under this assumption, the energy transfer to or from each object is reversible, and we may set Q Q r . The entropy change of the hot object is Sh Qr Th 8.00 J 373 K 0.021 4 J/K The cold object loses energy, and its entropy change is Sc Qr Tc 8.00 J 273 K 0.029 3 J/K This decrease in entropy of the Universe is in violation of the second law. That is, the spontaneous transfer of energy from a cold to a hot object cannot occur. In terms of disorder, let us consider the violation of the second law if energy were to continue to transfer spontaneously from a cold object to a hot object. Before the transfer, a certain degree of order is associated with the different temperatures of the objects. The hot object's molecules have a higher average energy than the cold object's molecules. If energy spontaneously flows from the cold object to the hot object, then, over a period of time, the cold object will become colder and the hot object will become hotter. The difference in average molecular energy will become even greater; this would represent an increase in order for the system and a violation of the second law. In comparison, the process that does occur naturally is the flow of energy from the hot object to the cold object. In this process, the difference in average molecular energy decreases; this represents a more random distribution of energy and an increase in disorder. Exercise Suppose that 8.00 J of energy is transferred from a hot object to a cold one. What is the net entropy change of the Universe? 0.007 9 J/K. We consider the two objects to be isolated from the rest of the Universe. Thus, the entropy change of the Universe is just Answer 22.7 Entropy Changes in Irreversible Processes 691 Entropy Change in a Free Expansion Let us again consider the adiabatic free expansion of a gas occupying an initial volume Vi (Fig. 22.16). A membrane separating the gas from an evacuated region is broken, and the gas expands (irreversibly) to a volume Vf . Let us find the changes in entropy of the gas and of the Universe during this process. The process is clearly neither reversible nor quasi-static. The work done by the gas against the vacuum is zero, and because the walls are insulating, no energy is transferred by heat during the expansion. That is, W 0 and Q 0. Using the first law, we see that the change in internal energy is zero. Because the gas is ideal, E int depends on temperature only, and we conclude that T 0 or Ti Tf . To apply Equation 22.9, we cannot use Q 0, the value for the irreversible process, but must instead find Q r ; that is, we must find an equivalent reversible path that shares the same initial and final states. A simple choice is an isothermal, reversible expansion in which the gas pushes slowly against a piston while energy enters the gas by heat from a reservoir to hold the temperature constant. Because T is constant in this process, Equation 22.9 gives f Insulating wall Vacuum Membrane Gas at Ti Figure 22.16 S i dQ r T 1 T f i dQ r f For an isothermal process, the first law of thermodynamics specifies that i dQ r is equal to the work done by the gas during the expansion from Vi to Vf , which is given by Equation 20.13. Using this result, we find that the entropy change for the gas is Adiabatic free expansion of a gas. When the membrane separating the gas from the evacuated region is ruptured, the gas expands freely and irreversibly. As a result, it occupies a greater final volume. The container is thermally insulated from its surroundings; thus, Q 0. S nR ln Vf Vi (22.13) Because Vf Vi , we conclude that S is positive. This positive result indicates that both the entropy and the disorder of the gas increase as a result of the irreversible, adiabatic expansion. Because the free expansion takes place in an insulated container, no energy is transferred by heat from the surroundings. (Remember that the isothermal, reversible expansion is only a replacement process that we use to calculate the entropy change for the gas; it is not the actual process.) Thus, the free expansion has no effect on the surroundings, and the entropy change of the surroundings is zero. Thus, the entropy change for the Universe is positive; this is consistent with the second law. EXAMPLE 22.8 Free Expansion of a Gas 18.3 J/K It is easy to see that the gas is more disordered after the expansion. Instead of being concentrated in a relatively small space, the molecules are scattered over a larger region. Calculate the change in entropy for a process in which 2.00 mol of an ideal gas undergoes a free expansion to three times its initial volume. Solution Vf S Using Equation 22.13 with n 3Vi , we find that nR ln Vf Vi 2.00 mol and (2.00 mol)(8.31 J/mol K) (ln 3) Entropy Change in Calorimetric Processes A substance of mass m 1 , specific heat c 1 , and initial temperature T1 is placed in thermal contact with a second substance of mass m 2 , specific heat c 2 , and initial 692 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics temperature T2 T1 . The two substances are contained in a calorimeter so that no energy is lost to the surroundings. The system of the two substances is allowed to reach therions involved in the free expansion. The instant after the partition is removed (and before the molecules have had a chance to rush into the other half of the container), all the molecules are in the initial volume. We assume that each molecule occupies some microscopic volume Vm . The total number of possible locations of a single molecule in a macroscopic initial volume Vi is the ratio wi Vi /Vm , which is a huge number. We use wi here to represent the number of ways that the molecule can be placed in the volume, or the number of microstates, which is equivalent to the number of available locations. We assume that the molecule's occupying each of these locations is equally probable. As more molecules are added to the system, the number of possible ways that the molecules can be positioned in the volume multiplies. For example, in considering two molecules, for every possible placement of the first, all possible placements of the second are available. Thus, there are w 1 ways of locating the first molecule, and for each of these, there are w 2 ways of locating the second molecule. The total number of ways of locating the two molecules is w 1 w 2 . Neglecting the very small probability of having two molecules occupy the same location, each molecule may go into any of the Vi /Vm locations, and so the number of ways of locating N molecules in the volume becomes W i wiN (Vi /Vm )N. (Wi is not to be confused with work.) Similarly, when the volume is increased to Vf , the number of ways of locating N molecules increases to W f wf N (Vf /Vm )N. The ratio of the number of ways of placing the molecules in the volume for the Vi Vacuum (a) Vf (b) Figure 22.17 In a free expansion, the gas is allowed to expand into a region that was previously a vacuum. 4 This section was adapted from A. Hudson and R. Nelson, University Physics, Philadelphia, Saunders College Publishing, 1990. 694 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics initial and final configurations is Wf Wi (Vf /Vm )N (Vi /Vm )N Vf Vi N If we now take the natural logarithm of this equation and multiply by Boltzmann's constant, we find that k B ln Wf Wi nNAk B ln Vf Vi where we have used the equality N nNA . We know from Equation 19.11 that NA k B is the universal gas constant R; thus, we can write this equation as k B ln W f k B ln W i nR ln Vf Vi (22.16) From Equation 22.13 we know that when n mol of a gas undergoes a free expansion from Vi to Vf , the change in entropy is Sf Si nR ln Vf Vi (22.17) Note that the right-hand sides of Equations 22.16 and 22.17 are identical. Thus, we make the following important connection between entropy and the number of microstates for a given macrostate: Entropy (microscopic definition) S k B ln W (22.18) The more microstates there are that correspond to a given macrostate, the greater is the entropy of that macrostate. As we have discussed previously, there are many more disordered microstates than ordered microstates. Thus, Equation 22.18 indicates mathematically that entropy is a measure of microscopic disorder. Although in our discussion we used the specific example of the free expansion of an ideal gas, a more rigorous development of the statistical interpretation of entropy would lead us to the same conclusion. Imagine the container of gas depicted in Figure 22.18a as having all of its molecules traveling at speeds greater than the mean value on the left side and all of its molecules traveling at speeds less than the mean value on the right side (an ordered microstate). Compare this with the uniform mixture of fast- and slow-mov- Faster molecules in this half Slower molecules in this half Nature tends toward this direction Fast and slow molecules intermixed (a) Ordered (b) Disordered Figure 22.18 A container of gas in two equally probable states of molecular motion. (a) An ordered arrangement, which is one of a few and therefore a collectively unlikely set. (b) A disordered arrangement, which is one of many and therefore a collectively likely set. 22.8 Entropy on a Microscopic Scale 695 Figure 22.19 By tossing a coin into a jar, the carnival-goer can win the fish in the jar. It is more likely that the coin will land in a jar containing a goldfish than in the one containing the black fish. ing molecules in Figure 22.18b (a disordered microstate). You might expect the ordered microstate to be very unlikely because random motions tend to mix the slow- and fast-moving molecules uniformly. Yet individually each of these microstates is equally probable. However, there are far more disordered microstates than ordered microstates, and so a macrostate corresponding to a large number of equivalent disordered microstates is much more probable than a macrostate corresponding to a small number of equivalent ordered microstates. Figure 22.19 shows a real-world example of this concept. There are two possible macrostates for the carnival game -- winning a goldfish and winning a black fish. Because only one jar in the array of jars contains a black fish, only one possible microstate corresponds to the macrostate of winning a black fish. A large number of microstates are described by the coin's falling into a jar containing a goldfish. Thus, for the macrostate of winning a goldfish, there are many equivalent microstates. As a result, the probability of winning a goldfish is much greater than the probability of winning a black fish. If there are 24 goldfish and 1 black fish, the probability of winning the black fish is 1 in 25. This assumes that all microstates have the same probability, a situation that may not be quite true for the situation shown in Figure 22.19. If you are an accurate coin tosser and you are aiming for the edge of the array of jars, then the probability of the coin's landing in a jar near the edge is likely to be greater than the probability of its landing in a jar near the center. Let us consider a similar type of probability problem for 100 molecules in a container. At any given moment, the probability of one molecule's being in the 1 left part of the container shown in Figure 22.20a as a result of random motion is 2. If there are two molecules, as shown in Figure 22.20b, the probability of both be1 ing in the left part is (2)2 or 1 in 4. If there are three molecules (Fig. 22.20c), the 1 probability of all of them being in the left portion at the same moment is (2)3, or 1 in 8. For 100 independently moving molecules, the probability that the 50 fastest 1 ones will be found in the left part at any moment is (2)50. Likewise, the probability that the remaining 50 slower molecules will be found in the right part at any moment is (1)50. Therefore, the probability of finding this fast-slow separation 2 as a result of random motion is the product (1 )50(1 )50 (1 )100, which corre2 2 2 sponds to about 1 in 1030. When this calculation is extrapolated from 100 molecules to the number in 1 mol of gas (6.02 1023), the ordered arrangement is found to be extremely improbable! QuickLab Roll a pair of dice 100 times and record the total number of spots appearing on the dice for each throw. Which total comes up most frequently? Is this expected? 696 CHAPTER 22 Heat Engines, Entropy, and the Second Law of Thermodynamics (a) (b) (c) Figure 22.20 (a) One molecule in a two-sided container has a 1-in-2 chance of being on the left side. (b) Two molecules have a 1-in-4 chance of being on the left side at the same time. (c) Three molecules have a 1-in-8 chance of being on the left side at the same time. EXAMPLE 22.10 Adiabatic Free Expansion -- One Last Time The number of microstates for all NA molecules in the final volume Vf 4Vi is Wf Vf Vm NA Let us verify that the macroscopic and microscopic approaches to the calculation of entropy lead to the same conclusion for the adiabatic free expansion of an ideal gas. Suppose that 1 mol of gas expands to four times its initial volume. As we have seen for this process, the initial and final temperatures are the same. (a) Using a macroscopic approach, calculate the entropy change for the gas. (b) Using statistical considerations, calculate the change in en . If P is far from the rod (a W ), then the in the denominator can be neglected, and E k eQ /a 2. This is just the form you would expect for a point charge. Therefore, at large values of a/ , the charge distribution appears to be a point charge of magnitude Q. The use of the limiting technique (a/ : ) often is a good method for checking a theoretical formula. y dq = dx dx x dE P a x Because every other element also produces a field in the negative x direction, the problem of summing their contributions is particularly simple in this case. The total field at P due to all segments of the rod, which are at different distances from P, is given by Equation 23.6, which in this case becomes3 a E a ke dx x2 Figure 23.16 where the limits on the integral extend from one end of the a). The constants ke and rod (x a) to the other (x can be removed from the integral to yield The electric field at P due to a uniformly charged rod lying along the x axis. The magnitude of the field at P due to the segment of charge dq is k e dq/x 2. The total field at P is the vector sum over all segments of the rod. EXAMPLE 23.8 The Electric Field of a Uniform Ring of Charge dE ke dq r2 A ring of radius a carries a uniformly distributed positive total charge Q . Calculate the electric field due to the ring at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring (Fig. 23.17a). Solution The magnitude of the electric field at P due to the segment of charge dq is 3 This field has an x component d E x d E cos along the axis and a component dE perpendicular to the axis. As we see in Figure 23.17b, however, the resultant field at P must lie along the x axis because the perpendicular components of all the It is important that you understand how to carry out integrations such as this. First, express the charge element dq in terms of the other variables in the integral (in this example, there is one variable, dx). The integral must be over scalar quantities; therefore, you x, and so we made the change dq must express the electric field in terms of components, if necessary. (In this example the field has only an x component, so we do not bother with this detail.) Then, reduce your expression to an integral over a single variable (or to multiple integrals, each over a single variable). In examples that have spherical or cylindrical symmetry, the single variable will be a radial coordinate. 23.5 Electric Field of a Continuous Charge Distribution various charge segments sum to zero. That is, the perpendicular component of the field created by any charge element is canceled by the perpendicular component created by an element on the opposite side of the ring. Because r (x 2 a 2 )1/2 and cos x/r, we find that d Ex d E cos ke dq r2 x r (x 2 kex dq a 2 )3/2 kex dq a 2 )3/2 kex Q a 2)3/2 kex a 2 )3/2 725 Ex (x 2 (x 2 dq (x 2 This result shows that the field is zero at x ing surprise you? 0. Does this find- All segments of the ring make the same contribution to the field at P because they are all equidistant from this point. Thus, we can integrate to obtain the total field at P : dq + + + Exercise Show that at great distances from the ring (x W a) the electric field along the axis shown in Figure 23.17 approaches that of a point charge of magnitude Q . 1 + + + + + + + + + + + r a + + + + + dE2 + + + + + x + P dEx + + + + 2 + + + dE (a) dE dE1 (b) Figure 23.17 A uniformly charged ring of radius a. (a) The field at P on the x axis due to an element of charge dq. (b) The total electric field at P is along the x axis. The perpendicular component of the field at P due to segment 1 is canceled by the perpendicular component due to segment 2. EXAMPLE 23.9 The Electric Field of a Uniformly Charged Disk butions of all rings making up the disk. By symmetry, the field at an axial point must be along the central axis. The ring of radius r and width dr shown in Figure 23.18 has a surface area equal to 2 r dr. The charge dq on this ring is equal to the area of the ring multiplied by trection at all points.) 17. A free electron and free proton are placed in an identical 733 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. electric field. Compare the electric forces on each particle. Compare their accelerations. Explain what happens to the magnitude of the electric field of a point charge as r approaches zero. A negative charge is placed in a region of space where the electric field is directed vertically upward. What is the direction of the electric force experienced by this charge? A charge 4q is a distance r from a charge q. Compare the number of electric field lines leaving the charge 4q with the number entering the charge q. In Figure 23.23, where do the extra lines leaving the charge 2q end? Consider two equal point charges separated by some distance d. At what point (other than ) would a third test charge experience no net force? A negative point charge q is placed at the point P near the positively charged ring shown in Figure 23.17. If x V a, describe the motion of the point charge if it is released from rest. Explain the differences between linear, surface, and volume charge densities, and give examples of when each would be used. If the electron in Figure 23.25 is projected into the electric field with an arbitrary velocity vi (at an angle to E), will its trajectory still be parabolic? Explain. It has been reported that in some instances people near where a lightning bolt strikes the Earth have had their clothes thrown off. Explain why this might happen. Why should a ground wire be connected to the metallic support rod for a television antenna? A light strip of aluminum foil is draped over a wooden rod. When a rod carrying a positive charge is brought close to the foil, the two parts of the foil stand apart. Why? What kind of charge is on the foil? Why is it more difficult to charge an object by rubbing on a humid day than on a dry day? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 23.1 Properties of Electric Charges Section 23.2 Insulators and Conductors Section 23.3 Coulomb's Law 1. (a) Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 10.0 g. Silver has 47 electrons per atom, and its molar mass is 107.87 g/mol. (b) Electrons are added to the pin until the net negative charge is 1.00 mC. How many electrons are added for every 109 electrons already present? 2. (a) Two protons in a molecule are separated by a distance of 3.80 10 10 m. Find the electric force exerted by one proton on the other. (b) How does the magnitude of this WEB force compare with the magnitude of the gravitational force between the two protons? (c) What must be the charge-to-mass ratio of a particle if the magnitude of the gravitational force between two of these particles equals the magnitude of the electric force between them? 3. Richard Feynman once said that if two persons stood at arm's length from each other and each person had 1% more electrons than protons, the force of repulsion between them would be enough to lift a "weight" equal to that of the entire Earth. Carry out an order-ofmagnitude calculation to substantiate this assertion. 4. Two small silver spheres, each with a mass of 10.0 g, are separated by 1.00 m. Calculate the fraction of the elec- 734 CHAPTER 23 Electric Fields 10. Review Problem. Two identical point charges each having charge q are fixed in space and separated by a distance d. A third point charge Q of mass m is free to move and lies initially at rest on a perpendicular bisector of the two fixed charges a distance x from the midpoint of the two fixed charges (Fig. P23.10). (a) Show that if x is small compared with d, the motion of Q is simple harmonic along the perpendicular bisector. Determine the period of that motion. (b) How fast will the charge Q be moving when it is at C. Find the magnitude and direction of the electric field at O, the center of the semicircle. 34. (a) Consider a uniformly charged right circular cylindrical shell having total charge Q , radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder, as shown in Figure P23.34. (Hint: Use the result of Example 23.8 and treat the cylinder as a collection of ring charges.) (b) Consider now a solid cylinder with the same dimensions and O Figure P23.33 h R d dx Figure P23.34 carrying the same charge, which is uniformly distributed through its volume. Use the result of Example 23.9 to find the field it creates at the same point. 35. A thin rod of length and uniform charge per unit length lies along the x axis, as shown in Figure P23.35. (a) Show that the electric field at P, a distance y from the rod, along the perpendicular bisector has no x component and is given by E 2k e sin 0/y. (b) Using your result to part (a), show that the field of a rod of infinite length is E 2k e /y. (Hint: First calculate the field at P due to an element of length dx, which has a charge dx. Then change variables from x to , using the facts that x y tan and dx y sec2 d , and integrate over .) y P 0 y x O dx Figure P23.35 36. Three solid plastic cylinders all have a radius of 2.50 cm and a length of 6.00 cm. One (a) carries charge with Problems uniform density 15.0 nC/m2 everywhere on its surface. Another (b) carries charge with the same uniform density on its curved lateral surface only. The third (c) carries charge with uniform density 500 nC/m3 throughout the plastic. Find the charge of each cylinder. 37. Eight solid plastic cubes, each 3.00 cm on each edge, are glued together to form each one of the objects (i, ii, iii, and iv) shown in Figure P23.37. (a) If each object carries charge with a uniform density of 400 nC/m3 throughout its volume, what is the charge of each object? (b) If each object is given charge with a uniform density of 15.0 nC/m2 everywhere on its exposed surface, what is the charge on each object? (c) If charge is placed only on the edges where perpendicular surfaces meet, with a uniform density of 80.0 pC/m, what is the charge of each object? 737 Section 23.7 Motion of Charged Particles in a Uniform Electric Field 41. An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48.0 ns after being released. 42. A proton is projected in the positive x direction into a region of uniform electric field E 6.00 10 5 i N/C. The proton travels 7.00 cm before coming to rest. Determine (a) the acceleration of the proton, (b) its initial speed, and (c) the time it takes the proton to come to rest. 43. A proton accelerates from rest in a uniform electric field of 640 N/C. At some later time, its speed has reached 1.20 106 m/s (nonrelativistic, since v is much less than the speed of light). (a) Find the acceleration of the proton. (b) How long does it take the proton to reach this speed? (c) How far has it moved in this time? (d) What is its kinetic energy at this time? 44. The electrons in a particle beam each have a kinetic energy of 1.60 10 17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm? 45. The electrons in a particle beam each have a kinetic energy K. What are the magnitude and direction of the electric field that stops these electrons in a distance d ? 46. A positively charged bead having a mass of 1.00 g falls from rest in a vacuum from a height of 5.00 m in a uniform vertical electric field with a magnitude of 1.00 104 N/C. The bead hits the ground at a speed of 21.0 m/s. Determine (a) the direction of the electric field (up or down) and (b) the charge on the bead. 47. A proton moves at 4.50 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 103 N/C. Ignoring any gravitational effects, find (a) the time it takes the proton to travel 5.00 cm horizontally, (b) its vertical displacement af4.8 A point charge located outside a closed surface. The number of lines entering the surface equals the number leaving the surface. The net electric flux through a closed surface is zero if there is no charge inside Note from Equation 24.5 that the net flux through the spherical surface is proportional to the charge inside. The flux is independent of the radius r because the area of the spherical surface is proportional to r 2, whereas the electric field is proportional to 1/r 2. Thus, in the product of area and electric field, the dependence on r cancels. Now consider several closed surfaces surrounding a charge q, as shown in Figure 24.7. Surface S 1 is spherical, but surfaces S 2 and S 3 are not. From Equation 24.5, the flux that passes through S 1 has the value q/ 0 . As we discussed in the previous section, flux is proportional to the number of electric field lines passing through a surface. The construction shown in Figure 24.7 shows that the number of lines through S 1 is equal to the number of lines through the nonspherical surfaces S 2 and S 3 . Therefore, we conclude that the net flux through any closed surface is independent of the shape of that surface. The net flux through any closed surface surrounding a point charge q is given by q/ 0 . Now consider a point charge located outside a closed surface of arbitrary shape, as shown in Figure 24.8. As you can see from this construction, any electric field line that enters the surface leaves the surface at another point. The number of electric field lines entering the surface equals the number leaving the surface. Therefore, we conclude that the net electric flux through a closed surface that surrounds no charge is zero. If we apply this result to Example 24.2, we can easily see that the net flux through the cube is zero because there is no charge inside the cube. Quick Quiz 24.1 Suppose that the charge in Example 24.1 is just outside the sphere, 1.01 m from its center. What is the total flux through the sphere? Let us extend these arguments to two generalized cases: (1) that of many point charges and (2) that of a continuous distribution of charge. We once again use the superposition principle, which states that the electric field due to many charges is the vector sum of the electric fields produced by the individual charges. Therefore, we can express the flux through any closed surface as E dA (E 1 E2 ) dA where E is the total electric field at any point on the surface produced by the vector addition of the electric fields at that point due to the individual charges. 24.2 Gauss's Law 749 Consider the system of charges shown in Figure 24.9. The surface S surrounds only one charge, q1 ; hence, the net flux through S is q1 / 0 . The flux through S due to charges q 2 and q 3 outside it is zero because each electric field line that enters S at one point leaves it at another. The surface S surrounds charges q 2 and q 3 ; hence, the net flux through it is (q 2 q 3 )/ 0. Finally, the net flux through surface S is zero because there is no charge inside this surface. That is, all the electric field lines that enter S at one point leave at another. Gauss's law, which is a generalization of what we have just described, states that the net flux through any closed surface is E dA q in 0 E (24.6) Gauss's law where q in represents the net charge inside the surface and E represents the electric field at any point on the surface. A formal proof of Gauss's law is presented in Section 24.6. When using Equation 24.6, you should note that although the charge q in is the net charge inside the gaussian surface, E represents the total electric field, which includes contributions from charges both inside and outside the surface. In principle, Gauss's law can be solved for E to determine the electric field due to a system of charges or a continuous distribution of charge. In practice, however, this type of solution is applicable only in a limited number of highly symmetric situations. As we shall see in the next section, Gauss's law can be used to evaluate the electric field for charge distributions that have spherical, cylindrical, or planar symmetry. If one chooses the gaussian surface surrounding the charge distribution carefully, the integral in Equation 24.6 can be simplified. You should also note that a gaussian surface is a mathematical construction and need not coincide with any real physical surface. Gauss's law is useful for evaluating E when the charge distribution has high symmetry S q2 q1 q3 S S Quick Quiz 24.2 For a gaussian surface through which the net flux is zero, the following four statements could be true. Which of the statements must be true? (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero everywhere on the surface. (d) The number of electric field lines entering the surface equals the number leaving the surface. The net electric flux through any closed surface depends only on the charge inside that surface. The net flux through surface S is q 1 / 0 , the net flux through surface S is (q 2 q 3 )/ 0 , and the net flux through surface S is zero. Figure 24.9 CONCEPTUAL EXAMPLE 24.3 A spherical gaussian surface surrounds a point charge q. Describe what happens to the total flux through the surface if (a) the charge is tripled, (b) the radius of the sphere is doubled, (c) the surface is changed to a cube, and (d) the charge is moved to another location inside the surface. lines from the charge pass through the sphere, regardless of its radius. (c) The flux does not change when the shape of the gaussian surface changes because all electric field lines from the charge pass through the surface, regardless of its shape. (d) The flux does not change when the charge is moved to another location inside that surface because Gauss's law refers to the total charge enclosed, regardless of where the charge is located inside the surface. Solution (a) The flux through the surface is tripled because flux is proportional to the amount of charge inside the surface. (b) The flux does not change because all electric field 750 CHAPTER 24 Gauss's Law 24.3 APPLICATION OF GAUSS'S LAW TO CHARGED INSULATORS As mentioned earlier, Gauss's law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry. The following examples demonstrate ways of choosing the gaussian surface over which the surface integral given by Equation 24.6 can be simplified and the electric field determined. In choosing the surface, we should always take advantage of the symmetry of the charge distribution so that we can remove E from the integral and solve for it. The goal in this type of calculation is to determine a surface that satisfies one or more of the following conditions: 1. The value of the electric field can be argued by symmetry to be constant over the surface. 2. The dot product in Equation 24.6 can be expressed as a simple algebraic product E dA because E and dA are parallel. 3. The dot product in Equation 24.6 is zero because E and dA are perpendicular. 4. The field can be argued to be zero over the surface. All four of these conditions are used in examples throughout the remainder of this chapter. EXAMPLE 24.4 The Electric Field Due to a Point Charge where we have used the fact that the surface area of a sphere is 4 r 2. Now, we solve for the electric field: E q 4 2 0r Starting with Gauss's law, calculate the electric field due to an isolated point charge q. A single charge represents the simplest possible charge distribution, and we use this familiar case to show how to solve for the electric field with Gauss's law. We choose a spherical gaussian surface of radius r centered on the point charge, as shown in Figure 24.10. The electric field due to a positive point charge is directed radially outward by symmetry and is therefore normal to the surface at every point. Thus, as in condition (2), E is parallel to d A at each point. Therefore, E d A E dA and Gauss's law gives E Solution ke q r2 This is the familiar electric field due to a point charge that we developed from Coulomb's law in Chapter 23. Gaussian surface r + q dA E E dA E dA q 0 By symmetry, E is constant everywhere on the surface, which satisfies condition (1), so it can be removed from the integral. Therefore, E dA E dA E(4 r 2 ) q 0 Figure 24.10 The point charge q is at the center of the spherical gaussian surface, and E is parallel to d A at every point on the surface. EXAMPLE 24.5 11.6 A Spherically Symmetric Charge Distribution An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q (Fig. 24.11). (a) Calculate the magnitude of the electric field at a point outside the sphere. Solution Because the charge distribution is spherically symmetric, we again select a spherical gaussian surface of radius r, concentric with the sphere, as shown in Figure 24.11a. For this choice, conditions (1) and (2) are satisfied, as they 24.3 Application of Gauss's Law to Charged Insulators were for the point charge in Example 24.4. Following the line of reasoning given in Example 24.4, we find that E ke Q r2 (for r a) 751 to the surface at each point -- both conditions (1) and (2) are satisfied. Therefore, Gauss's law in the region r a gives E dA Solving for E gives E dA E(4 r 2 ) q in 0 Note that this result is identical to the one we obtained for a point charge. Therefore, we conclude that, for a uniformly charged sphere, the field in the region external to the sphere is equivalent to that of a point charge located at the center of the sphere. (b) Find the magnitude of the electric field at a point inside the sphere. E q in 4 0r 2 4 3 4 r3 2 0r 3 r 0 Because Q / 4 a 3 by definition and since k e 3 this expression for E can be written as E Qr 3 0a k eQ r a3 (for r a) 1/(4 0 ), 4 Solution In this case we select a spherical gaussian surface having radius r a, concentric with the insulated sphere (Fig. 24.11b). Let us denote the volume of this smaller sphere by V . To apply Gauss's law in this situation, it is important to recognize that the charge q in within the gaussian surface of volume V is less than Q . To calculate q in , we use the fact that q in V : q in V (4 r 3 ) 3 By symmetry, the magnitude of the electric field is constant everywhere on the spherical gaussian surface and is normal Note that this result for E differs from the one we obtained in part (a). It shows that E : 0 as r : 0. Therefore, the result eliminates the problem that would exist at r 0 if E varied as 1/r 2 inside the sphere as it does outside the sphere. That is, if E 1/r 2 for r a, the field would be infinite at r 0, which is physically impossible. Note also that the expressions for parts (a) and (b) match when r a. A plot of E versus r is shown in Figure 24.12. a r a r Gaussian sphere a E E= k eQ r2 r (a) Gaussian sphere (b) a Figure 24.11 A uniformly charged insulating sphere of radius a and total charge Q. (a) The magnitude of the electric field at a point exterior to the sphere is k e Q /r 2. (b) The magnitude of the electric field inside the insulating sphere is due only to the charge within the gaussian sphere defined by the dashed circle and is k e Qr /a 3. A plot of E versus r for a uniformly charged insulating sphere. The electric field inside the sphere (r a) varies linearly with r. The field outside the sphere (r a) is the same as that of a point charge Q located at r 0. Figure 24.12 EXAMPLE 24.6 The Electric Field Due to a Thin Spherical Shell the shell is equivalent to that due to a point charge Q located at the center: E ke Q r2 (for r a) A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface (Fig. 24.13a). Find the electric field at points (a) outside and (b) inside the shell. Solution (a) The calculation for the field outside the shell is identical to that for the solid sphere shown in Example 24.5a. If we construct a spherical gaussian surface of radius r a concentric with the shell (Fig. 24.13b), the charge inside this surface is Q . Therefore, the field at a point outside (b) The electric field inside the spherical shell is zero. This follows from Gauss's law applied to a spherical surface of radius r a concentric with the shell (Fig. 24.13c). Because 752 CHAPTER 24 Gauss's Law We obtain the same results using Equation 23.6 and integrating over the charge distribution. This calculation is rather complicated. Gauss's law allows us to determine these results in a much simpler way. Gaussian surface r + + a + + + + + + + + a of the spherical symmetry of the charge distribution and because the net charge inside the surface is zero -- satisfaction of conditions (1) and (2) again -- application of Gauss's law shows that E 0 in the region r a. Gaussian surface + + + Ein = 0 + + + + a + E + + + + + + r + + + + + + (a) (b) (c) Figure 24.13 (a) The electric field inside a uniformly charged spherical shell is zero. The field outside is the same as that due to a point charge Q located at the center of the shell. (b) Gaussian surface for r a. (c) Gaussian surface for r a. EXAMPLE 24.7 11.7 A Cylindrically Symmetric Charge Distribution Gaussian surface + + + Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length (Fig. 24.14a). The symmetry of the charge distribution requires that E be perpendicular to the line charge and directed outward, as shown in Figure 24.14a and b. To reflect the symmetry of the charge distribution, we select a cylindrical gaussian surface of radius r and length that is coaxial with the line charge. For the curved part of this surface, E is constant in magnitude and perpendicular to the surface at each point -- satisfaction of conditions (1) and (2). Furthermore, the flux through the ends of the gaussian cylinder is zero because E is parallel to these surfaces -- the first application we have seen of condition (3). We take the surface integral in Gauss's law over the entire gaussian surface. Because of the zero value of E d A for the ends of the cylinder, however, we can restrict our attention to only the curved surface of the cylinder. The total charge inside our gaussian surface is . Applying Gauss's law and conditions (1) and (2), we find that for the curved surface E Solution r E dA + + + (a) E dA E dA EA q in 0 0 Figure 24.14 (a) An infinite line of charge surrounded by a cylindrical gaussian surface concentric with the line. (b) An end view shows that the electric field at the cylindrical surface is constant in magnitude and perpendicular to the surface. (b) + E + + + + + 24.3 Application of Gauss's Law to Charged Insulators The area of the curved surface is A E(2 r ) 0 753 2 r ; therefore, E 2 0r 2k e r (24.7) Thus, we see that the electric field due to a cylindrically symmetric charge distribution varies as 1/r, whereas the field external to a spherically symmetric charge distribution varies as 1/r 2. Equation 24.7 was also derived in Chapter 23 (see Problem 35[b]), by integration of the field of a point charge. If the line charge in this example were of finite length, the result for E would not be that given by Equation 24.7. A finite line charge does not possess sufficient symmetry for us to make use of Gauss's law. This is because the magnitude of the electric field is no longer constant over the surface of the gaussian cylinder -- the field near the ends of the line would be different from that far from the ends. Thus, condition (1) would not be satisfied in this situation. Furthermore, E is not perpendicular to the cylindrical surface at all points -- the field vectors near the ends would have a component parallel to the line. Thus, condition (2) would not be satisfied. When there is insufficient symmetry in the charge distribution, as in this situation, it is necessary to use Equation 23.6 to calculate E. For points close to a finite line charge and far from the ends, Equation 24.7 gives a good approximation of the value of the field. It is left for you to show (see Problem 29) that the electric field inside a uniformly charged rod of finite radius and infinite length is proportional to r. EXAMPLE 24.8 A Nonconducting Plane of Charge Because the distance from each flat end of the cylinder to the plane does not appear in Equation 24.8, we conclude that E /2 0 at any distance from the plane. That is, the field is uniform everywhere. An important charge configuration related to this example consists of two parallel planes, one positively charged and the other negatively charged, and each with a surface charge density (see Problem 58). In this situation, the electric fields due to the two planes add in the region between the planes, resulting in a field of magnitude / 0 , and cancel elsewhere to give a field of zero. + + + + + + + A Find the electric field due to a nonconducting, infinite plane of positive charge with uniform surface charge density . Solution By symmetry, E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane. The fact that the direction of E is away from positive charges indicates that the direction of E on one side of the plane must be opposite its direction on the other side, as shown in Figure 24.15. A gaussian surface that reflects the symmetry is a small cylinder whose axis is perpendicular to the plane and whose ends each have an area A and are equidistant from the plane. Because E is parallel to the curved surface -- and, therefore, perpendicular to d A everywhere on the surface -- condition (3) is satisfied and there is no contribution to the surface integral from this surface. For the flat ends of the cylinder, conditions (1) and (2) are satisfied. The flux through each end of the cylinder is EA; hence, the total flux through the entire gaussian surface is just that through the ends, E 2EA. Noting that the total charge inside the surface is q in A, we use Gauss's law and find that E + E + + + + + + + + + + + + + + + + + + + + + + + + + + + + E 2EA q in 0 A 0 Gaussian cylinder E 2 (24.8) 0 Figure 24.15 A cylindrical gaussian surface penetrating an infinite plane of charge. The flux is EA through each end of the gaussian surface and zero through its curved surface. CONCEPTUAL EXAMPLE 24.9 Explain why Gauss's law cannot be used to calculate the electric field near an electric dipole, a charged disk, or a triangle with a point charge at each corner. 754 CHAPTER 24 Gauss's Law Solution The charge distributions of all these configurations do not have sufficient symmetry to make the use of Gauss's law practical. We cannot find a closed surface surrounding any of these distributions that satisfies one or more of conditions (1) through (4) listed at the beginning of this section. 24.4 CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM As we learned in Section 23.2, a good electrical conductor contains charges (electrons) that are not bound to any atom and therefore are free to move about within the material. When there is no net motion of charge within a conductor, the conductor is in electrostatic equilibrium. As we shall see, a conductor in electrostatic equilibrium has the following properties: 1. The electric field is zero everywhere inside the conductor. 2. If an isolated conductor carries a charge, the charge resides on its surface. 3. The electric field just outside a charged conductor is perpendicular to the surface of the conductor and has a magnitude / 0 , where is the surface charge density at that point. 4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest. We verify the first three properties in the discussion that follows. The fourth property is presented here without further discussion so that we have a complete list of properties for conductors in electrostatic equilibrium. We can understand the first property by considering a conducting slab placed in an external field E (Fig. 24.16). We can argue that the electric field inside the conductor must be zero under the assumption that we have electrostatic equilibrium. If the field were not zero, free charges in the conductor would accelerate underd. QuickLab Wrap a radio or cordless telephone in aluminum foil and see if it still works. Does it matter if the foil touches the antenna? 2 The experiment is often referred to as Faraday's ice-pail experiment because Faraday, the first to perform it, used an ice pail for the hollow conductor. 758 CHAPTER 24 Gauss's Law Optional Section 24.6 FORMAL DERIVATION OF GAUSS'S LAW One way of deriving Gauss's law involves solid angles. Consider a spherical surface of radius r containing an area element A. The solid angle (uppercase Greek omega) subtended at the center of the sphere by this element is defined to be A r2 From this equation, we see that has no dimensions because A and r 2 both have dimensions L2. The dimensionless unit of a solid angle is the steradian. (You may want to compare this equation to Equation 10.1b, the definition of the radian.) Because the surface area of a sphere is 4 r 2, the total solid angle subtended by the sphere is 4 r2 4 steradians r2 Now consider a point charge q surrounded by a closed surface of arbitrary shape (Fig. 24.21). The total electric flux through this surface can be obtained by evaluating E A for each small area element A and summing over all elements. The flux through each element is A cos E A E A cos keq E r2 where r is the distance from the charge to the area element, is the angle between the electric field E and A for the element, and E k e q /r 2 for a point charge. In Figure 24.22, we see that the projection of the area element perpendicular to the radius vector is A cos . Thus, the quantity A cos /r 2 is equal to the solid angle that the surface element A subtends at the charge q. We also see that is equal to the solid angle subtended by the area element of a spherical surface of radius r. Because the total solid angle at a point is 4 steradians, the total flux E A q Figure 24.21 A closed surface of arbitrary shape surrounds a point charge q. The net electric flux through the surface is independent of the shape of the surface. A r A cos A E q Figure 24.22 charge q. The area element A subtends a solid angle ( A cos )/r 2 at the Summary 759 through the closed surface is E keq dA cos r2 keq d 4 keq q 0 Thus we have derived Gauss's law, Equation 24.6. Note that this result is independent of the shape of the closed surface and independent of the position of the charge within the surface. SUMMARY Electric flux is proportional to the number of electric field lines that penetrate a surface. If the electric field is uniform and makes an angle with the normal to a surface of area A, the electric flux through the surface is E EA cos (24.2) In general, the electric flux through a surface is E surface E dA (24.3) You need to be able to apply Equations 24.2 and 24.3 in a variety of situations, particularly those in which symmetry simplifies the calculation. Gauss's law says that the net electric flux E through any closed gaussian surface is equal to the net charge inside the surface divided by 0 : E E dA q in 0 (24.6) Using Gauss's law, you can calculate the electric field due to various symmetric charge distributions. Table 24.1 lists some typical results. TABLE 24.1 Typical Electric Field Calculations Using Gauss's Law Charge Distribution Insulating sphere of radius R, uniform charge density, and total charge Q Electric Field Q r2 Q r R3 Q r2 0 Line charge of infinite length and charge per unit length Nonconducting, infinite charged plane having surface charge density Conductor having surface charge density 0 2k e r Location ke ke r r R R Thin spherical shell of radius R and total charge Q ke r r R R Outside the line Everywhere outside the plane Just outside the conductor Inside the conductor 2 0 0 760 CHAPTER 24 Gauss's Law A conductor in electrostatic equilibrium has the following properties: 1. The electric field is zero everywhere inside the conductor. 2. Any net charge on the conductor resides entirely on its surface. 3. The electric field just outside the conductor is perpendicular to its surface and has a magnituniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Use approximate relationships to find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell. 29. Consider a long cylindrical charge distribution of radius R with a uniform charge density . Find the electric field at distance r from the axis where r R. L L Figure P24.19 Problems 19 and 20. 21. Consider an infinitely long line charge having uniform charge per unit length . Determine the total electric flux through a closed right circular cylinder of length L and radius R that is parallel to the line charge, if the distance between the axis of the cylinder and the line charge is d. (Hint: Consider both cases: when R d, and when R d.) 22. A 10.0- C charge located at the origin of a cartesian coordinate system is surrounded by a nonconducting hollow sphere of radius 10.0 cm. A drill with a radius of 1.00 mm is aligned along the z axis, and a hole is drilled in the sphere. Calculate the electric flux through the hole. WEB 764 CHAPTER 24 Gauss's Law the same charge is spread uniformly over the upper surface of an otherwise identical glass plate, compare the electric fields just above the center of the upper surface of each plate. A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find (a) the charge density of each face of the plate and (b) the total charge on each face. A hollow conducting sphere is surrounded by a larger concentric, spherical, conducting shell. The inner sphere has a charge Q , and the outer sphere has a charge 3Q. The charges are in electrostatic equilibrium. Using Gauss's law, find the charges and the electric fields everywhere. Two identical conducting spheres each having a radius of 0.500 cm are connected by a light 2.00-m-long conducting wire. Determine the tension in the wire if 60.0 C is placed on one of the conductors. (Hint: Assume that the surface distribution of charge on each sphere is uniform.) The electric field on the surface of an irregularly shaped conductor varies from 56.0 kN/C to 28.0 kN/C. Calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is (a) greatest and (b) smallest. A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of , and the cylinder has a net charge per unit length of 2 . From this information, use Gauss's law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis. A conducting spherical shell of radius 15.0 cm carries a net charge of 6.40 C uniformly distributed on its surface. Find the electric field at points (a) just outside the shell and (b) inside the shell. A thin conducting plate 50.0 cm on a side lies in the xy plane. If a total charge of 4.00 10 8 C is placed on the plate, find (a) the charge density on the plate, (b) the electric field just above the plate, and (c) the electric field just below the plate. A conducting spherical shell having an inner radius of a and an outer radius of b carries a net charge Q . If a point charge q is placed at the center of this shell, determine the surface charge density on (a) the inner surface of the shell and (b) the outer surface of the shell. A solid conducting sphere of radius 2.00 cm has a charge 8.00 C. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge 4.00 C. Find the electric field at (a) r 1.00 cm, (b) r 3.00 cm, (c) r 4.50 cm, and (d) r 7.00 cm from the center of this charge configuration. 30. A nonconducting wall carries a uniform charge density of 8.60 C/cm2. What is the electniform field. First, consider a uniform electric field directed along the negative y axis, as shown in Figure 25.1a. Let us calculate the potential difference between two points A and B separated by a distance d, where d is measured parallel to the field lines. Equation 25.3 gives B B B VB VA V A E ds A E cos 0 ds A E ds Because E is constant, we can remove it from the integral sign; this gives B V E A ds Ed (25.6) Potential difference in a uniform electric field The minus sign indicates that point B is at a lower electric potential than point A; that is, VB VA . Electric field lines always point in the direction of decreasing electric potential, as shown in Figure 25.1a. Now suppose that a test charge q 0 moves from A to B. We can calculate the change in its potential energy from Equations 25.3 and 25.6: U q0 V q 0 Ed (25.7) A A d q B B m d E g (a) (b) Figure 25.1 (a) When the electric field E is directed downward, point B is at a lower electric potential than point A. A positive test charge that moves from point A to point B loses electric potential energy. (b) A mass m moving downward in the direction of the gravitational field g loses gravitational potential energy. 772 CHAPTER 25 Electric Potential QuickLab It takes an electric field of about 30 000 V/cm to cause a spark in dry air. Shuffle across a rug and reach toward a doorknob. By estimating the length of the spark, determine the electric potential difference between your finger and the doorknob after shuffling your feet but before touching the knob. (If it is very humid on the day you attempt this, it may not work. Why?) E B From this result, we see that if q 0 is positive, then U is negative. We conclude that a positive charge loses electric potential energy when it moves in the direction of the electric field. This means that an electric field does work on a positive charge when the charge moves in the direction of the electric field. (This is analogous to the work done by the gravitational field on a falling mass, as shown in Figure 25.1b.) If a positive test charge is released from rest in this electric field, it experiences an electric force q 0 E in the direction of E (downward in Fig. 25.1a). Therefore, it accelerates downward, gaining kinetic energy. As the charged particle gains kinetic energy, it loses an equal amount of potential energy. If q 0 is negative, then U is positive and the situation is reversed: A negative charge gains electric potential energy when it moves in the direction of the electric field. If a negative charge is released from rest in the field E, it accelerates in a direction opposite the direction of the field. Now consider the more general case of a charged particle that is free to move between any two points in a uniform electric field directed along the x axis, as shown in Figure 25.2. (In this situation, the charge is not being moved by an external agent as before.) If s represents the displacement vector between points A and B, Equation 25.3 gives B B s A C V A E ds E A ds E s (25.8) A uniform electric field directed along the positive x axis. Point B is at a lower electric potential than point A. Points B and C are at the same electric potential. Figure 25.2 where again we are able to remove E from the integral because it is constant. The change in potential energy of the charge is U 11.9 q0 V q0 E s (25.9) An equipotential surface Finally, we conclude from Equation 25.8 that all points in a plane perpendicular to a uniform electric field are at the same electric potential. We can see this in Figure 25.2, where the potential difference VB VA is equal to the potential difference VC VA . (Prove this to yourself by working out the dot product E s for s A:B , where the angle between E and s is arbitrary as shown in Figure 25.2, and the dot product for s A:C , where 0.) Therefore, VB VC . The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential. Note that because U q 0 V, no work is done in moving a tent located a distance r from the charge, we begin with the general expression for potential difference: B VB B VA E ds A dr ds where A and B are the two arbitrary points shown in Figure 25.6. At any field r r point, the electric field due to the point charge is E k e q ^/r 2 (Eq. 23.4), where ^ is a unit vector directed from the charge toward the field point. The quantity E ds can be expressed as E ds ke q ^ ds r r2 r A rA rB ^ r q r r Because the magnitude of ^ is 1, the dot product ^ d s ds cos , where is r the angle between ^ and d s. Furthermore, d s cos is the projection of d s onto r; thus, ds cos dr. That is, any displacement d s along the path from point A to point B produces a change dr in the magnitude of r, the radial distance to the charge creating the field. Making these substitutions, we find that E ds (k e q/r 2 )dr; hence, the expression for the potential difference becomes rB The potential difference between points A and B due to a point charge q depends only on the initial and final radial coordinates rA and rB . The two dashed circles represent cross-sections of spherical equipotential surfaces. Figure 25.6 VB VB VA VA ke q E r dr 1 rB 1 rA keq rA dr r2 keq r rB rA (25.10) The integral of E ds is independent of the path between points A and B -- as it must be because the electric field of a point charge is conservative. Furthermore, Equation 25.10 expresses the important result that the potential difference between any two points A and B in a field created by a point charge depends only on the radial coordinates rA and rB . It is customary to choose the reference of electric potential . With this reference, the electric potential created by a point to be zero at r A charge at any distance r from the charge is V ke q r (25.11) Electric potential created by a point charge Electric potential is graphed in Figure 25.7 as a function of r, the radial distance from a positive charge in the xy plane. Consider the following analogy to gravitational potential: Imagine trying to roll a marble toward the top of a hill shaped like Figure 25.7a. The gravitational force experienced by the marble is analogous to the repulsive force experienced by a positively charged object as it approaches another positively charged object. Similarly, the electric potential graph of the region surrounding a negative charge is analogous to a "hole" with respect to any approaching positively charged objects. A charged object must be infinitely distant from another charge before the surface is "flat" and has an electric potential of zero. 25.3 Electric Potential and Potential Energy Due to Point Charges 775 2.5 2.0 Electric potential (V) 1.5 1.0 0.5 y 0 x (a) + (b) Figure 25.7 (a) The electric potential in the plane around a single positive charge is plotted on the vertical axis. (The electric potential function for a negative charge would look like a hole instead of a hill.) The red line shows the 1/r nature of the electric potential, as given by Equation 25.11. (b) View looking straight down the vertical axis of the graph in part (a), showing concentric circles where the electric potential is constant. These circles are cross sections of equipotential spheres having the charge at the center. 776 CHAPTER 25 Electric Potential Quick Quiz 25.3 A spherical balloon contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, does the electric potential at the surface of the balloon increase, decrease, or remain the same? How about the magnitude of the electric field? The electric flux? Electric potential due to several point charges We obtain the electric potential resulting from two or more point charges by applying the superposition principle. That is, the total electric potential at some point P due to several point charges is the sum of the potentials due to the individual charges. For a group of point charges, we can write the total electric potential at P in the form qi V ke (25.12) i ri where the potentlectric potential V to be zero at some convenient point. We illustrate both methods with several examples. EXAMPLE 25.5 Electric Potential Due to a Uniformly Charged Ring we can remove !x 2 V a 2 from the integral, and V reduces to ke dq k eQ (a) Find an expression for the electric potential at a point P located on the perpendicular central axis of a uniformly charged ring of radius a and total charge Q. Solution Let us orient the ring so that its plane is perpendicular to an x axis and its center is at the origin. We can then take point P to be at a distance x from the center of the ring, as shown in Figure 25.15. The charge element dq is at a distance !x 2 a 2 from point P. Hence, we can express V as V ke dq r ke dq ! x2 a2 !x 2 a2 (25.20) The only variable in this expression for V is x. This is not surprising because our calculation is valid only for points along the x axis, where y and z are both zero. (b) Find an expression for the magnitude of the electric field at point P. !x 2 a2 Because each element dq is at the same distance from point P, Solution From symmetry, we see that along the x axis E can have only an x component. Therefore, we can use Equa- 782 tion 25.16: Ex dV dx k eQ( (x 2 k eQ 1 2 2 )(x CHAPTER 25 Electric Potential ter, V has either a maximum or minimum value; it is, in fact, a maximum. d (x 2 dx a2) a2) 1/2 3/2(2x) dq k eQx a 2)3/2 (25.21) a x 2 + a 2 This result agrees with that obtained by direct integration (see Example 23.8). Note that E x 0 at x 0 (the center of the ring). Could you have guessed this from Coulomb's law? x P Exercise What is the electric potential at the center of the ring? What does the value of the field at the center tell you about the value of V at the center? V k eQ /a. Because E x dV/dx 0 at the cen- Answer Figure 25.15 A uniformly charged ring of radius a lies in a plane perpendicular to the x axis. All segments dq of the ring are the same distance from any point P lying on the x axis. EXAMPLE 25.6 Electric Potential Due to a Uniformly Charged Disk from the definition of surface charge density (see Section 23.5), we know that the charge on the ring is dq dA 2 r dr. Hence, the potential at the point P due to this ring is dV k e dq k e 2 r dr x2 Find (a) the electric potential and (b) the magnitude of the electric field along the perpendicular central axis of a uniformly charged disk of radius a and surface charge density . Solution (a) Again, we choose the point P to be at a distance x from the center of the disk and take the plane of the disk to be perpendicular to the x axis. We can simplify the problem by dividing the disk into a series of charged rings. The electric potential of each ring is given by Equation 25.20. Consider one such ring of radius r and width dr, as indicated in Figure 25.16. The surface area of the ring is dA 2 r dr ; ! r2 !r 2 x2 To find the total electric potential at P, we sum over all rings making up the disk. That is, we integrate dV from r 0 to r a: a V ke 2r dr a 0 !r 2 x2 ke (r 2 0 x2 ) 1/2 2r dr This integral is of the form u n du and has the value 1 r 2 x 2. This gives u n 1/(n 1), where n 2 and u a r x dA = 2rdr dr P r 2 + x2 V 2 k e [(x 2 a 2)1/2 x] (25.22) (b) As in Example 25.5, we can find the electric field at any axial point from Ex dV dx 2 ke 1 x ! x2 a2 (25.23) Figure 25.16 A uniformly charged disk of radius a lies in a plane perpendicular to the x axis. The calculation of the electric potential at any point P on the x axis is simplified by dividing the disk into many rings each of area 2 r dr. The calculation of V and E for an arbitrary point off the axis is more difficult to perform, and we do not treat this situation in this text. 25.5 Electric Potential Due to Continuous Charge Distributions 783 EXAMPLE 25.7 Electric Potential Due to a Finite Line of Charge Evaluating V, we find that V k eQ ln A rod of length located along the x axis has a total charge Q and a uniform linear charge density Q / . Find the electric potential at a point P located on th. The blue curves in the figure represent the cross-sections of the equipotential surfaces for this charge configuration. As usual, the field lines are perpendicular to the conducting surfaces at all points, and the equipotential surfaces are perpendicular to the field lines everywhere. Trying to move a positive charge in the region of these conductors would be like moving a marble on a hill that is flat on top (representing the conductor on the left) and has another flat area partway down the side of the hill (representing the conductor on the right). EXAMPLE 25.9 Two Connected Charged Spheres r1 q1 Two spherical conductors of radii r 1 and r 2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire, as shown in Figure 25.23. The charges on the spheres in equilibrium are q 1 and q 2 , respectively, and they are uniformly charged. Find the ratio of the magnitudes of the electric fields at the surfaces of the spheres. Solution Because the spheres are connected by a conducting wire, they must both be at the same electric potential: V ke q1 r1 ke q2 r2 r2 q2 Therefore, the ratio of charges is Figure 25.23 Two charged spherical conductors connected by a conducting wire. The spheres are at the same electric potential V. 25.6 Electric Potential Due to a Charged Conductor q1 q2 r1 r2 787 (1) Taking the ratio of these two fields and making use of Equation (1), we find that E1 E2 r2 r1 Because the spheres are very far apart and their surfaces uniformly charged, we can express the magnitude of the electric fields at their surfaces as E1 q ke 1 r 12 and E2 q ke 2 r 22 Hence, the field is more intense in the vicinity of the smaller sphere even though the electric potentials of both spheres are the same. A Cavity Within a Conductor Now consider a conductor of arbitrary shape containing a cavity as shown in Figure 25.24. Let us assume that no charges are inside the cavity. In this case, the electric field inside the cavity must be zero regardless of the charge distribution on the outside surface of the conductor. Furthermore, the field in the cavity is zero even if an electric field exists outside the conductor. To prove this point, we use the fact that every point on the conductor is at the same electric potential, and therefore any two points A and B on the surface of the cavity must be at the same potential. Now imagine that a field E exists in the cavity and evaluate the potential difference VB VA defined by Equation 25.3: B B A Figure 25.24 VB VA E ds A If E is nonzero, we can always find a path between A and B for which E ds is a positive number; thus, the integral must be positive. However, because VB VA 0, the integral of E ds must be zero for all paths between any two points on the conductor, which implies that E is zero everywhere. This contradiction can be reconciled only if E is zero inside the cavity. Thus, we conclude that a cavity surrounded by conducting walls is a field-free region as long as no charges are inside the cavity. A conductor in electrostatic equilibrium containing a cavity. The electric field in the cavity is zero, regardless of the charge on the conductor. Corona Discharge A phenomenon known as corona discharge is often observed near a conductor such as a high-voltage power line. When the electric field in the vicinity of the conductor is sufficiently strong, electrons are stripped from air molecules. This causes the molecules to be ionized, thereby increasing the air's ability to conduct. The observed glow (or corona discharge) results from the recombination of free electrons with the ionized air molecules. If a conductor has an irregular shape, the electric field can be very high near sharp points or edges of the conductor; consequently, the ionization process and corona discharge are most likely to occur around such points. Quick Quiz 25.4 (a) Is it possible for the magnitude of the electric field to be zero at a location where the electric potential is not zero? (b) Can the electric potential be zero where the olecules to form positive ions, electrons, and such negative ions as O2 . The air to be cleaned enters the duct and moves near the wire. As the electrons and negative ions created by the discharge are accelerated toward the outer wall by the electric field, the dirt particles in the air become charged by collisions and ion capture. Because most of the charged dirt particles are negative, they too are drawn to the duct walls by the electric field. When the duct is periodically shaken, the particles break loose and are collected at the bottom. Insulator Clean air out Dirty air in Weight Dirt out (a) (b) (c) Figure 25.28 (a) Schematic diagram of an electrostatic precipitator. The high negative electric potential maintained on the central coiled wire creates an electrical discharge in the vicinity of the wire. Compare the air pollution when the electrostatic precipitator is (b) operating and (c) turned off. 25.8 Applications of Electrostatics 791 In addition to reducing the level of particulate matter in the atmosphere (compare Figs. 25.28b and c), the electrostatic precipitator recovers valuable materials in the form of metal oxides. Xerography and Laser Printers The basic idea of xerography5 was developed by Chester Carlson, who was granted a patent for the xerographic process in 1940. The one feature of this process that makes it unique is the use of a photoconductive material to form an image. (A photoconductor is a material that is a poor electrical conductor in the dark but that becomes a good electrical conductor when exposed to light.) The xerographic process is illustrated in Figure 25.29a to d. First, the surface of a plate or drum that has been coated with a thin film of photoconductive material (usually selenium or some compound of selenium) is given a positive electrostatic charge in the dark. An image of the page to be copied is then focused by a lens onto the charged surface. The photoconducting surface becomes conducting only in areas where light strikes it. In these areas, the light produces charge carriers in the photoconductor that move the positive charge off the drum. However, positive Lens Light causes some areas of drum to become electrically conducting, removing positive charge Selenium-coated drum (a) Charging the drum (b) Imaging the document Negatively charged toner (c) Applying the toner Interlaced pattern of laser lines Laser beam (d) Transferring the toner to the paper (e) Laser printer drum The xerographic process: (a) The photoconductive surface of the drum is positively charged. (b) Through the use of a light source and lens, an image is formed on the surface in the form of positive charges. (c) The surface containing the image is covered with a negatively charged powder, which adheres only to the image area. (d) A piece of paper is placed over the surface and given a positive charge. This transfers the image to the paper as the negatively charged powder particles migrate to the paper. The paper is then heat-treated to "fix" the powder. (e) A laser printer operates similarly except the image is produced by turning a laser beam on and off as it sweeps across the selenium-coated drum. 5 Figure 25.29 The prefix xero- is from the Greek word meaning "dry." Note that no liquid ink is used anywhere in xerography. 792 CHAPTER 25 Electric Potential charges remain on those areas of the photoconductor not exposed to light, leaving a latent image of the object in the form of a positive surface charge distribution. Next, a negatively charged powder called a toner is dusted onto the photoconducting surface. The charged powder adheres only to those areas of the surface that contain the positively charged image. At this point, the image becomes visible. The toner (and hence the image) are then transferred to the surface of a sheet of positively charged paper. Finally, the toner is "fixed" to the surface of the paper as the toner melts while passing through high-temperature rollers. This results in a permanent copy of the original. A laser printer (Fig. 25.29e) operates by the saitor is initially uncharged, the battery establishes an electric field in the connecting wires when the connections are made. Let us focus on the plate connected to the negative terminal of the battery. The electric field applies a force on electrons in the wire just outside this plate; this force causes the electrons to move onto the plate. This movement continues until the plate, the wire, and the terminal are all at the same electric potential. Once this equilibrium point is attained, a potential difference no longer exists between the terminal and the plate, and as a result no electric field is present in the wire, and the movement of electrons stops. The plate now carries a negative charge. A similar process occurs at the other capacitor plate, with electrons moving from the plate to the wire, leaving the plate positively charged. In this final configuration, the potential difference across the capacitor plates is the same as that between the terminals of the battery. Suppose that we have a capacitor rated at 4 pF. This rating means that the capacitor can store 4 pC of charge for each volt of potential difference between the two conductors. If a 9-V battery is connected across this capacitor, one of the conductors ends up with a net charge of 36 pC and the other ends up with a net charge of 36 pC. Figure 26.2 A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge. 26.2 CALCULATING CAPACITANCE We can calculate the capacitance of a pair of oppositely charged conductors in the following manner: We assume a charge of magnitude Q , and we calculate the potential difference using the techniques described in the preceding chapter. We then use the expression C Q / V to evaluate the capacitance. As we might expect, we can perform this calculation relatively easily if the geometry of the capacitor is simple. We can calculate the capacitance of an isolated spherical conductor of radius R and charge Q if we assume that the second conductor making up the capacitor is a concentric hollow sphere of infinite radius. The electric potential of the sphere of radius R is simply ke Q /R, and setting V 0 at infinity as usual, we have C Q V Q k eQ /R R ke 4 0R QuickLab Roll some socks into balls and stuff them into a shoebox. What determines how many socks fit in the box? Relate how hard you push on the socks to V for a capacitor. How does the size of the box influence its "sock capacity"? (26.2) This expression shows that the capacitance of an isolated charged sphere is proportional to its radius and is independent of both the charge on the sphere and the potential difference. 806 CHAPTER 26 Capacitance and Dielectrics The capacitance of a pair of conductors depends on the geometry of the conductors. Let us illustrate this with three familiar geometries, namely, parallel plates, concentric cylinders, and concentric spheres. In these examples, we assume that the charged conductors are separated by a vacuum. The effect of a dielectric material placed between the conductors is treated in Section 26.5. Parallel-Plate Capacitors Two parallel metallic plates of equal area A are separated by a distance d, as shown in Figure 26.2. One plate carries a charge Q , and the other carries a charge Q . Let us consider how the geometry of these conductors influences the capacity of the combination to store charge. Recall that charges of like sign repel one another. As a capacitor is being charged by a battery, electrons flow into the negative plate and out of the positive plate. If the capacitor plates are large, the accumulated charges are able to distribute themselves over a substantial area, and the amount of charge that can be stored on a plate for a given potential difference increases as the plate area is increased. Thus, we expect the capacitance to be proportional to the plate area A. Now let us consider the region that separates teries, which combine to yield an equivalent capacitance of 4.0 F. Finally, the 2.0- F and 4.0- F capacitors in Figure 26.10c are in parallel and thus have an equivalent capacitance of 6.0 F. Exercise 1 1/2.0 F Consider three capacitors having capacitances of 3.0 F, 6.0 F, and 12 F. Find their equivalent capacitance when they are connected (a) in parallel and (b) in series. (a) 21 F; (b) 1.7 F. Answer 1.0 4.0 4.0 3.0 4.0 2.0 a 6.0 b 8.0 (a) a 8.0 (b) b a 4.0 (c) b a 6.0 b 2.0 8.0 (d) Figure 26.10 To find the equivalent capacitance of the capacitors in part (a), we reduce the various combinations in steps as indicated in parts (b), (c), and (d), using the series and parallel rules described in the text. 26.4 ENERGY STORED IN A CHARGED CAPACITOR 13.5 Almost everyone who works with electronic equipment has at some time verified that a capacitor can store energy. If the plates of a charged capacitor are connected by a conductor, such as a wire, charge moves between the plates and the connecting wire until the capacitor is uncharged. The discharge can often be observed as a visible spark. If you should accidentally touch the opposite plates of a charged capacitor, your fingers act as a pathway for discharge, and the result is an electric shock. The degree of shock you receive depends on the capacitance and on the voltage applied to the capacitor. Such a shock could be fatal if high voltages are present, such as in the power supply of a television set. Because the charges can be stored in a capacitor even when the set is turned off, unplugging the television does not make it safe to open the case and touch the components inside. Consider a parallel-plate capacitor that is initially uncharged, such that the initial potential difference across the plates is zero. Now imagine that the capacitor is connected to a battery and develops a maximum charge Q. (We assume that the capacitor is charged slowly so that the problem can be considered as an electrostatic system.) When the capacitor is connected to the battery, electrons in the wire just outside the plate connected to the negative terminal move into the plate to give it a negative charge. Electrons in the plate connected to the positive terminal move out of the plate into the wire to give the plate a positive charge. Thus, charges move only a small distance in the wires. To calculate the energy of the capacitor, we shall assume a different process -- one that does not actually occur but gives the same final result. We can make this 814 CHAPTER 26 Capacitance and Dielectrics QuickLab Here's how to find out whether your calculator has a capacitor to protect values or programs during battery changes: Store a number in your calculator's memory, remove the calculator battery for a moment, and then quickly replace it. Was the number that you stored preserved while the battery was out of the calculator? (You may want to write down any critical numbers or programs that are stored in the calculator before trying this!) assumption because the energy in the final configuration does not depend on the actual charge-transfer process. We imagine that we reach in and grab a small amount of positive charge on the plate connected to the negative terminal and apply a force that causes this positive charge to move over to the plate connected to the positive terminal. Thus, we do work on the charge as we transfer it from one plate to the other. At first, no work is required to transfer a small amount of charge dq from one plate to the other.3 However, once this charge has been transferred, a small potential difference exists between the plates. Therefore, work must be done to move additional charge through this potential difference. As more and more charge is transferred from one plate to the other, the potential difference increases in proportion, and more work is required. Suppose that q is the charge on the capacitor at some instant during the charging process. At the same instant, the potential difference across the capacitor is V q/C. From Section 25.2, we know that lose together without touching, thereby decreasing d and increasing C TABLE 26.1 Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature Material Air (dry) Bakelite Fused quartz Neoprene rubber Nylon Paper Polystyrene Polyvinyl chloride Porcelain Pyrex glass Silicone oil Strontium titanate Teflon Vacuum Water a Dielectric Constant 1.000 59 4.9 3.78 6.7 3.4 3.7 2.56 3.4 6 5.6 2.5 233 2.1 1.000 00 80 Dielectric Strengtha (V/m) 3 24 8 12 14 16 24 40 12 14 15 8 60 106 106 106 106 106 106 106 106 106 106 106 106 106 -- -- The dielectric strength equals the maximum electric field that can exist in a dielectric without electrical breakdown. Note that these values depend strongly on the presence of impurities and flaws in the materials. 820 CHAPTER 26 Capacitance and Dielectrics (a) (b) (a) Kirlian photograph created by dropping a steel ball into a high-energy electric field. Kirlian photography is also known as electrophotography. (b) Sparks from static electricity discharge between a fork and four electrodes. Many sparks were used to create this image because only one spark forms for a given discharge. Note that the bottom prong discharges to both electrodes at the bottom right. The light of each spark is created by the excitation of gas atoms along its path. Types of Capacitors Commercial capacitors are often made from metallic foil interlaced with thin sheets of either paraffin-impregnated paper or Mylar as the dielectric material. These alternate layers of metallic foil and dielectric are rolled into a cylinder to form a small package (Fig. 26.15a). High-voltage capacitors commonly consist of a number of interwoven metallic plates immersed in silicone oil (Fig. 26.15b). Small capacitors are often constructed from ceramic materials. Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets of metallic plates, one fixed and the other movable, and contain air as the dielectric. Often, an electrolytic capacitor is used to store large amounts of charge at relatively low voltages. This device, shown in Figure 26.15c, consists of a metallic foil in contact with an electrolyte -- a solution that conducts electricity by virtue of the motion of ions contained in the solution. When a voltage is applied between the foil and the electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, Metal foil Plates Case Electrolyte Contacts Oil Paper (a) (b) Metallic foil + oxide layer (c) Figure 26.15 Three commercial capacitor designs. (a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor. 26.5 Capacitors with Dielectrics 821 and this layer serves as the dielectric. Very large values of capacitance can be obtained in an electrolytic capacitor because the dielectric layer is very thin, and thus the plate separation is very small. Electrolytic capacitors are not reversible as are many other capacitors -- they have a polarity, which is indicated by positive and negative signs marked on the device. When electrolytic capacitors are used in circuits, the polarity must be aligned properly. If the polarity of the applied voltage is opposite that which is intended, the oxide layer is removed and the capacitor conducts electricity instead of storing charge. Quick Quiz 26.6 If you have ever tried to hang a picture, you know it can be difficult to locate a wooden stud in which to anchor your nail or screw. A carpenter's stud-finder is basically a capacitor with its plates arranged side by side instead of facing one another, as shown in Figure 26.16. When the device is moved over a stud, does the capacitance increase or decrease? Stud Capacitor plates Stud-finder Wall board (a) (b) Figure 26.16 A stud-finder. (a)The materials between the plates of the capacitor are the wallboard and air. (b) When the capacitor moves across a stud in the wall, the materials between the plates are the wallboard and thr does the stored energy change? 12. Why is it dangerous to touch the terminals of a highvoltage capacitor even after the applied voltage has been turned off? What can be done to make the capacitor safe to handle after the voltage source has been removed? 13. Describe how you can increase the maximum operating voltage of a parallel-plate capacitor for a fixed plate separation. 14. An air-filled capacitor is charged, disconnected from the power supply, and, finally, connected to a voltmeter. Explain how and why the voltage reading changes when a dielectric is inserted between the plates of the capacitor. 15. Using the polar molecule description of a dielectric, explain how a dielectric affects the electric field inside a capacitor. 16. Explain why a dielectric increases the maximum operating voltage of a capacitor even though the physical size of the capacitor does not change. 17. What is the difference between dielectric strength and the dielectric constant? 18. Explain why a water molecule is permanently polarized. What type of molecule has no permanent polarization? 19. If a dielectric-filled capacitor is heated, how does its capacitance change? (Neglect thermal expansion and assume that the dipole orientations are temperature dependent.) PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 26.1 Definition of Capacitance 1. (a) How much charge is on each plate of a 4.00- F capacitor when it is connected to a 12.0-V battery? (b) If this same capacitor is connected to a 1.50-V battery, what charge is stored? 2. Two conductors having net charges of 10.0 C and 10.0 C have a potential difference of 10.0 V. Determine (a) the capacitance of the system and (b) the potential difference between the two conductors if the charges on each are increased to 100 C and 100 C. 832 CHAPTER 26 Capacitance and Dielectrics Section 26.2 Calculating Capacitance 3. An isolated charged conducting sphere of radius 12.0 cm creates an electric field of 4.90 104 N/C at a distance 21.0 cm from its center. (a) What is its surface charge density? (b) What is its capacitance? 4. (a) If a drop of liquid has capacitance 1.00 pF, what is its radius? (b) If another drop has radius 2.00 mm, what is its capacitance? (c) What is the charge on the smaller drop if its potential is 100 V? 5. Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 7.00 C. (a) How is this total charge shared between the spheres? (Neglect any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be V 0 at r ? 6. Regarding the Earth and a cloud layer 800 m above the Earth as the "plates" of a capacitor, calculate the capacitance if the cloud layer has an area of 1.00 km2. Assume that the air between the cloud and the ground is pure and dry. Assume that charge builds up on the cloud and on the ground until a uniform electric field with a magnitude of 3.00 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold? 7. An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. If a 20.0-V potential difference is applied to these plates, calculate (a) the electric field between the plates, (b) the surface charge density, (c) the capacitance, and (d) the charge on each plate. 8. A 1-megabit computer memory chip contains many 60.0-fF capacitors. Each capacitor has a plate area of 21.0 10 12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The characteristic atomic diameter is 10 10 m 0.100 nm. Express the plate separation in nanometers.pherical gaussian surface outside and concentric with the capacitor, the net charge inside the surface is zero. Applying Gauss's law to this configuration, we find that E 0 at points outside the capacitor. 26.3 For a given voltage, the energy stored in a capacitor is proportional to C : U C( V )2/2. Thus, you want to maximize the equivalent capacitance. You do this by connecting the three capacitors in parallel, so that the capacitances add. 26.4 (a) C decreases (Eq. 26.3). (b) Q stays the same because there is no place for the charge to flow. (c) E remains constant (see Eq. 24.8 and the paragraph following it). (d) V increases because V Q /C, Q is constant (part b), and C decreases (part a). (e) The energy stored in the capacitor is proportional to both Q and V (Eq. 26.11) and thus increases. The additional energy comes from the work you do in pulling the two plates apart. 26.5 (a) C decreases (Eq. 26.3). (b) Q decreases. The battery supplies a constant potential difference V ; thus, charge must flow out of the capacitor if C Q / V is to decrease. (c) E decreases because the charge density on the plates decreases. (d) V remains constant because of the presence of the battery. (e) The energy stored in the capacitor decreases (Eq. 26.11). 26.6 It increases. The dielectric constant of wood (and of all other insulating materials, for that matter) is greater than 1; therefore, the capacitance increases (Eq. 26.14). This increase is sensed by the stud-finder's special circuitry, which causes an indicator on the device to light up. 26.7 (a) C increases (Eq. 26.14). (b) Q increases. Because the battery maintains a constant V, Q must increase if C ( Q / V ) increases. (c) E between the plates remains constant because V Ed and neither V nor d changes. The electric field due to the charges on the plates increases because more charge has flowed onto the plates. The induced surface charges on the dielectric create a field that opposes the increase in the field caused by the greater number of charges on the plates. (d) The battery maintains a constant V. (e) The energy stored in the capacitor increases (Eq. 26.11). You would have to push the dielectric into the capacitor, just as you would have to do positive work to raise a mass and increase its gravitational potential energy. P U Z Z L E R Electrical workers restoring power to the eastern Ontario town of St. Isadore, which was without power for several days in January 1998 because of a severe ice storm. It is very dangerous to touch fallen power transmission lines because of their high electric potential, which might be hundreds of thousands of volts relative to the ground. Why is such a high potential difference used in power transmission if it is so dangerous, and why aren't birds that perch on the wires electrocuted? (AP/Wide World Photos/Fred Chartrand) c h a p t e r Current and Resistance Chapter Outline 27.1 Electric Current 27.2 Resistance and Ohm's Law 27.3 A Model for Electrical Conduction 27.4 Resistance and Temperature 27.5 (Optional) Superconductors 27.6 Electrical Energy and Power 840 27.1 Electric Current 841 T hus far our treatment of electrical phenomena has been confined to the study of charges at rest, or electrostatics. We now consider situations involving electric charges in motion. We use the term electric current, or simply current, to describe the rate of flow of charge through some region of space. Most practical applications of electricity deal with electric currents. For example, the battery in a flashlight supplies current to the filament of the bulb when the switch is turned on. A variety of home appliances operate on alternating current. In these common situations, the charges flow through a conductor, such as a copper wire. It also is possible for currents to exist outside a conductor. For instance, a beam of electrons in a television picture tube constitutes a current. This chapter begins with the definitions of current and current density. A microscopic description of current is given, and some of the factors that contribute to the resistance to the flow of charge in conductors are discussed. A classical model is used to describe electrical conduction in metals, and some of the limitations of this model are cited. 27.1 13.2 ELECTRIC CURRENT It is instructive to draw an analogy between water flow and current. In many localities it is common practice to install low-flow showerheads in homes as a waterconservation measure. We quantify the flow of water from these and similar devices by specifying the amount of water that emerges during a given time interval, which is often measured in liters per minute. On a grander scale, we can characterize a river current by describing the rate at which the water flows past a particular location. For example, the flow over the brink at Niagara Falls is maintained at rates between 1 400 m3/s and 2 800 m3/s. Now consider a system of electric charges in motion. Whenever there is a net flow of charge through some region, a current is said to exist. To define current more precisely, suppose that the charges are moving perpendicular to a surface of area A, as shown in Figure 27.1. (This area could be the cross-sectional area of a wire, for example.) The current is the rate at which charge flows through this surface. If Q is the amount of charge that passes through this area in a time interval t, the average current I av is equal to the charge that passes through A per unit time: I av Q t (27.1) + + + + A I + Figure 27.1 Charges in motion through an area A. The time rate at which charge flows through the area is defined as the current I. The direction of the current is the direction in which positive charges flow when free to do so. If the rate at which charge flows varies in time, then the current varies in time; we define the instantaneous current I as the differential limit of average current: I The SI unit of current is the ampere (A): 1A 1C 1s (27.3) dQ dt (27.2) Electric current That is, 1 A of current is equivalent to 1 C of charge passing through the surface area in 1 s. The charges passing through the surface in Figure 27.1 can be positive or negative, or both. It is conventional to assign to the current the same direction as the flow of positive charge. In electrical conductors, such as copper or alu- The direction of the current 842 CHAPTER 27 Current and Resistance minum, the current is due to the motion of negatively charged electrons. Therefore, when we speak of current in an ordinary conductor, the direction of the current is opposite the direction of flow of electrons. However, if we are considering a beam of positively charged protons in an accelerator, the current is in the direction of motion of the protons. In some cases -- such as those involving gases and electrolytes, for instance -- the current is the result of the flow of both positive and negative charges. If the ends of a conducting wire are connected to form a loop, all points on the loop are at the same electric potential, and hence the electric field is zero within and at the surface of the conductor. Because the electric field is zero, there is no net transport of charge through the wire, and therefore there is no current. The current in the conductor is zero even if the conductor has an excess of charge on it. However, if the ends of the conducting wire are connected to a battery, all points on the loop are not at the same potential. The battery sets up a potential difference between the ends of the loop, creating an electric field within the wire. The electric field exerts forces on the conduction electrons in the wire, causing them to move around the loop and thus creating a current. It is common to refer to a moving charge (positive or negative) as a mobile charge carrier. For example, the mobile charge carriers in a metal are electrons. x vd A q Microscopic Model of Current We can relate current to the motion of the charge carriers by describing a microscopic model of conduction in a metal. Consider the current in a conductor of cross-sectional area A (Fig. 27.2). The volume of a section of the conductor of length x (the gray reg0 7 m2 4.6 /m Exercise (b) If a potential difference of 10 V is maintained across a 1.0-m length of the Nichrome wire, what is the current in the wire? Because a 1.0-m length of this wire has a resistance of 4.6 , Equation 27.8 gives I V R 10 V 4.6 2.2 A What is the resistance of a 6.0-m length of 22gauge Nichrome wire? How much current does the wire carry when connected to a 120-V source of potential difference? 28 ; 4.3 A. Answer Exercise Answer Solution Calculate the current density and electric field in the wire when it carries a current of 2.2 A. 6.8 106 A/m2; 10 N/C. EXAMPLE 27.4 The Radial Resistance of a Coaxial Cable completely filled with silicon, as shown in Figure 27.8a, and current leakage through the silicon is unwanted. (The cable is designed to conduct current along its length.) The radius Coaxial cables are used extensively for cable television and other electronic applications. A coaxial cable consists of two cylindrical conductors. The gap between the conductors is 850 CHAPTER 27 Current and Resistance of the inner conductor is a 0.500 cm, the radius of the outer one is b 1.75 cm, and the length of the cable is L 15.0 cm. Calculate the resistance of the silicon between the two conductors. dR 2 rL dr Solution In this type of problem, we must divide the object whose resistance we are calculating into concentric elements of infinitesimal thickness dr (Fig. 27.8b). We start by using the differential form of Equation 27.11, replacing with r for the distance variable: dR dr/A, where dR is the resistance of an element of silicon of thickness dr and surface area A. In this example, we take as our representative concentric element a hollow silicon cylinder of radius r, thickness dr, and length L, as shown in Figure 27.8. Any current that passes from the inner conductor to the outer one must pass radially through this concentric element, and the area through which this current passes is A 2 rL . (This is the curved surface area -- circumference multiplied by length -- of our hollow silicon cylinder of thickness dr.) Hence, we can write the resistance of our hollow cylinder of silicon as Because we wish to know the total resistance across the entire thickness of the silicon, we must integrate this expression from r a to r b : b b R a dR 2 L a dr r 2 L ln b a 640 m for Substituting in the values given, and using silicon, we obtain R 640 m 1.75 cm ln 2 (0.150 m ) 0.500 cm 851 Exercise If a potential difference of 12.0 V is applied between the inner and outer conductors, what is the value of the total current that passes between them? 14.1 mA. Answer L dr Silicon a b r Current direction Inner conductor Outer conductor (a) End view (b) Figure 27.8 A coaxial cable. (a) Silicon fills the gap between the two conductors. (b) End view, showing current leakage. 27.3 A MODEL FOR ELECTRICAL CONDUCTION In this section we describe a classical model of electrical conduction in metals that was first proposed by Paul Drude in 1900. This model leads to Ohm's law and shows that resistivity can be related to the motion of electrons in metals. Although the Drude model described here does have limitations, it nevertheless introduces concepts that are still applied in more elaborate treatments. Consider a conductor as a regular array of atoms plus a collection of free electrons, which are sometimes called conduction electrons. The conduction electrons, although bound to their respective atoms when the atoms are not part of a solid, gain mobility when the free atoms condense into a solid. In the absence of an electric field, the conduction electrons move in random directions through the con- 27.3 A Model for Electrical Conduction 851 ductor with average speeds of the order of 106 m/s. The situation is similar to the motion of gas molecules confined in a vessel. In fact, some scientists refer to conduction electrons in a metal as an electron gas. There is no current through the conductor in the absence of an electric field because the drift velocity of the free electrons is zero. That is, on V (27.22) 5 Power Note that once the current reaches its steady-state value, there is no change in the kinetic energy of the charge carriers creating the current. 27.6 Electrical Energy and Power 857 In this case, the power is supplied to a resistor by a battery. However, we can use Equation 27.22 to determine the power transferred to any device carrying a current I and having a potential difference V between its terminals. Using Equation 27.22 and the fact that V IR for a resistor, we can express the power delivered to the resistor in the alternative forms I 2R ( V )2 R (27.23) Power delivered to a resistor When I is expressed in amperes, V in volts, and R in ohms, the SI unit of power is the watt, as it was in Chapter 7 in our discussion of mechanical power. The power lost as internal energy in a conductor of resistance R is called joule heating 6; this transformation is also often referred to as an I 2R loss. A battery, a device that supplies electrical energy, is called either a source of electromotive force or, more commonly, an emf source. The concept of emf is discussed in greater detail in Chapter 28. (The phrase electromotive force is an unfortunate choice because it describes not a force but rather a potential difference in volts.) When the internal resistance of the battery is neglected, the potential difference between points a and b in Figure 27.14 is equal to the emf of the battery -- that is, V V b V a . This being true, we can state that the current in the circuit is I , the power supplied by the emf V/R /R. Because V source can be expressed as I , which equals the power delivered to the resistor, I 2R. When transporting electrical energy through power lines, such as those shown in Figure 27.15, utility companies seek to minimize the power transformed to internal energy in the lines and maximize the energy delivered to the consumer. Because I V, the same amount of power can be transported either at high currents and low potential differences or at low currents and high potential differences. Utility companies choose to transport electrical energy at low currents and high potential differences primarily for economic reasons. Copper wire is very expensive, and so it is cheaper to use high-resistance wire (that is, wire having a small cross-sectional area; see Eq. 27.11). Thus, in the expression for the power delivered to a resistor, I 2R , the resistance of the wire is fixed at a relatively high value for economic considerations. The I 2R loss can be reduced by keeping the current I as low as possible. In some instances, power is transported at potential differences as great as 765 kV. Once the electricity reaches your city, the potential difference is usually reduced to 4 kV by a device called a transformer. Another transformer drops the potential difference to 240 V before the electricity finally reaches your home. Of course, each time the potential difference decreases, the current increases by the same factor, and the power remains the same. We shall discuss transformers in greater detail in Chapter 33. Figure 27.15 Power companies transfer electrical energy at high potential differences. Quick Quiz 27.6 The same potential difference is applied to the two lightbulbs shown in Figure 27.16. Which one of the following statements is true? (a) The 30-W bulb carries the greater current and has the higher resistance. (b) The 30-W bulb carries the greater current, but the 60-W bulb has the higher resistance. QuickLab If you have access to an ohmmeter, verify your answer to Quick Quiz 27.6 by testing the resistance of a few lightbulbs. 6 It is called joule heating even though the process of heat does not occur. This is another example of incorrect usage of the word heat that has become entrenched in our language. 858 CHAPTER 27 Current and Resistance Figure 27.16 These lightbulbs operate at their rated power only when they are connected to a 120-V source. (c) The 30-W bulb has the higher resistance, but the 60-W bulb carries the greater current. (d) The 60-W bulb carries the greater current antity of charge q (in coulombs) passing through a surface of area 2.00 cm2 varies with time according to the equation q 4.00t 3 5.00t 6.00, where t is in seconds. (a) What is the instantaneous current through the surface at t 1.00 s? (b) What is the value of the current density? 8. An electric current is given by the expression I(t ) 100 sin(120 t), where I is in amperes and t is in seconds. What is the total charge carried by the current from t 0 to t 1/240 s? 9. Figure P27.9 represents a section of a circular conductor of nonuniform diameter carrying a current of 5.00 A. The radius of cross-section A1 is 0.400 cm. (a) What is the magnitude of the current density across A1 ? (b) If the current density across A2 is one-fourth the value across A1 , what is the radius of the conductor at A2 ? A2 WEB A1 I Figure P27.9 Problems 10. A Van de Graaff generator produces a beam of 2.00-MeV deuterons, which are heavy hydrogen nuclei containing a proton and a neutron. (a) If the beam current is 10.0 A, how far apart are the deuterons? (b) Is their electrostatic repulsion a factor in beam stability? Explain. 11. The electron beam emerging from a certain highenergy electron accelerator has a circular cross-section of radius 1.00 mm. (a) If the beam current is 8.00 A, what is the current density in the beam, assuming that it is uniform throughout? (b) The speed of the electrons is so close to the speed of light that their speed can be taken as c 3.00 10 8 m/s with negligible error. Find the electron density in the beam. (c) How long does it take for Avogadro's number of electrons to emerge from the accelerator? 12. An aluminum wire having a cross-sectional area of 4.00 10 6 m2 carries a current of 5.00 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2.70 g/cm3. (Assume that one electron is supplied by each atom.) 863 20. 21. 22. 23. 24. Section 27.2 Resistance and Ohm's Law 13. A lightbulb has a resistance of 240 when operating at a voltage of 120 V. What is the current through the lightbulb? 14. A resistor is constructed of a carbon rod that has a uniform cross-sectional area of 5.00 mm2. When a potential difference of 15.0 V is applied across the ends of the rod, there is a current of 4.00 10 3 A in the rod. Find (a) the resistance of the rod and (b) the rod's length. 15. A 0.900-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.600 mm2. What is the current in the wire? 16. A conductor of uniform radius 1.20 cm carries a current of 3.00 A produced by an electric field of 120 V/m. What is the resistivity of the material? 17. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance of R 0.500 , and if all of the copper is to be used, what will be (a) the length and (b) the diameter of this wire? 18. (a) Make an order-of-magnitude estimate of the resistance between the ends of a rubber band. (b) Make an order-of-magnitude estimate of the resistance between the `heads' and `tails' sides of a penny. In each case, state what quantities you take as data and the values you measure or estimate for them. (c) What would be the order of magnitude of the current that each carries if it were connected across a 120-V power supply? (WARNING! Do not try this at home!) 19. A solid cube of silver (density 10.5 g/cm3 ) has a mass of 90.0 g. (a) What is the resistance between opposite faces of the cube? (b) If there is one conduction electron for each silver atom, what is the average drift speed of electrons when a potential difference of 1.00 10 5 V is applied to opposite faces? (The atomic number of silver is 47, and its molar mass is 107.87 g/mol.) A metal wire of resistance R is cut into three equal pieces that are then connected side by side to form a new wire whose length is equal to one-third the original length. What is the resistance of this new wire? A wire with a resistance R is lengthened to 1.25 times its original length by being pulled through a small hole. Find the resistance battery, as described by Equation 28.1. Figure 28.2b is a graphical representation of the changes in electric potential as the circuit is traversed in the clockwise direction. By inspecting Figure 28.2a, we see that the terminal voltage V must equal the potential difference across the external resistance R, often called the load resistance. The load resistor might be a simple resistive circuit element, as in Figure 28.1, or it could be the resistance of some electrical device (such as a toaster, an electric heater, or a lightbulb) connected to the battery (or, in the case of household devices, to the wall outlet). The resistor represents a load on the battery because the battery must supply energy to operate the device. The potential difference across the load resistance is V IR . Combining this expression with Equation 28.1, we see that IR Ir (28.2) a b (b) c d Solving for the current gives Figure 28.2 (a) Circuit diagram of a source of emf (in this case, a battery), of internal resistance r, connected to an external resistor of resistance R. (b) Graphical representation showing how the electric potential changes as the circuit in part (a) is traversed clockwise. I R r (28.3) This equation shows that the current in this simple circuit depends on both the load resistance R external to the battery and the internal resistance r. If R is much greater than r, as it is in many real-world circuits, we can neglect r. If we multiply Equation 28.2 by the current I, we obtain I I 2R I 2r (28.4) This equation indicates that, because power I V (see Eq. 27.22), the total power output I of the battery is delivered to the external load resistance in the amount I 2R and to the internal resistance in the amount I 2r. Again, if r V R , then most of the power delivered by the battery is transferred to the load resistance. EXAMPLE 28.1 Terminal Voltage of a Battery (b) Calculate the power delivered to the load resistor, the power delivered to the internal resistance of the battery, and the power delivered by the battery. A battery has an emf of 12.0 V and an internal resistance of 0.05 . Its terminals are connected to a load resistance of 3.00 . (a) Find the current in the circuit and the terminal voltage of the battery. Using first Equation 28.3 and then Equation 28.1, we obtain I V R r Ir 12.0 V 3.05 12.0 V 3.93 A (3.93 A)(0.05 ) 11.8 V Solution Solution R The power delivered to the load resistor is I 2R (3.93 A)2 (3.00 ) 46.3 W The power delivered to the internal resistance is r I 2r (3.93 A)2 (0.05 ) 0.772 W To check this result, we can calculate the voltage across the load resistance R : V IR (3.93 A)(3.00 ) 11.8 V Hence, the power delivered by the battery is the sum of these quantities, or 47.1 W. You should check this result, using the expression I . 28.2 Resistors in Series and in Parallel 871 EXAMPLE 28.2 Matching the Load Show that the maximum power delivered to the load resistance R in Figure 28.2a occurs when the load resistance matches the internal resistance-- that is, when R r . max Solution The power delivered to the load resistance is equal to I 2R, where I is given by Equation 28.3: I 2R 2R (R r)2 r 2r 3r R When is plotted versus R as in Figure 28.3, we find that reaches a maximum value of 2/4r at R r. We can also prove this by differentiating with respect to R , setting the result equal to zero, and solving for R . The details are left as a problem for you to solve (Problem 57). Figure 28.3 Graph of the power delivered by a battery to a load resistor of resistance R as a function of R. The power delivered to the resistor is a maximum when the load resistance equals the internal resistance of the battery. 28.2 RESISTORS IN SERIES AND IN PARALLEL Suppose that you and your friends are at a crowded basketball game in a sports arena and decide to leave early. You have two choices: (1) your whole group can exit through a single door and walk down a long hallway containing several concession stands, each surrounded by a large crowd of people waiting to buy food or souvenirs; or (b) each me 2 ). Because the potential difference is Vbc across each of these resistors (since they are in parallel), we see that (6.0 )I 1 (3.0 )I 2 , or I 2 2I 1 . Using this result and the fact that I 1 I 2 3.0 A, we find that I 1 1.0 A and Solution 12 (b) a b 2.0 c 14 (c) a c Figure 28.6 28.2 Resistors in Series and in Parallel 875 EXAMPLE 28.4 Three Resistors in Parallel (c) Calculate the equivalent resistance of the circuit. Three resistors are connected in parallel as shown in Figure 28.7. A potential difference of 18 V is maintained between points a and b. (a) Find the current in each resistor. The resistors are in parallel, and so the potential difference across each must be 18 V. Applying the relationship V IR to each resistor gives I1 I2 I3 V R1 V R2 V R3 18 V 3.0 18 V 6.0 18 V 9.0 6.0 A Solution We can use Equation 28.8 to find R eq : 1 R eq 1 3.0 6 18 R eq 18 11 1 6.0 3 18 1.6 1 9.0 2 18 11 18 Solution 3.0 A Exercise 2.0 A the battery. Use R eq to calculate the total power delivered by Register to View AnswerCalculate the power delivered to each resistor and the total power delivered to the combination of resistors. We apply the relationship resistor and obtain 1 200 W. Solution ( V )2/R to each I1 I a I2 I3 V2 R1 V2 R2 V2 R3 (18 V )2 3.0 (18 V )2 6.0 (18 V )2 9.0 110 W 54 W 36 W 18 V 2 3.0 6.0 9.0 3 b This shows that the smallest resistor receives the most power. Summing the three quantities gives a total power of 200 W. Figure 28.7 Three resistors connected in parallel. The voltage across each resistor is 18 V. EXAMPLE 28.5 Finding Req by Symmetry Arguments Consider five resistors connected as shown in Figure 28.8a. Find the equivalent resistance between points a and b. 5 c 1 a 1 d (a) 1 5 1 a 1 1 b c,d 1 b 1 Solution In this type of problem, it is convenient to assume a current entering junction a and then apply symmetry 1/2 a c,d 1/2 b a 1 b (b) (c) (d) Figure 28.8 Because of the symmetry in this circuit, the 5- resistor does not contribute to the resistance between points a and b and therefore can be disregarded when we calculate the equivalent resistance. 876 CHAPTER 28 Direct Current Circuits be removed from the circuit and the remaining circuit then reduced as in Figures 28.8c and d. From this reduction we see that the equivalent resistance of the combination is 1 . Note that the result is 1 regardless of the value of the resistor connected between c and d. arguments. Because of the symmetry in the circuit (all 1- resistors in the outside loop), the currents in branches ac and ad must be equal; hence, the electric potentials at points c and d must be equal. This means that V cd 0 and, as a result, points c and d may be connected together without affecting the circuit, as in Figure 28.8b. Thus, the 5- resistor may CONCEPTUAL EXAMPLE 28.6 Operation of a Three-Way Lightbulb Figure 28.9 illustrates how a three-way lightbulb is constructed to provide three levels of light intensity. The socket of the lamp is equipped with a three-way switch for selecting different light intensities. The bulb contains two filaments. When the lamp is connected to a 120-V source, one filament receives 100 W of power, and the other receives 75 W. Explain how the two filaments are used to provide three different light intensities. Exercise Determine the resistances of the two filaments and their parallel equivalent resistance. Answer 144 , 192 , 82.3 . Solution The three light intensities are made possible by applying the 120 V to one filament alone, to the other filament alone, or to the two filaments in parallel. When switch S1 is closed and switch S2 is opened, current passes only through the 75-W filament. When switch S1 is open and switch S2 is closed, current passes only through the 100-W filament. When both switches are closed, current passes through both filaments, and the total power is 175 W. If the filaments were connected in series and one of them were to break, no current could pass through the bulb, and the bulb would give no illumination, regardless of the switch position. However, with the filaments connected in parallel, if one of them (for example, the 75-W filament) breaks, the bulb will still operate in two of the switch positions as current passes through the other (100-W) filament. 100-W filament 75-W filament S1 S2 120 V Figure 28.9 A three-way lightbulb. APPLICATION Strings of Lights In a parallel-wired string, each bulb operates at 120 V. By design, the bulbs are brighter and hotter than those on a series-wired string. As a result, these bulbs are inherently more dangerous (more likely to start a fire, for instance), but if one bulb in a parallel-wired string fails or is removed, the rest of the bulbs continue to glow. (A 25-bulb string of 4-W bulbs results in a power of 100 W; the total power becomes substantial when several strings are used.) A new design was developed for so-called "miniature" lights wired in series, to prevent the failure of one bulb from extinguishing the entire string. The solution is to create a connection (called a jumper) across the filament after it fails. (If an alternate connection existed across the filament before Strings of lights are used for many ornamental purposes, such as decorating Christmas trees. Over the years, both parallel and series connections have been used for multilight strings powered by 120 V.3 Series-wired bulbs are safer than parallel-wired bulbs for indoor Christmas-tree use because series-wired bulbs operate with less light per bulb and at a lower temperature. However, if the filament of a single bulb fails (or if the bulb is removed from its socket), all the lights on the string are extinguished. The popularity of series-wired light strings diminished because troubleshooting a failed bulb was a tedious, time-consuming chore that involved trialand-error substitution of a good bulb in each socket along the string until the defective bulb was found. 3 These and other household devices, such as the three-way lightbulb in Conceptual Example 28.6 and the kitchen appliances shown in this chapter's Puzzler, actually operate on alternating current (ac), to be introduced in Chapter 33. 28.3 Kirchhoff's Rules it failed, each bulb would represent a parallel circuit; in this circuit, the current would flow through the alternate connection, forming a short circuit, and the bulb would not glow.) When the filament breaks in one of these miniature lightbulbs, 120 V appears across the bulb because no current is present in the bulb and therefore no drop in potential occurs across the other bulbs. Inside the lightbulb, a small loop covered by an insulating material is wrapped around the filament leads. An arc burns the insulation and connects the filament leads when 120 V appears across the bulb -- that is, when the filament fails. This "short" now completes the circuit through the bulb even though the filament is no longer active (Fig. 28.10). 877 Suppose that all the bulbs in a 50-bulb miniature-light string are operating. A 2.4-V potential drop occurs across each bulb because the bulbs are in series. The power input to this style of bulb is 0.34 W, so the total power supplied to the string is only 17 W. We calculate the filament resistance at the operating temperature to be (2.4 V)2/(0.34 W) 17 . When the bulb fails, the resistance across its terminals is reduced to zero because of the alternate jumper connection mentioned in the preceding paragraph. All the other bulbs not only stay on but glow more brightly because the total resistance of the string is reduced and consequently the current in each bulb increases. Let us assume that the operating resistance of a bulb remains at 17 even though its temperature rises as a result of the increased current. If one bulb fails, the potential drop across each of the remaining bulbs increases to 2.45 V, the current increases from 0.142 A to 0.145 A, and the power increases to 0.354 W. As more lights fail, the current keeps rising, the filament of each bulb operates at a higher temperature, and the lifetime of the bulb is reduced. It is therefore a good idea to check for failed (nonglowingt 0, before the switch is closed. (c) Circuit diagram at time t 0, after the switch has been closed. difference across the resistor. We have used the sign conventions discussed earlier for the signs on and IR. For the capacitor, notice that we are traveling in the direction from the positive plate to the negative plate; this represents a decrease in potential. Thus, we use a negative sign for this voltage in Equation 28.11. Note that q and I are instantaneous values that depend on time (as opposed to steady-state values) as the capacitor is being charged. We can use Equation 28.11 to find the initial current in the circuit and the maximum charge on the capacitor. At the instant the switch is closed (t 0), the charge on the capacitor is zero, and from Equation 28.11 we find that the initial current in the circuit I 0 is a maximum and is equal to I0 R (current at t 0) (28.12) Maximum current At this time, the potential difference from the battery terminals appears entirely across the resistor. Later, when the capacitor is charged to its maximum value Q , charges cease to flow, the current in the circuit is zero, and the potential difference from the battery terminals appears entirely across the capacitor. Substituting I 0 into Equation 28.11 gives the charge on the capacitor at this time: Q C (maximum charge) (28.13) Maximum charge on the capacitor To determine analytical expressions for the time dependence of the charge and current, we must solve Equation 28.11 -- a single equation containing two variables, q and I. The current in all parts of the series circuit must be the same. Thus, the current in the resistance R must be the same as the current flowing out of and into the capacitor plates. This current is equal to the time rate of change of the charge on the capacitor plates. Thus, we substitute I dq /dt into Equation 28.11 and rearrange the equation: dq q dt R RC To find an expression for q, we first combine the terms on the right-hand side: dq dt C RC q RC q C RC 884 CHAPTER 28 Direct Current Circuits Now we multiply by dt and divide by q dq q C q 0 C to obtain 1 dt RC 0 at t t Integrating this expression, using the fact that q dq q C C C 1 RC t RC 0, we obtain dt 0 ln q From the definition of the natural logarithm, we can write this expression as Charge versus time for a capacitor being charged q(t ) C (1 e t/RC ) Q(1 e t /RC ) (28.14) where e is the base of the natural logarithm and we have made the substitution C Q from Equation 28.13. We can find an expression for the charging current by differentiating Equation 28.14 with respect to time. Using I dq /dt, we find that Current versus time for a charging capacitor I(t ) R e t /RC (28.15) Plots of capacitor charge and circuit current versus time are shown in Figure 28.17. Note that the charge is zero at t 0 and approaches the maximum value C as t : . The current has its maximum value I 0 /R at t 0 and decays exponentially to zero as t : . The quantity RC, which appears in the exponents of Equations 28.14 and 28.15, is called the time constant of the circuit. It represents the time it takes the current to decrease to 1/e of its initial value; that is, in a time , I e 1I 0 0.368I 0 . In a time 2 , I e 2I 0 0.135I 0 , and so forth. Likewise, in a time , the charge increases from zero to C (1 e 1) 0.632C . The following dimensional analysis shows that has the units of time: [ ] [RC] V I Q V Q Q/ t [ t] T q C 0.632 C I I0 = R =RC I0 0.368I0 (a) t (b) t (a) Plot of capacitor charge versus time for the circuit shown in Figure 28.16. After a time interval equal to one time constant has passed, the charge is 63.2% of the maximum value C . The charge approaches its maximum value as t approaches infinity. (b) Plot of current versus time for the circuit shown in Figure 28.16. The current has its maximum value I 0 /R at t 0 and decays to zero exponentially as t approaches infinity. After a time interval equal to one time constant has passed, the current is 36.8% of its initial value. Figure 28.17 28.4 RC Circuits 885 Because RCed with an ammeter connected in series with the resistor and battery of a circuit. An ideal ammeter has zero resistance. ries with other elements in the circuit, as shown in Figure 28.21. When using an ammeter to measure direct currents, you must be sure to connect it so that current enters the instrument at the positive terminal and exits at the negative terminal. Ideally, an ammeter should have zero resistance so that the current being measured is not altered. In the circuit shown in Figure 28.21, this condition requires that the resistance of the ammeter be much less than R 1 R 2 . Because any ammeter always has some internal resistance, the presence of the ammeter in the circuit slightly reduces the current from the value it would have in the meter's absence. The Voltmeter A device that measures potential difference is called a voltmeter. The potential difference between any two points in a circuit can be measured by attaching the terminals of the voltmeter between these points without breaking the circuit, as shown in Figure 28.22. The potential difference across resistor R 2 is measured by connecting the voltmeter in parallel with R 2 . Again, it is necessary to observe the polarity of the instrument. The positive terminal of the voltmeter must be connected to the end of the resistor that is at the higher potential, and the negative terminal to the end of the resistor at the lower potential. An ideal voltmeter has infinite resistance so that no current passes through it. In Figure 28.22, this condition requires that the voltmeter have a resistance much greater than R 2 . In practice, if this condition is not met, corrections should be made for the known resistance of the voltmeter. V R1 R2 Figure 28.22 The potential difference across a resistor can be measured with a voltmeter connected in parallel with the resistor. An ideal voltmeter has infinite resistance. The Galvanometer The galvanometer is the main component in analog ammeters and voltmeters. Figure 28.23a illustrates the essential features of a common type called the D'Arsonval galvanometer. It consists of a coil of wire mounted so that it is free to rotate on a pivot in a magnetic field provided by a permanent magnet. The basic op- Scale N S Spring Coil (a) (b) Figure 28.23 (a) The principal components of a D'Arsonval galvanometer. When the coil situated in a magnetic field carries a current, the magnetic torque causes the coil to twist. The angle through which the coil rotates is proportional to the current in the coil because of the counteracting torque of the spring. (b) A large-scale model of a galvanometer movement. Why does the coil rotate about the vertical axis after the switch is closed? 28.5 Electrical Instruments Galvanometer 889 60 Galvanometer Rs 60 Rp (a) (b) Figure 28.24 (a) When a galvanometer is to be used as an ammeter, a shunt resistor R p is connected in parallel with the galvanometer. (b) When the galvanometer is used as a voltmeter, a resistor R s is connected in series with the galvanometer. eration of the galvanometer makes use of the fact that a torque acts on a current loop in the presence of a magnetic field (Chapter 29). The torque experienced by the coil is proportional to the current through it: the larger the current, the greater the torque and the more the coil rotates before the spring tightens enough to stop the rotation. Hence, the deflection of a needle attached to the coil is proportional to the current. Once the instrument is properly calibrated, it can be used in conjunction with other circuit elements to measure either currents or potential differences. A typical off-the-shelf galvanometer is often not suitable for use as an ammeter, primarily because it has a resistance of about 60 . An ammeter resistance this great considerably alters the current in a circuit. You can understand this by considering the following example: The current in a simple series circuit containing a 3-V battery and a 3- resistor is 1 A. If you insert a 60- galvanometer in this circuit to measure the current, the tot8.33 A. The microwave oven, rated at 1 300 W, draws 10.8 A, and the coffee maker, rated at 800 W, draws 6.67 A. If the three appliances are operated simultaneously, they draw a total cur5 120 V Live Neutral Meter Circuit breaker R1 R2 R3 0V Figure 28.28 Live wire is a common expression for a conductor whose electric potential is above or below ground potential. Wiring diagram for a household circuit. The resistances represent appliances or other electrical devices that operate with an applied voltage of 120 V. 892 CHAPTER 28 Direct Current Circuits Figure 28.29 A power connection for a 240-V appliance. rent of 25.8 A. Therefore, the circuit should be wired to handle at least this much current. If the rating of the circuit breaker protecting the circuit is too small -- say, 20 A -- the breaker will be tripped when the third appliance is turned on, preventing all three appliances from operating. To avoid this situation, the toaster oven and coffee maker can be operated on one 20-A circuit and the microwave oven on a separate 20-A circuit. Many heavy-duty appliances, such as electric ranges and clothes dryers, require 240 V for their operation (Fig. 28.29). The power company supplies this voltage by providing a third wire that is 120 V below ground potential. The potential difference between this live wire and the other live wire (which is 120 V above ground potential) is 240 V. An appliance that operates from a 240-V line requires half the current of one operating from a 120-V line; therefore, smaller wires can be used in the higher-voltage circuit without overheating. Electrical Safety When the live wire of an electrical outlet is connected directly to ground, the circuit is completed and a short-circuit condition exists. A short circuit occurs when almost zero resistance exists between two points at different potentials; this results in a very large current. When this happens accidentally, a properly operating circuit breaker opens the circuit and no damage is done. However, a person in contact with ground can be electrocuted by touching the live wire of a frayed cord or other exposed conductor. An exceptionally good (although very dangerous) ground contact is made when the person either touches a water pipe (normally at ground potential) or stands on the ground with wet feet. The latter situation represents a good ground because normal, nondistilled water is a conductor because it contains a large number of ions associated with impurities. This situation should be avoided at all cost. Electric shock can result in fatal burns, or it can cause the muscles of vital organs, such as the heart, to malfunction. The degree of damage to the body depends on the magnitude of the current, the length of time it acts, the part of the body touched by the live wire, and the part of the body through which the current passes. Currents of 5 mA or less cause a sensation of shock but ordinarily do little or no damage. If the current is larger than about 10 mA, the muscles contract and the person may be unable to release the live wire. If a current of about 100 mA passes through the body for only a few seconds, the result can be fatal. Such a large current paralyzes the respiratory muscles and prevents breathing. In some cases, currents of about 1 A through the body can produce serious (and sometimes fatal) burns. In practice, no contact with live wires is regarded as safe whenever the voltage is greater than 24 V. Many 120-V outlets are designed to accept a three-pronged power cord such as the one shown in Figure 28.30. (This feature is required in all new electrical installations.) One of these prongs is the live wire at a nominal potential of 120 V. The second, called the "neutral," is nominally at 0 V and carries current to ground. The third, round prong is a safety ground wire that normally carries no current but is both grounded and connected directly to the casing of the appliance. If the live wire is accidentally shorted to the casing (which can occur if the wire insulation wears off), most of the current takes the low-resistance path through the appliance to ground. In contrast, if the casing of the appliance is not properly grounded and a short occurs, anyone in contact with the appliance experiences an electric shock because the body provides a low-resistance path to ground. Figure 28.30 A three-pronged power cord for a 120-V appliance. Summary 893 Special power outlets called ground-fault interrupters (GFIs) are now being used in kitchens, bathrooms, basements, exterior outlets, and other hazardous areas of new homes. These devices are designed to protect persons from electric shock by sensing small currents ( 5 mA) leaking to ground. (The principle of their operation is described in Chapter 31.) When an excessive leakage current is detected, the current is shut off in less than 1 ms. Quick Quiz 28.4 Is a circuit breaker wired in series or in parallel with the device it is protecting? SUMMARY The emf of a battery is equal to the voltage across its terminals when the current is zero. That is, the emf is equivalent to the open-circuit voltage of the battery. The equivalent resistance of a set of resistors connected in series is R eq R1 R2 R3 (28.6) The equivalent resistance of a set of resistors connected in parallel is 1 R eq 1 R1 1 R2 1 R3 (28.8) If it is possible to combine resistors into series or parallel equivalents, the preceding two equations make it easy to determine how the resistors influence the rest of the circuit. Circuits involving more than one loop are conveniently analyzed with the use of Kirchhoff 's rules: 1. The sum of the currents entering any junction in an electric circuit must equal the sum of the currents leaving that junction: I in I out (28.9) 2. The sum of the potential differences across all elements around any circuit loop must be zero: V closed loop 0 (28.10) The first rule is a statement of conservation of charge; the second is equivalent to a statement of conservation of energy. When a resistor is traversed in the direction of the current, the change in potential V across the resistor is IR. When a resistor is traversed in the direction IR . When a source of emf is traversed in the direcopposite the current, V tion of the emf (negative terminal to positive terminal), the change in potential is . When a source of emf is traversed opposite the emf (positive to negative), . The use of these rules together with Equations 28.9 the change in potential is and 28.10 allows you to analyze electric circuits. If a capacitor is charged with a battery through a resistor of resistance R, the charge on the capacitor and the current in the circuit vary in time according to 894 CHAPTER 28 Direct Current Circuits the expressions q(t) I(t) Q(1 R e e t /RC ) (28.14) (28.15) t /RC where Q C is the maximum charge on the capacitor. The product RC is called the time constant of the circuit. If a charged capacitor is discharged through a resistor of resistance R, the charge and current decrease exponentially in time according to the expressions q(t) I(t) Qe t/RC (28.17) t/RC Q e RC (28.18) where Q is the initial charge on the capacitor and Q /RC I 0 is the initial current in the circuit. Equations 28.14, 28.15, 28.17, and 28.18 permit you to analyze the current and potential differences in an RC circuit and the charge stored in the circuit's capacitor. QUESTIONS 1. Explain the difference between load resistance in a circuit and internal resistance in a battery. 2. Under what condition does the potential difference across the terminals of a battery equal its emf ? Can the terminal voltage ever exceed the emf ? Explain. 3. Is the direction of current through a battery always from the negative terminal to the positive one? Explain. 4. How would you connect resistors so that the equivalent resistance is greater than the greatest individual resistance? Give an example involving three resistors. 5. How would you connect resistors so that the equivalent resistance is less than the least individual resistance? Give an example involving three resistors. 6. Given three lightbulbs and a battery, sketch as 0 V. The terminal voltage of the battery is 11.6 V when it is delivering 20.0 W of power to an external load resistor R. (a) What is the value of R ? (b) What is the internal resistance of the battery? 2. (a) What is the current in a 5.60- resistor connected to a battery that has a 0.200- internal resistance if the terminal voltage of the battery is 10.0 V ? (b) What is the emf of the battery? 3. Two 1.50-V batteries -- with their positive terminals in the same direction -- are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.255 , the other an internal resistance of 0.153 . When the switch is closed, a current of 600 mA occurs in the lamp. (a) What is the lamp's resistance? (b) What percentage of the power from the batteries appears in the batteries themselves, as represented by an increase in temperature? 4. An automobile battery has an emf of 12.6 V and an internal resistance of 0.080 0 . The headlights have a total resistance of 5.00 (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor, which takes an additional 35.0 A from the battery, is operated? Section 28.2 Resistors in Series and in Parallel 5. The current in a loop circuit that has a resistance of R 1 is 2.00 A. The current is reduced to 1.60 A when an additional resistor R 2 3.00 is added in series with R 1 . What is the value of R 1 ? 6. (a) Find the equivalent resistance between points a and b in Figure P28.6. (b) Calculate the current in each resistor if a potential difference of 34.0 V is applied between points a and b. 7. A television repairman needs a 100- resistor to repair a malfunctioning set. He is temporarily out of resistors 896 7.00 4.00 CHAPTER 28 Direct Current Circuits the power delivered to each resistor? What is the total power delivered? 12. Using only three resistors -- 2.00 , 3.00 , and 4.00 -- find 17 resistance values that can be obtained with various combinations of one or more resistors. Tabulate the combinations in order of increasing resistance. 13. The current in a circuit is tripled by connecting a 500resistor in parallel with the resistance of the circuit. Determine the resistance of the circuit in the absence of the 500- resistor. 14. The power delivered to the top part of the circuit shown in Figure P28.14 does not depend on whether the switch is opened or closed. If R 1.00 , what is R ? Neglect the internal resistance of the voltage source. R 9.00 10.0 a b Figure P28.6 of this value. All he has in his toolbox are a 500- resistor and two 250- resistors. How can he obtain the desired resistance using the resistors he has on hand? 8. A lightbulb marked "75 W [at] 120 V" is screwed into a socket at one end of a long extension cord in which each of the two conductors has a resistance of 0.800 . The other end of the extension cord is plugged into a 120-V outlet. Draw a circuit diagram, and find the actual power delivered to the bulb in this circuit. 9. Consider the circuit shown in Figure P28.9. Find (a) the current in the 20.0- resistor and (b) the potential difference between points a and b. S R WEB R 10.0 a 10.0 25.0 V b Figure P28.14 15. Calculate the power delivered to each resistor in the circuit shown in Figure P28.15. 2.00 5.00 5.00 20.0 Figure P28.9 10. Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.00 cm2, 2.00 cm2, 3.00 cm2, and 5.00 cm2. If a voltage of 120 V is applied to the arrangement, what is the voltage across the 2.00-cm2 wire? 11. Three 100- resistors are connected as shown in Figure P28.11. The maximum power that can safely be delivered to any one resistor is 25.0 W. (a) What is the maximum voltage that can be applied to the terminals a and b? (b) For the voltage determined in part (a), what is 100 a 100 b 18.0 V 3.00 1.00 4.00 Figure P28.15 16. Two resistors connected in series have an equivalent resistance of 690 . When they are connected in parallel, their equivalent resisume that the resistivity and volume of the wire stay constant. 45. Consider the potentiometer circuit shown in Figure 28.27. If a standard battery with an emf of 1.018 6 V is used in the circuit and the resistance between a and d is 36.0 , the galvanometer reads zero. If the standard battery is replaced by an unknown emf, the galvanometer reads zero when the resistance is adjusted to 48.0 . What is the value of the emf ? 46. Meter loading. Work this problem to five-digit precision. Refer to Figure P28.46. (a) When a 180.00- resistor is put across a battery with an emf of 6.000 0 V and an internal resistance of 20.000 , what current flows in the resistor? What will be the potential difference across it? (b) Suppose now that an ammeter with a resistance of 0.500 00 and a voltmeter with a resistance of 6.000 0 V 20.000 V A V A 180.00 (a) (b) (c) Figure P28.46 900 CHAPTER 28 Direct Current Circuits 4.00 V a 20 000 are added to the circuit, as shown in Figure P28.46b. Find the reading of each. (c) One terminal of one wire is moved, as shown in Figure P28.46c. Find the new meter readings. (Optional) 12.0 V 2.00 Section 28.6 Household Wiring and Electrical Safety WEB 4.00 47. An electric heater is rated at 1 500 W, a toaster at 750 W, and an electric grill at 1 000 W. The three appliances are connected to a common 120-V circuit. (a) How much current does each draw? (b) Is a 25.0-A circuit breaker sufficient in this situation? Explain your answer. 48. An 8.00-ft extension cord has two 18-gauge copper wires, each with a diameter of 1.024 mm. What is the I 2R loss in this cord when it carries a current of (a) 1.00 A? (b) 10.0 A? 49. Sometimes aluminum wiring has been used instead of copper for economic reasons. According to the National Electrical Code, the maximum allowable current for 12-gauge copper wire with rubber insulation is 20 A. What should be the maximum allowable current in a 12-gauge aluminum wire if it is to have the same I 2R loss per unit length as the copper wire? 50. Turn on your desk lamp. Pick up the cord with your thumb and index finger spanning its width. (a) Compute an order-of-magnitude estimate for the current that flows through your hand. You may assume that at a typical instant the conductor inside the lamp cord next to your thumb is at potential 10 2 V and that the conductor next to your index finger is at ground potential (0 V). The resistance of your hand depends strongly on the thickness and moisture content of the outer layers of your skin. Assume that the resistance of your hand between fingertip and thumb tip is 10 4 . You may model the cord as having rubber insulation. State the other quantities you measure or estimate and their values. Explain your reasoning. (b) Suppose that your body is isolated from any other charges or currents. In order-of-magnitude terms, describe the potential of your thumb where it contacts the cord and the potential of your finger where it touches the cord. 10.0 b Figure P28.53 54. A 10.0- F capacitor is charged by a 10.0-V battery through a resistance R . The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. Find R . 55. When two unknown resistors are connected in series with a battery, 225 W is delivered to the combination with a total current of 5.00 A. For the same total current, 50.0 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors. 56. When two unknown resistors are connected in series with a battery, a total power s is delivered to the combination with a total current of I. For the same total current, a total power p is delivered when the resistors are connected in parallel. Determine the values of the two resistors. 57. A battery has an emf and internal resistance r. A variable resistor R is connected across the terminals of the battery. Determine the value of R such that (a) the potential difference across the terminals is a maximum, (b) the current in the circuit is a maximum, (c) the power delivered to the resistor is a maximum. 58. A pow 29.8 A wire segment of arbitrary shape carrying a current I in a magnetic field B experiences a magnetic force. The force on any segment d s is I ds B and is directed out of the page. You should use the right-hand rule to confirm this force direction. where d FB is directed out of the page for the directions assumed in Figure 29.8. We can consider Equation 29.4 as an alternative definition of B. That is, we can define the magnetic field B in terms of a measurable force exerted on a current element, where the force is a maximum when B is perpendicular to the element and zero when B is parallel to the element. To calculate the total force FB acting on the wire shown in Figure 29.8, we integrate Equation 29.4 over the length of the wire: b FB I a ds B (29.5) where a and b represent the end points of the wire. When this integration is carried out, the magnitude of the magnetic field and the direction the field makes with the vector ds (in other words, with the orientation of the element) may differ at different points. Now let us consider two special cases involving Equation 29.5. In both cases, the magnetic field is taken to be constant in magnitude and direction. Case 1 A curved wire carries a current I and is located in a uniform magnetic field B, as shown in Figure 29.9a. Because the field is uniform, we can take B outside the integral in Equation 29.5, and we obtain b FB I a ds B (29.6) I B b L I a ds (a) (b) B ds (a) A curved wire carrying a current I in a uniform magnetic field. The total magnetic force acting on the wire is equivalent to the force on a straight wire of length L running between the ends of the curved wire. (b) A current-carrying loop of arbitrary shape in a uniform magnetic field. The net magnetic force on the loop is zero. Figure 29.9 29.2 Magnetic Force Acting on a Current-Carrying Conductor 913 But the quantity b ds represents the vector sum of all the length elements from a to a b. From the law of vector addition, the sum equals the vector L , directed from a to b. Therefore, Equation 29.6 reduces to FB IL B (29.7) Case 2 An arbitrarily shaped closed loop carrying a current I is placed in a uniform magnetic field, as shown in Figure 29.9b. We can again express the force acting on the loop in the form of Equation 29.6, but this time we must take the vector sum of the length elements ds over the entire loop: FB I ds B Because the set of length elements forms a closed polygon, the vector sum must be zero. This follows from the graphical procedure for adding vectors by the polygon method. Because ds 0, we conclude that FB 0: The net magnetic force acting on any closed current loop in a uniform magnetic field is zero. EXAMPLE 29.2 Force on a Semicircular Conductor curved wire must also be into the page. Integrating our expression for dF2 over the limits 0 to (that is, the entire semicircle) gives F2 IRB 0 A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the xy plane, and a uniform magnetic field is directed along the positive y axis, as shown in Figure 29.10. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion. The force F1 acting on the straight portion has a magnitude F 1 ILB 2IRB because L 2R and the wire is oriented perpendicular to B. The direction of F1 is out of the page because L B is along the positive z axis. (That is, L is to the right, in the direction of the current; thus, according to the rule of cross products, L B is out of the page in Fig. 29.10.) To find the force F2 acting on the curved part, we first write an expression for the force d F2 on the length element d s shown in Figure 29.10. If is the angle between B and ds, then the magnitude of d F2 is dF2 I ds B IB sin ds sin d IRB cos 0) cos 0 Solution IRB(cos IRB( 1 1) 2IRB Because F2, with a magnitude of 2IRB , is directed into the page and because F1, with a magnitude of 2IRB , is directed out of the page, the net force on the closed loop is zero. This result is consias the same dimensions as the copper strip in this example and whose value for n 1.0 10 20 electrons/m3. Taking B 1.2 T and I 0.10 mA, we find that V H 7.5 mV. A potential difference of this magnitude is readily measured. 928 CHAPTER 29 Magnetic Fields SUMMARY The magnetic force that acts on a charge q moving with a velocity v in a magnetic field B is FB qv B (29.1) The direction of this magnetic force is perpendicular both to the velocity of the particle and to the magnetic field. The magnitude of this force is FB q vB sin (29.2) where is the smaller angle between v and B. The SI unit of B is the tesla (T), where 1 T 1 N/A m. When a charged particle moves in a magnetic field, the work done by the magnetic force on the particle is zero because the displacement is always perpendicular to the direction of the force. The magnetic field can alter the direction of the particle's velocity vector, but it cannot change its speed. If a straight conductor of length L carries a current I, the force exerted on that conductor when it is placed in a uniform magnetic field B is FB IL B (29.3) L. where the direction of L is in the direction of the current and L If an arbitrarily shaped wire carrying a current I is placed in a magnetic field, the magnetic force exerted on a very small segment ds is dFB I ds B (29.4) To determine the total magnetic force on the wire, one must integrate Equation 29.4, keeping in mind that both B and ds may vary at each point. Integration gives for the force exerted on a current-carrying conductor of arbitrary shape in a uniform magnetic field FB IL B (29.7) where L is a vector directed from one end of the conductor to the opposite end. Because integration of Equation 29.4 for a closed loop yields a zero result, the net magnetic force on any closed loop carrying a current in a uniform magnetic field is zero. The magnetic dipole moment of a loop carrying a current I is IA (29.10) where the area vector A is perpendicular to the plane of the loop and A is equal to the area of the loop. The SI unit of is A m2. The torque on a current loop placed in a uniform magnetic field B is B and the potential energy of a magnetic dipole in a magnetic field is U B (29.12) (29.11) If a charged particle moves in a uniform magnetic field so that its initial velocity is perpendicular to the field, the particle moves in a circle, the plane of which is perpendicular to the magnetic field. The radius of the circular path is r mv qB (29.13) Questions 929 where m is the mass of the particle and q is its charge. The angular speed of the charged particle is qB (29.14) m QUESTIONS 1. At a given instant, a proton moves in the positive x direction in a region where a magnetic field is directed in the negative z direction. What is the direction of the magnetic force? Does the proton continue to move in the positive x direction? Explain. 2. Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what can be said about them? 3. If a charged particle moves in a straight line through some region of space, can one say that the magnetic field in that region is zero? 4. Suppose an electron is chasing a proton up this page when suddenly a magnetic field directed perpendicular into the page is turned on. What happens to the particles? 5. How can the motion of a moving charged particle be used to distinguish between a magnetic field and an electric field? Give a specific example to justify your argument. 6. List several similarities and differences between electric and magnetic forces. 7. Justify the following statement: "It is impossible for a constant (in other words, a time-independent) magnetic field to alter the speed of a charged particle." 8. In view of the preceding statement, what is the role of a magnetic field in a cyclotron? 9. A current-carrying conductor experiences no magnetic force when placed in a certain manner in a uniform magnetic field. Explain. 10. Is it possible to orient a current loop in a unside of the loop sets up a perpendicular component of d B that cancels the perpendicular component set up by the current through the element diametrically opposite it. Therefore, the resultant field at P must be along the x axis and we can find it by integrating the components dB x dB cos . That is, B B x i, where Bx dB cos 0I Because the magnitude of the magnetic moment of the loop is defined as the product of current and loop area (see Eq. 29.10) -- I( R 2 ) for our circular loop -- we can express Equation 30.9 as B 0 2 x3 (30.10) This result is similar in form to the expression for the electric field due to an electric dipole, E k e(2qa/y 3 ) (see Example y 4 ds cos x2 R 2 ds and we must take the integral over the entire loop. Because , x, and R are constants for all elements of the loop and because cos R /(x 2 R 2 )1/2, we obtain Bx 0IR ^ r R z O x I I dBy r dB 4 (x 2 R 2 )3/2 ds 0IR 2 2(x 2 R 2)3/2 (30.7) P dBx x where we have used the fact that ds 2 R (the circumference of the loop). To find the magnetic field at the center of the loop, we set x 0 in Equation 30.7. At this special point, therefore, Geometry for calculating the magnetic field at a point P lying on the axis of a current loop. By symmetry, the total field B is along this axis. Figure 30.5 30.2 The Magnetic Force Between Two Parallel Conductors 23.6), where 2qa p is the electric dipole moment as defined in Equation 26.16. The pattern of the magnetic field lines for a circular current loop is shown in Figure 30.6a. For clarity, the lines are 943 drawn for only one plane -- one that contains the axis of the loop. Note that the field-line pattern is axially symmetric and looks like the pattern around a bar magnet, shown in Figure 30.6c. N N I S S (a) (b) (c) Figure 30.6 (a) Magnetic field lines surrounding a current loop. (b) Magnetic field lines surrounding a current loop, displayed with iron filings (Education Development Center, Newton, MA). (c) Magnetic field lines surrounding a bar magnet. Note the similarity between this line pattern and that of a current loop. 30.2 THE MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS In Chapter 29 we described the magnetic force that acts on a current-carrying conductor placed in an external magnetic field. Because a current in a conductor sets up its own magnetic field, it is easy to understand that two current-carrying conductors exert magnetic forces on each other. As we shall see, such forces can be used as the basis for defining the ampere and the coulomb. Consider two long, straight, parallel wires separated by a distance a and carrying currents I 1 and I 2 in the same direction, as illustrated in Figure 30.7. We can determine the force exerted on one wire due to the magnetic field set up by the other wire. Wire 2, which carries a current I2 , creates a magnetic field B2 at the location of wire 1. The direction of B2 is perpendicular to wire 1, as shown in Figure 30.7. According to Equation 29.3, the magnetic force on a length of wire 1 is F1 I 1 B2. Because is perpendicular to B2 in this situation, the magnitude of F1 is F 1 I 1 B 2 . Because the magnitude of B2 is given by Equation 30.5, we see that F1 I 1 B2 I1 0I2 0 I 1I 2 1 I1 B2 2 a F1 a I2 2 a 2 a (30.11) The direction of F1 is toward wire 2 because B2 is in that direction. If the field set up at wire 2 by wire 1 is calculated, the force F2 acting on wire 2 is found to be equal in magnitude and opposite in direction to F1 . This is what we expect be- Figure 30.7 Two parallel wires that each carry a steady current exert a force on each other. The field B2 due to the current in wire 2 exerts a force of magnitude I 1 B 2 on wire 1. The force is F1 attractive if the currents are parallel (as shown) and repulsive if the currents are antiparallel. 944 CHAPTER 30 Sources of the Magnetic Field cause Newton's third law must be obeyed.1 When the currents are in opposite directions (that is, when one of the currents is reversed in Fig. 30.7), the forces are reversed and the wires repel each other. Hence, we find mass of the electron with the mass of a proton or a neutron. Because the masses of the proton and neutron are much greater than that of the electron, their magnetic moments are on the order of 103 times smaller than that of the electron. TABLE 30.1 Magnetic Moments of Some Atoms and Ions Atom or Ion H He Ne Ce3 Yb3 Magnetic Moment (10 24 J/T) 9.27 0 0 19.8 37.1 Magnetization Vector and Magnetic Field Strength The magnetic state of a substance is described by a quantity called the magnetization vector M. The magnitude of this vector is defined as the magnetic moment per unit volume of the substance. As you might expect, the total magnetic field B at a point within a substance depends on both the applied (external) field B0 and the magnetization of the substance. To understand the problems involved in measuring the total magnetic field B in such situations, consider this: Scientists use small probes that utilize the Hall efMagnetization vector M 958 CHAPTER 30 Sources of the Magnetic Field fect (see Section 29.6) to measure magnetic fields. What would such a probe read if it were positioned inside the solenoid mentioned in the QuickLab on page 951 when you inserted the compass? Because the compass is a magnetic material, the probe would measure a total magnetic field B that is the sum of the solenoid (external) field B0 and the (magnetization) field Bm due to the compass. This tells us that we need a way to distinguish between magnetic fields originating from currents and those originating from magnetic materials. Consider a region in which a magnetic field B0 is produced by a current-carrying conductor. If we now fill that region with a magnetic substance, the total magnetic field B in the region is B B0 Bm , where Bm is the field produced by the magnetic substance. We can express this contribution in terms of the magnetization vector of the substance as Bm 0 M; hence, the total magnetic field in the region becomes B Magnetic field strength H B0 0M (30.29) When analyzing magnetic fields that arise from magnetization, it is convenient to introduce a field quantity, called the magnetic field strength H within the substance. The magnetic field strength represents the effect of the conduction currents in wires on a substance. To emphasize the distinction between the field strength H and the field B, the latter is often called the magnetic flux density or the magnetic induction. The magnetic field strength is a vector defined by the relationship H B0 / 0 (B/ 0 ) M. Thus, Equation 30.29 can be written B 0(H M) (30.30) The quantities H and M have the same units. In SI units, because M is magnetic moment per unit volume, the units are (ampere)(meter)2/(meter)3, or amperes per meter. To better understand these expressions, consider the torus region of a toroid that carries a current I. If this region is a vacuum, M 0 (because no magnetic material is present), the total magnetic field is that arising from the current alone, and B B0 0 H. Because B 0 0nI in the torus region, where n is the number of turns per unit length of the toroid, H B 0 / 0 0nI/ 0 , or H nI (30.31) In this case, the magnetic field B in the torus region is due only to the current in the windings of the toroid. If the torus is now made of some substance and the current I is kept constant, H in the torus region remains unchanged (because it depends on the current only) and has magnitude nI. The total field B, however, is different from that when the torus region was a vacuum. From Equation 30.30, we see that part of B arises from the term 0 H associated with the current in the toroid, and part arises from the term 0 M due to the magnetization of the substance of which the torus is made. Classification of Magnetic Substances Substances can be classified as belonging to one of three categories, depending on their magnetic properties. Paramagnetic and ferromagnetic materials are those made of atoms that have permanent magnetic moments. Diamagnetic materials are those made of atoms that do not have permanent magnetic moments. For paramagnetic and diamagnetic substances, the magnetization vector M is proportional to the magnetic field strength H. For these substances placed in an external magnetic field, we can write M H (30.32) Oxygen, a paramagnetic substance, is attracted to a magnetic field. The liquid oxygen in this photograph is suspended between the poles of the magnet. 30.8 Magnetism in Matter 959 TABLE 30.2 Magnetic Susceptibilities of Some Paramagnetic and Diamagnetic Substances at 300 K Paramagnetic Substance Aluminum Calcium Chromium Lithium Magnesium Niobium Oxygen Platinum Tungsten 2.3 1.9 2.7 2.1 1.2 2.6 2.1 2.9 6.8 10 10 10 10 10 10 10 10 10 5 5 4 5 5 4 6 4 5 Diamagnetic Substance Bismuth Copper Diamond Gold Lead Mercury Nitrogen Silver Silicon 1.66 9.8 2.2 3.6 1.7 2.9 5.0 2.6 4.2 10 10 10 10 10 10 10 10 10 5 6 5 5 5 5 9 5 6 where (Greek letter chi) is a dimensionless factor called the magnetic susceptibility. For paramagnetic substances, is positive and M is in the same direction as H. For diamagnetic substances, is negative and M is opposite H. (It is important to note that this linear relationship between M and H does not apply to ferromagnetic substances.) The susceptibilities of some substances are given in Table 30.2. Substituting Equation 30.32 for M into Equation 30.30 gives B or B mH 0(H Magnetic susceptibility M) 0(H H) 0(1 )H (30.33) where the constant m is called the magnetic permeability of the substance and is related to the susceptibility by m 0(1 ) (30.34) m Magnetic permeability m Substances may be classified in terms of how their magnetic permeability compares with 0 (the permeability of free space), as follows: Paramagnetic Diamagnetic m m 0 0 Because is very small for paramagnetic and diamagnetic substances (see Table 30.2), m is nearly equal to 0 for these substances. For ferromagnetic substances, however, m is typically several thousand times greater than 0 (meaning that is very great for ferromagnetic substances). Although Equation 30.33 provides a simple relationship between B and H, we must interpret it with care when dealing with ferromagnetic substances. As mentioned earlier, M is not a linear function of H for ferromagnetic substances. This is because the value of m is not only a characteristic of the ferromagnetic substance but also depends on the previous state of the substance and on the process it underwent as it moved from its previous state to its present one. We shall investigate this more deeply after the following example. 960 CHAPTER 30 Sources of the Magnetic Field EXAMPLE 30.10 An Iron-Filled Toroid B mH A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. The torus is iron, which has a magnetic permeability of m 5 000 0 under the given conditions. Find H and B inside the iron. 5 000 10 0H 7 5 000 4 T m A 300 A turns m 1.88 T This value of B is 5 000 times the value in the absence of iron! Solution H nI Using Equations 30.31 and 30.33, we obtain turns (5.00 A) m A turns m Exercise Answer Determine the magnitude of the magnetization vector inside the iron torus. M 1.5 10 6 A/m. 60.0 300 Quick Quiz 30.7 A current in a solenoid having air in the interior creates a magnetic field B 0 H. Describe qualitatively what happens to the magnitude of B as (a) aluminum, (b) copper, and (c) iron are placed in the interior. Ferromagnetism A small number of crystalline substances in which the atoms have permanent magnetic moments exhibit strong magnetic effects called ferromagnetism. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and dysprosium. These substances contain atomic magnetic moments that tend to align parallel to each other even in a weak external magnetic field. Once the moments are aligned, the substance remains magnetized after the external field is removed. This permanent alignment is due to a strong coupling between neighboring moments, a coupling that can be understood only in quantum-mechanical terms. All ferromagnetic materials are made up of microscopic regions called domains, regions within which all magnetic moments are aligned. These domains have volumes of about 10 12 to 10 8 m3 and contain 1017 to 1021 atoms. The boundaries between the various domains having different orientations are called domain walls. In an unmagnetized sample, the domains are randomly oriented so that the net magnetic moment is zero, as shown in Figure 30.28a. When the sample is placed in an external magnetic field, the magnetic moments of the atoms tend to align with the field, which results in a magnetized sample, as in Figure 30.28b. Observations show that domains initially oriented along the external field grow larger at the expense of the less favorably oriented domains. When the external field is removed, the sample may retain a net magnetization in the direction of the original field. At ordinary temperatures, thermal agitation is not sufficient to disrupt this preferred orientation of magnetic moments. A typical experimental arrangement that is used to measure the magnetic properties of a ferromagnetic material consists of a torus made of the material wound with N turns of wire, as shown in Figure 30.29, where the windings are represented in black and are referred to as the primary coil. This apparatus is sometimes referred to as a Rowland ring. A secondary coil (the red wires in Fig. 30.29) connected to a galvanometer is used to measure the total magnetic flux through the torus. The magnetic field B in the torus is measured by increasing the current in the toroid from zero to I . As the current changes, the magnetic flux through (a) B0 (b) Figure 30.28 (a) Random orientation of atomic magnetic moments in an unmagnetized substance. (b) When an external field B0 is applied, the atomic magnetic moments tend to align with the field, giving the sample a net magnetization vector M. 30.8 Magnetism in Matter 961 the secondary coil changes by an amount BA, where A is the cross-sectional area of the toroid. As we shall find in Chapter 31, because of this changing flux, an emf that is proportional to the rate of change in magnetic flux is induced in the secondary coil. If the galvanometer is properly calibrated, a value for B corresponding to any value of the current in the primary coil can be obtained. The magnetic field B is measured first in the absence of the torus and then with the torus in place. The magnetic properties of the torus material are then obtained from a comparison of the two measurements. Now consider a torus made of unmagnetized iron. If the current in the primary coil is increased from zero to some value I, the magnitude of the magnetic field strength H increases linearly with I according to the expression H nI. Furthermore, the magnitude of the total field B also increases with increasing current, as shown by the curve from point O to point a in Figure 30.30. At point O, the domains in the iron are randomly oriented, corresponding to B m 0. As the increasing current in the primary coil causes the external field B0 to increase, the domains become more aligned until all of them are nearly aligned at point a. At this point the iron core is approaching saturation, which is the condition in which all domains in the iron are aligned. Next, suppose that the current is reduced to zero, and the external field is consequently eliminated. The B versus H curve, called a magnetization curve, now follows the path ab in Figure 30.30. Note that at point b, B is not zero even though the external field is B0 0. The reason is that the iron is now magnetized due to the alignment of a large number of its domains (that is, B Bm ). At this point, the iron is said to have a remanent magnetization. If the current in the primary coil is reversed so that the direction of the external magnetic field is reversed, the domains reorient until the sample is again unmagnetized at point c, where B 0. An increase in the reverse current causes the iron to be magnetized in the opposite direction, approaching saturation at point d in Figure 30.30. A similar sequence of events occurs as the current is reduced to zero and then increased in the original (positive) direction. In this case the magnetization curve follows the path def. If the current is increased sufficiently, the magnetization curve returns to point a, where the sample again has its maximum magnetization. The effect just described, called magnetic hysteresis, shows that the magnetization of a ferromagnetic substance depends on the history of the substance as well as on the magnitude of the applied field. (The word hysteresis means "lagging behind.") It is often said that a ferromagnetic substance has a "memory" because it remains magnetized after the external field is removed. The closed loop in Figure 30.30 is referred to as a hysteresis loop. Its shape and size depend on the proper- QuickLab You've probably done this experiment before. Magnetize a nail by repeatedly dragging it across a bar magnet. Test the strength of the nail's magnetic field by picking up some paper clips. Now hit the nail several times with a hammer, and again test the strength of its magnetism. Explain what happens in terms of domains in the steel of the nail. S R G Figure 30.29 A toroidal winding arrangement used to measure the magnetic properties of a material. The torus is made of the material under study, and the circuit containing the galvanometer measures the magnetic flux. B a b c O f H e d Figure 30.30 material. Magnetization curve for a ferromagnetic 962 CHAPTER 30 Sources of the Magnetic Field B B H H (a) (b) Figure 30.31 netic material. B Hysteresis loops for (a) a hard ferromagnetic material and (b) a soft ferromag- H ties of the ferromagnetic substance and on the strength of the maximum applied field. The hysteresis loop for "hard" ferromagnetic materials is characteristically wide like the one shown in Figure 30.31a, corresponding to a large remanent magnetization. Such materials cannot be easily demagnetized by an external field. "Soft" ferromagnetic materials, such as iron, have a very narrow hysteresis loop and a small remanent magnetization (Fig. 30.31b.) Such materials are easily magnetized and demagnetized. An ideal soft ferromagnet would exhibit no hysteresis and hence would have no remanent magnetization. A ferromagnetic substance can be demagnetized by being carried through successive hysteresis loops, due to a decreasing applied magnetic field, as shown in Figure 30.32. Figure 30.32 Demagnetizing a ferromagnetic material by carrying it through successive hysteresis loops. Quick Quiz 30.8 Which material would make a better permanent magnet, one whose hysteresis loop looks like Figure 30.31a or one whose loop looks like Figure 30.31b? The magnetization curve is useful for another reason: The area enclosed by the magnetization curve represents the work required to take the material through the hysteresis cycle. The energy acquired by the material in the magnetization process originates from the source of the external field -- that is, the emf in the circuit of the toroidal coil. When the magnetization cycle is repeated, dissipative processes within the material due to realignment of the domains result in a transformation of magnetic energy into internal energy, which is evidenced by an increase in the temperature of the substance. For this reason, devices subjected to alternating fields (such as ac adapters for cell phones, power tools, and so on) use cores made of soft ferromagnetic substances, which have narrow hysteresis loops and correspondingly little energy loss per cycle. Magnetic computer disks store information by alternating the direction of B for portions of a thin layer of ferromagnetic material. Floppy disks have the layer on a circular sheet of plastic. Hard disks have several rigid platters with magnetic coatings on each side. Audio tapes and videotapes work the same way as floppy disks except that the ferromagnetic material is on a very long strip of plastic. Tiny coils of wire in a recording head are placed close to the magnetic material (which is moving rapidly past the head). Varying the current through the coils creates a magnetic field that magnetizes the recording material. To repelled by magnets! 964 CHAPTER 30 Sources of the Magnetic Field Figure 30.34 A small permanent magnet levitated above a disk of the superconductor Y Ba2Cu3O7 cooled to liquid nitrogen temperature (77 K). web For a more detailed description of the unusual properties of superconductors, visit www.saunderscollege.com/physics/ As you recall from Chapter 27, a superconductor is a substance in which the electrical resistance is zero below some critical temperature. Certain types of superconductors also exhibit perfect diamagnetism in the superconducting state. As a result, an applied magnetic field is expelled by the superconductor so that the field is zero in its interior. This phenomenon of flux expulsion is known as the Meissner effect. If a permanent magnet is brought near a superconductor, the two objects repel each other. This is illustrated in Figure 30.34, which shows a small permanent magnet levitated above a superconductor maintained at 77 K. EXAMPLE 30.11 Saturation Magnetization each atom contributes one Bohr magneton (due to one unpaired spin) to the magnetic moment, we obtain Ms 8.6 8.0 10 28 atoms m3 9.27 10 24 Estimate the saturation magnetization in a long cylinder of iron, assuming one unpaired electron spin per atom. The saturation magnetization is obtained when all the magnetic moments in the sample are aligned. If the sample contains n atoms per unit volume, then the saturation magnetization Ms has the value Ms n Solution A m2 atom 10 5 A/m where is the magnetic moment per atom. Because the molar mass of iron is 55 g/mol and its density is 7.9 g/cm3, the value of n for iron is 8.6 1028 atoms/m3. Assuming that This is about one-half the experimentally determined saturation magnetization for iron, which indicates that actually two unpaired electron spins are present per atom. Optional Section 30.9 THE MAGNETIC FIELD OF THE EARTH When we speak of a compass magnet having a north pole and a south pole, we should say more properly that it has a "north-seeking" pole and a "south-seeking" pole. By this we mean that one pole of the magnet seeks, or points to, the north geographic pole of the Earth. Because the north pole of a magnet is attracted toward the north geographic pole of the Earth, we conclude that the Earth's south magnetic pole is located near the north geographic pole, and the Earth's north magnetic pole is located near the south geographic pole. In fact, the configuration of the Earth's magnetic field, pictured in Figure 30.35, is very much like the one that would be achieved by burying a gigantic bar magnet deep in the interior of the Earth. 30.9 The Magnetic Field of the Earth South magnetic pole Geographic equator North geographic pole 965 QuickLab A gold ring is very weakly repelled by a magnet. To see this, suspend a 14or 18-karat gold ring on a long loop of thread, as shown in (a). Gently tap the ring and estimate its period of oscillation. Now bring the ring to rest, letting it hang for a few moments so that you can verify that it is not moving. Quickly bring a very strong magnet to within a few millimeters of the ring, taking care not to bump it, as shown in (b). Now pull the magnet away. Repeat this action many times, matching the oscillation period you estimated earlier. This is just like pushing a child on a swing. A small force applied at the resonant frequency results in a large-amplitude oscillation. If you have a platinum ring, you will be able to see a similar effect except that platinum is weakly attracted to a magnet because it is paramagnetic. S c equato r N Magneti South geographic pole North magnetic pole Figure 30.35 The Earth's magnetic field lines. Note that a south magnetic pole is near the north geographic pole, and a north magnetic pole is near the south geographic pole. If a compass needle is suspended in bearings that allow it to rotate in the vertical plane as well as in the horizontal plane, the needle is horizontal with respect to the Earth's surface only near the equator. As the compass is moved northward, the needle rotates so that it p. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is 1.00 A out of the page, and the current in the outer conductor is 3.00 A into the page. Determine the magnitude and direction of the magnetic field at points a and b. 22. The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00 T. (a) At what distance is it 0.100 T ? (b) At one instant, the two conductors in a long household extension cord carry equal WEB 25. 26. 27. 2.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 40.0 cm away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one-tenth as large? (d) The center wire in a coaxial cable carries current 2.00 A in one direction, and the sheath around it carries current 2.00 A in the opposite direction. What magnetic field does the cable create at points outside? The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radius of 1.30 m. If the toroid has 900 turns of large-diameter wire, each of which carries a current of 14.0 kA, find the magnitude of the magnetic field inside the toroid (a) along the inner radius and (b) along the outer radius. A cylindrical conductor of radius R 2.50 cm carries a current of I 2.50 A along its length; this current is uniformly distributed throughout the cross-section of the conductor. (a) Calculate the magnetic field midway along the radius of the wire (that is, at r R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field at r R/2. A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R 0.500 cm. (a) If each wire carries 2.00 A, what are the magnitude and direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (b) Would a wire on the outer edge of the bundle experience a force greater or less than the value calculated in part (a)? Niobium metal becomes a superconductor when cooled below 9 K. If superconductivity is destroyed when the surface magnetic field exceeds 0.100 T, determine the maximum current a 2.00-mm-diameter niobium wire can carry and remain superconducting, in the absence of any external magnetic field. A long, cylindrical conductor of radius R carries a current I, as shown in Figure P30.27. The current density J, however, is not uniform over the cross-section of the 972 I CHAPTER 30 Sources of the Magnetic Field Section 30.5 Magnetic Flux 33. A cube of edge length 2.50 cm is positioned as shown in Figure P30.33. A uniform magnetic field given by B (5.00 i 4.00 j 3.00 k) T exists throughout the region. (a) Calculate the flux through the shaded face. (b) What is the total flux through the six faces? y B r2 r1 R Figure P30.27 conductor but is a function of the radius according to J br, where b is a constant. Find an expression for the magnetic field B (a) at a distance r 1 R and (b) at a distance r 2 R , measured from the axis. 28. In Figure P30.28, both currents are in the negative x direction. (a) Sketch the magnetic field pattern in the yz plane. (b) At what distance d along the z axis is the magnetic field a maximum? z z x a a Figure P30.33 34. A solenoid 2.50 cm in diameter and 30.0 cm long has 300 turns and carries 12.0 A. (a) Calculate the flux through the surface of a disk of radius 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid, as in Figure P30.34a. (b) Figure P30.34b shows an enlarged end view of the same solenoid. Calculate the flux through the blue area, which is defined by an annulus that has an inner radius of 0.400 cm and outer radius of 0.800 cm. I x y I Figure P30.28 Section 30.4 The Magnetic Field of a Solenoid WEB 29. What current is required in the windings of a long soleic field at this location? 47. The magnetic moment of the Earth is approximately 8.00 1022 A m2. (a) If this were caused by the complete magnetization of a huge iron deposit, how many unpaired electrons would this correspond to? (b) At two unpaired electrons per iron atom, how many kilograms of iron would this correspond to? (Iron has a density of 7 900 kg/m3 and approximately 8.50 1028 atoms/m3.) Section 30.8 Magnetism in Matter 38. In Bohr's 1913 model of the hydrogen atom, the electron is in a circular orbit of radius 5.29 10 11 m, and its speed is 2.19 106 m/s. (a) What is the magnitude of the magnetic moment due to the electron's motion? (b) If the electron orbits counterclockwise in a horizontal circle, what is the direction of this magnetic moment vector? 39. A toroid with a mean radius of 20.0 cm and 630 turns (see Fig. 30.29) is filled with powdered steel whose magnetic susceptibility is 100. If the current in the windings is 3.00 A, find B (assumed uniform) inside the toroid. 40. A magnetic field of 1.30 T is to be set up in an iron-core toroid. The toroid has a mean radius of 10.0 cm and magnetic permeability of 5 000 0 . What current is re- ADDITIONAL PROBLEMS 48. A lightning bolt may carry a current of 1.00 104 A for a short period of time. What is the resultant magnetic 974 CHAPTER 30 Sources of the Magnetic Field z field 100 m from the bolt? Suppose that the bolt extends far above and below the point of observation. 49. The magnitude of the Earth's magnetic field at either pole is approximately 7.00 10 5 T. Suppose that the field fades away, before its next reversal. Scouts, sailors, and wire merchants around the world join together in a program to replace the field. One plan is to use a current loop around the equator, without relying on magnetization of any materials inside the Earth. Determine the current that would generate such a field if this plan were carried out. (Take the radius of the Earth as R E 6.37 10 6 m.) 50. Two parallel conductors carry current in opposite directions, as shown in Figure P30.50. One conductor carries a current of 10.0 A. Point A is at the midpoint between the wires, and point C is a distance d/2 to the right of the 10.0-A current. If d 18.0 cm and I is adjusted so that the magnetic field at C is zero, find (a) the value of the current I and (b) the value of the magnetic field at A. w I 0 b P y x Figure P30.53 in the plane of the strip at a distance b away from the strip. 54. For a research project, a student needs a solenoid that produces an interior magnetic field of 0.030 0 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 10.0 cm long. Determine the number of layers of wire she needs and the total length of the wire. WEB I 10.0 A A C d Figure P30.50 51. Suppose you install a compass on the center of the dashboard of a car. Compute an order-of-magnitude estimate for the magnetic field that is produced at this location by the current when you switch on the headlights. How does your estimate compare with the Earth's magnetic field? You may suppose the dashboard is made mostly of plastic. 52. Imagine a long, cylindrical wire of radius R that has a current density J (r) J 0(1 r 2/R 2 ) for r R and J(r) 0 for r R, where r is the distance from the axis of the wire. (a) Find the resulting magnetic field inside (r R) and outside (r R) the wire. (b) Plot the magnitude of the magnetic field as a function of r. (c) Find the location where the magnitude of the magnetic field is a maximum, and the value of that maximum field. 53. A very long, thin strip of metal of width w carries a current I along its length, as shown in Figure P30.53. Find the magnetic field at point P in the diagram. Point P is 55. A nonconducting ring with a radius of 10.0 cm is uniformly charged with a total positive charge of 10.0 C. The ring rotates at a constant angular speed of 20.0 rad/s about an axis through its center, perpendicular to the plane of the ring. What is the magnitude of thse the remanent magnetization at the point corresponding to point b in Figure 30.30 is greater. West to east. The lines of the Earth's magnetic field enter the planet in Hudson Bay and emerge from Antarctica; thus, the field lines resulting from the current would have to go in the opposite direction. Compare Figure 30.6a with Figure 30.35. 30.7 30.8 30.9 2.2 This is the Nearest One Head 979 P U Z Z L E R Before this vending machine will deliver its product, it conducts several tests on the coins being inserted. How can it determine what material the coins are made of without damaging them and without making the customer wait a long time for the results? (George Semple) c h a p t e r Faraday's Law Chapter Outline 31.1 31.2 31.3 31.4 Faraday's Law of Induction Motional emf Lenz's Law Induced emf and Electric Fields 31.5 (Optional) Generators and Motors 31.6 (Optional) Eddy Currents 31.7 Maxwell's Wonderful Equations 979 980 CHAPTER 31 Faraday's Law T he focus of our studies in electricity and magnetism so far has been the electric fields produced by stationary charges and the magnetic fields produced by moving charges. This chapter deals with electric fields produced by changing magnetic fields. Experiments conducted by Michael Faraday in England in 1831 and independently by Joseph Henry in the United States that same year showed that an emf can be induced in a circuit by a changing magnetic field. As we shall see, an emf (and therefore a current as well) can be induced in many ways -- for instance, by moving a closed loop of wire into a region where a magnetic field exists. The results of these experiments led to a very basic and important law of electromagnetism known as Faraday's law of induction. This law states that the magnitude of the emf induced in a circuit equals the time rate of change of the magnetic flux through the circuit. With the treatment of Faraday's law, we complete our introduction to the fundamental laws of electromagnetism. These laws can be summarized in a set of four equations called Maxwell's equations. Together with the Lorentz force law, which we discuss briefly, they represent a complete theory for describing the interaction of charged objects. Maxwell's equations relate electric and magnetic fields to each other and to their ultimate source, namely, electric charges. 31.1 12.6 & 12.7 FARADAY'S LAW OF INDUCTION A demonstration of electromagnetic induction. A changing potential difference is applied to the lower coil. An emf is induced in the upper coil as indicated by the illuminated lamp. What happens to the lamp's intensity as the upper coil is moved over the vertical tube? (Courtesy of Central Scientific Company) To see how an emf can be induced by a changing magnetic field, let us consider a loop of wire connected to a galvanometer, as illustrated in Figure 31.1. When a magnet is moved toward the loop, the galvanometer needle deflects in one direction, arbitrarily shown to the right in Figure 31.1a. When the magnet is moved away from the loop, the needle deflects in the opposite direction, as shown in Figure 31.1c. When the magnet is held stationary relative to the loop (Fig. 31.1b), no deflection is observed. Finally, if the magnet is held stationary and the loop is moved either toward or away from it, the needle deflects. From these observations, we conclude that the loop "knows" that the magnet is moving relative to it because it experiences a change in magnetic field. Thus, it seems that a relationship exists between current and changing magnetic field. These results are quite remarkable in view of the fact that a current is set up even though no batteries are present in the circuit! We call such a current an induced current and say that it is produced by an induced emf. Now let us describe an experiment conducted by Faraday 1 and illustrated in Figure 31.2. A primary coil is connected to a switch and a battery. The coil is wrapped around a ring, and a current in the coil produces a magnetic field when the switch is closed. A secondary coil also is wrapped arounderrupter. This electric range cooks food on the basis of the principle of induction. An oscillating current is passed through a coil placed below the cooking surface, which is made of a special glass. The current produces an oscillating magnetic field, which induces a current in the cooking utensil. Because the cooking utensil has some electrical resistance, the electrical energy associated with the induced current is transformed to internal energy, causing the utensil and its contents to become hot. (Courtesy of Corning, Inc.) 984 CHAPTER 31 Faraday's Law Pickup coil Magnet N N S Magnetized portion of string To amplifier Guitar string (a) (b) S Figure 31.5 (a) In an electric guitar, a vibrating string induces an emf in a pickup coil. (b) The circles beneath the metallic strings of this electric guitar detect the notes being played and send this information through an amplifier and into speakers. (A switch on the guitar allows the musician to select which set of six is used.) How does a guitar "pickup" sense what music is being played? (b, Charles D. Winters) the coil. When the string vibrates at some frequency, its magnetized segment produces a changing magnetic flux through the coil. The changing flux induces an emf in the coil that is fed to an amplifier. The output of the amplifier is sent to the loudspeakers, which produce the sound waves we hear. EXAMPLE 31.1 One Way to Induce an emf in a Coil is, from Equation 31.2, N t B A coil consists of 200 turns of wire having a total resistance of 2.0 . Each turn is a square of side 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s, what is the magnitude of the induced emf in the coil while the field is changing? The area of one turn of the coil is (0.18 m)2 0.032 4 m2. The magnetic flux through the coil at t 0 is zero because B 0 at that time. At t 0.80 s, the magnetic flux through one turn is B BA (0.50 T)(0.032 4 m2 ) 0.016 2 T m2. Therefore, the magnitude of the induced emf 200(0.016 2 T m2 0.80 s 4.1 V 0 T m2) 4.1 T m2/s Solution You should be able to show that 1 T m2/s 1 V. Exercise Answer What is the magnitude of the induced current in the coil while the field is changing? 2.0 A. EXAMPLE 31.2 An Exponentially Decaying B Field tially (Fig. 31.6). Find the induced emf in the loop as a function of time. A loop of wire enclosing an area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magnitude of B varies in time according to the expression B B maxe at, where a is some constant. That is, at t 0 the field is B max , and for t 0, the field decreases exponen- Solution Because B is perpendicular to the plane of the loop, the magnetic flux through the loop at time t 0 is 31.2 Motional EMF B Bmax 985 ABmaxe at B BA cos 0 Because AB max and a are constants, the induced emf calculated from Equation 31.1 is d B dt ABmax d e dt at aABmaxe at t This expression indicates that the induced emf decays exponentially in time. Note that the maximum emf occurs at t 0, where max aABmax . The plot of versus t is similar to the B-versus-t curve shown in Figure 31.6. Figure 31.6 Exponential decrease in the magnitude of the magnetic field with time. The induced emf and induced current vary with time in the same way. CONCEPTUAL EXAMPLE 31.3 What Is Connected to What? now the only resistance in the loop. As a result, the current in bulb 1 is greater than when bulb 2 was also in the loop. Once the switch is closed, bulb 2 is in the loop consisting of the wires attached to it and those connected to the switch. There is no changing magnetic flux through this loop and hence no induced emf. Two bulbs are connected to opposite sides of a loop of wire, as shown in Figure 31.7. A decreasing magnetic field (confined to the circular area shown in the figure) induces an emf in the loop that causes the two bulbs to light. What happens to the brightness of the bulbs when the switch is closed? Solution Bulb 1 glows brightet to left and so counteracts the decrease in the field produced by the solenoid. A metal ring is placed near a solenoid, as shown in Figure 31.15a. Find the direction of the induced current in the ring (a) at the instant the switch in the circuit containing the solenoid is thrown closed, (b) after the switch has been closed for several seconds, and (c) at the instant the switch is thrown open. Solution (a) At the instant the switch is thrown closed, the situation changes from one in which no magnetic flux passes through the ring to one in which flux passes through in the direction shown in Figure 31.15b. To counteract this change in the flux, the current induced in the ring must set up a magnetic field directed from left to right in Figure 31.15b. This requires a current directed as shown. (b) After the switch has been closed for several seconds, no change in the magnetic flux through the loop occurs; hence, the induced current in the ring is zero. (c) Opening the switch changes the situation from one in which magnetic flux passes through the ring to one in which there is no magnetic flux. The direction of the induced current is as shown in Figure 31.15c because current in this di- Switch (a) (b) (c) Figure 31.15 CONCEPTUAL EXAMPLE 31.7 A Loop Moving Through a Magnetic Field netic force experienced by charges in the right side of the loop. When the loop is entirely in the field, the change in magnetic flux is zero, and hence the motional emf vanishes. This happens because, once the left side of the loop enters the field, the motional emf induced in it cancels the motional emf present in the right side of the loop. As the right side of the loop leaves the field, the flux inward begins to decrease, a clockwise current is induced, and the induced emf is B v. As soon as the left side leaves the field, the emf decreases to zero. (c) The external force that must be applied to the loop to maintain this motion is plotted in Figure 31.16d. Before the loop enters the field, no magnetic force acts on it; hence, the applied force must be zero if v is constant. When the right side of the loop enters the field, the applied force necessary to maintain constant speed must be equal in magnitude and opposite in direction to the magnetic force exerted on that side: FB I B B 2 2v/R . When the loop is entirely in the field, the flux through the loop is not changing with time. Hence, the net emf induced in the loop is zero, and the current also is zero. Therefore, no external force is needed to maintain the motion. Finally, as the right side leaves the field, the applied force must be equal in magnitude and opposite A rectangular metallic loop of dimensions and w and resistance R moves with constant speed v to the right, as shown in Figure 31.16a, passing through a uniform magnetic field B directed into the page and extending a distance 3w along the x axis. Defining x as the position of the right side of the loop along the x axis, plot as functions of x (a) the magnetic flux through the area enclosed by the loop, (b) the induced motional emf, and (c) the external applied force necessary to counter the magnetic force and keep v constant. Solution (a) Figure 31.16b shows the flux through the area enclosed by the loop as a function x. Before the loop enters the field, the flux is zero. As the loop enters the field, the flux increases linearly with position until the left edge of the loop is just inside the field. Finally, the flux through the loop decreases linearly to zero as the loop leaves the field. (b) Before the loop enters the field, no motional emf is induced in it because no field is present (Fig. 31.16c). As the right side of the loop enters the field, the magnetic flux directed into the page increases. Hence, according to Lenz's law, the induced current is counterclockwise because it must produce a magnetic field directed out of the page. The motional emf B v (from Eq. 31.5) arises from the mag- 992 CHAPTER 31 Faraday's Law Furthermore, this example shows that the motional emf induced in the loop can be zero even whes the maximum current and is the angular frequency of the alternating current source (Fig. 31.18). (a) Determine the magnitude of the induced electric field outside the solenoid, a distance r R from its long central axis. First let us consider an external point and take the path for our line integral to be a circle of radius r centered on the solenoid, as illustrated in Figure 31.18. By sym- Solution Path of integration R The magnetic field inside a long solenoid is given by Equation 30.17, B Imax cos t into 0nI. When we substitute I this equation and then substitute the result into Equation (1), we find that E(2 r) R2 0nI max d (cos t) dt R2 R2 0nI max sin t (2) r E 0nI max 2r sin t (for r R) Hence, the electric field varies sinusoidally with time and its amplitude falls off as 1/r outside the solenoid. (b) What is the magnitude of the induced electric field inside the solenoid, a distance r from its axis? I max cos t Figure 31.18 A long solenoid carrying a time-varying current given by I I 0 cos t. An electric field is induced both inside and outside the solenoid. Solution For an interior point (r R), the flux threading an integration loop is given by B r 2. Using the same proce- 994 dure as in part (a), we find that E(2 r) r2 dB dt 0n Imax CHAPTER 31 Faraday's Law Exercise r2 0n I max sin t Show that Equations (2) and (3) for the exterior and interior regions of the solenoid match at the boundary, r R. (3) E 2 r sin t (for r R) Exercise Would the electric field be different if the solenoid had an iron core? Answer Yes, it could be much stronger because the maximum magnetic field (and thus the change in flux) through the solenoid could be thousands of times larger. (See Example 30.10.) This shows that the amplitude of the electric field induced inside the solenoid by the changing magnetic flux through the solenoid increases linearly with r and varies sinusoidally with time. Optional Section 31.5 GENERATORS AND MOTORS Electric generators are used to produce electrical energy. To understand how they work, let us consider the alternating current (ac) generator, a device that converts mechanical energy to electrical energy. In its simplest form, it consists of a loop of wire rotated by some external means in a magnetic field (Fig. 31.19a). In commercial power plants, the energy required to rotate the loop can be derived from a variety of sources. For example, in a hydroelectric plant, falling water directed against the blades of a turbine produces the rotary motion; in a coal-fired plant, the energy released by burning coal is used to convert water to steam, and this steam is directed against the turbine blades. As a loop rotates in a magnetic field, the magnetic flux through the area enclosed by the loop changes with time; this induces an emf and a current in the loop according to Faraday's law. The ends of the loop are connected to slip rings that rotate with the loop. Connections from these slip rings, which act as output terminals of the generator, to the external circuit are made by stationary brushes in contact with the slip rings. Loop Turbines turn generators at a hydroelectric power plant. (Luis Castaneda/The Image Bank) Slip rings N S max t External rotator Brushes (a) External circuit (b) Figure 31.19 (a) Schematic diagram of an ac generator. An emf is induced in a loop that rotates in a magnetic field. (b) The alternating emf induced in the loop plotted as a function of time. 31.5 Generators and Motors 995 B Suppose that, instead of a single turn, the loop has N turns (a more practical situation), all of the same area A, and rotates in a magnetic field with a constant angular speed . If is the angle between the magnetic field and the normal to the plane of the loop, as shown in Figure 31.20, then the magnetic flux through the loop at any time t is B Normal BA cos BA cos t t between angular displacement and anwhere we have used the relationship gular speed (see Eq. 10.3). (We have set the clock so that t 0 when 0.) Hence, the induced emf in the coil is 0 turns of wire rotates about a vertical axis at 1 500 rev/min, as indicated in Figure P31.38. The horizontal component of the Earth's magnetic field at the location of the coil is 2.00 10 5 T. Calculate the maximum emf induced in the coil by this field. N S WEB Figure P31.40 41. (a) What is the maximum torque delivered by an electric motor if it has 80 turns of wire wrapped on a rectangular coil of dimensions 2.50 cm by 4.00 cm? Assume that the motor uses 10.0 A of current and that a uniform 0.800-T magnetic field exists within the motor. (b) If the motor rotates at 3 600 rev/min, what is the peak power produced by the motor? 42. A semicircular conductor of radius R 0.250 m is rotated about the axis AC at a constant rate of 120 rev/min (Fig. P31.42). A uniform magnetic field in all of the lower half of the figure is directed out of the plane of rotation and has a magnitude of 1.30 T. (a) Calculate the maximum value of the emf induced in the conductor. (b) What is the value of the average induced emf for each complete rotation? (c) How would the answers to parts (a) and (b) change if B were allowed to extend a distance R above the axis of rotation? Sketch the emf versus time (d) when the field is as drawn in Figure P31.42 and (e) when the field is extended as described in part (c). 20.0 cm 20.0 cm Figure P31.38 1008 CHAPTER 31 Faraday's Law nal speed vt . (a) Show that A R C vt MgR B 2w 2 Bout Figure P31.42 43. The rotating loop in an ac generator is a square 10.0 cm on a side. It is rotated at 60.0 Hz in a uniform field of 0.800 T. Calculate (a) the flux through the loop as a function of time, (b) the emf induced in the loop, (c) the current induced in the loop for a loop resistance of 1.00 , (d) the power in the resistance of the loop, and (e) the torque that must be exerted to rotate the loop. (Optional) (b) Why is vt proportional to R ? (c) Why is it inversely proportional to B 2 ? 46. Figure P31.46 represents an electromagnetic brake that utilizes eddy currents. An electromagnet hangs from a railroad car near one rail. To stop the car, a large steady current is sent through the coils of the electromagnet. The moving electromagnet induces eddy currents in the rails, whose fields oppose the change in the field of the electromagnet. The magnetic fields of the eddy currents exert force on the current in the electromagnet, thereby slowing the car. The direction of the car's motion and the direction of the current in the electromagnet are shown correctly in the picture. Determine which of the eddy currents shown on the rails is correct. Explain your answer. Section 31.6 Eddy Currents 44. A 0.150-kg wire in the shape of a closed rectangle 1.00 m wide and 1.50 m long has a total resistance of 0.750 . The rectangle is allowed to fall through a magnetic field directed perpendicular to the direction of motion of the rectangle (Fig. P31.44). The rectangle accelerates downward as it approaches a terminal speed of 2.00 m/s, with its top not yet in the region of the field. Calculate the magnitude of B. N w I N v S S Figure P31.46 Section 31.7 Maxwell's Wonderful Equations Bout v Figure P31.44 Problems 44 and 45. 47. A proton moves through a uniform electric field E 50.0j V/m and a uniform magnetic field B (0.200i 0.300j 0.400k) T. Determine the acceleration of the proton when it has a velocity v 200i m/s. 48. An electron moves through a uniform electric field E (2.50i 5.00j) V/m and a uniform magnetic field B 0.400k T. Determine the acceleration of the electron when it has a velocity v 10.0i m/s. WEB 45. A conducting rectangular loop of mass M, resistance R, and dimensions w by falls from rest into a magnetic field B as in Figure P31.44. The loop approaches termi- ADDITIONAL PROBLEMS 49. A steel guitar string vibrates (see Fig. 31.5). The component of the magnetic field perpendicular to the area of Problems a pickup coil nearby is given by B 50.0 mT (3.20 mT) sin (2 523 t/s) 1009 The circular pickup coil has 30 turns and radius 2.70 mm. Find the emf induced in the coil as a function of time. 50. Fictor. Suppose that the switch S is thrown closed at t 0. The current in the circuit begins to increase, and a back emf that opposes the increasing current is induced in the inductor. The back emf is, from Equation 32.1, L b L Figure 32.3 A series RL circuit. As the current increases toward its maximum value, an emf that opposes the increasing current is induced in the inductor. dI dt Because the current is increasing, dI/dt is positive; thus, L is negative. This negative value reflects the decrease in electric potential that occurs in going from a to b across the inductor, as indicated by the positive and negative signs in Figure 32.3. With this in mind, we can apply Kirchhoff's loop rule to this circuit, traversing the circuit in the clockwise direction: IR L dI dt 0 (32.6) where IR is the voltage drop across the resistor. (We developed Kirchhoff's rules for circuits with steady currents, but we can apply them to a circuit in which the current is changing if we imagine them to represent the circuit at one instant of time.) We must now look for a solution to this differential equation, which is similar to that for the RC circuit (see Section 28.4). A mathematical solution of Equation 32.6 represents the current in the circuit as a function of time. To find this solution, we change variables for convenience, letting x tion 32.6 as x L dx R dt dx x Integrating this last expression, we have ln x x0 R t L 0 R dt L R I , so that dx dI. With these substitutions, we can write Equa- where we take the integrating constant to be ln x 0 and x 0 is the value of x at time t 0. Taking the antilogarithm of this result, we obtain x x 0e Rt /L 32.2 RL Circuits 1019 Because I 0 at t 0, we note from the definition of x that x 0 this last expression is equivalent to R I I R R e (1 Rt /L /R. Hence, I /R Rt /L ) e 0.63 R = L/R This expression shows the effect of the inductor. The current does not increase instantly to its final equilibrium value when the switch is closed but instead increases according to an exponential function. If we remove the inductance in the circuit, which we can do by letting L approach zero, the exponential term becomes zero and we see that there is no time dependence of the current in this case -- the current increases instantaneously to its final equilibrium value in the absence of the inductance. We can also write this expression as I where the constant (1 e t/ t Figure 32.4 Plot of the current versus time for the RL circuit shown in Figure 32.3. The switch is thrown closed at t 0, and the current increases toward its maximum value /R. The time constant is the time it takes I to reach 63% of its maximum value. R ) (32.7) is the time constant of the RL circuit: L/R (32.8) Time constant of an RL circuit Physically, is the time it takes the current in the circuit to reach (1 e 1 ) 0.63 of its final value /R . The time constant is a useful parameter for comparing the time responses of various circuits. Figure 32.4 shows a graph of the current versus time in the RL circuit. Note that the equilibrium value of the current, which occurs as t approaches infinity, is /R. We can see this by setting dI/dt equal to zero in Equation 32.6 and solving for the current I. (At equilibrium, the change in the current is zero.) Thus, we see that the current initially increases very rapidly and then gradually approaches the equilibrium value /R as t approaches infinity. Let us also investigate the time rate of change of the current in the circuit. Taking the first time derivative of Equation 32.7, we have dI dt L e t/ dI dt /L (32.9) From this result, we see that the time rate of change of the current is a maximum (equal to /L) at t 0 and falls off exponentially to zero as t approaches infinity (Fig. 32.5). Now let us consider the RL circuit shown in Figure 32.6. The circuit contains two switches that operate such that when one is closed, the other is opened. Suppose that S1 has been closed for a length of time sufficient to allow the current to reach its equilibrium value /R. In this situation, the circuit is described completely by the outer loop in Figure 32.6. If S2 is closed at the instant at which S1 is opened, the circuit changes so that it is described completely by just the upper loop in Figure 32.6. The lower loop no longer influences the behavior of the cir0). If we apply Kirchhoff's loop cuit. Thus, we have a circuit with no battery ( rule to the upper loop at the instant the switches are thrown, we obtain IR L dI dt 0 t Figure 32.5 Plot of dI/dt versus time for the RL circuit shown in Figure 32.3. The time rate of change of current is a maximum at t 0, which is the instant at which the switch is thrown closed. The rate decreases exponentially with time as I increases toward its maximum value. 1020 CHAPTER 32 Inductance R a L b I /R S2 - + S1 t Figure 32.6 An RL circuit containing two switches. When S1 is closed and S 2 open as shown, the battery is in the circuit. At the instant S 2 is closed, S1 is opened, and the battery is no longer part of the circuit. Figure 32.7 Current versus time for the upper loop of the circuit shown in Figure 32.6. For t 0, S1 is closed and S 2 is open. At t 0, S2 is closed as S1 is opened, and the current has its maximum value /R. It is left as a problem (Problem 18) to show that the solution of this differential equation is I R e t/ I 0e t/ (32.10) where is the emf of the battery and I 0 /R is the current at t 0, the instant at which S2 is closed as S1 is opened. If no inductor were present in the circuit, the current would immediately decrease to zero if the battery were removed. When the inductor is present, it acts to oppose the decrease in the current and to maintain the current. A graph of the current in the circuit versus time (Fig. 32.7) shows that the current is continuously decreasing with time. Note that the slope dI/dt is always negative and has its maximum value at t 0. The negative slope signifies that L L (dI/dt) is now positive; that is, point a in Figure 32.6 is at a lower electric potential than point b. Quick Quiz 32.2 Two circuits like the one shown in Figure 32.6 are identical except for the value of L. In circuit A the inductance of the inductor is L A , and in circuit B it is L B . Switch S 1 is thrown closed at t 0, while switch S 2 remains open. At t 10 s, switch S 1 is opened and switch S 2 is closed. The resulting time rates of change for the two currents are as graphed in Figure 32.8. If we assume that the time constant of each circuit is much less than 10 s, which of the following is true? (a) L A L B ; (b) L A L B ; (c) not enough information to tell. I A B t(s) 0 5 10 15 Figure 32.8 32.3 Energy in a Magnetic Field 1021 EXAMPLE 32.3 Time Constant of an RL Circuit 0. (a) Find A plot of Equation 32.7 for this circuit is given in Figure 32.9b. (c) Compare the potential difference across the resistor with that across the inductor. The switch in Figure 32.9a is thrown closed at t the time constant of the circuit. Solution The time constant is given by Equation 32.8: L R 30.0 10 6.00 3 H 5.00 ms 2.00 ms. (b) Calculate the current in the circuit at t Solution Using Equation 32.7 for the current as a function of time (with t and in milliseconds), we find that at t 2.00 ms I R (1 e t/ ) 12.0 V (1 6.00 30.0 mH e 0.400 ) 0.659 A Solution At the instant the switch is closed, there is no current and thus no potential difference across the resistor. At this instant, the battery voltage appears entirely across the inductor in the form of a back emf of 12.0 V as the inductor tries to maintain the zero-current condition. (The left end of the inductor is at a higher electric potential than the right end.) As time passes, the emf across the inductor decreases and the current through the resistor (and hence the potential difference across it) increases. The sum of the two potential differences at all times is 12.0 V, as shown in Figure 32.10. Exercise Calculate the current in the circuit and the voltage across the resistor after a time interval equal to one time constant has elapsed. 1.26 A, 7.56 V. 12.0 V 6.00 Answer V(V) 12 10 S (a) I (A) 2 8 6 VL 1 4 2 VR 0 2 4 6 (b) 8 10 t(ms) 0 t(ms) 2 4 6 8 10 Figure 32.9 t (a) The switch in this RL circuit is thrown closed at 0. (b) A graph of the current versus time for the circuit in part (a). Figure 32.10 The sum of the potential differences across the resistor and inductor in Figure 32.9a is 12.0 V (the battery emf) at all times. 32.3 13.6 ENERGY IN A MAGNETIC FIELD Because the emf induced in an inductor prevents a battery from establishing an instantaneous current, the battery must do work against the inductor to create a current. Part of the energy supplied by the battery appears as internal energy in the resistor, while the remaining energy is stored in the magnetic field of the inductor. If we multiply each term in Equation 32.6 by I and rearrange the expression, we have I I 2R LI dI dt (32.11) 1022 CHAPTER 32 Inductance This expression indicates that the rate at which energy is supplied by the battery (I ) equals the sum of the rate at which energy is delivered to the resistor, I 2R , and the rate at which energy is stored in the inductor, LI(dI /dt). Thus, Equation 32.11 is simply an expression of energy conservation. If we let U denote the energy stored in the inductor at any time, then we can write the rate dU/dt at which energy is stored as dU dI LI dt dt To find the total energy stored in the inductor, we can rewrite this expression as dU LI dI and integrate: I I U Energy stored in an inductor dU 0 1 2 2 LI LI dI L 0 I dI (32.12) U where L is constant and has been removed from the integral. This expression represents the energy stored in the magnetic field of the inductor when the current is I. Note that this equation is similar in form to Equation 26.11 for the energy stored in the electric field of a capacitor, U Q 2/2C . In either case, we see that energy is required to establish a field. We can also determine the energy density of a magnetic field. For simplicity, consider a solenoid whose inductance is given by Equation 32.5: L 0n 2A The magnetic field of a solenoid is given by Equation 30.17: B Substituting the expression for L and I U 1 2 2 LI 1 2 0n 2A 0nI B/ 0n into Equation 32.12 gives 2 B 0n B2 A 2 0 (32.13) Because A is the volume of the solenoid, the energy stored per unit volume in the magnetic field surrounding the inductor is Magnetic energy density uB U A B2 2 0 (32.14) Although this expression was derived for the special case of a solenoid, it is valid for any region of space in which a magnetic field exists. Note that Equation 32.14 is similar in form to Equation 26.13 for the energy per unit volume stored in an electric field, u E 1 0 E 2. In both cases, the energy density is proportional to 2 the square of the magnitude of the field. EXAMPLE 32.4 What Happens to the Energy in the Inductor? where I 0 /R is the initial current in the circuit and L/R is the time constant. Show that all the energy initially stored in the magnetic field of the inductor appears as internal energy in the resistor as the current decays to zero. Consider once again the RL circuit shown in Figure 32.6, in which switch S 2 is closed at the instant S1 is opened (at t 0). Recall that the current in the upper loop decays exponentially with time according to the expression I I 0e t / , 32.3 Energy in a Magnetic Field 1023 Solution The rate dU/dt at which energy is delivered to the resistor (which is the power) is equal to I 2R, where I is the instantaneous current: dU dt I 2R (I 0e Rt /L )2R The value of the definite integral is L/2R (this is left for the student to show in the exercise at the end of this example), and so U becomes U I 02R L 2R 1 LI 02 2 I 02Re 2Rt /L To find the total energy delivered to the resistor, we solve for dU and integrate this expression over the limits t 0 to t : (the upper limit is infinity because it takes an infinite amount of time for the current to reach zero): (1) U 0 Note that this is equal to the initial energy stored in the magnetic field of the inductor, given by Equation 32.13, as we set out to prters used were R 0.19 F. and C Plot of Q versus t for an overdamped RLC circuit, which occurs for values of R !4L/C . SUMMARY When the current in a coil changes with time, an emf is induced in the coil according to Faraday's law. The self-induced emf is L L dI dt (32.1) where L is the inductance of the coil. Inductance is a measure of how much opposition an electrical device offers to a change in current passing through the device. Inductance has the SI unit of henry (H), where 1 H 1 V s/A. The inductance of any coil is L N I B (32.2) where B is the magnetic flux through the coil and N is the total number of turns. The inductance of a device depends on its geometry. For example, the inductance of an air-core solenoid is L 0N 2A (32.4) where A is the cross-sectional area, and is the length of the solenoid. If a resistor and inductor are connected in series to a battery of emf , and if a switch in the circuit is thrown closed at t 0, then the current in the circuit varies in time according to the expression I R (1 e t/ ) (32.7) where L/R is the time constant of the RL circuit. That is, the current increases to an equilibrium value of /R after a time that is long compared with . If the battery in the circuit is replaced by a resistanceless wire, the current decays exponentially with time according to the expression I where R e t/ (32.10) /R is the initial current in the circuit. 1034 CHAPTER 32 Inductance The energy stored in the magnetic field of an inductor carrying a current I is U 1 2 2 LI (32.12) This energy is the magnetic counterpart to the energy stored in the electric field of a charged capacitor. The energy density at a point where the magnetic field is B is uB B2 2 0 (32.14) The mutual inductance of a system of two coils is given by M 12 N 2 12 I1 M 21 N 1 21 I2 M (32.15) This mutual inductance allows us to relate the induced emf in a coil to the changing source current in a nearby coil using the relationships 2 M 12 dI 1 dt and 1 M 21 dI 2 dt (32.16, 32.17) In an LC circuit that has zero resistance and does not radiate electromagnetically (an idealization), the values of the charge on the capacitor and the current in the circuit vary in time according to the expressions Q I dQ dt Q max cos ( t Q max sin( t ) ) (32.21) (32.23) where Q max is the maximum charge on the capacitor, is the angular frequency of oscillation: 1 !LC is a phase constant, and (32.22) The energy in an LC circuit continuously transfers between energy stored in the capacitor and energy stored in the inductor. The total energy of the LC circuit at any time t is Q2 LI 2 max max (32.26) U UC UL cos2 t sin2 t 2C 2 At t 0, all of the energy is stored in the electric field of the capacitor (U Q 2 /2C ). Eventually, all of this energy is transferred to the inductor max (U LI 2 /2). However, the total energy remains constant because energy transmax formations are neglected in the ideal LC circuit. QUESTIONS 1. Why is the induced emf that appears in an inductor called a "counter" or "back" emf? 2. The current in a circuit containing a coil, resistor, and battery reaches a constant value. Does the coil have an inductance? Does the coil affect the value of the current? 3. What parameters affect the inductance of a coil? Does the inductance of a coil depend on the current in the coil? 4. How can a long piece of wire be wound on a spool so that the wire has a negligible self-inductance? 5. A long, fine wire is wound as a solenoid with a selfinductance L. If it is connected across the terminals of a battery, how does the maximum current depend on L ? 6. For the series RL circuit shown in Figure Q32.6, can the back emf ever be greater than the battery emf? Explain. Problems R 1035 Switch L Figure Q32.6 7. Consider this thesis: "Joseph Henry, America's first professional physicist, changed the view of the Universe during a school vacation at the Albany Academy in 1830. Before that time, one could think of the Universe as consisting of just one thing: matter. In Henry's experiment, after a battery is removed from a coil, the energy that keeps the current flowing for a while does not belong to any piece of matter. This energy belongs to the magnetic field surrounding the coil. With Henry's discovery of self-induction, Nature forced us to admit that the Universe consists of fields as well as matter." What in your view constitutes the Universe? Argue for your answer. 8. Discuss the similarities and differences between the energy stored in the electric field of a charged capacitor and the energy stored in the magnetic field of a currentcarrying coil. 9. What is the inductance of two inductors connected in series? Does it matter if they are solenoids or toroids? 10. The centers of two circular loops are separated by a fixed distance. For what relative orientation of the loops is their mutual inductance a maximum? a minimum? Explain. 11. Two solenoids are connected in series so that each carries the same current at any instant. Is mutual induction present? Explain. 12. In the LC circuit shown in Figure 32.15, the charge on the capacitor is sometimes zero, even though current is in the circuit. How is this possible? 13. If the resistance of the wires in an LC circuit were not zero, would the oscillations persist? Explain. 14. How can you tell whether an RLC circuit is overdamped or underdamped? 15. What is the significance of critical damping in an RLC circuit? 16. Can an object exert a force on itself? When a coil induces an emf in itself, does it exert a force on itself? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 32.1 Self-Inductance 1. A coil has an inductance of 3.00 mH, and the current through it changes from 0.200 A to 1.50 A in a time of 0.200 s. Find the magnitude of the average induced emf in the coil during this time. 2. A coiled telephone cord forms a spiral with 70 turns, a diameter of 1.30 cm, and an unstretched length of 60.0 cm. Determine the self-inductance of one conductor in the unstretched cord. 3. A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is thrown open, the current is effectively zero in 10.0 ms. What is the average induced emf in the inductor during this time? 4. A small air-core solenoid has a length of 4.00 cm and a radius of 0.250 cm. If the inductance is to be 0.060 0 mH, how many turns per centimeter are required? 5. Calculate the magnetic flux through the area enclosed by a 300-turn, 7.20-mH coil when the current in the coil is 10.0 mA. 6. The current in a solenoid is increasing at a rate of 10.0 A/s. The cross-sectional area of the solenoid is cm2, and there are 300 turns on its 15.0-cm length. What is the induced emf opposing the increasing current? WEB 7. A 10.0-mH inductor carries a current I I max sin t, with I max 5.00 A and /2 60.0 Hz. What is the back emf as a function of time? 8. An emf of 24.0 mV is induced in a 500-turn coil at an instant when the current is 4.00 A and is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil? 9. An inductor in the form of a solenoid contains 420 turns, is 16.0 cm in length, and has a cross-sectional area of 3.00 cm2. What uniform rate of decrease of current through the inductor induces an emf of 175 V? 10. An inductor in the form of a solenoid contains N turns, has length , and has cross-sectional area A. What uniform rate of decrease of current through the inductor induces an emf ? 11. The current in a 90.0-mH inductor changes with time as I t 2 6.00t (in SI units). Find the magnitude of the induced emf at (a) t 1.00 s and (b) t 4.00 s. (c) At what time is the emf zero? 12. A 40.0-mA current is carried by a uniformly wound aircore solenoid with 450 turns, a 15.0-mm diameter, and 12.0-cm length. Compute (a) the magnetic field inside the solenoid, (b) the magnetic flux through each turn, 1036 CHAPTER 32 Inductance the induct these same inductors are now connected in parallel, is it necessarily true that they are equivalent to a single ideal inductor having 1/L eq 1/L 1 1/L 2 and 1/R eq 1/R 1 1/R 2 ? Explain your answer. Figure P32.27 28. One application of an RL circuit is the generation of time-varying high voltage from a low-voltage source, as shown in Figure P32.28. (a) What is the current in the circuit a long time after the switch has been in position A? (b) Now the switch is thrown quickly from A to B. Compute the initial voltage across each resistor and the inductor. (c) How much time elapses before the voltage across the inductor drops to 12.0 V? A S Section 32.3 Energy in a Magnetic Field 31. Calculate the energy associated with the magnetic field of a 200-turn solenoid in which a current of 1.75 A produces a flux of 3.70 10 4 T m2 in each turn. 32. The magnetic field inside a superconducting solenoid is 4.50 T. The solenoid has an inner diameter of 6.20 cm and a length of 26.0 cm. Determine (a) the magnetic energy density in the field and (b) the energy stored in the magnetic field within the solenoid. 33. An air-core solenoid with 68 turns is 8.00 cm long and has a diameter of 1.20 cm. How much energy is stored in its magnetic field when it carries a current of 0.770 A? 34. At t 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a resistance of 30.0 . (a) Find the energy stored in the magnetic field when the current reaches half its maximum value. (b) After the emf is connected, how long does it take the current to reach this value? 35. On a clear day there is a 100-V/m vertical electric field near the Earth's surface. At the same place, the Earth's magnetic field has a magnitude of 0.500 10 4 T. Compute the energy densities of the two fields. 36. An RL circuit in which L 4.00 H and R 5.00 is connected to a 22.0-V battery at t 0. (a) What energy is stored in the inductor when the current is 0.500 A? (b) At what rate is energy being stored in the inductor when I 1.00 A? (c) What power is being delivered to the circuit by the battery when I 0.500 A? 37. A 10.0-V battery, a 5.00- resistor, and a 10.0-H inductor are connected in series. After the current in the circuit B 12.0 V 1 200 2.00 H 12.0 Figure P32.28 WEB 29. A 140-mH inductor and a 4.90- resistor are connected with a switch to a 6.00-V battery, as shown in Figure P32.29. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from A to B. How much time elapses before the current falls to 160 mA? 30. Consider two ideal inductors, L 1 and L 2 , that have zero internal resistance and are far apart, so that their magnetic fields do not influence each other. (a) If these inductors are connected in series, show that they are equivalent to a single ideal inductor having L eq L 1 L 2 . (b) If these same two inductors are connected in parallel, show that they are equivalent to a WEB 1038 CHAPTER 32 Inductance h 0.400 mm, w 1.30 mm, and L 2.70 mm, what is their mutual inductance? 47. Two inductors having self-inductances L 1 and L 2 are connected in parallel, as shown in Figure P32.47a. The mutual inductance between the two inductors is M. Determine the equivalent self-inductance L eq for the system (Fig. P32.47b). has reached its maximum value, calculate (a) the power being supplied by the battery, (b) the power being delivered to the resistor, (c) the power being delivered to the inductor, and (d) the energy stored in the magnetic field of the inductor. 38. A uniform electric field with a magnitude of 680 kV/m throughout a cylindrical volume results in a total energy of 3.40 J. What magnetic field over this same region stores the same total energy? 39. Assume that the magnitude of the magnetic field outside a sphere of radius R is B B 0(R/r)2, where B0 is a constant. Determine the total energy stored in the magnetic field outside the sphere and evaluate your resul energy stored in the capacitor first exceeds that in the inductor. 62. An inductor having inductance L and a capacitor having capacitance C are connected in series. The current in the circuit increases linearly in time as described by I Kt . The capacitor is initially uncharged. Determine (a) the voltage across the inductor as a function of time, (b) the voltage across the capacitor as a function of time, and (c) the time when the energy stored in the capacitor first exceeds that in the inductor. 63. A capacitor in a series LC circuit has an initial charge Q and is being discharged. Find, in terms of L and C, the flux through each of the N turns in the coil, when the charge on the capacitor is Q /2. 64. The toroid in Figure P32.64 consists of N turns and has a rectangular cross-section. Its inner and outer radii are a and b, respectively. (a) Show that L 0N 2h Figure P32.54 WEB 55. An LC circuit like that illustrated in Figure 32.14 consists of a 3.30-H inductor and an 840-pF capacitor, initially carrying a 105- C charge. At t 0 the switch is thrown closed. Compute the following quantities at t 2.00 ms: (a) the energy stored in the capacitor; (b) the energy stored in the inductor; (c) the total energy in the circuit. 2 ln b a (Optional) Section 32.6 The RLC Circuit 56. In Figure 32.19, let R 7.60 , L 2.20 mH, and C 1.80 F. (a) Calculate the frequency of the damped oscillation of the circuit. (b) What is the critical resistance? 57. Consider an LC circuit in which L 500 mH and C 0.100 F. (a) What is the resonant frequency 0 ? (b) If a resistance of 1.00 k is introduced into this circuit, what is the frequency of the (damped) oscillations? (c) What is the percent difference between the two frequencies? 58. Show that Equation 32.29 in the text is Kirchhoff's loop rule as applied to Figure 32.19. 59. Electrical oscillations are initiated in a series circuit containing a capacitance C, inductance L, and resistance R. (a) If R V !4L/C (weak damping), how much time elapses before the amplitude of the current oscillation falls off to 50.0% of its initial value? (b) How long does it take the energy to decrease to 50.0% of its initial value? (b) Using this result, compute the self-inductance of a 500-turn toroid for which a 10.0 cm, b 12.0 cm, and h 1.00 cm. (c) In Problem 14, an approximate formula for the inductance of a toroid with R W r was derived. To get a feel for the accuracy of that result, use the expression in Problem 14 to compute the approximate inductance of the toroid described in part (b). Compare the result with the answer to part (b). h a b Figure P32.64 65. (a) A flat circular coil does not really produce a uniform magnetic field in the area it encloses, but estimate the self-inductance of a flat circular coil, with radius R and N turns, by supposing that the field at its center is uniform over its area. (b) A circuit on a laboratory table consists of a 1.5-V battery, a 270- resistor, a switch, and three 30cm-long cords connecting them. Suppose that the circuit is arranged to be circular. Think of it as a flat coil with one turn. Compute the order of magnitude of its selfinductance and (c) of the time constant describing how fast the current increases when you close the switch. 66. A soft iron rod ( m 800 0 ) is used as the core of a solenoid. The rod has a diameter of 24.0 mm and is ADDITIONAL PROBLEMS 60. Initially, the capacitor in a series LC circuit is charged. A switch is closed, allowing the capacitor to discharge, and after time t the energy stored in the capacitor is onefourth its initial value. Determine L if C is known. 61. A 1.00-mH inductor and a 1.00- F capacitor are connected in series. The current in the circuit is described by I 20.0t, where t is in seconds and I is in amperes. 1040 CHAPTER 32 Inductance R1 S 10.0 cm long. A 10.0-m piece of 22-gauge copper wire (diameter 0.644 mm) is wrapped around the rod in a single uniform layer, except for a 10.0-cm length at each end, which is to be used for connections. (a) How many turns of this wire can wrap around the rod? (Hint: The difabricated. It would carry a maximum current of 50.0 kA through each winding of a 150-turn Nb3Sn solenoid. (a) If the inductance of this huge coil were 50.0 H, what would be the total energy stored? (b) What would be the compressive force per meter length acting between two adjacent windings 0.250 m apart? 78. Review Problem. Superconducting Power Transmission. The use of superconductors has been proposed for the manufacture of power transmission lines. A single coaxial cable (Fig. P32.78) could carry 1.00 103 MW (the output of a large power plant) at 200 kV, dc, over a distance of 1 000 km without loss. An inner wire with a radius of 2.00 cm, made from the superconductor Nb3Sn, carries the current I in one direction. A surrounding superconducting cylinder, of radius 5.00 cm, would carry the return current I. In such a system, what is the magnetic field (a) at the surface of the inner conductor and (b) at the inner surface of the outer conductor? (c) How much energy would be stored in the space between the conductors in a 1 000-km superconducting line? (d) What is the pressure exerted on the outer conductor? 7.50 R 12.0 V 450 mH 10.0 V Figure P32.73 74. An air-core solenoid 0.500 m in length contains 1 000 turns and has a cross-sectional area of 1.00 cm2. (a) If end effects are neglected, what is the self-inductance? (b) A secondary winding wrapped around the center of the solenoid has 100 turns. What is the mutual inductance? (c) The secondary winding carries a constant current of 1.00 A, and the solenoid is connected to a load of 1.00 k . The constant current is suddenly stopped. How much charge flows through the load resistor? 75. The lead-in wires from a television antenna are often constructed in the form of two parallel wires (Fig. P32.75). (a) Why does this configuration of conductors have an inductance? (b) What constitutes the flux loop for this configuration? (c) Neglecting any magnetic flux inside the wires, show that the inductance of a length x I TV set I TV antenna I a b I a = 2.00 cm b = 5.00 cm Figure P32.75 Figure P32.78 1042 CHAPTER 32 Inductance 79. Review Problem. The Meissner Effect. Compare this problem with Problem 63 in Chapter 26 on the force attracting a perfect dielectric into a strong electric field. A fundamental property of a Type I superconducting material is perfect diamagnetism, or demonstration of the Meissner effect, illustrated in the photograph on page 855 and again in Figure 30.34, and described as follows: The superconducting material has B 0 everywhere inside it. If a sample of the material is placed into an externally produced magnetic field, or if it is cooled to become superconducting while it is in a magnetic field, electric currents appear on the surface of the sample. The currents have precisely the strength and orientation required to make the total magnetic field zero throughout the interior of the sample. The following problem will help you to understand the magnetic force that can then act on the superconducting sample. Consider a vertical solenoid with a length of 120 cm and a diameter of 2.50 cm consisting of 1 400 turns of copper wire carrying a counterclockwise current of 2.00 A, as shown in Figure P32.79a. (a) Find the magnetic field in the vacuum inside the solenoid. (b) Find the energy density of the magnetic field, and note that the units J/m3 of energy density are the same as the units N/m2( Pa) of pressure. (c) A superconducting bar 2.20 cm in diameter is inserted partway into the solenoid. Its upper end is far outside the solenoid, where the magnetic field is small. The lower end of the bar is deep inside the solenoid. Identify the direction required for the current on the curved surface of the bar so that the total magnetic field is zero within the bar. The field created by the supercurrents is sketched in Figure P32.79b, and the total field is sketched in Figure Btot B0 I (a) (b) (c) Figure P32.79 P32.79c. (d) The field of the solenoid exerts a force on the current in the superconductor. Identify the direction of the force on the bar. (e) Calculate the magnitude of the force by multiplying the energy density of the solenoid field by the area of the bottom end of the superconducting bar. ANSWERS TO QUICK QUIZZES 32.1 When it is being opened. When the switch is initially open, there is no current in the circuit; when the switch is then closed, the inductor tends to maintain the nocurrent condition, and as a result there is very little chance of sparking. When the switch is initially closed, there is current in the circuit; when the switch is then opened, the current decreases. An induced emf is set up across the inductor, and this emf tends to maintain the original current. Sparking can occur as the current bridges the air gap between the poles of the switch. 32.2 (b). Figure 32.8 shows that circuit B has the greater time constant because in this circuit it takes longer for the current to reach its maximum value and then longer for this current to decrease to zero after switch S 2 is closed. Equation 32.8 indicates that, for equal resistances R A and R B , the condition B A means that LA LB. 32.3 (a) M12 increases because the magnetic flux through coil 2 increases. (b) M12 decreases because rotation of coil 1 decreases its flux through coil 2. 32.4 (a) No. Mutual inductance requires a system of coils, and each coil has self-inductance. (b) Yes. A single coil has self-inductance but no mutual inductance because it does not interact with any other coils. 32.5 From Equation 32.25, I max Q max . Thus, the amplitude of the I - t graph is times the amplitude of the Q - t graph. 32.6 Equation 32.31 without the cosine factor. The dashed lines represent the positive and negative amplitudes (maximum values) for each oscillation period, and it is the Q Q maxe Rt /2L part of Equation 32.31 that gives the value of the ever-decreasing amplitude. P U Z Z L E R Small "black boxes" like this one are commonly used to supply power to electronic devices such as CD players and tape players. Whereas these devices need only about 12 V to operate, wall outlets provide an output of 120 V. What do the black boxes do, and how do they work? (George Semple) c h a p t e r Alternating-Current Circuits Chapter Outline 33.1 33.2 33.3 33.4 33.5 33.6 ac Sources and Phasors Resistors in an ac Circuit Inductors in an ac Circuit Capacitors in an ac Circuit The RLC Series Circuit Power in an ac Circuit 33.7 Resonance in a Series RLC Circuit 33.8 The Transformer and Power Transmission 33.9 (Optional) Rectifiers and Filters 1043 1044 CHAPTER 33 Alternating-Current Circuits I n this chapter we describe alternating-current (ac) circuits. Every time we turn on a television set, a stereo, or any of a multitude of other electrical appliances, we are calling on alternating currents to provide the power to operate them. We begin our study by investigating the characteristics of simple series circuits that contain resistors, inductors, and capacitors and that are driven by a sinusoidal voltage. We shall find that the maximum alternating current in each element is proportional to the maximum alternating voltage across the element. We shall also find that when the applied voltage is sinusoidal, the current in each element is sinusoidal, too, but not necessarily in phase with the applied voltage. We conclude the chapter with two sections concerning transformers, power transmission, and RC filters. 33.1 AC SOURCES AND PHASORS An ac circuit consists of circuit elements and a generator that provides the alternating current. As you recall from Section 31.5, the basic principle of the ac generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency , a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage v is v V max sin t where Vmax is the maximum output voltage of the ac generator, or the voltage amplitude. From Equation 13.6, the angular frequency is 2 f 2 T where f is the frequency of the generator (the voltage source) and T is the period. The generator determines the frequency of ecalling that the rate at which electrical energy is converted to internal energy in a resistor is the power i 2R, where i is the instantaneous current in the resistor. Because this rate is proportional to the square of the current, it makes no difference whether the current is direct or alternating -- that is, whether the sign associated with the current is positive or negative. However, the temperature increase produced by an alternating 1 The lowercase symbols v and i are used to indicate the instantaneous values of the voltage and the current. The current in a resistor is in phase with the voltage vR R v = Vmax sin t Figure 33.1 A circuit consisting of a resistor of resistance R connected to an ac generator, designated by the symbol . 1046 CHAPTER 33 Alternating-Current Circuits iR , vR a iR vR b T iR , vR Imax Vmax t iR vR Imax Vmax c (a) t (b) Figure 33.2 (a) Plots of the instantaneous current iR and instantaneous voltage vR across a resistor as functions of time. The current is in phase with the voltage, which means that the current is zero when the voltage is zero, maximum when the voltage is maximum, and minimum when the voltage is minimum. At time t T, one cycle of the time-varying voltage and current has been completed. (b) Phasor diagram for the resistive circuit showing that the current is in phase with the voltage. current having a maximum value I max is not the same as that produced by a direct current equal to I max . This is because the alternating current is at this maximum value for only an instant during each cycle (Fig. 33.3a). What is of importance in an ac circuit is an average value of current, referred to as the rms current. As we learned in Section 21.1, the notation rms stands for root mean square, which in this case means the square root of the mean (average) value of the square of the current: I rms i 2. Because i 2 varies as sin2 t and because the average value of i 2 is 1 2 2 2 I max (see Fig. 33.3b), the rms current is rms current I rms I max 2 0.707I max (33.4) This equation states that an alternating current whose maximum value is 2.00 A delivers to a resistor the same power as a direct current that has a value of (0.707) (2.00 A) 1.41 A. Thus, we can say that the average power delivered to a resistor that carries an alternating current is Average power delivered to a resistor av I2 R rms 2 That the square root of the average value of i 2 is equal to I max/2 can be shown as follows: The current in the circuit varies with time according to the expression i I max sin t, so i 2 I 2 sin2 t. Therefore, we can find the average value of i 2 by calculating the average value of sin2 t. A max graph of cos2 t versus time is identical to a graph of sin2 t versus time, except that the points are shifted on the time axis. Thus, the time average of sin2 t is equal to the time average of cos2 t when taken over one or more complete cycles. That is, (sin2 t)av Using this fact and the trigonometric identity sin2 (sin2 t )av (cos2 t)av cos2 1, we obtain 1 (cos2 t )av (sin2 t )av 2(sin2 t )av 1 2 When we substitute this result in the expression i 2 I 2 sin2 t, we obtain (i 2 )av i 2 I 2 max rms I 2 /2, or I rms I max/2. The factor 1/2 is valid only for sinusoidally varying currents. Other wavemax forms, such as sawtooth variations, have different factors. 33.2 Resistors in an ac Circuit i Imax t 1047 (a) i2 I2 max i2 = 1 I 2 2 max t (b) Figure 33.3 (a) Graph of the current in a resistor as a function of time. (b) Graph of the current squared in a resistor as a function of time. Notice that the gray shaded regions under the curve and above the dashed line for I 2 /2 have the same area as the gray shaded regions above max the curve and below the dashed line for I 2 /2. Thus, the average value of i 2 is I 2 /2. max max Alternating voltage also is best discussed in terms of rms voltage, and the relationship is identical to that for current: V rms V max 0.707 V max (33.5) rms voltage 2 When we speak of measuring a 120-V alternating voltage from an electrical outlet, we are reower points has the value R/L, so Q 0L R QuickLab Tune a radio to your favorite station. Can you determine what the product of LC must be for the radio's tuning circuitry? (33.36) The curves plotted in Figure 33.16 show that a high-Q circuit responds to only a very narrow range of frequencies, whereas a low-Q circuit can detect a much broader range of frequencies. Typical values of Q in electronic circuits range from 10 to 100. The receiving circuit of a radio is an important application of a resonant circuit. One tunes the radio to a particular station (which transmits a specific electromagnetic wave or signal) by varying a capacitor, which changes the resonant frequency of the receiving circuit. When the resonance frequency of the circuit matches that of the incoming electromagnetic wave, the current in the receiving circuit increases. This signal caused by the incoming wave is then amplified and fed to a speaker. Because many signals are often present over a range of frequencies, it is important to design a high-Q circuit to eliminate unwanted signals. In this manner, stations whose frequencies are near but not equal to the resonance frequency give signals at the receiver that are negligibly small relative to the signal that matches the resonance frequency. av Small R, high Q Large R, low Q 0 Figure 33.16 4 The quality factor is also defined as the ratio 2 E/ E, where E is the energy stored in the oscillating system and E is the energy lost per cycle of oscillation. The quality factor for a mechanical system can also be defined, as noted in Section 13.7. Average power versus frequency for a series RLC cirof each curve is cuit. The width measured between the two points where the power is half its maximum value. The power is a maximum at the resonance frequency 0. 1060 CHAPTER 33 Alternating-Current Circuits Quick Quiz 33.5 An airport metal detector (Fig. 33.17) is essentially a resonant circuit. The portal you step through is an inductor (a large loop of conducting wire) that is part of the circuit. The frequency of the circuit is tuned to the resonant frequency of the circuit when there is no metal in the inductor. Any metal on your body increases the effective inductance of the loop and changes the current in it. If you want the detector to be able to detect a small metallic object, should the circuit have a high quality factor or a low one? Signal Circuit A Circuit B C Figure 33.17 When you pass through a metal detector, you become part of a resonant circuit. As you step through the detector, the inductance of the circuit changes, and thus the current in the circuit changes. (Terry Qing/FPG International) EXAMPLE 33.8 A Resonating Series RLC Circuit C 1 0 2L Consider a series RLC circuit for which R 150 , L 20.0 mH, V rms 20.0 V, and 5 000 s 1. Determine the value of the capacitance for which the current is a maximum. The current has its maximum value at the resonance frequency 0 , which should be made to match the "driving" frequency of 5 000 s 1: 0 1 (25.0 10 6 s 2 )(20.0 10 3 H) 2.00 F Solution Exercise Answer Calculate the maximum value of the rms current in the circuit as the frequency is varied. 0.133 A. 5.00 10 3 s 1 1 LC 33.8 THE TRANSFORMER AND POWER TRANSMISSION When electric power is transmitted over great distances, it is economical to use a high voltage and a low current to minimize the I 2R loss in the transmission lines. 33.8 The Transformer and Power Transmission 1061 Soft iron V1 N1 N2 Z1 Primary (input) S R Z2 Secondary (output) Consequently, 350-kV lines are common, and in many areas even higher-voltage (765-kV) lines are under construction. At the receiving end of such lines, the consumer requires power at a low voltage (for safety and for efficiency in design). Therefore, a device is required that can change the alternating voltage and current without causing appreciable changes in the power delivered. The ac transformer is that device. In its simplest form, the ac transformer consists of two coils of wire wound around a core of iron, as illustra I rms V rms (X L X C )2 R2 (33.32) A series RLC circuit is in resonance when the inductive reactance equals the capacitive reactance. When this condition is met, the current given by Equation 33.32 reaches its maximum value. When X L X C in a circuit, the resonance frequency 0 of the circuit is 1 (33.33) 0 LC The current in a series RLC circuit reaches its maximum value when the frequency of the generator equals 0 -- that is, when the "driving" frequency matches the resonance frequency. Transformers allow for easy changes in alternating voltage. Because energy (and therefore power) are conserved, we can write I 1 V1 I 2 V2 (33.40) to relate the currents and voltages in the primary and secondary windings of a transformer. QUESTIONS 1. Fluorescent lights flicker on and off 120 times every second. Explain what causes this. Why can't you see it happening? 2. Why does a capacitor act as a short circuit at high frequencies? Why does it act as an open circuit at low frequencies? 3. Explain how the acronyms in the mnemonic "ELI the ICE man" can be used to recall whether current leads voltage or voltage leads current in RLC circuits. (Note that "E" represents voltage.) 4. Why is the sum of the maximum voltages across the elements in a series RLC circuit usually greater than the maximum applied voltage? Doesn't this violate Kirchhoff's second rule? 5. Does the phase angle depend on frequency? What is the phase angle when the inductive reactance equals the capacitive reactance? 6. Energy is delivered to a series RLC circuit by a generator. This energy appears as internal energy in the resistor. What is the source of this energy? 1068 CHAPTER 33 Alternating-Current Circuits 15. Why are the primary and secondary windings of a transformer wrapped on an iron core that passes through both coils? 16. With reference to Figure Q33.16, explain why the capacitor prevents a dc signal from passing between circuits A and B, yet allows an ac signal to pass from circuit A to circuit B. (The circuits are said to be capacitively coupled.) C 7. Explain why the average power delivered to an RLC circuit by the generator depends on the phase between the current and the applied voltage. 8. A particular experiment requires a beam of light of very stable intensity. Why would an ac voltage be unsuitable for powering the light source? 9. Consider a series RLC circuit in which R is an incandescent lamp, C is some fixed capacitor, and L is a variable inductance. The source is 120-V ac. Explain why the lamp glows brightly for some values of L and does not glow at all for other values. 10. What determines the maximum voltage that can be used on a transmission line? 11. Will a transformer operate if a battery is used for the input voltage across the primary? Explain. 12. How can the average value of a current be zero and yet the square root of the average squared current not be zero? 13. What is the time average of the "square-wave" voltage shown in Figure Q33.13? What is its rms voltage? V Vmax 0 t Circuit A Circuit B Signal Figure Q33.16 17. With reference to Figure Q33.17, if C is made sufficiently large, an ac signal passes from circuit A to ground rather than from circuit A to circuit B. Hence, the capacitor acts as a filter. Explain. Signal Circuit A Circuit B C Figure Q33.13 14. Explain how the quality factor is related to the response characteristics of a radio receiver. Which variable most strongly determines the quality factor? Figure Q33.17 PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Note: Assume that all ac voltages and currents are sinusoidal unless stated otherwise. Note that an ideal ammeter has zero resistance and that an ideal voltmeter has infinite resistance. V max = 100 V Section 33.1 ac Sources and Phasors Section 33.2 Resistors in an ac Circuit 1. The rms output voltage of an ac generator is 200 V, and the operating frequency is 100 Hz. Write the equation giving the output voltage as a function of time. 2. (a) What is the resistance of a lightbulb that uses an average power of 75.0 W when connected to a 60.0-Hz power source having a maximum voltage of 170 V? (b) What is the resistance of a 100-W bulb? 3. An ac power supply produces a maximum voltage V max 100 V. This power supply is connected to a 24.0- resistor, and the current and resistor voltage are measured with an ideal ac ammeter and voltmeter, as shown in Figure P33.3. What does each meter read? A R = 24.0 V Figure P33.3 Problems 4. In the simple ac circuit shown in Figure 33.1, R 70.0 and v V max sin t. (a) If v R 0.250 V max for the first time at t 0.010 0 s, what is the angular frequency of the generator? (b) What is the next value of t for which v R 0.250 V max? 5. The current in the circuit shown in Figure 33.1 equals 60.0% of the peak current at t 7.00 ms. What is the smallest frequency of the generator that gives this current? 6. Figure P33.6 shows three lamps connected to a 120-V ac (rms) household supply voltage. Lamps 1 and 2 have 150-W bulbs; lamp 3 has a 100-W bulb. Find the rms current and the resistance of each bulb. Lamp Lamp Lamp 1 2 3 1069 WEB 11. For the circuit shown in Figure 33.4, V max 80.0 V, 65.0 rad/s, and L 70.0 mH. Calculate the current in the inductor at t 15.5 ms. 12. A 20.0-mH inductor is connected to a standard outlet ( V rms 120 V, f 60.0 Hz). Determine the energy stored in the inductor at t (1/180) s, assuming that this energy is zero at t 0. 13. Review Problem. Determine the maximum magnetic flux through an inductor connected to a standard outlet ( V rms 120 V, f 60.0 Hz). Section 33.4 Capacitors in an ac Circuit 14. (a) For what frequencies does a 22.0- F capacitor have a reactance below 175 ? (b) Over this same frequency range, what is the reactance of a 44.0- F capacitor? 15. What maximum current is delivered by a 2.20- F capacitor when it is connected across (a) a North American outlet having V rms 120 V and f 60.0 Hz? (b) a European outlet having V rms 240 V and f 50.0 Hz? 16. A capacitor C is connected to a power supply that operates at a frequency f and produces an rms voltage V. What is the maximum charge that appears on either of the capacitor plates? 17. What maximum current is delivered by an ac generator with V max 48.0 V and f 90.0 Hz when it is connected across a 3.70- F capacitor? 18. A 1.00-mF capacitor is connected to a standard outlet ( V rms 120 V, f 60.0 Hz). Determine the current in the capacitor at t (1/180) s, assuming that at t 0 the energy stored in the capacitor is zero. 120 V Figure P33.6 7. An audio amplifier, represented by the ac source and resistor in Figure P33.7, delivers to the speaker alternating voltage at audio frequencies. If the source voltage has an amplitude of 15.0 V, R 8.20 , and the speaker is equivalent to a resistance of 10.4 , what time-averaged power is transferred to it? R Section 33.5 The RLC Series Circuit 19. An inductor (L 400 mH), a capacitor (C 4.43 F), and a resistor (R 500 ) are connected in series. A 50.0-Hz ac generator produces a peak current of 250 mA in the circuit. (a) Calculate the required peak voltage V max . (b) Determine the phase angle by which the current leads or lags the applied voltage. 20. At what frequency does the inductive reactance of a 57.0- H inductor equal the capacitive reactance of a 57.0- F capacitor? 21. A series ac circuit contains the following components: R 150 , L 250 mH, C 2.00 F, and a generator with V max 210 V operating at 50.0 Hz. Calculate the (a) inductive reactance, (b) capacitive reactance, (c) impedance, (d) maximum current, and (e) phase angle between current and generator voltage. 22. A sinusoidal voltage v(t ) (40.0 V) sin(100t) is applied to a series RLC circuit with L 160 mH, C 99.0 F, and R 68.0 . (a) What is the impedance of the circuit? (b) What is the maximum current? (c) Determine the numerical values for I max , , and in the equation i(t ) I max sin( t ). 23. An RLC circc series circuit for which R 300 , C 11.0 F, L 0.200 H, and f (500/ ) Hz. 27. A coil of resistance 35.0 and inductance 20.5 H is in series with a capacitor and a 200-V (rms), 100-Hz source. The rms current in the circuit is 4.00 A. (a) Calculate the capacitance in the circuit. (b) What is V rms across the coil? Section 33.6 Power in an ac Circuit 28. The voltage source in Figure P33.28 has an output V rms 100 V at 1 000 rad/s. Determine (a) the current in the circuit and (b) the power supplied by the source. (c) Show that the power delivered to the resistor is equal to the power supplied by the source. Problems R, the average power delivered to the diode circuit shown in Figure P33.35. Diode 2R R R Diode V R 1071 Figure P33.35 Section 33.7 Resonance in a Series RLC Circuit 36. The tuning circuit of an AM radio contains an LC combination. The inductance is 0.200 mH, and the capacitor is variable, so the circuit can resonate at any frequency between 550 kHz and 1 650 kHz. Find the range of values required for C. 37. An RLC circuit is used in a radio to tune in to an FM station broadcasting at 99.7 MHz. The resistance in the circuit is 12.0 , and the inductance is 1.40 H. What capacitance should be used? 38. A series RLC circuit has the following values: L 20.0 mH, C 100 nF, R 20.0 , and V max 100 V, V max sin t. Find (a) the resonant frequency, with v (b) the amplitude of the current at the resonant frequency, (c) the Q of the circuit, and (d) the amplitude of the voltage across the inductor at resonance. 39. A 10.0- resistor, a 10.0-mH inductor, and a 100- F capacitor are connected in series to a 50.0-V (rms) source having variable frequency. What is the energy delivered to the circuit during one period if the operating frequency is twice the resonance frequency? 40. A resistor R, an inductor L, and a capacitor C are connected in series to an ac source of rms voltage V and variable frequency. What is the energy delivered to the circuit during one period if the operating frequency is twice the resonance frequency? 41. Compute the quality factor for the circuits described in Problems 22 and 23. Which circuit has the sharper resonance? 43. A transformer has N 1 350 turns and N 2 2 000 turns. If the input voltage is v(t ) (170 V) cos t, what rms voltage is developed across the secondary coil? 44. A step-up transformer is designed to have an output voltage of 2 200 V (rms) when the primary is connected across a 110-V (rms) source. (a) If there are 80 turns on the primary winding, how many turns are required on the secondary? (b) If a load resistor across the secondary draws a current of 1.50 A, what is the current in the primary under ideal conditions? (c) If the transformer actually has an efficiency of 95.0%, what is the current in the primary when the secondary current is 1.20 A? 45. In the transformer shown in Figure P33.45, the load resistor is 50.0 . The turns ratio N 1 :N 2 is 5 : 2, and the source voltage is 80.0 V (rms). If a voltmeter across the load measures 25.0 V (rms), what is the source resistance R s ? Rs Vs N1 N2 RL Figure P33.45 46. The secondary voltage of an ignition transformer in a furnace is 10.0 kV. When the primary operates at an rms voltage of 120 V, the primary impedance is 24.0 and the transformer is 90.0% efficient. (a) What turns ratio is required? What are (b) the current in the secondary and (c) the impedance in the secondary? 47. A transmission line that has a resistance per unit length of 4.50 10 4 /m is to be used to transmit 5.00 MW over 400 mi (6.44 105 m). The output voltage of the generator is 4.50 kV. (a) What is the power loss if a transformer is used to step up the voltage to 500 kV? (b) What fraction of the input power is lost to the line under these circumstances? (c) What difficulties would be encountered on attempting to transmit the 5.00 MW at the generator voltage of 4.50 kV? (Optional) Section 33.9 Rectifiers and Filters Section 33.8 The Transformer and Power Transmission 42. A step-down transformer is used for recharging the batteries of portable devices such as tapound this rectangle. The contributions from the top and bottom of the rectangle are zero because E is perpendicular to ds for these paths. We can express the electric field on the right side of the rectangle as x z B E(x dx, t) E(x, t) Figure 34.5 As a plane wave passes through a rectangular path of width dx lying in the xy plane, the electric field in the y direction varies from E to E d E. This spatial variation in E gives rise to a time-varying magnetic field along the z direction, according to Equation 34.6. dE dx dx t constant E(x, t) E dx x while the field on the left side is simply E(x, t).3 Therefore, the line integral over this rectangle is approximately E ds E(x dx, t) E(x, t) ( E/ x) dx (34.14) Because the magnetic field is in the z direction, the magnetic flux through the rectangle of area dx is approximately B B dx. (This assumes that dx is very small compared with the wavelength of the wave.) Taking the time derivative of 3 Because dE/ dx in this equation is expressed as the change in E with x at a given instant t, dE/ is dx equivalent to the partial derivative E / x . Likewise, dB/ means the change in B with time at a particudt lar position x, so in Equation 34.15 we can replace dB/ with B/ t . dt 34.3 Energy Carried by Electromagnetic Waves 1083 y the magnetic flux gives d B dt dx dB dt dx x constant B t (34.15) E Substituting Equations 34.14 and 34.15 into Equation 34.3, we obtain E x dx E x dx B t B t z dx B + dB B x This expression is Equation 34.6. In a similar manner, we can verify Equation 34.7 by starting with Maxwell's fourth equation in empty space (Eq. 34.5). In this case, we evaluate the line integral of B ds around a rectangle lying in the xz plane and having width dx and length , as shown in Figure 34.6. Noting that the magnitude of the magnetic field changes from B(x, t) to B(x dx , t) over the width dx, we find the line integral over this rectangle to be approximately B ds B(x, t) B(x dx, t) E Figure 34.6 As a plane wave passes through a rectangular path of width dx lying in the xz plane, the magnetic field in the z direction varies from B to B d B. This spatial variation in B gives rise to a time-varying electric field along the y direction, according to Equation 34.7. ( B/ x) dx (34.16) The electric flux through the rectangle is with respect to time, gives E E dx, which, when differentiated E t t dx (34.17) Substituting Equations 34.16 and 34.17 into Equation 34.5 gives ( B/ x) dx B x which is Equation 34.7. 0 0 dx( E/ t) E t 0 0 34.3 ENERGY CARRIED BY ELECTROMAGNETIC WAVES Poynting vector Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects placed in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector, which is defined by the expression S 1 0 E B (34.18) The magnitude of the Poynting vector represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation. Thus, the magnitude of the Poynting vector represents power per unit area. The direction of the vector is along the direction of wave propagation (Fig. 34.7). The SI units of the Poynting vector are J/s m2 W/m2. Magnitude of the Poynting vector for a plane wave 1084 CHAPTER 34 Electromagnetic Waves As an example, let us evaluate the magnitude of S for a plane electromagnetic wave where E B EB. In this case, S Because B EB 0 (34.19) E /c, we can also express this as S E2 0c c 0 B2 Wave intensity y These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area. What is of greater interest for a sinusoidal plane electromagnetic wave is the time average of S over one or more cycles, which is called the wave intensity I. (We discussed the intensity of sound waves in Chapter 17.) When this average is taken, we obtain an expression involving the time average of cos2(kx t), which equals 1 . Hence, the average value of S (in other words, the intensity of the wave) is 2 I S av E max B max 2 0 E2 max 2 0c c 2 0 E B2 max (34.20) S z B c x Recall that the energy per unit volume, which is the instantaneous energy density uE associated with an electric field, is given by Equation 26.13, uE 1 2 0E 2 Figure 34.7 The Poynting vector S for a plane electromagnetic wave is along the direction of wave propagation. and that the instantaneous energy density uB associated with a magnetic field is given by Equation 32.14: B2 uB 2 0 Because E and B vary with time for an electromagnetic wave, the energy densities also vary with time. When we use the relationships B E /c and c 1/ 0 0 , Equation 32.14 becomes uB (E /c)2 2 0 0 0 2 E2 0 1 2 0E 2 Comparing this result with the expression for uE , we see that uB uE 1 2 0E 2 B2 2 0 Total instantaneous energy density That is, for an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field. Hence, in a given volume the energy is equally shared by the two fields. The total instantaneous energy density u is equal to the sum of the energy densities associated with the electric and magnetic fields: u uE uB 0E 2 B2 0 Average energy density of an electromagnetic wave When this total instantaneous energy density is averaged over one or more cycles of an electromagnetic wave, we again obtain a factor of 1 . Hence, for any electro2 magnetic wave, the total average energy per unit volume is 34.4 Momentum and Radiation Pressure 1085 u av 0(E 2) av 1 2 2 0 E max B2 max 2 0 (34.21) EXAMPLE 34.2 Fields on the Page E max Estimate the maximum magnitudes of the electric and magnetic fields of the light that is incident on this page because of the visible light coming from your desk lamp. Treat the bulb as a point source of electromagnetic radiation that is about 5% efficient at converting electrical energy to visible light. Recall from Equation 17.8 that the wave intensity 2 I a distance r from a point source is I av /4 r , where av is the average power output of the source and 4 r 2 is the area of a sphere of radius r centered on the source. Because the intensity of an electromagnetic wave is also given by Equation 34.20, we have I 4 av r2 0c av 7 2 r2 (4 10 T m/A)(3.00 10 8 m/s)(3.0 W) 2 (0.30 m )2 Solution 45 V/m From Equation 34.13, B max E max c 45 V/m 3.00 10 8 m/s 1.5 10 7 T E2 max 2 0c This value is two orders of magnitude smaller than the Earth's magnetic field, which, unlike the magnetic field in the light wave from your desk lamp, is not oscillating. We must now make some assumptions about numbers to enter in this equation. If we have a 60-W lightbulb, its output at 5% efficiency is approximately 3.0 W in the form of visible light. (The remaining energy transfers out of the bulb by conduction and invisible radiation.) A reasonable distance from the bulb to the page might be 0.30 m. Thus, we have Exercise Answer Estimate the energy density of the light wave just before it strikes this page. 9.0 10 9 J/m3. Comparing this result with Equation 34.20 for the average value of S, we see that I S av cu av (34.22) In other words, the intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light. 34.4 MOMENTUM AND RADIATION PRESSURE Momentum transported to a perfectly absorbing surface Electromagnetic waves transport linear momentum as well as energy. It follows that, as this momentum is absorbed by some surface, pressure is exerted on the surface. We shall assume in this discussion that the electromagnetic wave strikes the surface at normal incidence and transports a total energy U to the surface in a time t. Maxwell showed that, if the surface absorbs all the incident energy U in this time (as does a black body, introduced in Chapter 20), the total momentum p transported to the surface has a magnitude p U c (complete absorption) (34.23) The pressure exerted on the surface is defined as force per unit area F/A. Let us combine this with Newton's second law: 1086 CHAPTER 34 Electromagnetic Waves P F A 1 dp A dt Radiation pressure exerted on a perfectly absorbing surface If we now replace p, the momentum transported to the surface by light, from Equation 34.23, we have P 1 dp A dt 1 d A dt U c 1 (dU/dt) c A QuickLab Using Example 34.2 as a starting point, estimate the total force exerted on this page by the light from your desk lamp. Does it make a difference if the page contains large, dark photographs instead of mostly white space? We recognize (dU/dt)/A as the rate at which energy is arriving at the surface per unit area, which is the magnitude of the Poynting vector. Thus, the radiation pressure P exerted on the perfectly absorbing surface is P S c (34.24) Radiation pressure exerted on a perfectly reflecting surface Note that Equation 34.24 is an expression for uppercase P, the pressure, while Equation 34.23 is an expression for lowercase p, linear momentum. If the surface is a perfect reflector (such as a mirror) and incidence is normal, then the momentum transported to the surface in a time t is twice that given by Equation 34.23. That is, the momentum transferred to the surface by the incoming light is p U/c, and that transferred by the reflected light also is p U/c. Therefore, 2U p (complete reflection) (34.25) c The momentum delivered to a surface having a reflectivity somewhere between these two extremes has a value between U/c and 2U/c, depending on the properties of the surface. Finally, the radiation pressure exerted on a perfectly reflecting surface for normal incidence of the wave is4 P 2S c (34.26) web Visit http://pds.jpl.nasa.gov for more information about missions to the planets. You may also want to read Arthur C. Clarke's 1963 science fiction story The Wind from the Sun about a solar yacht race. Although radiation pressures are very small (about 5 10 6 N/m2 for direct sunlight), they have been measured with torsion balances such as the one shown in Figure 34.8. A mirror (a perfect reflector) and a black disk (a perfect absorber) are connected by a horizontal rod suspended from a fine fiber. Normal-incidence light striking the black disk is completely absorbed, so all of the momentum of the Light Mirror Black disk Figure 34.8 An apparatus for measuring the pressure exerted by light. In practice, the system is contained in a high vacuum. 4 Figure 34.9 Mariner 10 used its solar panels to "sail on sunlight." For oblique incidence on a perfectly reflecting surface, the momentum transferred is (2U cos )/c and the pressure is P (2S cos2 ) /c, where is the angle between the normal to the surface and the direction of wave propagation. 34.4 Momentum and Radiation Pressure 1087 light is transferred to the disk. Normal-incidence light striking the mirror is totally reflected, and hence the momentum transferred to the mirror is twice as great as that transferred to the disk. The radiation pressure is determined by measuring the angle through which the horizontal connecting rod rotates. The apparatus CONCEPTUAL EXAMPLE 34.3 Sweeping the Solar System cube of the radius of a spherical dust particle because it is proportional to the mass and therefore to the volume 4 r 3/3 of the particle. The radiation pressure is proportional to the square of the radius because it depends on the planar crosssection of the particle. For large particles, the gravitational force is greater than the force from radiation pressure. For particles having radii less than about 0.2 m, the radiationpressure force is greater than the gravitational force, and as a result these particles are swept out of the Solar System. A great amount of dust exists in interplanetary space. Although in theory these dust particles can vary in size from molecular size to much larger, very little of the dust in our solar system is smaller than about 0.2 m. Why? Solution The dust particles are subject to two significant forces -- the gravitational force that draws them toward the Sun and the radiation-pressure force that pushes them away from the Sun. The gravitational force is proportional to the EXAMPLE 34.4 Pressure from a Laser Pointer flected beam would apply a pressure of P 2S /c. We can model the actual reflection as follows: Imagine that the surface absorbs the beam, resulting in pressure P S /c. Then the surface emits the beam, resulting in additional pressure P S /c. If the surface emits only a fraction f of the beam (so that f is the amount of the incident beam reflected), then the pressure due to the emitted beam is P f S /c. Thus, the total pressure on the surface due to absorption and re-emission (reflection) is P S c f S c (1 f) S c Many people giving presentations use a laser pointer to direct the attention of the audience. If a 3.0-mW pointer creates a spot that is 2.0 mm in diameter, determine the radiation pressure on a screen that reflects 70% of the light that strikes it. The power 3.0 mW is a time-averaged value. Solution We certainly do not expect the pressure to be very large. Before we can calculate it, we must determine the Poynting vector of the beam by dividing the time-averaged power delivered via the electromagnetic wave by the crosssectional area of the beam: S 3.0 A r2 2.0 10 10 2 3W 3 m 2 955 W/m2 Notice that if f 1, which represents complete reflection, this equation reduces to Equation 34.26. For a beam that is 70% reflected, the pressure is P (1 0.70) 955 W/m2 3.0 10 8 m/s 5.4 10 6 This is about the same as the intensity of sunlight at the Earth's surface. (Thus, it is not safe to shine the beam of a laser pointer into a person's eyes; that may be more dangerous than looking directly at the Sun.) Now we can determine the radiation pressure from the laser beam. Equation 34.26 indicates that a completely re- N/m2 This is an extremely small value, as expected. (Recall from Section 15.2 that atmospheric pressure is approximately 105 N/m2.) EXAMPLE 34.5 Solar Energy represents the power per unit area, or the light intensity. Assuming that the radiation is incident normal to the roof, we obtain SA 1.60 (1 000 W/m2 )(8.00 10 5 W 20.0 m2 ) As noted in the preceding example, the Sun delivers about 1 000 W/m2 of energy to the Earth's surface via electromagnetic radiation. (a) Calculate the total power that is incident on a roof of dimensions 8.00 m 20.0 m. Solution The magnitude of the Poynting vector for solar radiation at the surface of the Earth is S 1 000 W/m2 ; this 1088 CHAPTER 34 Electromagnetic Waves Using Equation 34.24 with S find that the radiation pressure is P S c 1 000 W/m2 3.00 10 8 m/s 3.33 If all of this power could be converted to electrical energy, it would provide more than enough power for the average home. However, solar energy is not easily harnessed, and the prospects for large-scale conversion are not as bright as may appear from this calculation. For example, the efficiency of conversion from solar to electrical energy is typically 10% for photovoltaic cells. Roof systems for converting solar energy to internal energy are approximately 50% efficient; however, solar energy is associated with other practical problems, such as overcast days, geographic location, and methods of energy storage. (b) Determine the radiation pressure and the radiation force exerted on the roof, assuming that the roof covering is a perfect absorber. Solution 1 000 W/m2, we 10 6 N/m2 Because pressure equals force per unit area, this corresponds to a radiation force of F PA (3.33 10 6 N/m2 )(160 m2 ) 5.33 10 4 N Exercise 1 h? How much solar energy is incident on the roof in 5.76 108 J. Answer y Js must be placed in a high vacuum to eliminate the effects of air currents. NASA is exploring the possibility of solar sailing as a low-cost means of sending spacecraft to the planets. Large reflective sheets would be used in much the way canvas sheets are used on earthbound sailboats. In 1973 NASA engineers took advantage of the momentum of the sunlight striking the solar panels of Mariner 10 (Fig. 34.9) to make small course corrections when the spacecraft's fuel supply was running low. (This procedure was carried out when the spacecraft was in the vicinity of the planet Mercury. Would it have worked as well near Pluto?) Optional Section 34.5 RADIATION FROM AN INFINITE CURRENT SHEET z x A portion of an infinite current sheet lying in the yz plane. The current density is sinusoidal and is given by the expression Js J max cos t. The magnetic field is everywhere parallel to the sheet and lies along z. Figure 34.10 In this section, we describe the electric and magnetic fields radiated by a flat conductor carrying a time-varying current. In the symmetric plane geometry employed here, the mathematics is less complex than that required in lower-symmetry situations. Consider an infinite conducting sheet lying in the yz plane and carrying a surface current in the y direction, as shown in Figure 34.10. The current is distributed across the z direction such that the current per unit length is Js . Let us assume that Js varies sinusoidally with time as Js J max cos t where J max is the amplitude of the current variation and is the angular frequency of the variation. A similar problem concerning the case of a steady current was treated in Example 30.6, in which we found that the magnetic field outside the sheet is everywhere parallel to the sheet and lies along the z axis. The magnetic field was found to have a magnitude Bz 0 Radiated magnetic field Js 2 5 Note that the solution could also be written in the form cos( t t ). That is, cos is an even function, which means that cos( cos(kx kx), which is equivalent to ) cos . 34.5 Radiation from an Infinite Current Sheet y Js B 1089 c z x E Figure 34.11 Representation of the plane electromagnetic wave radiated by an infinite current sheet lying in the yz plane. The vector B is in the z direction, the vector E is in the y direction, and the direction of wave motion is along x. Both vector B and vector E behave according to the t ). Compare this drawing with Figure 34.3a. expression cos(kx In the present situation, where Js varies with time, this equation for B z is valid only for distances close to the sheet. Substituting the expression for Js , we have Bz 0 2 J max cos t (for small values of x) To obtain the expression valid for B z for arbitrary values of x, we can investigate the solution:5 Bz 0 J max 2 cos(kx t) (34.27) Radiated electric field You should note two things about this solution, which is unique to the geometry under consideration. First, when x is very small, it agrees with our original solution. Second, it satisfies the wave equation as expressed in Equation 34.9. We conclude that the magnetic field lies along the z axis, varies with time, and is characand an terized by a transverse traveling wave having an angular frequency angular wave number k 2 / . We can obtain the electric field radiating from our infinite current sheet by using Equation 34.13: 0 J maxc (34.28) E y cB z cos(kx t) 2 That is, the electric field is in the y direction, perpendicular to B, and has the same space and time dependencies. These expressions for B z and E y show that the radiation field of an infinite current sheet carrying a sinusoidal current is a plane electromagnetic wave propagating with a speed c along the x axis, as shown in Figure 34.11. We can calculate the Poynting vector for this wave from Equations 34.19, EXAMPLE 34.6 An Infinite Sheet Carrying a Sinusoidal Current An infinite current sheet lying in the yz plane carries a sinusoidal current that has a maximum density of 5.00 A/m. (a) Find the maximum values of the radiated magnetic and electric fields. Solution From Equations 34.27 and 34.28, we see that the maximum values of B z and E y are B max 0 J max 2 and E max 0 J maxc 2 1090 Using the values 0 4 10 and c 3.00 10 8 m/s, we get B max (4 10 7 7 CHAPTER 34 T m/A, J max Electromagnetic Waves 5.00 A/m, Solution The intensity, or power per unit area, radiated in each direction by the current sheet is given by Equation 34.30: I 0 T m/A)(5.00 A/m ) 2 6 J2 c max 8 10 7 (4 3.14 (4 10 7 10 T 10 8 m/s) T m/A)(5.00 A /m)2(3.00 8 10 8 m/s) E max T m/A)(5.00 A/m )(3.00 2 1.18 10 3 W/m2 Mulld an infrared photograph of a person look different from a photograph of that person taken with visible light? 22. Suppose a creature from another planet had eyes that were sensitive to infrared radiation. Describe what the creature would see if it looked around the room you are now in. That is, what would be bright and what would be dim? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 34.1 Maxwell's Equations and Hertz's Discoveries Section 34.2 Plane Electromagnetic Waves Note: Assume that the medium is vacuum unless specified otherwise. 1. If the North Star, Polaris, were to burn out today, in what year would it disappear from our vision? Take the distance from the Earth to Polaris as 6.44 1018 m. 2. The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is v 1/ 0 0 , where is the dielectric constant of the substance. Determine the speed of light in water, which has a dielectric constant at optical frequencies of 1.78. 3. An electromagnetic wave in vacuum has an electric field amplitude of 220 V/m. Calculate the amplitude of the corresponding magnetic field. 4. Calculate the maximum value of the magnetic field of an electromagnetic wave in a medium where the speed of light is two thirds of the speed of light in vacuum and where the electric field amplitude is 7.60 mV/m. 1098 CHAPTER 34 Electromagnetic Waves is induced by this signal between the ends of the receiving antenna. What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically with an average power of 250 kW? A monochromatic light source emits 100 W of electromagnetic power uniformly in all directions. (a) Calculate the average electric-field energy density 1.00 m from the source. (b) Calculate the average magneticfield energy density at the same distance from the source. (c) Find the wave intensity at this location. A community plans to build a facility to convert solar radiation to electric power. They require 1.00 MW of power, and the system to be installed has an efficiency of 30.0% (that is, 30.0% of the solar energy incident on the surface is converted to electrical energy). What must be the effective area of a perfectly absorbing surface used in such an installation, assuming a constant intensity of 1 000 W/m2 ? Assuming that the antenna of a 10.0-kW radio station radiates spherical electromagnetic waves, compute the maximum value of the magnetic field 5.00 km from the antenna, and compare this value with the surface magnetic field of the Earth. The filament of an incandescent lamp has a 150- resistance and carries a direct current of 1.00 A. The filament is 8.00 cm long and 0.900 mm in radius. (a) Calculate the Poynting vector at the surface of the filament. (b) Find the magnitude of the electric and magnetic fields at the surface of the filament. In a region of free space the electric field at an instant of time is E (80.0 i 32.0 j 64.0 k) N/C and the magnetic field is B (0.200 i 0.080 0 j 0.290 k) T. (a) Show that the two fields are perpendicular to each other. (b) Determine the Poynting vector for these fields. A lightbulb filament has a resistance of 110 . The bulb is plugged into a standard 120-V (rms) outlet and emits 1.00% of the electric power delivered to it as electromagnetic radiation of frequency f. Assuming that the bulb is covered with a filter that absorbs all other frequencies, find the amplitude of the magnetic field 1.00 m from the bulb. A certain microwave oven contains a magnetron that has an output of 700 W of microwave power for an electrical input power of 1.40 kW. The microwaves are entirely transferred from the magnetron into the oven chamber through a waveguide, which is a metal tube of rectangular cross-section with a width of 6.83 cm and a height of 3.81 cm. (a) What is the efficiency of the magnetron? (b) Assuming that the food is absorbing all the microwaves produced by the magnetron and that no energy is reflected back into the waveguide, find the direction and magnitude of the Poynting vector, averaged over time, in the waveguide near the entrance to the oven chamber. (c) What is the maximum electric field magnitude at this point? WEB 5. Figure 34.3a shows a plane electromagnetic sinusoidal wave propagating in what we choose as the x direction. Suppose that the wavelength is 50.0 m, and the electric field vibrates in the xy plane with an amplitude of 22.0 V/m. Calculate (a) the frequency of the wave and (b) the magnitude and direction of B when the electric field has its maximum value in the negative y direction. (c) Write an expression for B in the form B B max cos(kx t) WEB 13. 14. with numerical values for B max , k, and . 6. Write down expressions for the electric and magnetic fields of a sinusoidal plane electromagnetic wave having a frequency of 3.00 GHz and traveling in the positive x direction. The amplitude of the electric field is 300 V/m. 7. In SI units, the electric field in an electromagnetic wave is described by Ey 100 sin(1.00 10 7x t) 15. 16. Find (a) the amplitude of the corresponding magnetic field, (b) the wavelength , and (c) the frequency f. 8. Verify by substitution that the following equations are solutions to Equations 34.8 and 34.9, respectively: WEB 17. E B E max cos(kx B max cos(kx t) t) 9. Review Problem. A standing-wave interference pattern is set up by radio waves between two metal sheets 2.00 m apart. This is the shortest distance between the plates that will produce a standing-wave pattern. What is the fundamental frequency? 10. A microwave oven is powered by an electron tube called a magnetron, which generates electromagnetic waves of frequency 2.45 GHz. The microwaves enter the oven and are reflected by the walls. The standing-wave pattern produced in the oven can cook food unevenly, with hot spots in the food at antinodes and cool spots at nodes, so a turntable is often used to rotate the food and distribute the energy. If a microwave oven intended for use with a turntable is instead used with a cooking dish in a fixed position, the antinodes can appear as burn marks on foods such as carrot strips or cheese. The separation distance between the burns is measured to be 6 cm 5%. From these data, calculate the speed of the microwaves. 18. 19. 20. Section 34.3 Energy Carried by Electromagnetic Waves 11. How much electromagnetic energy per cubic meter is contained in sunlight, if the intensity of sunlight at the Earth's surface under a fairly clear sky is 1 000 W/m2 ? 12. An AM radio station broadcasts isotropically (equally in all directions) with an average power of 4.00 kW. A dipole receiving antenna 65.0 cm long is at a location 4.00 miles from the transmitter. Compute the emf that Problems 21. High-power lasers in factories are used to cut through cloth and metal (Fig. P34.21). One such laser has a beam diameter of 1.00 mm and generates an electric field with an amplitude of 0.700 MV/m at the target. Find (a) the amplitude of the magnetic field produced, (b) the intensity of the laser, and (c) the power delivered by the laser. 1099 27. 28. WEB 29. 30. Figure P34.21 A laser cutting device mounted on a robot arm is being used to cut through a metallic plate. (Philippe Plailly/SPL/Photo Researchers) 31. 22. At what distance from a 100-W electromagnetic-wave point source does E max 15.0 V/m? 23. A 10.0-mW laser has a beam diameter of 1.60 mm. (a) What is the intensity of the light, assuming it is uniform across the circular beam? (b) What is the average energy density of the beam? 24. At one location on the Earth, the rms value of the magnetic field caused by solar radiation is 1.80 T. From this value, calculate (a) the average electric field due to solar radiation, (b) the average energy density of the solar component of electromagnetic radiation at this location, and (c) the magnitude of the Poynting vector for the Sun's radiation. and a night setting. The night setting greatly diminishes the intensity of the image in order that lights from trailing vehicles do not blind the driver. How does such a mirror work? Solution Figure 36.7 shows a cross-sectional view of a rearview mirror for each setting. The unit consists of a reflective coating on the back of a wedge of glass. In the day setting (Fig. 36.7a), the light from an object behind the car strikes the glass wedge at point 1. Most of the light enters the wedge, refracting as it crosses the front surface, and reflects Reflecting side of mirror B B 1 D Incident light Daytime setting (a) D Incident light Nighttime setting (b) Figure 36.7 Cross-sectional views of a rearview mirror. (a) With the day setting, the silvered back surface of the mirror reflects a bright ray B into the driver's eyes. (b) With the night setting, the glass of the unsilvered front surface of the mirror reflects a dim ray D into the driver's eyes. 36.2 IMAGES FORMED BY SPHERICAL MIRRORS Concave Mirrors 14.7 A spherical mirror, as its name implies, has the shape of a section of a sphere. This type of mirror focuses incoming parallel rays to a point, as demonstrated by the colored light rays in Figure 36.8. Figure 36.9a shows a cross-section of a spherical mirror, with its surface represented by the solid, curved black line. (The blue band represents the structural support for the mirrored surface, such as a curved piece of glass on which the silvered surface is deposited.) Such a mirror, in which light is reflected from the inner, concave surface, is called a concave mirror. The mirror has a radius of curvature R , and its center of curvature is point C. Point V is the center of the spherical section, and a line through C and V is called the principal axis of the mirror. Now consider a point source of light placed at point O in Figure 36.9b, where O is any point on the principal axis to the left of C. Two diverging rays that originate at O are shown. After reflecting from the mirror, these rays converge (come together) at the image point I. They then continue to diverge from I as if an object were there. As a result, we have at point I a real image of the light source at O. We shall consider in this section only rays that diverge from the object and make a small angle with the principal axis. Such rays are called paraxial rays. All 1144 CHAPTER 36 Geometric Optics Mirror Center of curvature Mirror R C C Principal axis V O I V (a) (b) (a) A concave mirror of radius R. The center of curvature C is located on the principal axis. (b) A point object placed at O in front of a concave spherical mirror of radius R, where O is any point on the principal axis farther than R from the mirror surface, forms a real image at I. If the rays diverge from O at small angles, they all reflect through the same image point. Figure 36.9 Figure 36.8 Red, blue, and green light rays are reflected by a curved mirror. Note that the point where the three colors meet is white. such rays reflect through the image point, as shown in Figure 36.9b. Rays that are far from the principal axis, such as those shown in Figure 36.10, converge to other points on the principal axis, producing a blurred image. This effect, which is called spherical aberration, is present to some extent for any spherical mirror and is discussed in Section 36.5. We can use Figure 36.11 to calculate the image distance q from a knowledge of the object distance p and radius of curvature R . By convention, these distances are measured from point V. Figure 36.11 shows two rays leaving the tip of the object. One of these rays passes through the center of curvature C of the mirror, hitting the mirror perpendicular to the mirror surface and reflecting back on itself. The second ray strikes the mirror at its center (point V ) and reflects as shown, obeying the law of reflection. The image of the tip of the arrow is located at the point where these two rays intersect. From the gold right triangle in Figure 36.11, we see that tan h/p, and from the blue right triangle we see that tan u h /q. The negative sign is introduced because the image is inverted, so h is taken to be negative. Thus, from Equation 36.1 and these results, we find that the magnification of the mirror is M h h q p (36.2) h O C Principal axis I h V Figure 36.10 Rays diverging from the object at large angles from the principal axis reflect from a spherical concave mirror to intersect the principal axis at different points, resulting in a blurred image. This condition is called spherical aberration. q R p Figure 36.11 The image formed by a spherical concave mirror when the object O lies outside the center of curvature C. 36.2 Images Formed by Spherical Mirrors 1145 We also note from the two triangles in Figure 36.11 that have that tan from which we find that h h R p q R h p R and tan h R q as one angle (36.3) If we compare Equations 36.2 and 36.3, we see that R p Simple algebra reduces this to 1 p 1 q 2 R (36.4) Mirror equation in terms of R q R q p This expression is called the mirror equation. It is applicable only to paraxial rays. If the object is very far from the mirror -- that is, if p is so much greater than R that p can be said to approach infinity -- then 1/p 0, and we see from Equation 36.4 that q R/2. That is, when the object is very far from the mirror, the image point is halfway between the center of curvature and the center point on the mirror, as shown in Figure 36.12a. The incoming rays from the object are essentially parallel in this figure because the source is assumed to be very far from the mirror. We call the image point in this special case the focal point F and the image distance the focal length f, where f R 2 (36.5) Focal length C F f R (a) (b) Figure 36.12 (a) Light rays from a distant object ( p ) reflect from a concave mirror through the focal point F. In this case, the image distance q R/2 f, where f is the focal length of the mirror. (b) Reflection of parallel rays from a concave mirror. 1146 CHAPTER 36 Geometric Optics Front Back O p q I F C Figure 36.13 Formation of an image by a spherical convex mirror. The image formed by the real object is virtual and upright. Focal length is a parameter particular to a given mirror and therefore can be used to compare one mirror with another. The mirror equation can be expressed in terms of the focal length: 1 p 1 q 1 f Mirror equation in terms of f (36.6) Notice that the focal length of a mirror depends only on the curvature of the mirror and not on the material from which the mirror is made. This is because the formation of the image results from rays reflected from the surface of the material. We shall find in Section 36.4 that the situation is different for lenses; in that case the light actually passes through the material. Convex Mirrors Figure 36.13 shows the formation of an image by a convex mirror -- that is, one silvered so that light is reflected from the outer, convex surface. This is sometimes called a diverging mirror because the rays from any point on an object diverge after reflection as though they were coming from some point behind the mirror. The image in Figure 36.13 is virtual because the reflected rays only appear to originate at the image point, as indicated by the dashed lines. Furthermore, the image is always upright and smaller than the object. This type of mirror is often used in stores to foil shoplifters. A single mirror can be used to survey a large field of view because it forms a smaller image of the interior of the store. We do not derive any equations for convex spherical mirrors because we can use Equations 36.2, 36.4, and 36.6 for either concave or convex mirrors if we adhere to the following procedure. Let us refer to the region in which light rays move toward the mirror as the front side of the mirror, and the other side as the back side. For example, in Figures 36.10 and 36.12, the side to the left of the mirrors is the front side, and the side to the right of the mirrors is the back side. Figure 36.14 states the sign conventions for object and image distances, and Table Image from a Mirror q which means that rays originating from an object positioned at the focal point of a mirror are reflected so that the image is formed at an infinite distance from the mirror; that is, the rays travel parallel to one another after reflection. This is the situation in a flashlight, where the bulb filament is placed at the focal point of a reflector, producing a parallel beam of light. (c) When the object is at p 5.00 cm, it lies between the focal point and the mirror surface, as shown in Figure 36.15b. Thus, we expect a magnified, virtual, upright image. In this case, the mirror equation gives 1 5.00 cm 1 q q 1 10.0 cm 10.0 cm Assume that a certain spherical mirror has a focal length of 10.0 cm. Locate and describe the image for object distances of (a) 25.0 cm, (b) 10.0 cm, and (c) 5.00 cm. Solution Because the focal length is positive, we know that this is a concave mirror (see Table 36.1). (a) This situation is analogous to that in Figure 36.15a; hence, we expect the image to be real and closer to the mirror than the object. According to the figure, it should also be inverted and reduced in size. We find the image distance by using the Equation 36.6 form of the mirror equation: 1 p 1 25.0 cm 1 q 1 q q 1 f 1 10.0 cm 16.7 cm The magnification is given by Equation 36.2: M q p 16.7 cm 25.0 cm 0.668 The image is virtual because it is located behind the mirror, as expected. The magnification is M q p 10.0 cm 5.00 cm 2.00 The fact that the absolute value of M is less than unity tells us that the image is smaller than the object, and the negative sign for M tells us that the image is inverted. Because q is positive, the image is located on the front side of the mirror and is real. Thus, we see that our predictions were correct. (b) When the object distance is 10.0 cm, the object is located at the focal point. Now we find that 1 10.0 cm 1 q 1 10.0 cm The image is twice as large as the object, and the positive sign for M indicates that the image is upright (see Fig. 36.15b). Exercise 1.00? At what object distance is the magnification Answer 20.0 cm. EXAMPLE 36.5 The Image from a Convex Mirror A woman who is 1.5 m tall is located 3.0 m from an antishoplifting mirror, as shown in Figure 36.16. The focal length of the mirror is 0.25 m. Find (a) the position of her image and (b) the magnification. Solution (a) This situation is depicted in Figure 36.15c. We should expect to find an upright, reduced, virtual image. To find the image position, we use Equation 36.6: 1 p 1 q 1 f 1 q q 1 0.25 m 1 0.25 m 0.23 m 1 3.0 m Figure 36.16 Convex mirrors, often used for security in department stores, provide wide-angle viewing. 1150 CHAPTER 36 Geometric Optics The image is much smaller than the woman, and it is upright because M is positive. The negative value of q indicates that her image is virtual, or behind the mirror, as shown in Figure 36.15c. (b) The magnification is M q p 0.23 m 3.0 m 0.077 Exercise Answer Find the height of the image. 0.12 m. 36.3 IMAGES FORMED BY REFRACTION In this section we describe how images are formed when light rays are refracted at the boundary between two transparent materials. Consider two transparent media having indices of refraction n 1 and n 2 , where the boundary between the two media is a spherical surface of radius R (Fig. 36.17). We assume that the object at O is in the medium for which the index of refraction is n 1 , where n 1 n 2 . Let us consider the paraxial rays leaving O. As we shall see, all such rays are refracted at the spherical surface and focus at a single point I, the image point. Figure 36.18 shows a single ray leaving point O and focusing at point I. Snell's law of refraction applied to this refracted ray gives n 1 sin Because tion sin 1 1 n 2 sin 2 and 2 are assumed to be small, we can use the small-angle approxima(angles in radians) and say that n1 1 n2 2 Now we use the fact that an exterior angle of any triangle equals the sum of the two opposite interior angles. Applying this rule to triangles OPC and PIC in Figure 36.18 gives 1 2 vely. Because the cornea and water have almost identical indices of refraction, very little refraction occurs when a person under water views objects with the naked eye. In this case, light rays from an object focus behind the retina, resulting in a blurred image. When a mask is used, the air space between the eye and the mask surface provides the normal Solution Solution If a lens prescription is ground into the glass of the mask so that the wearer can see without eyeglasses, only the inside surface is curved. In this way the prescription is accurate whether the mask is used under water or in air. If the curvature were on the outer surface, the refraction at the outer surface of the glass would change depending on whether air or water were present on the outside of the mask. EXAMPLE 36.7 Gaze into the Crystal Ball n1 p 1.50 1.0 cm n2 q 1 q q n2 R 1.00 1.50 3.0 cm 0.75 cm n1 A dandelion seed ball 4.0 cm in diameter is embedded in the center of a spherical plastic paperweight having a diameter of 6.0 cm (Fig. 36.20a). The index of refraction of the plastic is n 1 1.50. Find the position of the image of the near edge of the seed ball. Because n 1 n 2 , where n 2 1.00 is the index of refraction for air, the rays originating from the seed ball are refracted away from the normal at the surface and diverge outward, as shown in Figure 36.20b. Hence, the image is formed inside the paperweight and is virtual. From the given dimensions, we know that the near edge of the seed ball is 1.0 cm beneath the surface of the paperweight. Applying Equation 36.8 and noting from Table 36.2 that R is negative, we obtain Solution The negative sign for q indicates that the image is in front of the surface -- in other words, in the same medium as the object, as shown in Figure 36.20b. Being in the same medium as the object, the image must be virtual (see Table 36.2). The surface of the seed ball appears to be closer to the paperweight surface than it actually is. 36.3 Images Formed by Refraction 1153 n1>n2 3.0 cm 2.0 cm q n1 n2 (a) 1.0 cm (b) Figure 36.20 (a) An object embedded in a plastic sphere forms a virtual image between the surface of the object and the sphere surface. All rays are assumed paraxial. Because the object is inside the sphere, the front of the refracting surface is the interior of the sphere. (b) Rays from the surface of the object form an image that is still inside the plastic sphere but closer to the plastic surface. EXAMPLE 36.8 The One That Got Away Because q is negative, the image is virtual, as indicated by the dashed lines in Figure 36.21. The apparent depth is threefourths the actual depth. A small fish is swimming at a depth d below the surface of a pond (Fig. 36.21). What is the apparent depth of the fish, as viewed from directly overhead? Solution Because the refracting surface is flat, R is infinite. Hence, we can use Equation 36.9 to determine the location of the image with p d. Using the indices of refraction given in Figure 36.21, we obtain q n2 p n1 1.00 d 1.33 0.752d n 2 = 1.00 q d n 1 = 1.33 Figure 36.21 The apparent depth q of the fish is less than the true depth d . All rays are assumed to be paraxial. 1154 CHAPTER 36 Geometric Optics 36.4 14.8 THIN LENSES Lenses are commonly used to form images by refraction in optical instruments, such as cameras, telescopes, and microscopes. We can use what we just learned about images formed by refracting surfaces to help us locate the image formed by a lens. We recognize that light passing through a lens experiences refraction at two surfaces. The development we shall follow is based on the notion that the image formed by one refracting surface serves as the object for the second surface. We shall analyze a thick lens first and then let the thickness of the lens be approximately zero. Consider a lens having an index of refraction n and two spherical surfaces with radii of curvature R 1 and R 2 , as in Figure 36.22. (Note that R 1 is the radius of curvature of the lens surface that the light leaving the object reaches first and that R 2 is the radius of curvature of the other surface of the lens.) An object is placed at point O at a distance p 1 in front of surface 1. If the object were far from surface 1, the light rays from the object that struck the surface would be almost parallel to each other. The refraction at the surface would focus these rays, forming a real image to the right of surface 1 in Figure 36.22 (as in Fig. 36.17). If the object is placed close to surface 1, as shown in Figure 36.22, the rays diverging from the object and striking the surface cover a wide range of angles and are not parallel to each other. In this case, the refraction at the surface is not sufficient to cause the rays to converge on the right side of the surface. They still diverge, although they are closer to parallel than they were before they struck the surface. This results in a virtual image of the object at I 1 to the left of the surface, as shown in Figure 36.22. This image is then used as the object for surface 2, which results in a real image I 2 to the right of the lens. Let us begin with the virtual image formed by surface 1. Using Equation 36.8 and assuming that n 1 1 because the lens is surrounded by air, we find that the image I 1 formed by surface 1 satisfies the equation (1) 1 p1 n q1 n R1 1 where q 1 is a negative number because it represents a virtual image formed on the front side of surface 1. Now we apply Equation 36.8 to surface 2, taking n 1 n and n 2 1. (We make this switch in index because the light rays from I 1 approaching surface 2 are in the material of the lens, and this material has index n. We could also imagine removing the object at O, filling all of the space to the left of surface 1 with the mate- n1 = 1 I1 O p1 q1 p2 q2 t I2 R1 Surface 1 n R2 Surface 2 Figure 36.22 To locate the image formed by a lens, we use the virtual image at I1 formed by surface 1 as the object for the image formed by surface 2. The final image is real and is located at I 2 . 36.4 Thin Lenses 1155 rial of the lens, and placing the object at I 1 ; the light rays approaching surface 2 would be the same as in the actual situation in Fig. 36.22.) Taking p 2 as the object distance for surface 2 and q 2 as the image distance gives (2) n p2 1 q2 1 R2 n We now introduce mathematically the fact that the image formed by the first surface acts as the object for the second surface. We do this by noting from Figure 36.22 that p 2 is the sum of q 1 and t and by setting p2 q 1 t, where t is the thickness of the lens. (Remember that q 1 is a negative number and that p 2 must be positive by our sign convention -- thus, we must introduce a negative sign for q 1 .) For a thin lens (for which the thickness is small compared to the radii of curvature), we can neglect t. In this approximation, we see that p 2 q 1 . Hence, Equation (2) becomes (3) n q1 1 q2 1 R2 n Adding Equations (1) and (3), we find that (4) 1 p1 1 q2 (n 1) 1 R1 1 R2 O R2 C2 R1 C1 I For a thin lens, we can omit the subscripts on p 1 and q 2 in Equation (4) and call the object distance p and the image distance q, as in Figure 36.23. Hence, we can write Equation (4) in the form 1 p 1 q (n 1) 1 R1 1 R2 (36.10) This expression relates the image distance q of the image formed by a thin lens to the object distance p and to the thin-lens properties (index of refraction and radii of curvature). It is valid only for paraxial rays and only when the lens thickness is much less than R 1 and R 2 . The focal length f of a thin lens is the image distance that corresponds to an infinite object distance, just as with mirrors. Letting p approach and q approach f in Equation 36.10, we see that the inverse of the focal length for a thin lens is 1 f (n 1) 1 R1 1 R2 (36.11) p q Figure 36.23 Simplified geometry for a thin lens. Lens makers' equation This relationship is called the lens makers' equation because it can be used to determine the values of R 1 and R 2 that are needed for a given index of refraction and a desired focal length f. Conversely, if the index of refraction and the radii of curvature of a lens are gaced 30.0 cm in front of the lens. Locate the image. Exercise Solution p Using the thin-lens equation (Eq. 36.12) with 20.0 cm, we obtain 30.0 cm and f 1 30.0 cm 1 q q 1 20.0 cm 12.0 cm Determine both the magnification and the height of the image. M 0.400; h 0.800 cm. Answer EXAMPLE 36.10 An Image Formed by a Converging Lens sign for M means that the image is inverted. The situation is like that pictured in Figure 36.27a. (b) No calculation is necessary for this case because we know that, when the object is placed at the focal point, the image is formed at infinity. We can readily verify this by substituting p 10.0 cm into the thin-lens equation. (c) We now move inside the focal point, to an object distance of 5.00 cm: 1 5.00 cm 1 q q 1 10.0 cm 10.0 cm q p 10.0 cm 5.00 cm A converging lens of focal length 10.0 cm forms an image of each of three objects placed (a) 30.0 cm, (b) 10.0 cm, and (c) 5.00 cm in front of the lens. In each case, find the image distance and describe the image. Solution (a) The thin-lens equation can be used again: 1 p 1 30.0 cm 1 q 1 q q 1 f 1 10.0 cm 15.0 cm The positive sign indicates that the image is in back of the lens and real. The magnification is M q p 15.0 cm 30.0 cm 0.500 M 2.00 The image is reduced in size by one half, and the negative The negative image distance indicates that the image is in front of the lens and virtual. The image is enlarged, and the positive sign for M tells us that the image is upright, as shown in Figure 36.27b. EXAMPLE 36.11 A Lens Under Water A converging glass lens (n 1.52) has a focal length of 40.0 cm in air. Find its focal length when it is immersed in water, which has an index of refraction of 1.33. Solution We can use the lens makers' equation (Eq. 36.11) in both cases, noting that R 1 and R 2 remain the same in air and water: 1160 1 f air 1 f water CHAPTER 36 1 R1 1 R1 1 R2 1 R2 Geometric Optics f water f air Because f air f water n n 1 1 1.52 1.14 1 1 (n (n 1) 1) 3.71 40.0 cm, we find that 3.71f air 3.71(40.0 cm ) 148 cm where n is the ratio of the index of refraction of glass to that of water: n 1.52/1.33 1.14. Dividing the first equation by the second gives The focal length of any glass lens is increased by a factor (n 1)/(n 1) when the lens is immersed in water. Combination of Thin Lenses If two thin lenses are used to form an image, the system can be treated in the following manner. First, the image formed by the first lens is located as if the second lens were not present. Then a ray diagram is drawn for the second lens, with the image formed by the first lens now serving as the object for the second lens. The second image formed is the final image of the system. One configuration is particularly straightforward; that is, if the image formed by the first lens lies on the back side of the second lens, then that image is treated as a virtual object for the second lens (that is, p is negative). The same procedure can be extended to a system of three or more lenses. The overall magnification of a system of thin lenses equals the product of the magnifications of the separate lenses. Let us consider the special case of a system of two lenses in contact. Suppose two thin lenses of focal lengths f 1 and f 2 are placed in contact with each other. If p is the object distance for the combination, application of the thin-lens equation (Eq. 36.12) to the first lens gives 1 p 1 q1 1 f1 Light from a distant object brought into focus by two converging lenses. where q 1 is the image distance for the first lens. Treating this image as the object for the second lens, we see that the object distance for the second lens must be q 1 (negative because the object is virtual). Therefore, for the second lens, 1 q1 1 q 1 f2 where q is the final image distance from the second lens. Adding these equations eliminates q 1 and gives 1 p Focal length of two thin lenses in contact 1 q 1 f 1 f1 1 f1 1 f2 1 f2 (36.13) Because the two thin lenses are touching, q is also the distance of the final image from the first lens. Therefore, two thin lenses in contact with each other are equivalent to a single thin lens having a focal length given by Equation 36.13. 36.4 Thin Lenses 1161 EXAMPLE 36.12 Where Is the Final Image? 1 10.0 cm 1 q2 q2 1 20.0 cm 6.67 cm Even when the conditions just described do not apply, the lens equations yield image position and magnification. For example, two thin converging lenses of focal lengths f 1 10.0 cm and f 2 20.0 cm are separated by 20.0 cm, as illustrated in Figure 36.29. An object is placed 15.0 cm to the left of lens 1. Find the position of the final image and the magnification of the system. Solution First we locate the image formed by lens 1 while ignoring lens 2: 1 p1 1 15.0 cm 1 q1 1 q1 q1 1 f1 1 10.0 cm 30.0 cm The final image lies 6.67 cm to the right of lens 2. The individual magnifications of the images are q1 30.0 cm M1 2.00 p1 15.0 cm M2 q2 p2 6.67 cm 10.0 cm 0.667 The total magnification M is equal to the product M 1M 2 ( 2.00)(0.667) 1.33. The final image is real because q 2 is positive. The image is also inverted and enlarged. 15.0 cm 20.0 cm where q 1 is measured from lens 1. A positive value for q 1 means that this first image is in back of lens 1. Because q 1 is greater than the separation between the two lenses, this image formed by lens 1 lies 10.0 cm to the right of lens 2. We take this as the object distance for the second lens, so p 2 10.0 cm, where distances are now measured from lens 2: 1 p2 1 q2 1 f2 O f1 = 10.0 cm f2 = 20.0 cm Figure 36.29 A combination of two converging lenses. CONCEPTUAL EXAMPLE 36.13 Watch Your p's and q's! near side of the lens. At this point, the rays leaving the lens are parallel, making the image infinitely far away. This is described in the graph by the asymptotic approach of the curve to the line p f 10 cm. As the object moves inside the focal point, the image becomes virtual and located near q . We are now following the curve in the lower left portion of Figure 36.30a. As the object moves closer to the lens, the virtual image also moves closer to the lens. As p : 0, the image distance q also approaches 0. Now imagine that we bring the object to the back side of the lens, where p 0. The object is now a virtual object, so it must have been formed by some other lens. For all locations of the virtual object, the image distance is positive and less than the focal length. The final image is real, and its position approaches the focal point as p gets more and more negative. The f 10 cm graph shows that a distant real object forms an image at the focal point on the front side of the lens. As the object approaches the lens, the image remains Use a spreadsheet or a similar tool to create two graphs of image distance as a function of object distance -- one for a lens for which the focal length is 10 cm and one for a lens for which the focal length is 10 cm. Solution The graphs are shown in Figure 36.30. In each graph a gap occurs where p f, which we shall discuss. Note the similarity in the shapes -- a result of the fact that image and object distances for both lenses are related according to the same equation -- the thin-lens equation. 10 cm The curve in the upper right portion of the f graph corresponds to an object on the front side of a lens, which we have drawn as the left side of the lens in our previous diagrams. When the object is at positive infinity, a real image forms at the focal point on the back side (the positive side) of the lens, q f. (The incoming rays are parallel in this case.) As the object gets closer to the lens, the image moves farther from the lens, corresponding to the upward path of the curve. This continues until the object is located at the focal point on the 1162 CHAPTER 36 Geometric Optics the image shifts from a location at positive infinity to one at negative infinity. Finally, as the virtual object continues moving away from the lens, the image is virtual, starts moving in from negative infinity, and approaches the focal point. q(cm) virtual and moves closer to the lens. But as we continue toward the left end of the p axis, the objn in Fig. 36.32) have focal points intermediate between those of red and violet. Chromatic aberration for a diverging lens also results in a shorter focal length for violet light than for red light, but on the front side of the lens. Chromatic aberration can be greatly reduced by combining a converging lens made of one type of glass and a diverging lens made of another type of glass. Optional Section Violet Red FR FV Violet Red Figure 36.32 Chromatic aberration caused by a converging lens. Rays of different wavelengths focus at different points. 36.6 THE CAMERA The photographic camera is a simple optical instrument whose essential features are shown in Figure 36.33. It consists of a light-tight box, a converging lens that produces a real image, and a film behind the lens to receive the image. One focuses the camera by varying the distance between lens and film. This is accom- 1164 CHAPTER 36 Geometric Optics Shutter Lens Film Image p Aperture q Figure 36.33 Cross-sectional view of a simple camera. Note that in reality, p W q. plished with an adjustable bellows in older-style cameras and with some other mechanical arrangement in modern cameras. For proper focusing -- which is necessary for the formation of sharp images -- the lens-to-film distance depends on the object distance as well as on the focal length of the lens. The shutter, positioned behind the lens, is a mechanical device that is opened for selected time intervals, called exposure times. One can photograph moving objects by using short exposure times, or photograph dark scenes (with low light levels) by using long exposure times. If this adjustment were not available, it would be impossible to take stop-action photographs. For example, a rapidly moving vehicle could move enough in the time that the shutter was open to produce a blurred image. Another major cause of blurred images is the movement of the camera while the shutter is open. To prevent such movement, either short exposure times or a tripod should be used, even for stationary objects. Typical shutter speeds (that is, exposure times) are 1/30, 1/60, 1/125, and 1/250 s. For handheld cameras, the use of slower speeds can result in blurred images (due to movement), but the use of faster speeds reduces the gathered light intensity. In practice, stationary objects are normally shot with an intermediate shutter speed of 1/60 s. More expensive cameras have an aperture of adjustable diameter to further control the intensity of the light reaching the film. As noted earlier, when an aperture of small diameter is used, only light from the central portion of the lens reaches the film; in this way spherical aberration is reduced. The intensity I of the light reaching the film is proportional to the area of the lens. Because this area is proportional to the square of the diameter D, we conclude that I is also proportional to D 2. Light intensity is a measure of the rate at which energy is received by the film per unit area of the image. Because the area of the image is proportional to q 2, and q f (when p W f, so p can be approximated as infinite), we conclude that the intensity is also proportional to 1/f 2, and thus I D 2/f 2. The brightness of the image formed on the film depends on the light intensity, so we see that the image brightness depends on both the focal length and the diameter of the lens. The ratio f/D is called the f-number of a lens: f -number f D (36.14) Hence, the intensity of light incident on the film can be expressed as I 1 ( f/D)2 1 ( f -number)2 (36.15) The f-number is often given as a description of the lens "speed." The lower the f-number, the wider the aperture and the higher the rate at which energy from the light exposes the film -- thus, a lens with a low f-number is a "fast" lens. The conventional notation for an f-number is "f/" followed by the actual number. For example, "f/4" means an f-number of 4 -- it does not mean to divide f by 4! Extremely fast lenses, which have f-numbers as low as approximately f/1.2, are expensive because it is very difficult to keep aberightedness f 2.5 m A particular nearsighted person is unable to see objects clearly when they are beyond 2.5 m away (the far point of this particular eye). What should the focal length be in a lens prescribed to correct this problem? Solution The purpose of the lens in this instance is to "move" an object from infinity to a distance where it can be seen clearly. This is accomplished by having the lens produce an image at the far point. From the thin-lens equation, we have 1 p 1 q 1 1 2.5 m 1 f Why did we use a negative sign for the image distance? As you should have suspected, the lens must be a diverging lens (one with a negative focal length) to correct nearsightedness. Exercise Answer What is the power of this lens? 0.40 diopter. Optional Section 36.8 THE SIMPLE MAGNIFIER p Figure 36.39 The size of the image formed on the retina depends on the angle subtended at the eye. The simple magnifier consists of a single converging lens. As the name implies, this device increases the apparent size of an object. Suppose an object is viewed at some distance p from the eye, as illustrated in Figure 36.39. The size of the image formed at the retina depends on the angle subtended by the object at the eye. As the object moves closer to the eye, increases and a larger image is observed. However, an average normal eye cannot focus on an object closer than about 25 cm, the near point (Fig. 36.40a). Therefore, is maximum at the near point. To further increase the apparent angular size of an object, a converging lens can be placed in front of the eye as in Figure 36.40b, with the object located at point O, just inside the focal point of the lens. At this location, the lens forms a virtual, upright, enlarged image. We define angular magnification m as the ratio of the angle subtended by an object with a lens in use (angle in Fig. 36.40b) to the angle subtended by the object placed at the near point with no lens in use (angle 1 The word lens comes from lentil, the name of an Italian legume. (You may have eaten lentil soup.) Early eyeglasses were called "glass lentils" because the biconvex shape of their lenses resembled the shape of a lentil. The first lenses for farsightedness and presbyopia appeared around 1280; concave eyeglasses for correcting nearsightedness did not appear for more than 100 years after that. 36.8 The Simple Magnifier 1171 h 0 25 cm (a) 25 cm h' I h F O (b) p (a) An object placed at the near point of the eye (p 25 cm) subtends an angle 0 h/25 at the eye. (b) An object placed near the focal point of a converging lens produces a magnified image that subtends an angle h /25 at the eye. Figure 36.40 0 in Fig. 36.40a): m 0 (36.16) Angular magnification with the object at the near point The angular magnification is a maximum when the image is at the near point of the eye -- that is, when q 25 cm. The object distance corresponding to this image distance can be calculated from the thin-lens equation: 1 p 1 25 cm p 1 f 25f 25 f where f is the focal length of the magnifier in centimeters. If we make the smallangle approximations tan 0 0 h 25 and tan h p (36.17) Equation 36.16 becomes m max 0 h/p h/25 25 cm f 25 p 25 25f /(25 f) (36.18) m max 1 Although the eye can focus on an image formed anywhere between the near point and infinity, it is most relaxed when the image is at infinity. For the image formed by the magnifying lens to appear at infinity, the object has to be at the focal point of the lens. In this case, Equations 36.17 become 0 h 25 and h f 1172 CHAPTER 36 Geometric Optics and the magnification is m min 0 25 cm f (36.19) With a single lens, it is possible to obtain angular magnifications up to about 4 without serious aberrations. Magnifications up to about 20 can be achieved by using one or two additional lenses to correct for aberrations. EXAMPLE 36.16 Maximum Magnification of a Lens When the eye is relaxed, the image is at infinity. In this case, we use Equation 36.19: m min 25 cm f 25 cm 10 cm 2.5 What is the maximum magnification that is possible with a lens having a focal length of 10 cm, and what is the magnification of this lens when the eye is relaxed? The maximum magnification occurs when the image is located at the near point of the eye. Under these circumstances, Equation 36.18 gives m max 1 25 cm f 1 25 cm 10 cm 3.5 Solution Optional Section 36.9 THE COMPOUND MICROSCOPE A simple magnifier provides only limited assistance in inspecting minute details of an object. Greater magnification can be achieved by combining two lenses in a device called a compound microscope, a schematic diagram of which is shown in Figure 36.41a. It consists of one lens, the objective, that has a very short focal length Objective O I2 Fo fo Fe Il Eyepiece fe p1 q1 L (a) (b) Figure 36.41 (a) Diagram of a compound microscope, which consists of an objective lens and an eyepiece lens. (b) A compound microscope. The three-objective turret allows the user to choose from several powers of magnification. Combinations of eyepieces with different focal lengths and different objectives can produce a wide range of magnifications. 36.9 The Compound Microscope 1173 f o 1 cm and a second lens, the eyepiece, that has a focal length f e of a few centimeters. The two lenses are separated by a distance L that is much greater than either f o or f e . The object, which is placed just outside the focal point of the objective, forms a real, inverted image at I 1 , and this image is located at or close to the focal point of the eyepiece. The eyepiece, which serves as a simple magnifier, produces at I 2 a virtual, inverted image of I 1 . The lateral magnification M 1 of the first image is q 1 /p 1 . Note from Figure 36.41a that q 1 is approximately equal to L and that the object is very close to the focal point of the objective: p 1 f o . Thus, the lateral magnification by the objective is M1 L fo The angular magnification by the eyepiece for an object (corresponding to the image at I 1 ) placed at the focal point of the eyepiece is, from Equation 36.19, me 25 cm fe The overall magnification of the compound microscope is defined as the product of the lateral and angular magnifications: M M 1m e L fo 25 cm fe (36.20) The negative sign indicates that the image is inverted. The microscope has extended human vision to the point where we can view previously unknown details of incredibly small objects. The capabilities of this instrument have steadily increased with improved techniques for precision grinding of lenses. An often-asked question about microscopes is: "If one were extremely patient and careful, would it be possible to construct a microscope that would enable the human eye to see an atom?" The answer is no, as long as light is used to illuminate the object. The reason is that, for an object under an optical microscope (one that uses visible light) to be seen, the object must be at least as large as a wavelength of light. Because the diameter of any atom is many times smaller than the wavelengths of visible light, the mysteries of the atom must be probed using other types of "microscopes." The ability to use other types of waves to "see" objects also depends on wavelength. We can illustrate this with water waves in a bathtub. Suppose you vibrate your hand in the water until waves having a wavelength of about 15 cm are moving along the surface. If you hold a small object, such as a toothpick, so that it lies in the path of the waves, it does not appreciably disturb the waves; they continue along their path "oblivious" to it. Now suppose you hold a larger object, such as a toy sailboat, in the path of the 15-cm waves. In this case, the waves are considerably disturbed by the object. Because the toothpick was smaller than the wavelength of the waves, the waves did not "see" it (the intensity of the scattered waves was low). Because it is about the same size as the wavelength of the waves, however, the boat creates a disturbance. In other words, the object acts as the source of scattered waves that appear to come from it. Light waves behave in this same general way. The ability of an optical microscope to view an object depends on the size of the object relative to the wavelength of the light used to observe it. Hence, we can never observe atoms with an optical 1174 CHAPTER 36 Geometric Optics microscope2 because their dimensions are small ( 0.1 nm) relative to the wavelength of the light ( 500 nm). Optional Section 36.10 14.1 & 14.9 THE TELESCOPE Two fundamentally different types of telescopes exist; both are designed to aid in viewing distant objects, such as the planets in our Solar System. The refracting telescope uses a combination of lenses to form an image, and the reflecting telescope uses a curved mirror and a lens. The lens combination shown in Figure 36.42a is that of a refracting telescope. Like the compound microscope, this telescope has an objective and an eyepiece. The two lenses are arranged so that the objective forms a real, inverted image of the distant object very near the focal point of the eyepiece. Because the object is essentially at infinity, this point at which I 1 forms is the focal point of the objective. Hence, the two lenses are separated by a distance f o f e , which corresponds to the length of the telescope tube. The eyepiece then forms, at I 2 , an enlarged, inverted image of the image at I 1 . The angular magnification of the telescope is given by / o , where o is the angle subtended by the object at the objective and is the angle subtended by the final image at the viewer's eye. Consider Figure 36.42a, in which the object is a very great distance to the left of the figure. The angle o (to the left of the objective) subtended by the object at the objective is the same as the angle (to the right of the objective) subtended by the first image at the objective. Thus, h fo where the negative sign indicates that the image is inverted. The angle subtended by the final image at the eye is the same as the angle that a ray coming from the tip of I 1 and traveling parallel to the principal axis makes with the principal axis after it passes through the lens. Thus, tan o o h fe We have not used a negative sign in this equation because the final image is not inverted; the object creating this final image I 2 is I 1 , and both it and I 2 point in the same direction. To see why the adjacent side of the triangle containing angle is f e and not 2f e , note that we must use only the bent length of the refracted ray. Hence, the angular magnification of the telescope can be expressed as tan h /f e fo (36.21) h /f o fe o and we see that the angular magnification of a telescope equals the ratio of the objective focal length to the eyepiece focal length. The negative sign indicates that the image is inverted. m Quick Quiz 36.6 Why isn't the lateral magnification given by Equation 36.1 a useful concept for telescopes? Single-molecule near-field optic studies are routinely performed with visible light having wavelengths of about 500 nm. The technique uses very small apertures to produce images having resolution as small as 10 nm. 2 36.10 The Telescope 1175 Eyepiece lens Objective lens o Fe Fo h' I1 Fe' o fe fo I2 (a) fe (b) Figure 36.42 (a) Lens arrangement in a refracting telescope, with the object at infinity. (b) A refracting telescope. When we look through a telescope at such relatively nearby objects as the Moon and the planets, magnification is important. However, stars are so far away that they always appear as small points of light no matter how great the magnification. A large research telescope that is used to study very distant objects must have a great diameter to gather as much light as possible. It is difficult and expensive to manufacture large lenses for refracting telescopes. Another difficulty with large lenses is that their weight leads to sagging, which is an additional source of aberration. These problems can be partially overcome by replacing the objective with a concave mirror, which results in a reflecting telescope. Because light is reflected from the mirror and does not pass through a lens, the mirror can have rigid supports on the back side. Such supports eliminate the problem of sagging. Figure 36.43 shows the design for a typical reflecting telescope. Incoming light rays pass down the barrel of the telescope and are reflected by a parabolic mirror at the base. These rays converge toward point A in the figure, where an image would be formed. However, before this image is formed, a small, flat mirror M reflects the light toward an opening in the side of the tube that passes into an eyepiece. This particular design is said to have a Newtonian focus because Newton developed it. Note that in the reflecting telescope the light never passes through glass (except through the small eyepiece). As a result, problems associated with chromatic aberration are virtually eliminated. The largest reflecting telescopes in the world are at the Keck Observatory on Mauna Kea, Hawaii. The site includes two telescopes with diameters of 10 m, each containing 36 hexagonally shaped, computer-controlled mirrors that work together to form a large reflecting surface. In contrast, the largest refracting telescope in the world, at the Yerkes Observatory in Williams Bay, Wisconsin, has a diameter of only 1 m. A M Eyepiece Parabolic mirror Figure 36.43 A Newtonian-focus reflecting telescope. web For more information on the Keck telescopes, visit http://www2.keck.hawaii.edu:3636/ 1176 CHAPTER 36 Geometric Optics SUMMARY The lateral magnification M of a mirror or lens is defined as the ratio of the image height h to the object height h: M h h (36.1) In the paraxial ray approximation, the object distance p and image distance q for a spherical mirror of radius R are related by the mirror equation: 1 p 1 q 2 R 1 f (36.4, 36.6) where f R /2 is the focal length of the mirror. An image can be formed by refraction from a spherical surface of radius R. The object and image distances for refraction from such a surface are related by n1 p n2 q n2 R n1 (36.8) where the light is incident in the medium for which the index of refraction is n 1 and is refracted in the medium for which the index of refraction is n 2 . The inverse of the focal length f of a thin lens surrounded by air is given by the lens makers' equation: 1 f (n 1) 1 R1 1 R2 (36.11) Converging lenses have positive focal lengths, and diverging lenses have negative focal lengths. For a thin lens, and in the paraxial ray approximation, the object and image distances are related by the thin-lens equation: 1 p 1 q 1 f (36.12) QUESTIONS 1. What is wrong with the caption of the cartoon shown in Figure Q36.1? 2. Using a simple ray diagram, such as the one shown in Figure 36.2, show that a flat mirror whose top is at eye level need not be as long as you are for you to see your entire body in it. 3. Consider a concave spherical mirror with a real object. Is the image always inverted? Is the image always real? Give conditions for your answers. 4. Repeat the preceding question for a convex spherical mirror. 5. Why does a clear stream of water, such as a creek, always appear to be shallower than it actually is? By how much is its depth apparently reduced? 6. Consider the image formed by a thin converging lens. Under what conditions is the image (a) inverted, (b) upright, (c) real, (d) virtual, (e) larger than the object, and (f) smaller than the object? 7. Repeat Question 6 for a thin diverging lens. 8. Use the lens makers' equation to verify the sign of the focal length of each of the lenses in Figure 36.26. "Most mirrors reverse left and right. This one reverses top and bottom." Figure Q36.1 Questions 9. If a cylinder of solid glass or clear plastic is placed above the words LEAD OXIDE and viewed from the side as shown in Figure Q36.9, the LEAD appears inverted but the OXIDE does not. Explain. 1177 Figure Q36.9 10. If the camera "sees" a movie actor's reflection in a mirror, what does the actor see in the mirror? 11. Explain why a mirror cannot give rise to chromatic aberration. 12. Why do some automobile mirrors have printed on them the statement "Objects in mirror are closer than they appear" ? (See Fig. Q36.12.) 14. Explain why a fish in a spherical goldfish bowl appears larger than it really is. 15. Lenses used in eyeglasses, whether converging or diverging, are always designed such that the middle of the lens curves away from the eye, like the center lenses of Figure 36.26a and b. Why? 16. A mirage is formed when the air gets gradually cooler with increasing altitude. What might happen if the air grew gradually warmer with altitude? This often happens over bodies of water or snow-covered ground; the effect is called looming. 17. Consider a spherical concave mirror, with an object positioned to the left of the mirror beyond the focal point. Using ray diagrams, show that the image moves to the left as the object approaches the focal point. 18. In a Jules Verne novel, a piece of ice is shaped into a magnifying lens to focus sunlight to start a fire. Is this possible? 19. The f-number of a camera is the focal length of the lens divided by its aperture (or diameter). How can the f-number of the lens be changed? How does changing this number affect the required exposure time? 20. A solar furnace can be constructed through the use of a concave mirror to reflect and focus sunlight into a furnace enclosure. What factors in the design of the reflecting mirror would guarantee very high temperatures? 21. One method for determining the position of an image, either real or virtual, is by means of parallax. If a finger or another object is placed at the position of the image, as shown in Figure Q36.21, and the finger and the image are viewed simultaneously (the image is viewed through the lens if it is virtual), the finger and image have the same parallax; that is, if the image is viewed from different positions, it will appear to move along with the finger. Use this method to locate the image formed by a lens. Explain why the method works. Finger Image Figure Q36.21 22. Figure Q36.22 shows a lithograph by M. C. Escher titled Hand with Reflection Sphere (Self-Portrait in Spherical Mirror). Escher had this to say about the work: "The picture shows a spherical mirror, resting on a left hand. But as a print is the reverse of the original drawing on stone, it was my right hand that you see depicted. (Being left-handed, I needed my left hand to make the drawing.) Such a globe reflection collects almost one's whole surroundings in one disk-shaped image. The whole room, four walls, the Figure Q36.12 13. Why do some emergency vehicles have the symbol ECNALUBMA written on the front? 1178 CHAPTER 36 Geometric Optics floor, and the ceiling, everything, albeit distorted, is compressed into that one small circle. Your own head, or more exactly the point between your eyes, is the absolute center. No matter how you turn or twist yourself, you can't get out of that central point. You are immovably the focus, the unshakable core, of your world." Comment on the accuracy of Escher's description. 23. You can make a corner reflector by placing three flat mirrors in the corner of a room where the ceiling meets the walls. Show that no matter where you are in the room, you can see yourself reflected in the mirrors -- upside down. Figure Q36.22 PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide WEB = solution posted at http://www.saunderscollege.com/physics/ = Computer useful in solving problem = Interactive Physics = paired numerical/symbolic problems Section 36.1 Images Formed by Flat Mirrors 1. Does your bathroom mirror show you older or younger than you actually are? Compute an order-of-magnitude estimate for the age difference, based on data that you specify. 2. In a church choir loft, two parallel walls are 5.30 m apart. The singers stand against the north wall. The organist faces the south wall, sitting 0.800 m away from it. To enable her to see the choir, a flat mirror 0.600 m wide is mounted on the south wall, straight in front of her. What width of the north wall can she see? Hint: Draw a top-view diagram to justify your answer. 3. Determine the minimum height of a vertical flat mirror in which a person 5 10 in height can see his or her full image. (A ray diagram would be helpful.) 4. Two flat mirrors have their reflecting surfaces facing each other, with an edge of one mirror in contact with an edge of the other, so that the angle between the mirrors is . When an object is placed between the mirrors, a number of images are formed. In general, if the angle is such that n 360 , where n is an integer, the number of images formed is n 1. Graphically, find all the image positions for the case n 6 when a point object is between the mirrors (but not on the angle bisector). 5. A person walks into a room with two flat mirrors on opposite walls, which produce multiple images. When the person is 5.00 ft from the mirror on the left wall and 10.0 ft from the mirror on the right wall, find the distances from that person to the first three images seen in the mirror on the left. Section 36.2 Images Formed by Spherical Mirrors 6. A concave spherical mirror has a radius of curvature of 20.0 cm. Find the location of the image for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification. 7. At an intersection of hospital hallways, a convex mirror is mounted high on a wall to help people avoid collisions. The mirror has a radius of curvature of 0.550 m. Locate and describe the image of a patient 10.0 m from the mirror. Determine the magnification. 8. A large church has a niche in one wall. On the floor plan it appears as a semicircular indentation of radius 2.50 m. A worshiper stands on the center line of the niche, 2.00 m out from its deepest point, and whispers a prayer. Where is the sound concentrated after reflection from the back wall of the niche? Problems 1179 WEB 9. A spherical convex mirror has a radius of curvature of 40.0 cm. Determine the position of the virtual image and the magnification (a) for an object distance of 30.0 cm and (b) for an object distance of 60.0 cm. (c) Are the images upright or inverted? 10. The height of the real image formed by a concave mirror is four times the object height when the object is 30.0 cm in front of the mirror. (a) What is the radius of curvature of the mirror? (b) Use a ray diagram to locate this image. 11. A concave mirror has a radius of curvature of 60.0 cm. Calculate the image position and magnification of an object placed in front of the mirror (a) at a distance of 90.0 cm and (b) at a distance of 20.0 cm. (c) In each case, draw ray diagrams to obtain the image characteristics. 12. A concave mirror has a focal length of 40.0 cm. Determine the object position for which the resulting image is upright and four times the size of the object. 13. A spherical mirror is to be used to form, on a screen 5.00 m from the object, an image five times the size of the object. (a) Describe the type of mirror required. (b) Where should the mirror be positioned relative to the object? 14. A rectangle 10.0 cm 20.0 cm is placed so that its right edge is 40.0 cm to the left of a concave spherical mirror, as in Figure P36.14. The radius of curvature of the mirror is 20.0 cm. (a) Draw the image formed by this mirror. (b) What is the area of the image? Section 36.3 Images Formed by Refraction 18. A flint-glass plate (n 1.66) rests on the bottom of an aquarium tank. The plate is 8.00 cm thick (vertical dimension) and covered with water (n 1.33) to a depth of 12.0 cm. Calculate the apparent thickness of the plate as viewed from above the water. (Assume nearly normal incidence.) 19. A cubical block of ice 50.0 cm on a side is placed on a level floor over a speck of dust. Find the location of the image of the speck if the index of refraction of ice is 1.309. 20. A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the transparent cornea. Assume that this outer surface has a 6.00-mm radius of curvature, and assume that the eyeball contains just one fluid with an index of refraction of 1.40. Prove that a very distant object will be imaged on the retina, 21.0 mm behind the cornea. Describe the image. 21. A glass sphere (n 1.50) with a radius of 15.0 cm has a tiny air bubble 5.00 cm above its center. The sphere is viewed looking down along the extended radius containing the bubble. What is the apparent depth of the bubble below the surface of the sphere? 22. A transparent sphere of unknown composition is observed to form an image of the Sun on the surface of the sphere opposite the Sun. What is the refractive index of the sphere material? 23. One end of a long glass rod (n 1.50) is formed into a convex surface of radius 6.00 cm. An object is positioned in air along the axis of the rod. Find the image positions corresponding to object distances of (a) 20.0 cm, (b) 10.0 cm, and (c) 3.00 cm from the end of the rod. 24. A goldfish is swimming at 2.00 cm/s toward the front wall of a rectangular aquarium. What is the apparent speed of the fish as measured by an observer looking in from outside the front wall of the tank? The index of refraction of water is 1.33. 25. A goldfish is swimming inside a spherical plastic bowl of water, with an index of refraction of 1.33. If the goldfish is 10.0 cm from the wall of the 15.0-cm-radius bowl, where does it appear to an observer outside the bowl? 20.0 cm 10.0 cm C 40.0 cm Figure P36.14 15. A dedicated sports-car enthusiast polishes the inside and outside surfaces of a hubcap that is a section of a sphere. When she looks into one side of the hubcap, she sees an image of her face 30.0 cm in back of the hubcap. She then flips the hubcap over and sees another image of her face 10.0 cm in back of the hubcap. (a) How far is her face from the hubcap? (b) What is the radius of curvature of the hubcap? 16. An object is 15.0 cm from the surface of a reflective spherical Christmas-tree ornament 6.00 cm in diameter. What are the magnification and position of the image? 17. A ball is dropped from rest 3.00 m directly above the vertex of a concave mirror that has a radius of 1.00 m and lies in a horizontal plane. (a) Describe the motion of the ball's image in the mirror. (b) At what time do the ball and its image coincide? Section 36.4 Thin Lenses 26. A contact lens is made of plastic with an index of refraction of 1.50. The lens has an outer radius of curvature of 2.00 cm and an inner radius of curvature of 2.50 cm. What is the focal length of the lens? 27. The left face of a biconvex lens has a radius of curvature of magnitude 12.0 cm, and the right face has a radius of curvature of magnitude 18.0 cm. The index of refraction of the glass is 1.44. (a) Calculate the focal length of the lens. (b) Calculate the focal length if the radii of curvature of the two faces are interchanged. WEB 1180 CHAPTER 36 Geometric Optics location and (b) the magnification of the image. (c) Construct a ray diagram for this arrangement. 38. Figure P36.38 shows a thin glass (n 1.50) converging lens for which the radii of curvature are R 1 15.0 cm and R 2 12.0 cm. To the left of the lens is a cube with a face area of 100 cm2. The base of the cube is on the axis of the lens, and the right face is 20.0 cm to the left of the lens. (a) Determine the focal length of the lens. (b) Draw the image of the square face formed by the lens. What type of geometric figure is this? (c) Determine the area of the image. WEB 28. A converging lens has a focal length of 20.0 cm. Locate the image for object distances of (a) 40.0 cm, (b) 20.0 cm, and (c) 10.0 cm. For each case, state whether the image is real or virtual and upright or inverted. Find the magnification in each case. 29. A thin lens has a focal length of 25.0 cm. Locate and describe the image when the object is placed (a) 26.0 cm and (b) 24.0 cm in front of the lens. 30. An object positioned 32.0 cm in front of a lens forms an image on a screen 8.00 cm behind the lens. (a) Find the focal length of the lens. (b) Determine the magnification. (c) Is the lens converging or diverging? 31. The nickel's image in Figure P36.31 has twice the diameter of the nickel and is 2.84 cm from the lenn 36.8 The Simple Magnifier Section 36.9 The Compound Microscope Section 36.10 The Telescope 47. A philatelist examines the printing detail on a stamp, using a biconvex lens with a focal length of 10.0 cm as a simple magnifier. The lens is held close to the eye, and the lens-to-object distance is adjusted so that the virtual image is formed at the normal near point (25.0 cm). Calculate the magnification. 48. A lens that has a focal length of 5.00 cm is used as a magnifying glass. (a) Where should the object be placed to obtain maximum magnification? (b) What is the magnification? 49. The distance between the eyepiece and the objective lens in a certain compound microscope is 23.0 cm. The focal length of the eyepiece is 2.50 cm, and that of the objective is 0.400 cm. What is the overall magnification of the microscope? The desired overall magnification of a compound microscope is 140 . The objective alone produces a lateral magnification of 12.0 . Determine the required focal length of the eyepiece. The Yerkes refracting telescope has a 1.00-m-diameter objective lens with a focal length of 20.0 m. Assume that it is used with an eyepiece that has a focal length of 2.50 cm. (a) Determine the magnification of the planet Mars as seen through this telescope. (b) Are the Martian polar caps seen right side up or upside down? Astronomers often take photographs with the objective lens or the mirror of a telescope alone, without an eyepiece. (a) Show that the image size h for this telescope is given by h f h/( f p), where h is the object size, f is the objective focal length, and p is the object distance. (b) Simplify the expression in part (a) for the case in which the object distance is much greater than objective focal length. (c) The "wingspan" of the International Space Station is 108.6 m, the overall width of its solar-panel configuration. Find the width of the image formed by a telescope objective of focal length 4.00 m when the station is orbiting at an altitude of 407 km. Galileo devised a simple terrestrial telescope that produces an upright image. It consists of a converging objective lens and a diverging eyepiece at opposite ends of the telescope tube. For distant objects, the tube length is the objective focal length less the absolute value of the eyepiece focal length. (a) Does the user of the telescope see a real or virtual image? (b) Where is the final image? (c) If a telescope is to be constructed with a tube 10.0 cm long and a magnification of 3.00, what are the focal lengths of the objective and eyepiece? A certain telescope has an objective mirror with an aperture diameter of 200 mm and a focal length of 2 000 mm. It captures the image of a nebula on photographic film at its prime focus with an exposure time of 1.50 min. To produce the same light energy per unit area on the film, what is the required exposure time to photograph the same nebula with a smaller telescope, which has an objective lens with an aperture diameter of 60.0 mm and a focal length of 900 mm? ADDITIONAL PROBLEMS 55. The distance between an object and its upright image is 20.0 cm. If the magnification is 0.500, what is the focal length of the lens that is being used to form the image? 56. The distance between an object and its upright image is d. If the magnification is M, what is the focal length of the lens that is being used to form the image? 57. The lens and mirror in Figure P36.57 have focal lengths of 80.0 cm and 50.0 cm, respectively. An object is 1182 Object CHAPTER 36 Geometric Optics mirror, what is the position of the image? Is the latter image real or virtual? 61. A parallel beam of light enters a glass hemisphere perpendicular to the flat face, as shown in Figure P36.61. The radius is R 6.00 cm, and the index of refraction is n 1.560. Determine the point at which the beam is focused. (Assume paraxial rays.) Lens Mirror WEB 1.00 m 1.00 m Air n I Figure P36.57 placed 1.00 m to the left of the lens, as shown. Locate the final image, which is formed by light that has gone through the lens twice. State whether the image is bject is 3.40 mm from the objective, what is the magnification? (Hint: Use the lens equation for the objective.) Two converging lenses with focal lengths of 10.0 cm and 20.0 cm are positioned 50.0 cm apart, as shown in Figure P36.70. The final image is to be located between the lenses, at the position indicated. (a) How far to the left f 1 (10.0 cm) Final image f 2 (20.0 cm) 67. like device, and by the requirement that the implant provide for correct distant vision. (a) If the distance from lens to retina is 22.4 mm, calculate the power of the implanted lens in diopters. (b) Since no accommodation occurs and the implant allows for correct distant vision, a corrective lens for close work or reading must be used. Assume a reading distance of 33.0 cm, and calculate the power of the lens in the reading glasses. 72. A floating strawberry illusion consists of two parabolic mirrors, each with a focal length of 7.50 cm, facing each other so that their centers are 7.50 cm apart (Fig. P36.72). If a strawberry is placed on the lower mirror, an image of the strawberry is formed at the small opening at the center of the top mirror. Show that the final image is formed at that location, and describe its characteristics. (Note: A very startling effect is to shine a flashlight beam on these images. Even at a glancing angle, the incoming light beam is seemingly reflected off the images! Do you understand why?) Small hole 68. 69. Strawberry 70. Object p 31.0 cm 50.0 cm Figure P36.72 73. An object 2.00 cm high is placed 40.0 cm to the left of a converging lens with a focal length of 30.0 cm. A diverging lens with a focal length of 20.0 cm is placed 110 cm to the right of the converging lens. (a) Determine the final position and magnification of the final image. (b) Is the image upright or inverted? (c) Repeat parts (a) and (b) for the case in which the second lens is a converging lens with a focal length of 20.0 cm. Figure P36.70 of the first lens should the object be? (b) What is the overall magnification? (c) Is the final image upright or inverted? 71. A cataract-impaired lens in an eye may be surgically removed and replaced by a manufactured lens. The focal length required for the new lens is determined by the lens-to-retina distance, which is measured by a sonar- 1184 CHAPTER 36 Geometric Optics ANSWERS TO QUICK QUIZZES 36.1 At C. A ray traced from the stone to the mirror and then to observer 2 looks like this: 1 2 3 F F Front Back Extension of lens 2 1 Figure QQA36.2 rays, but the remaining ones still come from all parts of the object. 36.5 The eyeglasses on the left are diverging lenses, which correct for nearsightedness. If you look carefully at the edge of the person's face through the lens, you will see that everything viewed through these glasses is reduced in size. The eyeglasses on the right are converging lenses, which correct for farsightedness. These lenses make everything that is viewed through them look larger. 36.6 The lateral magnification of a telescope is not well defined. For viewing with the eye relaxed, the user may slightly adjust the position of the eyepiece to place the final image I 2 in Figure 36.42a at infinity. Then, its height and its lateral magnification also are infinite. The angular magnification of a telescope as we define it is the factor by which the telescope increases in the diameter -- on the retina of the viewer's eye -- of the real image of an extended object. Figure QQA36.1 36.2 The focal length is infinite. Because the flat surfaces of the pane have infinite radii of curvature, Equation 36.11 indicates that the focal length is also infinite. Parallel rays striking the pane focus at infinity, which means that they remain parallel after passing through the glass. 36.3 An infinite number. In general, an infinite number of rays leave each point of any object and travel outward in all directions. (The three principal rays that we use to locate an image make up a selected subset of the infinite number of rays.) When an object is taller than a lens, we merely extend the plane conining two small openings (usually in the shape of slits). The light emerging from the two slits is coherent because a single source produces the original light beam and the two slits serve only to separate the original beam into two parts (which, after all, is what was done to the sound signal from the side-by-side loudspeakers). Any random change in the light 37.2 Young's Double-Slit Experiment 1187 emitted by the source occurs in both beams at the same time, and as a result interference effects can be observed when the light from the two slits arrives at a viewing screen. 37.2 YOUNG'S DOUBLE-SLIT EXPERIMENT Interference in light waves from two sources was first demonstrated by Thomas Young in 1801. A schematic diagram of the apparatus that Young used is shown in Figure 37.1a. Light is incident on a first barrier in which there is a slit S0 . The waves emerging from this slit arrive at a second barrier that contains two parallel slits S 1 and S 2 . These two slits serve as a pair of coherent light sources because waves emerging from them originate from the same wave front and therefore maintain a constant phase relationship. The light from S 1 and S 2 produces on a viewing screen a visible pattern of bright and dark parallel bands called fringes (Fig. 37.1b). When the light from S 1 and that from S 2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears. When the light from the two slits combines destructively at any location on the screen, a dark fringe results. Figure 37.2 is a photograph of an interference pattern produced by two coherent vibrating sources in a water tank. max min max S1 S0 S2 First barrier Second barrier min max min max min A B max (a) Viewing screen (b) Figure 37.1 (a) Schematic diagram of Young's double-slit experiment. Slits S1 and S 2 behave as coherent sources of light waves that produce an interference pattern on the viewing screen (drawing not to scale). (b) An enlargement of the center of a fringe pattern formed on the viewing screen with many slits could look like this. Figure 37.2 An interference pattern involving water waves is produced by two vibrating sources at the water's surface. The pattern is analogous to that observed in Young's double-slit experiment. Note the regions of constructive (A) and destructive (B) interference. 1188 CHAPTER 37 Interference of Light Waves Quick Quiz 37.1 If you were to blow smoke into the space between the second barrier and the viewing screen of Figure 37.1a, what would you see? QuickLab Look through the fabric of an umbrella at a distant streetlight. Can you explain what you see? (The fringe pattern in Figure 37.1b is from rectangular slits. The fabric of the umbrella creates a two-dimensional set of square holes.) Quick Quiz 37.2 Figure 37.2 is an overhead view of a shallow water tank. If you wanted to use a small ruler to measure the water's depth, would this be easier to do at location A or at location B? Figure 37.3 shows some of the ways in which two waves can combine at the screen. In Figure 37.3a, the two waves, which leave the two slits in phase, strike the screen at the central point P. Because both waves travel the same distance, they arrive at P in phase. As a result, constructive interference occurs at this location, and a bright fringe is observed. In Figure 37.3b, the two waves also start in phase, but in this case the upper wave has to travel one wavelength farther than the lower wave to reach point Q . Because the upper wave falls behind the lower one by exactly one wavelength, they still arrive in phase at Q , and so a second bright fringe appears at this location. At point R in Figure 37.3c, however, midway between points P and Q , the upper wave has fallen half a wavelength behind the lower wave. This means that a trough of the lower wave overlaps a crest of the upper wave; this gives rise to destructive interference at point R. For this reason, a dark fringe is observed at this location. We can describe Young's experiment quantitatively with where I max is the maximum intensity on the screen and the expression represents the time average. Substituting the value for given by Equation 37.8 into this expression, we find that I Alternatively, because sin Equation 37.12 in the form I max cos2 d sin (37.12) in Figure 37.4, we can write y/L for small values of I max cos2 d y L I (37.13) Constructive interference, which produces light intensity maxima, occurs when the quantity dy/ L is an integral multiple of , corresponding to y ( L/d )m. This is consistent with Equation 37.5. A plot of light intensity versus d sin is given in Figure 37.6. Note that the interference pattern consists of equally spaced fringes of equal intensity. Remember, however, that this result is valid only if the slit-to-screen distance L is much greater than the slit separation, and only for small values of . We have seen that the interference phenomena arising from two sources depend on the relative phase of the waves at a given point. Furthermore, the phase difference at a given point depends on the path difference between the two waves. The resultant light intensity at a point is proportional to the square of the resultant electric field at that point. That is, the light intensity is proportional to (E 1 E 2 )2. It would be incorrect to calculate the light intensity by adding the intensities of the individual waves. This procedure would give E 12 E 22, which of course is not the same as (E 1 E 2 )2. Note, however, that (E 1 E 2)2 has the same average value as E 12 E 22 when the time average is taken over all values of the 37.4 Phasor Addition of Waves 1193 phase difference between E 1 and E 2 . Hence, the law of conservation of energy is not violated. E1 37.4 PHASOR ADDITION OF WAVES t E0 In the preceding section, we combined two waves algebraically to obtain the resultant wave amplitude at some point on a screen. Unfortunately, this analytical procedure becomes cumbersome when we must add several wave amplitudes. Because we shall eventually be interested in combining a large number of waves, we now describe a graphical procedure for this purpose. Let us again consider a sinusoidal wave whose electric field component is given by E1 E 0 sin t (a) E2 E0 t + where E 0 is the wave amplitude and is the angular frequency. This wave can be represented graphically by a phasor of magnitude E 0 rotating about the origin counterclockwise with an angular frequency , as shown in Figure 37.7a. Note that the phasor makes an angle t with the horizontal axis. The projection of the phasor on the vertical axis represents E 1 , the magnitude of the wave disturbance at some time t. Hence, as the phasor rotates in a circle, the projection E 1 oscillates along the vertical axis about the origin. Now consider a second sinusoidal wave whose electric field component is given by E2 E 0 sin( t ) (b) E0 EP E2 E1 ER t E 0 This wave has the same amplitude and frequency as E 1 , but its phase is with respect to E 1 . The phasor representing E 2 is shown in Figure 37.7b. We can obtain the resultant wave, which is the sum of E 1 and E 2 , graphically by redrawing the phasors as shown in Figure 37.7c, in which the tail of the second phasor is placed at the tip of the first. As with vector addition, the resultant phasor ER runs from the tail of the first phasor to the tip of the second. Furthermore, ER rotates along with the two individual phasors at the same angular frequency . The projection of ER along the vertical axis equals the sum of the projections of the two other phasors: E P E 1 E 2 . It is convenient to construct the phasors at t 0 as in Figure 37.8. From the geometry of one of the right triangles, we see that cos which gives ER 2E 0 cos , E R /2 E0 (c) Figure 37.7 (a) Phasor diagram for the wave disturbance E 1 E 0 sin t. The phasor is a vector of length E 0 rotating counterclockwise. (b) Phasor diagram for the wave E 2 ). (c) The disturE 0 sin( t bance ER is the resultant phasor formed from the phasors of parts (a) and (b). Because the sum of the two opposite interior angles equals th Sketch a phasor diagram to illustrate the resultant of E 1 E 01 sin t and E 2 E 02 sin( t ), where /3. Use the sketch and E 02 1.50E 01 and /6 the law of cosines to show that, for two coherent waves, the resultant intensity can be written in the form I R I 1 I 2 2 !I 1I 2 cos . 27. Consider N coherent sources described by E 1 E 0 sin( t ), E 2 E 0 sin( t 2 ), E 3 E 0 sin( t 3 ), . . . , E N E 0 sin( t N ). Find the minimum value of for which E R E 1 E 2 E N is zero. E3 . . . to coat a piece of glass (n 1.50). What should be the minimum thickness of this film if it is to minimize reflection of 500-nm light? 33. A film of MgF2 (n 1.38) having a thickness of 1.00 10 5 cm is used to coat a camera lens. Are any wavelengths in the visible spectrum intensified in the reflected light? 34. Astronomers observe the chromosphere of the Sun with a filter that passes the red hydrogen spectral line of wavelength 656.3 nm, called the H line. The filter consists of a transparent dielectric of thickness d held between two partially aluminized glass plates. The filter is held at a constant temperature. (a) Find the minimum value of d that produces maximum transmission of perpendicular H light, if the dielectric has an index of refraction of 1.378. (b) Assume that the temperature of the filter increases above its normal value and that its index of refraction does not change significantly. What happens to the transmitted wavelength? (c) The dielectric will also pass what near-visible wavelength? One of the glass plates is colored red to absorb this light. 35. A beam of 580-nm light passes through two closely spaced glass plates, as shown in Figure P37.35. For what minimum nonzero value of the plate separation d is the transmitted light bright? d Section 37.5 Change of Phase Due to Reflection Section 37.6 Interference in Thin Films 28. A soap bubble (n 1.33) is floating in air. If the thickness of the bubble wall is 115 nm, what is the wavelength of the light that is most strongly reflected? 29. An oil film (n 1.45) floating on water is illuminated by white light at normal incidence. The film is 280 nm thick. Find (a) the dominant observed color in the reflected light and (b) the dominant color in the transmitted light. Explain your reasoning. 30. A thin film of oil (n 1.25) is located on a smooth, wet pavement. When viewed perpendicular to the pavement, the film appears to be predominantly red (640 nm) and has no blue color (512 nm). How thick is the oil film? 31. A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n 1.50, how thick would you make the coating? 32. A material having an index of refraction of 1.30 is used Figure P37.35 36. When a liquid is introduced into the air space between the lens and the plate in a Newton's-rings apparatus, the diameter of the tenth ring changes from 1.50 to 1.31 cm. Find the index of refraction of the liquid. 37. An air wedge is formed between two glass plates separated at one edge by a very fine wire, as shown in Figure P37.37. When the wedge is illuminated from above by 600-nm light, 30 dark fringes are observed. Calculate the radius of the wire. WEB Figure P37.37 Problems 37 and 38. 1208 CHAPTER 37 Interference of Light Waves cent zones of constructive interference. (b) To make the angles in the interference pattern easy to measure with a plastic protractor, you should use an electromagnetic wave with frequency of what order of magnitude? How is this wave classified on the electromagnetic spectrum? In a Young's double-slit experiment using light of wavelength , a thin piece of Plexiglas having index of refraction n covers one of the slits. If the center point on the screen is a dark spot instead of a bright spot, what is the minimum thickness of the Plexiglas? Review Problem. A flat piece of glass is held stationary and horizontal above the flat top end of a 10.0-cm-long vertical metal rod that has its lower end rigidly fixed. The thin film rence pattern moves upward a distance y . Find y . 62. Waves broadcast by a 1 500-kHz radio station arrive at a home receiver by two paths. One is a direct path, and the other is from reflection off an airplane directly above the receiver. The airplane is approximately 100 m above the receiver, and the direct distance from station to home is 20.0 km. What is the precise height of the airplane if destructive interference is occurring? (Assume that no phase change occurs on reflection.) 63. In a Newton's-rings experiment, a plano-convex glass (n 1.52) lens having a diameter of 10.0 cm is placed on a flat plate, as shown in Figure 37.18a. When 650-nm light is incident normally, 55 bright rings are observed, with the last ring right on the edge of the lens. (a) What is the radius of curvature of the convex surface of the lens? (b) What is the focal length of the lens? 64. A piece of transparent material having an index of re- 1210 CHAPTER 37 Interference of Light Waves fraction n is cut into the shape of a wedge, as shown in Figure P37.64. The angle of the wedge is small, and monochromatic light of wavelength is normally incident from above. If the height of the wedge is h and the width is , show that bright fringes occur at the positions x (m 1 )/2hn and that dark fringes occur at 2 the positions x m/2hn, where m 0, 1, 2, . . . and x is measured as shown. r R h Figure P37.66 r !m R /n film x Figure P37.64 65. Use phasor addition to find the resultant amplitude and phase constant when the following three harmonic functions are combined: E 1 sin( t /6), E 2 3.0 sin( t 7 /2), E 3 6.0 sin( t 4 /3). 66. A plano-convex lens having a radius of curvature of r 4.00 m is placed on a concave reflecting surface whose radius of curvature is R 12.0 m, as shown in Figure P37.66. Determine the radius of the 100th bright ring if 500-nm light is incident normal to the flat surface of the lens. 67. A plano-convex lens has index of refraction n. The curved side of the lens has radius of curvature R and rests on a flat glass surface of the same index of refraction, with a film of index n film between them. The lens is illuminated from above by light of wavelength . Show that the dark Newton's rings have radii given approximately by where m is an integer and r is much less than R . 68. A soap film (n 1.33) is contained within a rectangular wire frame. The frame is held vertically so that the film drains downward and becomes thicker at the bottom than at the top, where the thickness is essentially zero. The film is viewed in white light with near-normal incidence, and the first violet ( 420 nm ) interference band is observed 3.00 cm from the top edge of the film. (a) Locate the first red ( 680 nm ) interference band. (b) Determine the film thickness at the positions of the violet and red bands. (c) What is the wedge angle of the film? 69. Interference fringes are produced using Lloyd's mirror and a 606-nm source, as shown in Figure 37.14. Fringes 1.20 mm apart are formed on a screen 2.00 m from the real source S. Find the vertical distance h of the source above the reflecting surface. 70. Slit 1 of a double slit is wider than slit 2, so that the light from slit 1 has an amplitude 3.00 times that of the light from slit 2. Show that Equation 37.11 is replaced by the equation I (4I max /9)(1 3 cos2 /2) for this situation. ANSWERS TO QUICK QUIZZES 37.1 Bands of light along the orange lines interspersed with dark bands running along the dashed black lines. 37.2 At location B. At A, which is on a line of constructive interference, the water surface undulates so much that you probably could not determine the depth. Because B is on a line of destructive interference, the water level does not change, and you should be able to read the ruler easily. 37.3 The graph is shown in Figure QQA37.1. The width of the primary maxima is slightly narrower than the N 5 primary width but wider than the N 10 primary width. 1 Because N 6, the secondary maxima are 36 as intense as the primary maxima. 37.4 The greater the variation in thickness, the narrower t from overlapping, as shown in Figure 38.12a, their images can be distinguished and are said to be resolved. If the sources are close together, however, as shown in Figure 38.12b, the two central maxima overlap, and the images are not resolved. In determining whether two images are resolved, the following condition is often used: When the central maximum of one image falls on the first minimum of the other image, the images are said to be just resolved. This limiting condition of resolution is known as Rayleigh's criterion. Figure 38.13 shows diffraction patterns for three situations. When the objects are far apart, their images are well resolved (Fig. 38.13a). When the angular sepa- S1 S1 S2 S2 Slit Viewing screen Slit Viewing screen (a) (b) Figure 38.12 Two point sources far from a narrow slit each produce a diffraction pattern. (a) The angle subtended by the sources at the slit is large enough for the diffraction patterns to be distinguishable. (b) The angle subtended by the sources is so small that their diffraction patterns overlap, and the images are not well resolved. (Note that the angles are greatly exaggerated. The drawing is not to scale.) 38.3 Resolution of Single-Slit and Circular Apertures 1221 (a) (b) (c) Figure 38.13 Individual diffraction patterns of two point sources (solid curves) and the resultant patterns (dashed curves) for various angular separations of the sources. In each case, the dashed curve is the sum of the two solid curves. (a) The sources are far apart, and the patterns are well resolved. (b) The sources are closer together such that the angular separation just satisfies Rayleigh's criterion, and the patterns are just resolved. (c) The sources are so close together that the patterns are not resolved. ration of the objects satisfies Rayleigh's criterion (Fig. 38.13b), the images are just resolved. Finally, when the objects are close together, the images are not resolved (Fig. 38.13c). From Rayleigh's criterion, we can determine the minimum angular separation min subtended by the sources at the slit for which the images are just resolved. Equation 38.1 indicates that the first minimum in a single-slit diffraction pattern occurs at the angle for which sin a where a is the width of the slit. According to Rayleigh's criterion, this expression gives the smallest angular separation for which the two images are resolved. Because V a in most situations, sin is small, and we can use the approximation sin . Therefore, the limiting angle of resolution for a slit of width a is Figure 38.14 min a (38.8) where min is expressed in radians. Hence, the angle subtended by the two sources at the slit must be greater than /a if the images are to be resolved. Many optical systems use circular apertures rather than slits. The diffraction pattern of a circular aperture, shown in Figure 38.14, consists of a central circular The diffraction pattern of a circular aperture consists of a central bright disk surrounded by concentric bright and dark rings. 1222 CHAPTER 38 Diffraction and Polarization bright disk surrounded by progressively fainter bright and dark rings. Analysis shows that the limiting angle of resolution of the circular aperture is Limiting angle of resolution for a circular aperture min 1.22 D (38.9) where D is the diameter of the aperture. Note that this expression is similar to Equation 38.8 except for the factor 1.22, which arises from a complex mathematical analysis of diffraction from the circular aperture. EXAMPLE 38.3 Limiting Resolution of a Microscope Violet light (400 nm) gives a limiting angle of resolution of min Light of wavelength 589 nm is used to view an object under a microscope. If the aperture of the objective has a diameter of 0.900 cm, (a) what is the limiting angle of resolution? 1.22 400 0.900 10 9 m 10 2 m 5.42 10 5 rad Solution min (a) Using Equation 38.9, we find that the limiting angle of resolution is 1.22 589 0.900 10 9 m 10 2 m 7.98 10 5 (c) Suppose that water (n 1.33) fills the space between the object and the objective. What effect does this have on resolving power when 589-nm light is used? rad This means that any two points on the object subtending an angle smaller than this at the objective cannot be distinguished in the image. (b) If it were possible to use visible light of any wavelength, what would be the maximum limit of resolution for this microscope? To obtain the smallest limiting angle, we have to use the shortest wavelength available in the visible spectrum. Solution We find the wavelength of the 589-nm light in the water using Equation 35.7: air water n water 589 nm 1.33 443 nm The limiting angle of resolution at this wavelength is now smaller than that calculated in part (a): min Solution 1.22 443 0.900 10 9 m 10 2 m 6.00 10 5 rad EXAMPLE 38.4 Resolution of a Telescope mospheric blurring. This seeing limit is usually about 1 s of arc and is never smaller than about 0.1 s of arc. (This is one of the reasons for the superiority of photographs from the Hubble Space Telescope, which views celestial objects from an orbital position above the atmosphere.) The Hale telescope at Mount Palomar has a diameter of 200 in. What is its limiting angle of resolution for 600-nm light? Solution 10 7 Because D 200 in. m, Equation 38.9 gives min 5.08 m and 6.00 10 7 m 5.08 m 0.03 s of arc 6.00 1.22 D 1.22 10 7 Exercise 1.44 rad Any two stars that subtend an angle greater than or equal to this value are resolved (if atmospheric conditions are ideal). The Hale telescope can never reach its diffraction limit because the limiting angle of resolution is always set by at- The large radio telescope at Arecibo, Puerto Rico, has a diameter of 305 m and is designed to detect 0.75-m radio waves. Calculate the minimum angle of resolution for this telescope and compare your answer with that for the Hale telescope. 3.0 10 3 rad (10 min of arc), more than 10 000 times larger (that is, worse) than the Hale minimum. Answer EXAMPLE 38.5 Resolution of the Eye Estimate the limiting angle of resolution for the human eye, assuming its resolution is limited only by diffraction. Solution Let us choose a wavelength of 500 nm, near the center of the visible spectrum. Although pupil diameter 38.3 Resolution of Single-Slit and Circular Apertures varies from person to person, we estimate a diameter of 2 mm. We use Equation 38.9, taking 500 nm and D 2 mm: min 1223 S1 d S2 1.22 3 D 10 1.22 4 5.00 10 7 m 2 10 3 m 1 min of arc min rad L We can use this result to determine the minimum separation distance d between two point sources that the eye can distinguish if they are a distance L from the observer (Fig. 38.15). Because min is small, we see that sin min min Figure 38.15 Two point sources separated by a distance d as observed by the eye. d L Exercise d L min For example, if the point sources are 25 cm from the eye (the near point), then d (25 cm)(3 10 4 Suppose that the pupil is dilated to a diameter of 5.0 mm and that two point sources 3.0 m away are being viewed. How far apart must the sources be if the eye is to resolve them? 0.037 cm. rad) 8 10 3 cm Answer This is approximately equal to the thickness of a human hair. APPLICATION Loudspeaker Design The three-way speaker system shown in Figure 38.16 contains a woofer, a midrange speaker, and a tweeter. The smalldiameter tweeter is for high frequencies, and the largediameter woofer is for low frequencies. The midrange speaker, of intermediate diameter, is used for the frequency band above the high-frequency cutoff of the woofer and below the low-frequency cutoff of the tweeter. Circuits known as crossover networks include low-pass, midrange, and high-pass filters that direct the electrical signal to the appropriate speaker. The effective aperture size of a speaker is approximately its diameter. Because the wavelengths of sound waves are comparable to the typical sizes of the speakers, diffraction effects determine the angular radiation pattern. To be most useful, a speaker should radiate sound over a broad range of angles so that the listener does not haverder maximum for each wavelength occurs at a specific angle. All wavelengths are seen at 0, corresponding to m 0, the zeroth-order maximum. The first-order maximum (m 1) is observed at an angle that satisfies the relationship sin /d; the second-order maximum (m 2) is observed at a larger angle , and so on. The intensity distribution for a diffraction grating obtained with the use of a monochromatic source is shown in Figure 38.19. Note the sharpness of the principal maxima and the broadness of the dark areas. This is in contrast to the broad bright fringes characteristic of the two-slit interference pattern (see Fig. 37.6). Because the principal maxima are so sharp, they are very much brighter than two-slit 1226 CHAPTER 38 Diffraction and Polarization (a) Figure 38.20 (b) (a) Addition of two wave fronts from two slits. (b) Addition of ten wave fronts from ten slits. The resultant wave is much stronger in part (b) than in part (a). QuickLab Stand a couple of meters from a lightbulb. Facing away from the light, hold a compact disc about 10 cm from your eye and tilt it until the reflection of the bulb is located in the hole at the disc's center. You should see spectra radiating out from the center, with violet on the inside and red on the outside. Now move the disc away from your eye until the violet band is at the outer edge. Carefully measure the distance from your eye to the center of the disc and also determine the radius of the disc. Use this information to find the angle to the first-order maximum for violet light. Now use Equation 38.10 to determine the spacing between the grooves on the disc. The industry standard is 1.6 m. How close did you come? interference maxima. The reason for this is illus