Physics 53 Summer Final Exam. Solutions

Transcription

1 Final Exam Solutions In questions or problems not requiring numerical answers, express the answers in terms of the symbols given, and standard constants such as g. If numbers are required, use g = 10 m/s 2. Part A: Multiple choice questions. Check the best answer. Each question has a value of 4 points. 1. Which is a correct statement of a fundamental law of mechanics? If all the external forces on a system are conservative, total mechanical energy is conserved. If the total external force on a system is zero the total angular momentum about any point is conserved. If only conservative forces do work, the total linear momentum of the system is conserved. If any component of the total external force is zero, that component of the linear momentum is conserved. 2. In simple harmonic motion in one dimension: The acceleration and the displacement are always in opposite directions. The acceleration and the velocity are always in opposite directions. The velocity and the acceleration have their greatest magnitudes at the same time. None of the above is true. 3. A billiard ball (a solid sphere of radius R and mass m) moves to the right across a rough table. It started without rotating, so it is still skidding, although eventually it will roll without slipping. Which of the following is NOT true? The torque about its CM is clockwise. Both its CM speed and its angular speed of rotation are decreasing. [Angular speed is increasing] It angular momentum about a point on the table is constant. Its total kinetic energy is decreasing. 1

2 4. Three objects (1,2,3) start from rest at the same time from the same height on an incline, and roll to the bottom without slipping. Their moments of inertia obey I 1 > I 2 > I 3. The order of speeds with which they will reach the bottom is: 1,2,3 3,2,1 1,3,2 2,3,1 5. Shown is an auto with the normal forces on the front and rear wheels indicated. When the auto is at rest N 1 is somewhat larger than N 2 because the CM of the auto is closer to the front wheels. When the auto accelerates forward: Both normal forces remain the same. Both normal forces increase. N 1 increases while N 2 decreases. N 2 increases while N 1 decreases. [Consider torques about the CM.] N 1 N 2 6. Concerning Kepler s three laws, which of the following is NOT true? One law can be used to determine the mass of a planet if it has a moon. One law says that the quantity T 2 /a 3 (T is the period, a the semi-major axis) is the same for the earth s orbit around the sun and moon s orbit around the earth. [Must orbit the same object] One law says the planetary orbits are ellipses with the sun at a focus. One law is equivalent to conservation of angular momentum of the orbit. 7. The change in internal energy of an ideal gas is given by!u = nc V!T : Only for processes at constant volume. Only for adiabatic processes. Only for isothermal processes. For any process. 2

3 8. A heat engine operates between reservoirs at 300 K and 250 K. In each cycle it takes heat 300 J from the hot reservoir. The net entropy change of the engine itself is 1 J/K per cycle. [Zero] The engine s efficiency is 1/5. [No greater than 1/6] At least 250 J of heat must be expelled to the cold reservoir. None of the above is true. Part B: True-false questions. Check the correct answer. Each question has a value of 3 points. 1. A particle acted upon only by a conservative force and released from rest will move toward lower potential energy. True False 2. When you unscrew the top of a bottle, your thumb and fingers exert forces in opposite directions but torques in the same direction. True False 3. Galileo s finding, that an object dropped from rest in a gravitational field has an acceleration independent of its mass, is only valid as an approximation for objects near the earth s surface. True False [Always exactly true] 4. Spring tides, when the sun and moon combine to make especially high tides, occur only once per lunar month, when the sun and moon are on the same side of the earth. True False [Twice, also when they are on opposite sides] 5. Nights in the desert can be quite cold, even though the days are very hot, because the clear sky allows rapid energy loss by radiation at night. True False 6. Any kind of heat flow from higher to lower temperature is an irreversible process. True False 3

4 Part C: Problems. Work problems in the space provided, indicating your method clearly. Problems have the point values shown. 1. A hunter with a blow-gun intends to shoot a monkey hanging from a tree limb at height h above the ground. The hunter is at distance 3h from the base of the tree. Having studied physics, the hunter knows that if he aims his blow gun directly at the monkey, and shoots at the instant the monkey releases its grip on the limb to drop to the ground, then the dart will strike the falling monkey. But the dart must travel fast enough to hit the monkey before it reaches the ground. a. How long does it take the monkey to reach the ground? b. What must be the minimum horizontal component of the dart s initial velocity? c. What must be the minimum initial speed of the dart? Give all answers in terms of h and g. [Draw a picture.] [15 points] a. The time is given by h = 1 2 gt2, so t = 2h/ g. b. It must go 3h in this time or less, so 3h = v 0x t, or v 0x = 3h! g/2h = 3 gh/2. c. To shoot at the monkey, we must have v 0y /v 0x = h/3h = 1/3, so v 0y = 1 3 v 0 = gh/2. The speed is given by x v 2 = v 2 0x + v 2 0y = 9gh/2 + gh/2 = 5gh, or v = 5gh. 3h h 4

5 2. A ball of mass m attached to a massless string of length R is moving as shown in a vertical circle. We are interested in the situation at the three points indicated. a. The ball has the minimum speed necessary to stay in the circle at point A. What is that speed? Explain. b. What is its speed at point B? c. Suppose when it reaches point C the string breaks. To what height h above C will it rise before it falls again? Give answers in terms of R and g. [15 points] A B v C a. The minimum downward acceleration is g (when the tension in the string is zero) so the minimum speed is given by g = v 2 A /R, or v A = Rg. b. By conservation of energy, 1 2 mv A 2 + mg(2r) = 1 2 mv B 2. This gives (using the previous answer) v B = 5Rg. c. The speed at C is given by 1 2 mv A 2 + mg(2r) = 1 2 mv C 2 + mgr, so v C = 3Rg. It is moving vertically so it is at rest at height h.. Using conservaton of energy again we find mgh = 1 2 mv C 2, or h = 3 2 R. 5

6 3. A crate is resting on the bed of a truck as shown. The truck is moving to the left with speed v 0 when the driver applies the brakes to avoid hitting a stopped car in front of him. He wants to stop quickly, but not to have the crate either slide forward on the bed or tip over. Assume the braking acceleration is constant. The line from the CM of the crate to the front corner makes the angle θ shown. The coefficient of static friction between the crate and the truck bed is µ s. a. If the crate does not tip over before it slides, what is the smallest distance x in which the truck can stop without making the crate slide? b. If the crate tips over before it slides, what is the smallest distance x in which the truck can stop without making the crate tip over? [Consider torques about the CM in an inertial frame, or else use the frame of the truck and g eff.] c. For what value of µ s are these distances the same? [15 points] v 0 θ a. The only horizontal force on the crate is friction, so f s = ma. Using the maximum friciton force we find a max = µ s mg/m = µ s g. The stopping distance is given by v 0 2 = 2ax, so x min = v 0 2 /2a max = v 0 2 /2µ s g. b. Torques about CM: Friction acts to the right along the bed surface, with magnitude ma. Its torque is counter-clockwise, with moment arm h, the height of the CM. The normal force has magnitude mg and is upward. It can act effectively at any point where the crate contacts the bed. Its maximum clockwise torque occurs when it acts at the left corner, when its moment arm is w/2, half the width of the crate. Just before the crate tips over the torques are mg! w/2 = ma! h, or a = g! w/2h = g tan". Frame of the truck: Take torques about the left lower corner. Friction has no moment arm, and if the normal force acts at that corner it has no moment arm. The only torque would come from gravity, given by g eff. If this vector is along the line from the CM to the corner, there is no moment arm, so the total torque is zero. This happens when a = g tan!. If a is larger, there is a counter-clockwise torque about the corner and the crate rotates that way. In either case, the minimum stopping distance is x min = v 0 2 /2a max = v 0 2 / gtan!. c. The two distances are the same if µ s = tan!. 6

7 4. Shown is a bicycle plus rider of total mass M accelerating to the right at rate a, both wheels rolling without slipping. The wheels have the same radius R and moment of inertia I about their axles. A clockwise torque! 0 is applied by the rider to the rear wheel through the pedal mechanism. You are to show that the increase in the kinetic energy of the system comes from the power supplied by this torque, according to the formula P =!", where ω is the angular velocity of the object the torque τ is applied to (in this case, the rear wheel). a. Call the friction forces by the road on the front and rear wheels f F and f R. Show their directions on the drawing and write the equation relating these forces to a. [What causes the angular acceleration of the front wheel?] b. Write the two equations relating the total torque on each wheel to its angular acceleration α, in terms of f F, f R and given quantities. c. Eliminate f F and f R to find! 0 in terms of a, α and given quantities. d. Finally, write the expression for the total kinetic energy K of the system and show that dk /dt =! 0 ". [Use the fact that the wheels are rolling.] [20 points] a a. The rear wheel will tend to slip backwards on the road because of the applied torque! 0, so friction is forward. The wheels have a clockwise angular acceleration! = a/r (rolling). For the front wheel, the torque to produce! can only be caused by friction, which must therefore be backward. Friction is the only external horizontal force, so f R! f F = ma. b. For the rear wheel! 0 " f R R = I#. For the front wheel f F R = I!. c. Substituting for the friction forces in the equation in (a) we find! 0 = mar + 2I". d. We see that P =! 0 " = mar" + 2I#w = mav + 2I#" (using the rolling condition v = R! ). Now K = 1 2 mv2 + 2! 1 2 I" 2, so dk /dt = mav + 2I!", which is equal to P as claimed. [The momentum of the system is provided by the friction forces, but the energy comes from the work done by the torque! 0, which of course comes indirectly from the muscles of the rider.] 7

8 5. A child of mass m runs at speed v 0 and jumps onto the rim of a playground turntable at rest, as shown from above. The turntable has radius R and moment of inertia I = 9mR 2 about its axle. The child s velocity is tangent to the rim when she jumps on. Give all answers in terms of m, R and v 0. a. What is the angular speed of the rotation when the child comes to rest relative to the turntable? b. What fraction of the child s initial energy is lost in bringing her to rest relative to the turntable? c. Later the child jumps off, but it such a way that her velocity relative to the ground is zero as she leaves the turntable. How much energy do her muscles contribute in making this jump? [What is the final angular speed of the turntable?] [15 points] a. Angular momentum about the axle is conserved (nothing else is) so we have mrv 0 = I tot! = (9mR 2 + mr 2 )! = 10mR 2!, or! = v 0 /10R. b. The final kinetic energy is K = 1 2 I tot! 2 = 5mR 2 "(v 0 /10R) 2 = 1 20 mv 0 2, which is 1/10 of the original kinetic energy 1 2 mv 0 2. The collision lost 9/10 of the kinetic energy. c. Angular momentum is conserved again, and the child s final angular momentum is zero, so we have I tot! = I!", or 10mR 2!(v 0 /10R) = (9mR 2 )# ", which gives "! = v 0 /9R. The final kinetic energy is K f = 1 2 I "! 2 = 1 2 # 9mR2 #(v 0 /9R) 2 = 1 18 mv 0 2. She has added energy equal to 1 18 mv 0 2! 1 20 mv 0 2 = mv 0 2 by jumping off. 8

9 6. Two questions about fluids, 5 points each. a. A bubble forms at the bottom of a glass of champagne and rises toward the surface. As it rises it occupies more volume, and its upward acceleration becomes larger. Explain both of these facts. The volume increases because the pressure from the liquid decreases as the bubble rises, while the temperature stays constant. Because the volume increases, so does the buoyant force, and the upward acceleration also increases. b. If the long arm of a siphon exceeds about 10 m in length the siphon will not work. Explain why. L Compare points at the top and bottom of the long arm. The speed of the fluid is the same at both points because the area of the tube is the same (continuity). At the bottom the pressure is air pressure. We have from Bernoulli P top + 1 2!v2 +!gl = P !v2, or P top = P 0! "gl. If L is more than about 10 m, P top will become negative and the flow will form cavities and stop. 9

10 7. A block of mass m is attached to a spring of stiffness k as shown. a. If the surface the block slides on is frictionless and the block is set into oscillation, what is the angular frequency ω? b. Suppose the surface has friction coefficients µ k = 0.25 and µ s = 0.6. The block is pulled out from the wall, stretching the spring by A = mg/k, and released from rest. How far will the block slide past the equilibrium point of the spring before coming momentarily to rest? [Consider the change in total mechanical energy, and recall that a 2! b 2 = (a + b)(a! b).] c. Will it stay at rest at that point? Explain with a free body diagram. Give answers in terms of m, g and k. [15 points] a. The usual formula,! = k/m. b. Let the distance beyond the equilibrium point be x. The work done by friction is W f =! f k (A + x) =!µ k mg "(A + x). Use the fact that the change in total mechanical energy is the work done by non-conservative forces. The initial and final kinetic energies are zero, so we have 1 2 kx2! 1 2 ka2 = W f =!µ k mg "(x + A). Using the hint this becomes 1 2 k(x! A) =!µ k mg, so x = A! 2(m/k)µ k g. Using the valus of A and µ k we find x = 0.5(mg/k). c. The spring force is kx = 0.5mg, while the maximum static friction force is f s = µ s mg = 0.6mg. The block will stay at rest. 10

11 8. Two questions concerning waves. a. The frequencies of the 7 notes in the diatonic scale in music are such that harmonics 2-6 of the lowest note in the scale also produce notes in the scale, in higher octaves. But the 7 th harmonic is not a note in the scale; for that reason the piano is designed to suppress it. How far from one end should the hammer strike the string (of length L, fixed at both ends) so no energy goes into the 7 th harmonic? Explain with a picture. [The point where the hammer strikes cannot be a node for any harmonic.] [5 points] The 7 th harmonic will have a node at distance L/7 from each end. The hammer should strike at one of those points, forcing an antinode and eliminating the 7 th harmonic from the sound. b. Two waves of equal wavelength and frequency and moving in the same direction interfere. The waves have separate intensities I 1 and I 2. i. Show that the maximum resultant intensity is ( I 1 + I 2 ) 2. ii. Show that the minimum resultant intensity is ( I 1! I 2 ) 2. [Consider the relation between intensity and amplitude.] [10 points] For any wave the intensity is related to the amplitude by I = K! A 2, where K is a constant dependent on the type of wave. Thus A 1 = I 1 /K and A 2 = I 2 /K. The resultant wave will have amplitude A = I /K. For constructive interference, A = A 1 + A 2, and for destructive interference A = A 1! A 2. Thus for constructive interference I /K = I 1 /K + I 2 /K, so I = ( I 1 + I 2 ) 2. This is the maximum intensity claimed. A similar argument gives the minimum intensity claimed. 11

12 9. An engine using one mole of an ideal monatomic gas (c V = 3 R ) operates in the 2 reversible cycle shown as a circle on the P-V diagram. The cycle starts at a and proceeds clockwise. a. In which steps (denoted as a! b, etc.) is heat taken into the engine and in which steps is it expelled? Explain how you know for each step. [Find the temperatures at the four points shown.] b. The net work output per cycle (the area inside the circle) is!p 0 V 0. Find the total heat taken in per cycle, in terms of P 0 and V 0 only. [Each quadrant of the circle represents ¼ of the total work represented by the circle.] c. Find the efficiency of the engine. [15 points] 3P 0 P 0 a V 0 b d c 3V 0 a. Point a: 2P 0, V 0, so T a = 2P 0 V 0 /R. Point b: 3P 0, 2V 0, so T b = 6P 0 V 0 /R. Point c: 2P 0, 3V 0, so T c = 6P 0 V 0 /R. Point d: P 0, 2V 0, so T d = 2P 0 V 0 /R. a! b : W > 0,!U > 0 (temperature increases) so heat is taken in. b! c : W > 0,!U = 0, so heat is taken in. c! d : W < 0,!U < 0 (temperature decreases), so heat is expelled. d! a : W < 0,!U = 0, so heat is expelled. b. a! c : W = 1 2!P 0 V 0 + 4P 0 V 0,!U = c V!T = 3 2 R(T c " T a ) = 6P 0 V 0, so Q in = 10P 0 V !P 0 V 0. c. The efficiency is e = W tot Q in =!P 0 V 0 10P 0 V !P 0 V 0 = 2!! + 20 "

13 10. Two questions about devices. a. The rate at which heat flows by conduction into a house on a hot day is proportional to the difference between outdoor and indoor temperatures. Call this amount A(T o! T i ), where A is some constant. In order to keep the indoor temperature constant the air conditioner must remove heat from the house at the same rate. Assume it runs on a Carnot cycle and show that the power supplied to the air conditioner (from the electric power grid) in order to keep the indoor temperature constant is proportional to (T o! T i ) 2. [10 points] The heat removed by the air conditioner is Q C, where W = Q H! Q C. Using the Carnot condition we have Q H = Q C!T o /T i, so W = Q C T o! T i T i and the power required to remove this heat is dw dt = dq C dt! T o " T i T i. Setting dq C dt equal to the rate at which heat enters the house by conduction, we find dw dt = A (T o! T i )2 T i as claimed. b. When a refrigerator turns water into ice, the entropy of the freezer and its contents decreases. Why is this not a violation of the 2 nd law of thermodynamics? [5 points] The 2 nd law says the total entropy of a closed system cannot decrease. The freezer compartment is not a closed system, since it interacts with the rest of the refrigerator and with the external world, e.g., the kitchen. 13