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Chapter14 - CHAPTER ENERGY AND STABILITY I do not consider...

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Unformatted text preview: CHAPTER ENERGY AND STABILITY I do not consider these principles to be certain mysterious qualities feigned as arising from characteristic forms of things, but as universal laws of Nature, by the influence of which these very things have been created. For the phenomena of Nature show that these principles do indeed exist, although their nature has not yet been elucidated. To assert that each and every species is endowed with a mysterious property characteristic to it, due to which it has a definite mode in action, is really equivalent to saying nothing at all. On the other hand, to derive from the phenomena of Nature two or three general principles, and then to explain how the properties and actions of all corporate things follow from those principles, this would indeed be a mighty advance in philosophy, even if the causes of those principles had not at the time been discovered. Roger Boscovich, A Theory of Natural Philosophy (1763) 14.1 FORMS OF ENERGY The concept of energy, as we saw in Chapter 13, is subtle, elegant, and rich. It describes a dynamic property of the universe which is strictly and absolutely conserved; energy can neither be created nor destroyed. Not even in the presence of friction is energy ever lost; it is simply transformed into other forms. Nevertheless, the universe is winding down. Energy tends to be transformed from well-organized forms into more disorganized forms, until it becomes completely useless. 267 268 ENERGY AND STABILITY The organized forms of energy we discussed are potential and kinetic energy. In order for one object, or part of the universe, to receive such energy, another part must lose an equal amount of energy. To keep track of exchanges of energy between different parts of the universe, we have the bookkeeping device known as work. Let’s reexamine the ideas of work and energy, in part to tidy up a few loose ends, but mainly to generate deeper insights. Recall that work is the displacement of an object multiplied by the component of the force in the same direction. If the force F varies in magnitude but always has the same direction, then the work done by F (x) in moving a particle from x = a to x = b along that direction is the integral we introduced in Chapter 13: b W = f F(x)dx. (13-2) Consider now a force F, which may vary both in magnitude and direction, and suppose it acts on a particle moving along a curve from A to B, as illustrated in Fig. 14.1. The problem of determining the work done by this force is now more complicated. In fact, the real problem here is to decide on a reasonable definition of work. Again, we shall use an integral to define work. This integral must take into account both the force F and the curve along which the particle is moved. F Figure 14.1 A variable force F acting along a curve. Here’s an intuitive procedure for arriving at such an integral. Describe the curve by its position vector, say r=xi+yj+zk. It seems reasonable to say that for a force F that produces a small displacement dr the amount of work done is the dot product dW=F-dr. The total work done in moving the particle from point A to point B is obtained by adding all these small amounts of work. We indicate the summation process by the integral symbol and write B W=J F-dr. (14.1) A This notation resembles that in (13.2), but the symbol in (14.1) is a new kind of integral. It is called a line integral or a contour integral because the integration takes place along a curve joining A and B. Such an integral can be defined in terms of ordinary integrals 14.1 FORMS OF ENERGY 269 of the type we are familiar with. To do this, we consider the position vector as a function of time t and write r = r(t). The initial point A of the curve corresponds to some value of I, say t = a, so A = r(a). Similarly, B = r(b) for some time t = b. We can interpret the dot product F 'dr to mean (F ' dr/dt) dt. This suggests that we simply define the line integral IRF ' dr by the equation B [7 dl' F- = I F ' - dt. 14.2 [A (11' a dt ( ) The integral on the right-hand side is our common garden variety of integral over an interval from t = a to t = b. It incorporates the curve as well as the force because in the integrand on the right-hand side the derivative dr/dt is to be evaluated at time t and the force F is to be evaluated at the position r(t). Thus the integrand is the following function of t: d F[r(t)] - 3:. Equation (14.2) is taken to be the definition of the line integral f2 F - dr, and then Eq. (14.1) is used to define the work done by F. If the force causes the acceleration dv/dt we can use this integral to prove that the work done is equal to the change in kinetic energy, the same result we obtained in Chapter 13 for linear motion. Using Newton’s second law we replace F by m dv/dt in the integrand in (14.2) and dr/dt by v to obtain B b IF-dr=Jm‘~1!-vdt. A a d! The dot product dv/dt ~ v is related to the speed v. Recalling that v2 = v - v we know that d 2 d dv dv dv __ :_ . : ._+__.. =2 .__., dt(v) dz(v V) v dt dt v v dt andhence dv _1d 2 dt V‘zdzm' Substituting this into the last integral and denoting the work done by WAB, we get b d l WAB = J; gt (2 m'Uz) (1’. Since the integrand is the derivative of %mv2(t) we can evaluate the integral by the second fundamental theorem of calculus to obtain WAB = émv2(b) — émv2(a). 270 ENERGY AND STABILITY The scalar quantity K = émvzm is called the kinetic energy of the particle at time t, and we have just shown that the total work done by F is the change in kinetic energy, B WAB = [A F ' dl‘ : KB _ KA' (14.3) Another way of performing work on the same particle is to apply a force that is equal and opposite to another force (gravity, for example) acting on the particle. The object is displaced without accelerating since there is no net force. The work done on it goes entirely into potential energy, the energy that results from the position of the object. So, if a force F acts on a particle and if we apply an equal and opposite force —F to it, the work we do goes into a change of the object’s potential energy: B UB — UA = IA —F -dr. (14.4) We now have two expressions relating the work done on an object, one to kinetic energy, the other to potential energy. They show how external forces can increase or decrease the energy of a system. To give a body potential energy, we apply a force to balance one that already exists (like gravity). If we move the particle from point B to point A, we give it potential energy U A — UB. Suppose we then remove our applied force (e. g., drop the body we were holding), leaving only the preexisting force. It gives the body kinetic energy KB — K A as it moves the body from A to B. From (14.3) and (14.4) we see that UA—UB:KB‘KA‘ The potential energy we had given the body is turned into kinetic energy. Rearranging terms, we find the conservation law In other words, once we stop doing work that transfers energy into our system, the total energy of the system, potential plus kinetic, is conserved. The integral in (14.4) describes the change in potential energy, not the potential energy itself. The same type of integral can sometimes be used to define the potential energy itself (as will be done later), but there is a small technical difficulty that must be faced. Suppose we see a stone of mass m on a table and ask for its potential energy. One may reply that it is mgh. But from what level is h measured? From the floor? From the street level outside? Or from the center of the earth? In practice, potential energy is always calculated with respect to some arbitrary level of reference, usually the lowest level a body attains during a given discussion. The actual choice of the original level is often irrelevant because we are interested in difi’erences or changes in the potential energy between two locations and the difference is independent of the choice of origin. 14.1 FORMS OF ENERGY 271 Example 1 A small block of mass m slides down a frictionless incline, starting from rest at a height h, and around a circular track of radius a. Find the speed of the block at point C. A To find the speed 1; of the block at point C, we shall equate the total energy at points A and C. At A, the total energy is stored as potential energy, EA = mgh. At point C, the block has both kinetic and potential energy: EC = mg(2a) 1+- émvz. Setting E A = EC and solving for v, we get '0 = \/2g(h — 2a). Even in the absence of work transferring energy from or to another part of the universe, the total potential and kinetic energy of a system may not be conserved. There are forms of energy that are hidden from ordinary view. One example is heat, the disorganized motions of atoms and molecules. Heat is a kind of hidden kinetic energy. There can also be hidden potential energy. For example, the potential energy of a carbon atom in a piece of coal is far greater than that in a molecule of carbon dioxide, producedwhen the coal burns. This is chemical energy, our name for the electrical potential energy built into the structure of molecules, crystals, and so on. When coal burns, of course, the released potential energy turns directly into heat. Heat itself is not necessarily useless. For example, the intense heat in a jet engine is used to propel an airplane. The dilute heat dispersed by a ball bouncing on the floor, on the other hand, is not easy to recover. Such is the future of the energy of the universe. Eventually the world will run down, not run out of energy, as organized energy is transformed into useless thermal energy. Example 2 A 0.03-kg marble is dropped into a cylinder containing glycerin. Starting from a height of 0.15 m, the marble reaches the bottom of the cylinder with a speed of 1.4 m/s. What fraction of the marble’s initial energy is dissipated by the viscous force of the glycerin? 272 ENERGY AND STABILITY The marble starts out with potential energy mgh (h = 0.15 m) and ends up with kinetic energy émvz, where v 1.4 m/s. The difference between these is the energy dissipated: energy dissipated = mgh — émv2 = 1.5 X 10‘7- J. To find the fraction of the initial energy this amount represents, we divide the energy dissipated by mgh. The result is 0.33. Not only can we follow the transformation of energy into the future, but we also can also trace its evolution backward in time. For example, the energy you use to lift a book was, at an earlier stage, chemical energy stored in muscles. And it was whatever brand of hamburger you had for lunch that provided that chemical energy. For jet engines, the energy to fly you to a vacation resort comes from potential energy stored in the jet fuel. The fossil fuel, in turn, came from creatures that inhabited primeval forests that covered the earth millions of years ago. Their energy ultimately came from sunlight. Every last joule of energy on earth came originally from the sun, with the sole exception of that generated by nuclear reactions. But the sun itself is a cosmic nuclear reactor. Most energy starts out as nuclear energy, and all the energy that exists in the universe is a legacy to us from the instant in which the universe originated — the Big Bang. Further back than that, we can’t trace it. But the amount of primordial energy from the Big Bang is still the same as it was at that first instant, 20 billion years ago, but that energy has been converted into various forms. Questions 1. A spring is compressed and dropped into a vat of nitric acid which dissolves it. What happens to the potential energy of the spring? 2. Trace the following forms of energy back to their sources insofar as you possibly can: (a) the calories in an apple, (b) a lightbulb, (c) a wind-up wristwatch. 3. A pendulum bob is pulled aside and released. A peg is located as shown in the sketch. How high above the peg will the bob rise? 14.1 FORMS OF ENERGY 273 4. A dockworker is loading lOO-kg drums onto a truck by rolling them up a ramp. The bed of the truck is 1 m above the ground and the ramp is 2 m long. What is the smallest force that must be exerted on a drum to get it up the ramp? 5. The chemical energy stored in a certain amount of gasoline is converted into ki- netic energy as a car speeds up from 0 to 50 km/h. How does the energy to accel- erate from 50 to 100 km/h compare to that used to go from 0 to 50 km/h? The answer is not “the same amount.” 6. A child sits on a hemispherical mound of ice. Assuming the mound to be friction- less, if the child is given a slight nudge and slides off, find the angle from the horizontal at which the child leaves the mound. (Him: The child leaves the mound when the normal force is zero.) 7. A penny given an initial speed of 1.5 m/s slides a distance of 25 cm across a hori- zontal table before stopping. Find the coefficient of friction between the penny and tabletop. 8. For the track in Example 1, determine the minimum height h can be (in terms of the radius a) so that the block barely makes it to point C. (Hint: Use the condition that if the block barely makes it to point C, the normal force exerted on the block by the track at that point will be zero.) 9. A toy car of mass 0.02 kg travels along the frictionless track shown below. The spring (k = 3.0 X 102 N/m) is initially compressed 0.05 m. i... o D (a) How much work is done by the normal force as the car moves from B to C? (b) What is the speed of the car at C? (c) If at D a small parachute is released, bringing the car to rest, how much energy is dissipated by air resistance? 10. Suppose you are driving along a fog—shrouded road at a speed 11, and suddenly a brick wall appears a distance R in front of you. Would it be better to slam on the 274 ENERGY AND STABILITY brakes (and hope that you stop before reaching the wall) or swerve in a circular arc of radius R to avoid the wall? To help you decide, compare the force neces- sary to change all the car’s kinetic energy entirely into work done against friction with that necessary to turn it in a circle. 14.2 GRAVITATIONAL POTENTIAL ENERGY A swinging pendulum, a plunging roller coaster, and a vibrating guitar string are all examples of potential energy changing into kinetic energy, and back into potential energy, and so on in an energy volley. In each of these examples, the system is endowed with potential energy by changing its position — pulling aside a pendulum, raising a roller coaster, plucking a guitar string. And once the system is released, the volley begins as potential energy is converted back and forth into kinetic energy. In all cases, all the energy is eventually transformed into heat, but it hasn’t been totally useless; it may have told someone the time, provided a thrill, or pleased an ear. It is also possible to get a system started by giving it kinetic energy. A baseball thrown by a major league outfielder, an arrow fired from a hunter’s bow, and a planetary probe launched by NASA are examples of giving a system kinetic energy and allowing that energy to change into gravitational potential energy. For a rocket, however, we need to think more about its potential energy; it isn’t mgh. Why? In Chapter 13 we assumed that the force of gravity on an object of mass m is simply mg, but we know (from Chapter 8) that this is only true near the surface of the earth. Newton’s universal law of gravity expresses the general case: mM r2 where all distances are measured from the center of the earth. So we need to consider the change in the force of gravity as the rocket moves away from the earth. Let’s calculate the work done in sending a rocket from the surface of the earth; where r = Re, to some distance Rf far away. First, to lift the rocket we must apply a force equal and opposite to the force of gravity given by Eq. (8.1). For simplicity, let’s assume that we send the rocket straight out, in a radial direction. The force — F and displacement will be in the same direction as shown in Fig. 14.2. Since this is linear motion the work we do is F=—G e r, (8.1) Figure 14.2 Work done to move a mass m from Re to Rf. 14.2 GRAVITATIONAL POTENTIAL ENERGY 275 rr Rf W =J’ —F - dr =f —Fdr. (13.3) Substituting for F, which is a function of r, we have W— RmeMed — G M mg _ Rc r2 r _ m 6 Re r2. Knowing that an antiderivative of Ur2 is — l/r, we can evaluate the integral and obtain 1 1 W = GmMe — — — . (Re Rf) Since l/Re is bigger than l/Rf, the work done is a positive quantity, which it had to be because we needed to supply energy to get the rocket to Rf. This last result holds if the motion is along any path from the surface of the earth to any point in space a distance Rf away. We shall prove this in Example 3, but to get an idea of why this is so, consider a special path from distance Re to distance Rf which consists of steps which alternate along radial lines and circular arcs (perpendicular to the radial lines) as shown in Fig. 14.3. Figure 14.3 A path consisting of radial lines and circular arcs from the surface of the earth to any point in space. Since the force F is always radial, no work is done along the circular arcs because F is perpendicular to the displacement there. So the total work done is obtained by adding work done in the radial directions alone, and this is the same result we got before, 1 l W = GmMe <— — —>. Re Rf According to our bookkeeping, the work we do on the rocket increases its potential energy. The potential energy is higher at Rf than it would be on the surface of the earth by an amount AU = W, or (14.5) 276 ENERGY AND STABILITY By conservation of energy, if the rocket stopped at Rf and fell back to the earth, it would arrive with a kinetic energy émvz equal to AU. We’ve ignored one thing, though: in actual rocket propulsion, the mass of a rocket changes as it burns up fuel. Example 3 Prove that 'the work done on the rocket is independent of the path from the surface of the earth to any point in space at distance Rf. Let’s say the rocket moves along a path with position vector r(t) at time t. At time t = 0 it is at the surface of the earth, so R6 = r(0) = |r(0)|. At some later time II it is at a distance Rf, so Rf = r(t1) = Ir(t1)|. The work we do to send the rocket from r(0) to r(t1) is the line integral 1‘01) I: d W: -$-m=J-4-im r(0) where F is given by (8.1). The position vector satisfies 1' = ri‘, so its derivative is dr df' dr _ = _ + _ m rm- wr Taking the dot product of this with —F we get d_r di' dr A —F' 617 ——rF'E—EF'I‘. But I", being a unit vector, has constant length so it is always perpendicular to its derivative df/dt. Since —F has the same direction as I", it, too, is perpendicular to df/dt, so F - df/dt = 0 and the foregoing equation becomes d_r dr _GmM d_r —F- =——F- d_r dt r2 dt’ where we have used (8.1) for F. Therefore the integral for work becomes " GmMe dr ’1 d —GmM W j _W_ [— _._e r2 0dt< r > The integrand is now a derivative so we can evaluate the integral by the second fundamental theorem to obtain 1 1 1 1 W‘ GmMe<n<TW> ‘ “Mia—1;) the same formula we got for linear motion. Example 4 A ballistic missile is fired vertically from the earth with a speed of 9.0 km/s. Neglecting atmospheric friction, how far above the earth’s surface will it rise? (Use R8 = 6400 km, Me = 6.0 x 1024 kg.) 14.2 GRAVITATIONAL POTENTIAL ENERGY 277 As the projectile travels upward, its kinetic energy is converted into potential energy. Using Eq. (14.5) for the potential energy it gains, energy conservation implies lmvz=GmM i—i 2 6Re R,‘ Solving for Rf, we get —1 1 v2 R=—- =l.8X104k. ‘ (Re 20M) m The distance above the earth’s surface, Rf — Re, is 1.2 X 104 km, almost two Earth radii. In early science-fiction stories, like those of Jules Verne and H. G. Wells, human beings ventured into space, not in sleek rocket ships, but in shells fired out of colossal cannons. You might wonder, as these writers certainly did, is it really possible to release humanity from the shackles of the earth’s gravity in such a way? How fast would a shell need to be fired to escape from the earth and ne...
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