Chs 13 & 14

Chs 13 & 14 - I said, “Why not let him see if...

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Unformatted text preview: I said, “Why not let him see if hrough a large angle?“ I may rot believe that they would be, vas a very fast massive particle, 1 you could show that if the tted effect of a number of small article’s being scattered back- :membered two or three days It excitement and saying, “We the a-particles coming back- incredible event that has ever : almost as incredible as if you tissue paper and it came back [ realized that this scattering at single collision, and when I as impossible to get anything 'ou took a system in which the atom was concentrated in a had the idea of an atom with charge. I worked out mathe- should obey, and I found that through a given angle should the scattering foil, the square 1y proportional to the fourth JCthflS were later verified by beautiful experimentsl luced some excerpts from rsden. It is interesting to :attering (> 90°) on the )mic model of 1910. This 1e negative electrons were a sphere of uniform posi- A passing alpha particle : repulsion of the positive mass of the atom.2 The :ounter was quite small. ;everal atomsrmi-ght occur :ctures by various scientists at nd W. Pagel, eds.), Cambridge to light compared to the alpha aside in a collision between PROBLEMS 617 in sufficiently thick foils, producing a net deflection which is large. For a gold foil 10"4 cm thick such as Geiger and Marsden used for some of their experiments, the Thomson theory pre- dicted that the fraction of alpha particles scattered at angles greater than 90° would be about one out of every 10””! That is tantamount to saying that it would never happen. (Recall, for the purposes of comparison, that the total number of all the electrons, protons, and neutrons in all the galaxies of the observ- able universe is only about 1080.) No wonder Rutherford was astonished when Geiger and Marsden observed for a foil of this thickness that approximately one out of every to4 alpha particles was deflected at angles greater than 90°. 13—] The circular orbits under the action of a certain central force F(r) are found all to have the same rate of sweeping out area by the radius vector, independent of the orbital radius. Determine how F varies with r. 13—2 In the Bohr model of the hydrogen atom an electron (mass m) moves in a circular orbit around an effectively stationary proton, under the central Coulomb force F(r) = —ke2/r2. (a) Obtain an expression for the speed c of the electron as a function of r. (b) Obtain an expression for the orbital angular momentum l as a function of r. (c) Introduce Bohr’s postulate (of the so-called “old quantum theory,” now superseded) that the angular momentum in a circular orbit is equal to nit/2w, where h is Planck’s constant. Obtain an expression for the permitted orbital radii. (d) Calculate the potential energy of the system from the equation I U(r) = —-/ F(r)dr no Hence find an expression for the total energy of the quantized system as a function of n. (e) For the lowest energy state of the atom (corresponding to n = 1) calculate the numerical values of the orbital radius and the energy, measured in electron volts, needed to ionize the atom. (k = 9 x109N—m2/C2;e = 1.6 x 10'19C;m = 9.1 x1o—3‘ kg; h = h/21r = 1.05 x 10'34 J-sec.) 13—3 A mass m is joined to a fixed point 0 by a string of length l. Initially the string is slack and the mass is moving with constant speed [)0 along a straight line. At its closest approach the distance of the mass from 0 is It. When the mass reaches a distance 1 from 0, the string becomes taut and the mass goes into a circular path around 0. Find the ratio of the final kinetic energy of the mass to its initial kinetic energy. (Neglect any effects of gravity.) Where did the energy go? 13—4 A particle A, of mass m, is acted on by the gravitational force from a second particle, B, which remains fixed at the origin. Initially, when A is very far from B (r = 00), A has a velocity v0 directed along the line shown in the figure. The perpendicular distance between B and this line is D. The particle A is deflected from its initial course by B and moves along the trajectory shown in the figure. The shortest distance betWeen this trajectory and B is found to be d. Deduce the mass of B in terms of the quantities given and the gravitational con- stant G. i Trajectory 13—5 A particle of mass m moves in the field of a repulsive central force Ant/r”, where A is a constant. At a very large distance from the force center the particle has speed 1:0 and its impact parameter is b. Show that the closest m comes to the center of force is given by ‘) ‘) rtnin = (b2 'l‘ A/UO")l/' 13—6 A nonrotating, spherical planet with no atmosphere has mass M and radius R. A particle is fired off from the surfaCe with a speed equal to three quarters of the escape speed. By considering conserva— tion of total energy and angular momentum, calculate the farthest distance that it reaches (measured from the center of the planet) il~ it is fired oil (a) radially and (b) tangentially. Sketch the ell‘eetive po— tential-energy curve, given by for case (b). Draw the line representing the total energy of the motion, and thus verify your result. 13—7 imagine a spherical, nonrotating planet of mass M, radius R, that has no atmosphere. A satellite is tired from the surface ol‘ the planet with speed (In at 30° to the local vertical. In its subsequent orbit the satellite reaches a maximum distance of 5Ry'2 from the center ig with constant speed :h the distance of the listance I from 0, the 'cular path around 0. he mass to its initial Where did the energy he gravitational force t the origin. Initially, tcity v0 directed along r distance between B 'rom its initial course e figure. The shortest to be d. Deduce the ,he gravitational con- 'ajec tory )f a repulsive central 'ge distance from the pact parameter is b. rce is given by tmosphere has mass surface with a speed onsidering conserva— Ilculate the farthest r of the planet) if it tch the effective po- 1ergy of the motion, mass M, radius R, l the surface of the In its subsequent R, '2 from the center 619 of the planet. Using the principles of conservation of energy and angular momentum, show that 00 = (SGM/4R)“2 13—8 A particle moves under the influence of a central attractive force, —k/r3. At a very large (effectively infinite) distance away, it has a nonzero velocity that does not point toward the center. Con- struct the effective potential-energy diagram for the radial component of the motion. What conclusions can you draw about the dependence on r of the radial component of velocity? 13—9 A satellite in a circular orbit around the earth fires a small rocket. Without going into detailedcalculations, consider how the orbit is changed according to whether the rocket is fired (a) forward; (b) backward; (c) toward the earth; and (d) perpendicular to the plane of the orbit. 13—10 Two spacecraft are coasting in exactly the same circular orbit around the earth, but one is a few hundred yards behind the other. An astronaut in the rear wants to throw a ham sandwich to his partner in the other craft. How can he do it? Qualitatively describe the various possible paths of transfer open to him. (This question was posed by Dr. Lee DuBridge in an after-dinner speech to the American Physical Society on April 27, 1960.) 13—] I The elliptical orbit of an earth satellite has major axis 2a and minor axis 2b. The distance between the earth’s center and the other focus is 2c. The period is T. (a) Verify that b = (a2 — 02)“2. (b) Consider the satellite at perigee (r; a — c) and apogee (r2 = a + c). At these two points its velocity vector and its radius vector are at right angles. Verify that conservation of energy implies that 2 GMm lmv] 2 GMm 2 ._ 1 =—mt:2 —————=E (1-0 2 a+c Verify also that conservation of angular momentum implies that b E = %(a - c)vi = %(a + c)02 T (c) From the above relationships, deduce the following results, corresponding to Eqs. (13—36) and (13—39) in the text: T2 = 41r2a3/GM and = -—GMm/2a 13—12 A satellite of mass m is in an elliptical orbit about the earth. When the satellite is at its perigee, a distance R0 from the center of Eriroitéiems the earth, it is traveling with a speed v0. The mass of the earth, M, is much greater than m. (a) If the length of the major axis of the elliptical orbit is 4R0, what is the speed of the satellite at its apogee (the maximum distance from the earth) in terms of G, M, and R0? (b) Show that the length of the minor axis of the elliptical orbit is 2V5 R0, and find the period of the satellite in terms of Do and R0. 13—13 A satellite of mass m is traveling at speed we in a circular orbit of radius ro under the gravitational force of a fixed mass at 0. ' (a) Taking the potential energy to be zero at r = co , show that the total mechanical energy of the satellite is -—%mvo2. (b) At a certain point B in the orbit (see the figure) the direction of motion of the satellite is suddenly changed without any change in the magnitude of the velocity. As a result the satellite goes into an elliptic orbit. Its closest distance of approach to 0 (at point P) is now ro/5. What is the speed of the satellite at P, expressed as a multiple of 00? (c) Through what angle a (see the figure) was the velocity of the satellite turned at the point B? 13—14 A small satellite is in a circular orbit of radius r1 around the earth. The direction of the satellite‘s velocity is now changed, causing it to move in an elliptical orbit around the earth. The change in velocity is made in such a manner that the satellite loses half its orbital angular momentum, but its total energy remains unchanged. Cal- culate, in terms of r1, the perigee and apogee distances of the new orbit (measured with respect to the earth's center). 13—15 An experimental rocket is fired from Cape Kennedy with an initial speed v0 and angle 0 to the horizontal (see the figure). Neglect- ing air friction and the earth’s rotational motion, calculate the maxi- mum distance from the center of the earth that the rocket achieves in terms of the earth’s mass and radius (M and R), the gravitational constant G, and 6 and v0. 13—16 A satellite of mass m is traveling in a perfectly circular orbit of radius r about the earth (mass M). An explosion breaks up the satellite into two equal fragments, each of mass m/2. Immediately mass of the earth, M, is ie elliptical orbit is 4R0, (the maximum distance xis of the elliptical orbit in terms of 00 and R0. ed 120 in a circular orbit fixed mass at 0. r0 at r = 00 , show that -%mv02. > V the figure) the direction without any change in satellite goes into an ) 0 (at point P) is now xpressed as a multiple 'e) was the velocity of ' radius r] around the now changed, causing arth. The change in te loses half its orbital ins unchanged. Cal- distances of the new r). tpe Kennedy with an : the figure). Neglect- n, calculate the maxi- the rocket achieves in R), the gravitational :rfectly circular orbit losion breaks up the s m/2. Immediately after the explosion the two fragments have radial components of velocity equal to 120/2, where no is the orbital speed of the satellite prior to the explosion; in the reference frame of the satellite at the instant of the explosion the fragments appear to separate along the line joining the satellite to the center of the earth. (a) In terms of G, M, m, and r, what are the energy and the angular momentum (with respect to the earth’s center) of each frag- ment? (b) Make a sketch showing the original circular orbit and the orbits of the two fragments. In making the sketch, use the fact that the major axis of the elliptic orbit of a satellite is inversely proportional to the total energy. 13—17 A spaceship is in an elliptical orbit around the earth. It has a certain amount of fuel for orbit alteration. Where in the orbit should this fuel be used to attain the greatest distance from earth? Do you notice any similarity between this problem and the one con— cerning a rocket ignited after falling down a chute (Problem 10~1 3)? 13—18 The commander of a spaceship that has shut down its engines and is coasting near a strange-appearing gas cloud notes that the ship is following a circular path that will lead directly into the cloud (see the figure). He also deduces from the ship’s motion that its angular momentum with respect to the cloud is not changing. What attractive (central) force could account for such an orbit? Spaceship \ 13—19 (a) Make an analysis of an earth-to-Mars orbit transfer similar to that carried out in the text for the transfer to Venus. Assume that earth and Mars are in circular orbits of radii l and 1.52 AU, re- spectively. (b) In part (a), and in the discussion in the text, the gravita-' tional fields of the planets are neglected. (The problem was taken to be simply that of shifting from one orbit to another, not from the surface of one planet to the surface of the other.) At what distance from the earth is the earth’s field equal in magnitude to that of the sun? Similarly, at what distance from Mars is the sun’s field equaled by that of the planet? Further, compare the work done against the sun’s gravity in the transfer with that done against the earth’s gravity, and with the energy gained from the gravitational field of Mars. 13—20 The problem of dropping a spacecraft into the sun from the earth’s orbit with the application of minimum possible impulse (given to the spacecraft by firing a rocket engine) is not solved by firing the rocket in a direction opposite to the earth's orbital motion so as to reduce the velocity of the spacecraft to zero. A two-step process can accomplish the goal with a smaller rocket. Assume the initial orbit to be a circle of radius r; with the sun at the center (see the figure). By means of a brief rocket burn the spacecraft is Speeded up tangentially in the direction of the orbit velocity, so that it assumes an elliptical orbit whose perihelion coincides with the firing point. At the aphelion of this orbit the spacecraft is given a backward impulse sufficient to reduce its space velocity to zero, so that it will subsequently fall into the sun. (As in the previous transfer problem, the effects of the earth‘s gravity are neglected.) (a) For a given value of the aphelion distance, r2 of the space- craft, calculate the required increment of speed given to it at first firing. (b) Find the speed of the spacecraft at its aphelion distance, and so find the sum of the speed increments that must be given to the spacecraft in the two steps to make it fall into the sun. This sum pro- vides a measure of the total impulse that the-rocket engine must be able to supply. Compare this sum with the speed of the spacecraft in its initial earth orbit for the case r2 = 10r1. [Notes This problem is discussed by E. Feenberg, “Orbit to the Sun," Am. J. Phys., 28, 497 (1960).] 13—2] The sun loses mass at the rate of about 4 X 106 tons/sec. What change in the length of the year should this have produced within the span of recorded history (~5000 yr)? Note that the equa- tion for circular motion can be employed (even though the earth spirals away from the sun) because the fractional yearly change in radius is so small. The other condition needed to describe the gradual shift is the over-all conservation of angular momentum about the CM of the system. (This problem was given in a simplified form as Prob- lem 8—19.) ,he sun’s field equaled done against the sun’s 1e earth’s gravity, and 1d of Mars. nto the sun from the ossible impulse (given at solved by firing the 'bital motion so as to . two—step process can me the initial orbit to er (see the figure). By peeded up tangentially t assumes an elliptical Joint. At the aphelion l impulse sufficient to subsequently fall into 1e effects of the earth’s tance, r2 of the space- iven to it at first firing. its aphelion distance, t must be given to the 1e sun. This sum pro- ‘ocket engine must be ed of the spacecraft in g, “Orbit to the Sun,” out 4 X 106tons/sec. d this have produced ? Note that the equa— ven though the earth anal yearly change in :0 describe the gradual nentum ab0ut the CM nplified form as Prob- 623 13—22 A particle of mass m moves about a massive center of force C, with the attraction given by —f(r)e,, where r is the position of the particle as measured from C. If the particle is also subjected to a retarding force —)\v, and initially has angular momentum Lo about C, find its angular momentum as a function of time. 13—23 Consider a central force in a horizontal plane given by F(r) = —kr, where k is a constant. (This provides a good description, for example, of the pendulum encountered in the laboratory. Rarely is a pendulum physically confined to swing in only one vertical plane.) (a) A particle of mass m is moving under the influence of such a force. Initially the particle has position vector re and velocity vo as measured from the stationary force center. Set up a Cartesian co- ordinate system with the xy plane containing re and v0, and find the time dependence of the position (x, y) of the particle. Does the orbit correspond to any particular geometric curve? (Keep in mind the differences between this interaction and the gravitational problem.) What physical quantities are conserved? (b) Suppose the particle is originally in a circular orbit of radius R. What is its orbital speed? If at some point its velocity is doubled, what will be the maximum value of r in its subsequent motion? 13—24 According to general relativity theory, the gravitational po- tential energy of a mass m orbiting about a mass M is modified by the addition of a term —GMmC2/c2r3, where C = r2 dB/dt and c is the speed of light. Thus the period of a circular orbit of radius r is slightly smaller than would be predicted by Newtonian theory. (a) Show that the fractional change in the period of a circular orbit of radius r due to this relativistic term is —(127r2r2/C2To2), where To is the period predicted by Newtonian theory. (b) Since, by Kepler's third law, we have T 02 ~ r3, the effect of this relativistic correction is greatest for the planet closest to the sun, i.e., Mercury. Consider the effect of the relativistic term on the radial and angular periods, and see if you can thereby arrive at the famous result that the perihelion of Mercury’s orbit precesses at the rate of about 43 seconds of are per century. You may find it useful to refer back to Problem 8—20, which also deals with this question. 13—25 A beam of atoms traveling in the positive x direction and passing through a medium containing n particles per unit volume suffers an attenuation given by dN(x) dx = -—'AnN(x) where A is the cross section for scattering of an atom in the beam by ii’méréienzs an atom of the medium. Therefore, if the beam contains No atoms at x = 0, the number still traveling in the beam at x is just N(x) = Noe—An: (a) Set up a simple model of beam attenuation that gives the results stated above. (b) The graph summarizes a set of measurements of the at- tenuation of a beam of potassium atoms by argon gas at various pressures (the pressures are given in millimeters of mercury; the temperature is 0°C throughout). (These data are from the film “The Size of Atoms from an Atomic Beam Experiment,” by J. G. King, Education Development Center, Newton, Mass.I 1961.) Deduce the cross section for the scattering of a potassium atom by an argon atom. (1 cm3 of a perfect gas at STP contains 2.7 X 1019 molecules.) Check Whether theresults for difl‘erent values of 'the'pressure'a’gre'e.’ (c) If the potassium and argon atoms are visualized simply as hard spheres of radii rK and rA, respectively, what is implied about rK and rA by the result of part (b)? 13—26 (a) In the Rutherford scattering problem one can calculate a distance of closest approach do for alpha particles of a given energy approaching a nucleus head on. Verify that do is given by do = quiqz/mvoz. (b) The force of repulsion between two protons, separated by 10"14 m, is 2.3 N. Use this to deduce the value of do for alpha par- ticles (charge 2e) of kinetic energy 5 MeV approaching nuclei of gold (charge 79e). (c) By introducing do, the expression for the fraction of in- cident alpha particles scattered into d9 at go becomes seam contains No atoms at beam at x is just N(x) = attenuation that gives the f measurements of the at- ; by argon gas at various llimeters of mercury; the ata are from the film “The periment,” by J. G. King, Mass., 1961.) Deduce the lm atom by an argon atom. X 1019 molecules.) Check ie pressure agree. ns are visualized simply as ely, what is implied about oblem one can calculate a particles of a given energy that do is given by do = two protons, separated by value of do for alpha par- approaching nuclei of gold ion for the fraction of in- , becomes d9 Sin“ (so/2) where n is the number of nuclei per unit volume and As is the length of the path through the foil. Putting d9 = 21r sin g0 dgo, show that the fraction of alpha particles scattered through angles _>_ 900 is given by _ 1 2 df—16(nAs)d0 1r — n As (1'02 cot2 $2 > = f(_s0o) 4 2 (d) A foil of gold leaf 10—4 cm thick is bombarded with alpha particles of energy 5 MeV. Out of 1 million alpha particles, incident normally on the foil, how many would be deflected through 90° or more? ODensity of gold = 1.9 X 10" kg/m3; atomic weight = 197.) PROBLEMS 6 x 10"""’m3 4 x 1033m3 M... m 7 x 1022 kg, r...3 ~ 2 x 1030 kg, r? 22 S 2 22 Hence 9 z (2.5 X 10—9)(l.7 X 10—3) z 4 X lO—lzsec—l and so T = Z5 z 1.5 X 1012 sec z 50,000 years We could try to trim this result a little~for example, we have someWhat underestimated the mean density of the earth’s crust and (by treating the earth as a uniform sphere) we have some- what overestimated the moment of inertia. Both of these would cause us to underestimate the precessional rate and obtain too large a value for the precessional period. But in view of the gross assumptions we have made elsewhere in the calculation, we should not set any great store on making small refinements of this type. The important thing is that, by quite simple means, we have verified that the precession of the equinoxes can indeed be understood in terms of Newtonian dynamical principles. But Newton got there first! 14—] (a) Devise a criterion for whether there is external force acting on a system of two particles. Use this criterion on the following one- dimensional system. A particle of mass m is observed to follow the path x(t) = A sin (wt) + L + 1;! The other particle, of mass M, follows X(t) = Bsin (wt) + Vt The different constants are arbitrary except that mA = — MB. (b) Try it on the system with x(t) = A sin (wt) and X(t) = Bsin (wt + (p) where A and B are related as before, and go # 0. >< 1025m3 X 1033 m3 4 X 10~12 sec”1 )0 years .ittle—for example, we have density of the earth’s crust "orm sphere) we have some- nertia. Both of these would :ssional rate and obtain too )eriod. But in view of the lsewhere in the calculation, n making small refinements that, by quite simple means, of the equinoxes can indeed 1 dynamical principles. But r there is external force acting 'riterion on the following one- s m is observed to follow the :pt that mA = —MB. = Bsin (0)! + (p) l<p5£0. 14—2 Consider a system of three particles, each of mass m, which remain always in the same plane. The particles interact among them- selves, always in a manner consistent with Newton’s third law. If the particles A, B, and C have positions at various times as given in the table, determine whether any external forces are acting on the system. Time A B C 0 (1, 1) (2, 2) (3, 3) 1 (1, 0) (0, 1) (3, 3) 2 (O, 1) (1, 2) (2, 0) 14—3 Two skaters, each of mass 70 kg, skate at speeds of 4 m/sec in opposite directions along parallel lines 1.5 m apart. As they are about to pass one another they join hands and go into circular paths about their common center of mass. (a) What is their total angular momentum? (b) A third skater is skating at 2 m/sec along a line parallel to the initial directions of the other two and 6 m off to the side of the track of the nearer one. From his standpoint, what is the total angular momentum of the other two skaters as they rotate? 14—4 A molecule of carbon monoxide (CO) is moving along in a straight line with a kinetic energy equal to the value of AT at room temperature (A- = Boltzmann’s constant = 1.38 X l0‘23 J/°K). The molecule is also rotating about its center of mass with a total angular momentum equal to h (= 1.05 X 10—34 J-sec). The internuclear distance in the C0 molecule is 1.1 A. Compare the kinetic energy of its rotational motion with its kinetic energy of translation. What does this result suggest about the ease or difficulty of exciting such rotational motion in a gas of C0 molecules at room temperature? 14—5 A uniform disk of mass Mand radius R is rotating freely about a vertical axis with initial angular velocity w”. Then sand is poured onto the disk in a thin stream so that it piles up on the disk at the radius r (< R). The sand is added at the constant rate [.1 (mass per unit time). (a) At what rate are the angular velocity and the rotational kinetic energy varying with time at a given instant? (b) After what length of time is the rotational kinetic energy reduced to half of its initial value? What has happened to this energy? 14—6 Two men, each of mass 100 kg, stand at opposite ends of the diameter of a rotating turntable of mass 200 kg and radius 3 m. In- itially the turntable makes one revolution every 2 sec. The two men make their way to the middle of the turntable at equal rates. (a) Calculate the final rate of revolution and the factor by which the kinetic energy of rotation has been increased. (b) Analyze, at least qualitatively, the means by which the in- crease of rotational kinetic energy occurs. (c) At what radial distance from the axis of rotation do the men experience the greatest centrifugal force as they make their way to the center? 14—7 Estimate the kinetic energy in a hurricane. Take the density of air as 1 kg/m3. 14—8 A useful way of calculating the approximate value of the moment of inertia of a continuous object is to consider the object as if it were built up of concentrated masses, and to calculate the value of Zmr2_ As an example, take the case of a long uniform bar of mass M and lengthrL (with its transverse dimensions much less than L). We know that its moment of inertia about one end is ML2/3. (a) The most primitive approximation is to consider the total mass M to be concentrated at the midpoint, distant L/2 from the end. You will not be surprised to find that this is a poor approximation. (b) Next, treat the bar as being made up of two masses, each equal to M/2, at distances L/4 and 3L/4 from one end. (c) Examine the improvements obtained from finer subdivisions —e.g., 3 parts, 5 parts, 10 parts. 14—9 (a) Calculate the moment of inertia of a thin-walled spherical shell, of mass M and radius R, about an axis passing through its center. Consider the shell as a set of rings defined by the amounts of material lying within angular ranges dB at the various angles 0 to the axis (see the figure). (b) Verify the formula I = 2MR2/5 for the moment of inertia of a solid sphere of uniform density about an axis through its center. You can proceed just as in part (a), except that the system is to be regarded as a stack of circular disks instead of rings. 14—10 (a) Calculate the moment of inertia of a thin square plate about an axis through its center perpendicular to its plane. (Use the per- pendicular-axis theorem.) - (b) Making appropriate use of the theorems of parallel and perpendicular axes, calculate the moment of inertia of a hollow cubical box about an axis passing through the centers of two opposite faces. (c) Using the result of (a), deduce the moment of inertia of a uniform, solid cube about an axis passing through the midpoints of tw0 opposite faces. (d) For a cube of mass M and edge a, you should have obtained the result Ma2/6. It is noteworthy that the moment of inertia has this same value about any axis passing through the center of the cube. See how far you can go toward verifying this result, perhaps by con- sidering other special axes—e.g., an axis through diagonally opposite corners of the cube or an axis through the midpoints of opposite edges. increased. y, the means by which the in- rs. n the axis of rotation do the force as they make their way iurricane. Take the density of proximate value of the moment onsider the object as if it were u calculate the value of Zmrr". ; uniform bar of mass M and much less than L). We know id is ML2/3. 1ation is to consider the total )int, distant L/2 from the end. this is a poor approximation. made up of two masses, each ‘4 from one end. )tained from finer subdivisions rtia of a thin-walled spherical axis passing through its center. ed by the amounts of material various angles 0 to the axis 2/5 for the moment of inertia am an axis through its center. :cept that the system is to be :ead of rings. ia of a thin square plate about r to its plane. (Use the per- the theorems of parallel and t of inertia of a hollow cubical centers of two opposite faces. :e the moment of inertia of a ing through the midpoints of ;e a, you should have obtained the moment of inertia has this ough the center of the cube. g this result, perhaps by con- ; through diagonally opposite e midpoints of opposite edges. 14—11 Refer to Fig. 11—19(a), which shows the variation of density with radial distance inside the earth. Using this graph, compare the moment of inertia of the earth about its axis with the moment of inertia of a sphere of the same mass and radius but of uniform density. You can do quite well by considering the earth to be made up of a central core and two thick concentric shells, each of approximately uniform density. The boundaries between these three regions correspond to the abrupt changes of density shown in the graph. [Alternatively, consider the earth as built up of three superposed solid spheres—a basic one, occupying the whole volume of the earth, with the density of the outer- most region (r > 0.54RE) and two other spheres with densities corre- sponding to the mean density differences between the successive regions] 14—12 (a) A hoop of mass WM and radius R rolls down a slope that makes an angle 0 with the horizontal. This means that when the linear velocity of its center is 1; its angular velocity is v/R. Show that the kinetic energy of the rolling hoop is M122. (b) There is a traditional story about the camper—physicist who has a can of bouillon and a can of beans, but the labels have come off, so he lets them roll down a board to discover which is which. What would you expect to happen? Does the method work? (Try it!) 14—13 A skier is enjoying the mountain air while standing on a 30° snow slope when he suddenly notices a huge snowball rolling down at him. By the time he notices the ball, it is only 100 m away and is traveling at 25 m/sec. The skier gives himself a speed of 10 m/sec almost instantaneously and proceeds to accelerate down the slope at g sin 0 (= g/2). Does he get away? (Assume that the snowball has a constant acceleration corresponding to that of a sphere of given radius rolling, without slipping, down the slope. Assume that the moment of inertia of the snowball about an axis through its center is 2MR2/5.) 14—14 The preceding problem suggests another one. If an object is rolling down a slope, gathering material as it goes, how‘does its acceler- ation compare, in fact, with a similar object that is not adding material in this way? To give yourself a relatively straightforward situation to consider, take the case of a cylinder, rather than a sphere, that grows in size as it rolls. Make whatever assumptions seem reasonable about the way in which the rate of increase in radius depends on the existing radius, R, and on the instantaneous speed, 12. 14—15 Two masses, of 9 kg and 1 kg, hang from the ends of a string that passes around a pulley of mass 40 kg and radius 0.5 m (I = lMR2) as shown in the diagram. The system is released from rest and the 9-kg mass drops, starting the pulley rotating. (a) What is the acceleration of the 9-kg mass? - (b) What is the angular velocity of 'the pulley after the 9-kg mass has dropped 2 m? (c) What is the tension in the part of the string which is between the pulley and the 9-kg mass? Between the pulley and the l-kg mass? (d) If the coefificient of friction between the string and the pulley is 0.2, what is the least number of turns that the string must make around the pulley to prevent slipping? (Cf. Problem 5—14.) 14—16 An amusement park has a downhill racetrack in which the competitors ride down a 30° slope on small carts. Each cart has four wheels, each of mass 20 kg and diameter 1 m. The frame of each cart has a mass of 20 kg. (a) What is the acceleration of a cart if its rider has a mass of 50 kg? (Assume that the moment of inertia of a wheel is given by 0.8MR3, where R is its radius.) (b) If two riders, of masses 50 and 60 kg, respectively, start ofl’ simultaneously, what is the distance between them when the winner passes the finishing line 60 m down the slope? 14—17 A uniform rod of length 31) swings as a pendulum about a pivot a distance x from one end. For what value(s) of x does this pendulum have the same period as a simple pendulum of length 21;? 14—18 (a) A piece of putty of mass m is stuck very near the rim of a uniform disk of mass 2m and radius R. The disk is set on edge on a table on which it can roll without slipping. The equilibrium position is obviously that in which the piece of putty is closest to the table. Find the period of small-amplitude oscillations about this position and the length of the equivalent simple pendulum. (b) A circular hoop hangs over a nail on a wall. Find the period of its small—amplitude oscillations and the length of the equiva- lent simple pendulum. (In these and similar problems, use the equation of conservation of energy as a starting point. The more complicated the system, the greater is the advantage that this method has over a direct application of Newton’s law.) 14—19 A uniform cylinder of mass M and radius R can rotate about a shaft but is restrained by a spiral spring (like the balance wheel of your watch). When the cylinder is turned through an angle 0 from its equilibrium position, the spring exerts a restoring torque M equal to —c0. Set up an equation for the angular oscillations of this system and find the period, T. 14—20 Assuming that you let your legs swing more or less like rigid pendulums, estimate the approximate time of one stride. Hence estimate your comfortable walking speed in miles per hour. How does it compare with your actual pace? 14—2] A torsion balance to measure the momentum of electrons con- sists of a rectangular vane of thin aluminum foil, 10 by 2 by 0.005 cm, attached to a very thin vertical fiber, as shown. The period of torsional oscillation is 20 sec, and the density of aluminum is 2.7 times that of water. :n the pulley and the l-kg mass? between the string and the pulley urns that the string must make .’ (Cf. Problem 5—14.) 3wnhill racetrack in which the small carts. Each cart has four ter 1 m. The frame of each cart ‘a cart if its rider has a mass of f inertia of a wheel is given by and 60 kg, respectively, start off between them when the winner e slope? :wings as a pendulum about a )r what value(s) of x does this simple pendulum of length 2b? is stuck very near the rim of a E. The disk is set on edge on a ping. The equilibrium position 3f putty is closest to the table. oscillations about this position e pendulum. :r a nail on a wall. Find the ns and the length of the equiva- he equation of conservation of ‘e complicated the system, the 0d has over a direct application and radius R can rotate about ring (like the balance wheel of 1ed through an angle 0 from its a restoring torque M equal to .ular oscillations of this system 5 swing more or less like rigid : time of one stride. Hence :d in miles per hour. How does 1e momentum of electrons con- inum foil, 10 by 2 by 0.005 cm, shown. The period of torsional ' aluminum is 2.7 times that of Electron— beam spot (a) What is the torsion constant of the suspension, in m-N/rad ? (b) What horizontal force, applied perpendicular to the surface of the vane at a point 3 cm from the axis, will produce an angular deflectipn of, 10°? , , ,, (c) A beam of 1 mA of electrons accelerated through 500 V strikes the vane perpendicularly at a point 4 cm from the axis. What steady angular deflection is produced, assuming that the electrons are stopped in the vane? 14—22 The torsion constant of a wire or fiber of length I, and of circular cross section of radius a, is given by c = Esra4/21, where E, is an elastic constant of the material known as the shear modulus, measured in N/m2. The maximum load that can be supported by such a fiber is given by its cross-sectional area, 7ra2, multiplied by the ultimate tensile strength of the material, also measured in N/m2. For glass fibers the value of E, is about 2.5 X 10lo N/mz, and the ultimate tensile strength is about 109 N/mz. (a) Calculate the diameter of the thinnest glass fiber that can safely support two lead spheres, each of mass 20 g, in a gravity torsion balance. Allow a safety factor of about 3. (b) If the spheres are at the ends of a light bar of length 20 cm, and the length of the suspending fiber is also 20 cm, what is the period of torsional oscillation of this system? (The measurement of this period is the practical way of inferring the torsion constant of the suspension.) (c) What angular deflection of this system is produced by placing lead spheres of mass 2 kg with their centers 5 cm from the centers of the small suspended spheres? What linear displacement would this give in a spot of light reflected from a mirror on the torsion arm to a scale 5 m away? Compare this result with the figures used in Problem 5—3 on a Cavendish experiment. 14—23 A wheel of uniform thickness, of mass 10 kg and radius 10 cm, is driven by a motor through a belt (see the figure).- The drive wheel on the motor is 2 cm in radius, and the motor is capable of delivering a torque of 5 m-N. (a) Assuming that the belt does not slip on the wheel, how long does it take to accelerate the large wheel from rest up to 100 rpm? (b) If the coeflicient of friction between belt and wheel is Path of CM Drive wheel T. T2 Driven wheel 0.3, what are the tensions in the belt on the two sides of the wheel? (Assume that the belt touches the wheel over half its circumference.) . 14—24 A possible scheme for stopping the rotation of a spacecraft of radius R is to let two small masses, m, swing out at the ends of strings of length l, which are attached to the spacecraft at the points P and I" (see the figure). Initially, the masses are held at the positions shown and are rotating with the body of the spacecraft. When the masses have swung out to their maximum distance, with the strings extending radially straight out, the ends P and P’ of the strings are released from the spacecraft. For given values of m, R, and I (the moment of inertia of the spacecraft), what value of 1 will leave the spacecraft in a non- rotating state as a result of this operation? Apply the result to a spacecraft that can be regarded as a uniform disk of mass M and radius R. (Put in some numbers, too, maybe.) P P, 14—25 The technique of “pumping” a playground swing in order to increase the amplitude of its motion can be learned by example or (less easily) by trial and error. The mechanics of the procedure are not trivial. According to one model of the process, the pumping is taken to consist of a sudden elevation of the rider’s center of mass at each passing of the vertical, or low point (the rider lifts and holds himself above the seat), and a subsequent return to resting on the swing seat at each turning point (see the figure). The support ropes are assumed to be always straight, and the instantaneous changes of effective length of the “pendulum” allow conservation of angular momentum to be applied not only to the low-point pumping motions but also to those at the turning points. (a) Carry out the analysis as indicated above and show that increase of amplitude can be achieved. Note that the result agrees with the qualitative experience that any given amount of increase is more easily achieved as the amplitude increases. 3%, E: my; {been in the two sides of the wheel? el over half its circumference.) . the rotation of a spacecraft of :wing out at the ends of strings racecraft at the points P and P’ re held at the positions shown spacecraft. When the masses .nce, with the strings extending 3f the strings are released from t, and I (the moment of inertia leave the spacecraft in a non- ttion? Apply the result to a uniform disk of mass M and maybe.) I7! P, playground swing in order to an be learned by example or echanics of the procedure are f the process, the pumping is «f the rider’s center of mass at lint (the rider lifts and holds lent return to resting on the Ie figure). The support ropes the instantaneous changes of low conservation of angular e low-point pumping motions :licated above and show that Note that the result agrees ' given amount of increase is creases. 707 (b) Consider in what ways the analysis indicated above may be imperfect. Also, how well does this idealized technique match the actual pumping method that children utilize every day? [The analysis suggested above may be found in an article by P. L. Tea, Jr., and H. Falk, “Pumping on a Swing,” Am. J. Phys., 36, 1165 (1968).] 14—26 Two gear wheels, A and B, of radii RA and R B, and of moments of inertia IA and 13, respectively, are mounted on parallel shafts so that they are not quite in contact (see the figure). Both wheels can rotate completely freely on their shafts. Initially, A is rotating with angular velocity mo, and B is stationary. At a certain instant, one shaft is moved slightly so that the gear wheels engage. Find the resulting angular velocity of each in terms of the given quantities. (Warning: Do not be tempted into a glib use of angular momentum conservation. Consider the forces and torques resulting from the contact.) 14—27 A section of steel pipe of large diameter and relatively thin wall is mounted as shown on a flat—bed truck. The driver of the truck, not realizing that the pipe has not been lashed in place, starts up the truck with a constant acceleration of 0.5 g. As a result, the pipe rolls back- ward (relative to the truck bed) without slipping, and falls to the ground. The length of the truck bed is 5 m. (a) With what horizontal velocity does the pipe strike the ground? (b) What is its angular velocity at this instant? (c) How far does it skid before beginning to roll without slip- ping, if the coefficient of friction between pipe and ground is 0.3? (d) What is its linear velocity when its motion changes to rolling without slipping? 14—28 (a) How far above the center of a billiard ball or pool ball should the ball be struck (horizontally) by the cue so that it will be sure to begin rolling without slipping? (b) Analyze the consequences of striking the ball at the level of its center if the coefficient of friction between the ball and the table is p. 14—29 A man kicks sharply at the bottom end of a vertical uniform post which is stuck in the ground so that 6 ft of it are above ground. 2? z" {:3 isle 2'; Unfortunately for him the post has rotted where it enters the ground and breaks off at this point. To appreciate why “unfortunately” is the appropriate word, consider the subsequent motion of the top end of the post. 14—30 Refer to page 673 for the discussion of a rectangular board rotating about an axis in its plane. Using the notation and the method of attack of that discussion, show that the angular momentum com- ponent L’ about an axis in the plane of the board and perpendicular to w is given by combining the resolved parts of 1,0), and Iyw” in this direction; i.e., L’ = Izwz sin 6 — Iva), cos 6 14—31 A flywheel in the form of a uniform'disk of radius 5 cm is mounted on an axle that just fits along the diameter of a gimbal ring of diameter 12 cm. The flywheel is set rotating at 1000 rpm and the gimbal ring is supported at the point where one end of the axle meets it. Calculate the rate of precession in rpm. 14—32 In most cars the engine has its axis of rotation pointing fore and aft along the car. The gyroscopic properties of the engine when rotating at high speed are not negligible. Consider the tendency of this gyroscopic property to make the front end of the car rise or fall as the car follows a curve in the road. What about the corresponding effects for a car with its engine mounted transversely? Try to make some quantitative estimates of the importance of such effects. Con- sider whether a left-hand curve or a right-hand curve might involve the greater risk of losing control over the steering of the car. 14—33 See if you can pick up the challenge, given in the text, of making a more respectable calculation of the precession of the equinoxes. ...
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This note was uploaded on 04/29/2008 for the course PHYS 230 taught by Professor Harris during the Fall '07 term at McGill.

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Chs 13 &amp; 14 - I said, “Why not let him see if...

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