t = g = 10 m/s 2 = 2 s T = 2π g

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1 Annotated Answers to the 1984 AP Physics C Mechanics Multiple Choice 1. D. Torque is the rotational analogue of force; F net = ma corresponds to τ net = Iα. 2. C. The horizontal speed does not affect the time to fall; use y = 1 2 gt2 or t = 2y 20 m g = 10 m/s 2 = 2 s 3. C. The formula for the period of a pendulum is l T = 2π g and is independent of the mass. 4. A. The cars were at the same spot at t = 0; we might as well call this the origin. The distance traveled by each during the next 20 seconds is the integral of v with respect to t; it is the area under the curve. Car X has traveled twice as far as Car Y in this time, because the rectangle has twice the area of the triangle. 5. B. The areas are the same, so each car has traveled the same distance in the 40 seconds. But Car Y is traveling much faster than Car X, so it is passing Car X. 6. C. When the skater draws her arms in, her rotational inertia I is decreased. Angular momentum is conserved because there are no external torques acting on the skater. However, her kinetic energy increases, because she has to do work to draw in her arms. Alternatively, L = Iω stays the same, but the kinetic energy goes like 1 2 Iω2 which is effectively a constant times ω, which increases since I decreases. 7. E. The velocity is proportional to t 2. Then the acceleration, the derivative of the velocity, will be proportional to the derivative of t 2, or t. The force is proportional to the acceleration, so it will be a linear function of time, t. 8. B. Power is work done over time, or in this case (mgh)/t. 9. E. The acceleration of the entire system is a = F/3 kg. The net force on the 1 kg block is the tension in the cord. That is, ( ) F F net = T = ma = (1 kg) = 1 3 kg 3 F 10. A. The (constant) angular acceleration equals ω/ t = (1 rad/s 0 rad/s)/2 s = 0.5 rad/s D. The angular momentum is given by L = Iω = (4 kg-m 2 )(1 rad/s) = 4 kg-m 2 /s. 12. B. The kinetic energy is due to rotational motion only, and is equal to 1 2 Iω2 = 1 2 (4 kg-m2 )(1 rad/s) 2 = 2 J 13. B. The momentum of the heavier mass is m 2 v, down; the momentum of the lighter is m 1 v, up. As momentum is a vector, the net has a magnitude of (m 2 m 1 )v. 14. A. Gravitational potential energy is U = mgh. The mass m 2 drops a distance h, but necessarily the mass m 1 must rise the same distance, so the net loss in potential energy is (m 2 m 1 )gh. 15. E. Use the Work-Energy Theorem; W = K = (area under the F vs x graph). The area under the graph in E is larger than any other; the areas for A, B and D are all equal, and C is the least.

2 16. A. To be in equilibrium, the net force must be equal to zero, and the net torque must be equal to zero. The only two choices that would make the net force equal to zero (remember, force is a vector) are A and D. Take torques around the center of mass of the square. The 10 N force produces no torque about the center (the angle between r and F is 180, and the torque involves the sine of this angle). The 5 N force produces a clockwise torque; an equal counter-clocwise torque is needed to offset it. The force at D produces no torque (same argument as for the 10 N force); the appropriate answer is A. 17. B. Before the collision, the total momentum is m(v o cos 60, v o sin 60) + m(v o cos 60, v o sin 60) = (2mv o cos 60, 0) = (mv o, 0) After the collision, a single object of mass 2m has a velocity of v ; that is, so that v = ( 1 2 v o, 0). 2mv = (mv o, 0) 18. D. As the mass passes through the equilibrium point, the potential energy is zero and the kinetic energy is maximum: C is wrong. As the mass reaches the greatest stretch or the greatest compression, the kinetic energy is zero and the potential energy is maximum: E is wrong. Of course if there is no friction the total energy is constant. So both the kinetic and the potential energies vary from zero to a maximum of all of the energy, making A and B wrong, and D the right answer. 19. D. Since the object is moving in a circle, you know v = ωr. The angular frequency ω is the derivative of the angle, θ; ω = d ( 3t 2 + 2t ) = 6t + 2 dt At t = 4 s, ω = 26 rad/s, and the radius r = 2 m, so v = 52 m/s. 20. B. Let M = the mass of the earth, and R = the radius of the earth. The gravitational acceleration g X of Planet X is given by g X = GM X RX 2 = G 1 10 M ( 1 = 4 GM. 2 = 4 m/s 2 R)2 10 R C. Power is equal to force times velocity, if the velocity is constant, as it is here. The force provided by the person equals the friction (or the box would accelerate.) The problem was not well-worded because if the person pushes at any angle other than horizontal, we cannot find the normal force, and hence the friction. So, assuming the person is pushing horizontally, the friction is given by f k = µ k N = (0.25)(mg). = (0.25)(40 kg)(10 m/s 2 ). = 100 N Then P = f k v = (100 N)(0.5 m/s) = 50 W. 22. B. Draw a diagram: Let P be the force exerted by the hinge at O, and let the length of the bar be l. The angle θ is drawn with the correct orientation, but you don t have to assume it is correct; the physics is going to give us the angle momentarily. Take torques about the hinge at O. If the bar is not to rotate, the net torque must be zero: τ = mg 1 2 l sin 90 + F l sin 30 = 0 = 1 2 mgl F l

3 so that F = mg. If the bar is not to accelerate, the net force must be zero: Substituting F = mg gives Dividing, F net,x = P cos θ F cos 30 = 0 F net,y = P sin θ + F sin 30 mg = 0 P cos θ = mg cos 30 P sin θ = mg(1 sin 30) = mg sin 30 P cos θ P sin θ = mg cos 30 mg sin 30 so that tan θ = tan 30; or θ = 30 (and also P = mg.) In retrospect this can be solved by inspection. Because of the angle of 30, it follows quickly from the torque that F = mg, and from the F y equation that F sin 30 + P sinθ = mg. But if F = mg, then F sin 30 = 1 2mg, which means that P sin θ must also equal 1 2 mg = F sin 30. The F x equation requires P cos θ = F cos 30. So P = F, and θ = 30, by inspection. 23. A. In the diagram shown, the mass is initially in the unstretched position, and then allowed to fall to a new position a distance of y below the original point. At this lowest point, the mass is momentarily at rest, and all the energy is potential. Suppose we call the initial height y = 0. Then the total energy at the start is zero, and the energy at the lowest point the energy is still zero. unstretched maximum stretch At the lowest point, there are two sorts of potential energy, gravitational mgy and elastic ky2 ; that is, E = mgy ky2 = 0 y = 2mg k If there is no loss in energy, the spring will oscillate between its original position and the lowest point at y. The midpoint of the oscillation is at 1 2y, or mg/k. This is also the amplitude; A = mg k. = (0.1 kg)(10 m/s2 ) = 1 40 N/m 40 m 24. C. The time for the block to return to its original position is just the period, T. This is given by the formula m T = 2π k = 2π 0.1 kg 40 N/m = π 10 s 25. E. From the graph, the amplitude of the oscillation is 4 (meters, one presumes) and the period is 2 (seconds?). That means ω = 2π/T = π, and so the position function is given by x = 4 cos πt The velocity is given by the derivative of this function; v = d(4 cos πt)/dt = 4π sin πt

4 26. B. The angular momentum L = mvr sin φ. See the illustration. Note that r sin φ = a, so L = mva. 27. A. By definition, I = m i r 2 i The distance of the mass M from the center of the stick is 1 2L, so I = I o + M( 1 2 L)2 = I o ML2 28. C. At t = 0, the object is at the origin. At t = 1, the object is at (24 m/s)(1 s) = 24 m. The initial speed is 24 m/s. The object is now given an acceleration of 6 m/s 2. At t = 11, this is ten seconds after the acceleration begins. That is, reset the clock to zero at t = 11. Then x = x o + v o t at2 = (10) 3(10) 2 = = 36 m 29. B. The wire is made of three identical pieces of length L, so each has a mass m. Let the left edge be located on the y axis. Then the center of mass of the assembly is (x cm, y cm ) = m ( 0, 1 2 L) + m ( 1 2 L, 0) + m(l, 1 2 L) ( = 1 3 m + m + m 3 2 L, L) = ( 1 2 L, 1 3 L) Point B is above 1 4 L, but below 1 2 L, so it s the best answer. (All the points are located at x = 1 2 L, which is obviously correct by symmetry. The center of mass calculation was carried out for both x and y above as a check; had we not found x cm = 1 2L, we d have known the calculation was wrong.) 30. A. Because the velocity is given as perpendicular to the acceleration, the speed of the car is constant. Then the acceleration must be toward the center of the circle. The only place toward the center is directed due south is when the car is at the northernmost point of the circle. See the illustration. The velocity is directed east, so the car is moving clockwise. The acceleration a = v 2 /r, so v = ar = 300 m/s 2 3 m = 30 m/s. 31. E. The definition of the center of mass velocity is the total momentum divided by the total mass; mi v i v cm = = M 1v mi M 1 + M D. The weight hangs at equilibrium, so calling the left cord T 2 and the right cord (the subject of the question) T 1, it follows ( T 2, 0) + (T 1 cos 30, T 1 sin 30) + (0, 100 N) = F net = (0, 0) Looking only at the y-component, T 1 sin 30 = 100 N or T 1 = 100 N/ sin 30 = 200 N.

5 33. C. If the position is proportional to t 3/2, then the velocity is proportional to the derivative of this, or t 1/2, and the kinetic energy, proportional to the square of the velocity, is proportional to t (to the first power.) 34. C. There s no reason not to call the ground y = 0. The starting energy of the ball is E = K + U = 1 2 mv2 + mgy = 1 2 (1 kg)(10 m/s)2 + (1 kg)(10 m/s 2 )(70 m) = 750 J and the final energy E = K + U = 1 2 (1 kg)(30 m/s)2 + 0 = 450 J, so the loss is 300 J. 35. A. We have to assume that the pivot is frictionless. The counterclockwise torque is M o g(2l) = 2M o gl. The clockwise torque is 3M o gl. The net torque is the difference, or M o gl. The moment of inertia is Discussion The instantaneous angular acceleration α is I = m i r 2 i = (3M o )(l 2 ) + (M o )(2l) 2 = 7M o l 2 α = τ I = M ogl 7M o l 2 = g 7l A nice test, though perhaps too many plug in problems, and not enough problems requiring insight. Some of the problems were a little repetitive; 7 isn t that different from 19 or from 33. Here and there a problem is missing some important information: in 21 you need to assume the push is horizontal, in 35 you need to assume the pivot is frictionless, and there were no units given in 25. On the other hand, some problems needed a little thought; 22 and 23; and 35. The problems missed most frequently were 24, 26 and 29, not at all hard but perhaps requiring topics not reinforced (the angular momentum of a point mass often gets overlooked, and the center of mass is sometimes neglected.) David Derbes, loki@uchicago.edu University of Chicago Laboratory Schools 25 August 2009

Physics 111. Tuesday, November 2, Rotational Dynamics Torque Angular Momentum Rotational Kinetic Energy

Physics 111. Tuesday, November 2, Rotational Dynamics Torque Angular Momentum Rotational Kinetic Energy ics Tuesday, ember 2, 2002 Ch 11: Rotational Dynamics Torque Angular Momentum Rotational Kinetic Energy Announcements Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468 Announcements This