Name Student ID Score Last First. I = 2mR 2 /5 around the sphere s center of mass?

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Buck McCoy
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NOTE: ignore air resistance in all Questions. In all Questions choose the answer that is the closest!! Question I. (15 pts) Rotation 1. (5 pts) A bowling ball that has an 11 cm radius and a 7.

It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill.

1 NOTE: ignore air resistance in all Questions. In all Questions choose the answer that is the closest!! Question I. (15 pts) Rotation 1. (5 pts) A bowling ball that has an 11 cm radius and a 7.2 kg mass is rolling without slipping at 2 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill. What is the value of h (in cm) if you model the bowling ball as a uniform sphere with I = 2mR 2 /5 around the sphere s center of mass? (A) 29 (B) 40 (C) 58 (D) 70 (E) none of A-D Conserve energy: mv 2 /2 + I 2 /2 = mgh, where I = 2mR 2 /5 and =v/r Thus, mv 2 /2 + (2mR 2 /5) (v/r) 2 /2 = mgh h = 7v 2 /(10g) =.29 m = 29 cm. The equations were setup in class 2. (5 pts) A rigid body is in equilibrium if A. F net ext = 0 B. net ext = 0 C. neither A nor B. D. either A or B. E. both A and B. 3. (5 pts) Determine the moment of inertia of a uniform solid sphere (about its CM, I = 2MR 2 /5) of mass M and radius R about an axis that is tangent to the surface of the sphere as shown. (A) MR 2 /2 (B) 7MR 2 /5 (C) 2MR 2 /3 (D) 35MR 2 /108 (E) 29MR 2 /110 Use the Parallel Axis Thm: I = I cm + MR 2 = 2MR 2 /5 +MR 2 = 7MR 2 /5 WebAssign homework

2 Question II. (25 pts) Torque and Angular Momentum 4. (5 pts) Rank in order, from largest to smallest, the five torques a thru e. The rods all have the same length and are pivoted at the dot. 2N 2N 2N 4N (a) (b) (c) (d) (e) N (A) e > a = d > b > c (B) d e a b c (C) d e a b c (D) d c a b c (E e a d b c only if one chooses the moment arm for (d) to be less than half the distance compared to case (a) 5. (5 pts) Two buckets spin around in a horizontal circle on frictionless bearings as shown. Suddenly, it starts to rain with vertical drops. As a result, A. The buckets slow down because the angular momentum of the bucket + rain system is conserved. B. The buckets continue to rotate at constant angular velocity because the rain is falling vertically while the buckets move in a horizontal plane. C. The buckets continue to rotate at constant angular velocity because the total mechanical energy of the bucket + rain system is conserved. D. The buckets speed up because the potential energy of the rain is transformed into kinetic energy. E. None of the above. 6. (5 pts) A uniform disc in the x-y plane of radius 20 m and mass 1200 kg rotates at 0.75 rad/s about its axis, which is also the z-axis, as shown. When viewed from a point on the positive z-axis, the disc rotates counterclockwise. What is the magnitude (in kg m 2 /s) and direction of the angular momentum? I disc = mr 2 /2.

3 (A) 1.8 x 10 5 k (B) 3.6 x 10 5 k (C) 1.8 x 10 5 j (D) 3.6 x 10 5 i (E) 14.4 x 10 5 k L = I = mr 2 /2 = 1.8x10 5 kg-m 2 /s in the positive z direction 7. (5 pts) If an object were to suddenly shrink and decrease its moment of inertia by a factor of 2 without any external force, what is the ratio of the final to initial rotational kinetic energies? (A) 3 (B) 9 (C) 8 (D) 2 (E) 1 8. (5 pts) Disc 1 is rotating freely and has an angular velocity i = 0.5 rev/s about an axis that coincides with its symmetry axis as shown. Its moment of inertia is I 1 = 20 kg m 2. It drops onto disc 2, of moment of inertia I 2 = 20 kg m 2, that is initially at rest. Disc 2 is centered on the same axis as disc 1 and is free to rotate about the axis. Because of kinetic friction, the two discs eventually attain a common angular velocity f. What is f (in rev/s)? (A) 1.5 (B) 1.0 (C) 0.5 (D) 0.25 (E) Worked out in class. Conserve angular momentum L i = L f I 1 i = (I 1 + I 2 ) f So w f = I 1 w i /[I 1 + I 2 ) = 0.25 rev/s

Question III. (15 pts) Rotational Dynamics 9. (5 pts) Consider the inclined plane shown (not to scale). A wheel of radius 0.

0 kg that slides on a frictionless surface inclined at = 20 0 with the horizontal, as shown. The object accelerates down the incline at 2.0 m/s 2.

6 For the sliding block, mgsin(20 0 ) T = ma and for the pulley, Tr = I or T = Ia/r 2 Substituting for T and solving for I gives I = [mgsin(20 0 ) ma]r 2 /a = 0.054kg-m 2 10.

One point mass, m 1 = 20 kg, is placed at one end of the board and the other point mass, m 2 = 40 kg, is placed on the other end of the board 3 m away.

4 Question III. (15 pts) Rotational Dynamics 9. (5 pts) Consider the inclined plane shown (not to scale). A wheel of radius 0.20 m is mounted on a frictionless horizontal axis through its geometric center. A massless cord is wrapped around the wheel and attached to an object of mass m = 2.0 kg that slides on a frictionless surface inclined at = 20 0 with the horizontal, as shown. The object accelerates down the incline at 2.0 m/s 2. What is the value, in units of kgm 2, of the rotational inertia of the wheel about its axis of rotation? (A) 5 (B) 2 (C) (D) (E) 14.6 For the sliding block, mgsin(20 0 ) T = ma and for the pulley, Tr = I or T = Ia/r 2 Substituting for T and solving for I gives I = [mgsin(20 0 ) ma]r 2 /a = 0.054kg-m (5 pts) A see-saw is comprised of a massless stiff board of length L = 3 m with two point masses placed on it. One point mass, m 1 = 20 kg, is placed at one end of the board and the other point mass, m 2 = 40 kg, is placed on the other end of the board 3 m away. The fulcrum is located 1 m from the end where m 2 is located. Where must a third mass, M = 10 kg, be placed to balance the see-saw? (A) 1.5 m from m 1 (B) 0.5 m from m 2 (C) on fulcrum (D) 0.2 m between fulcrum and m 1 (E) 0.2 m between fulcrum and m 2 The torques clockwise must equal the torques counterclockwise. Choose the pivot point to be at the fulcrum. So, m 2 g(1m) =Mgx + m 1 g(2m) x = (5 pts) If m 2 is changed from 40 kg to 60 kg in question 9, where should mass M be placed to balance the see-saw? (A) on top of m 1 (B) 0.2 m from fulcrum (C) 0.5 m from m 1 (D) on top of m 2 (E) close to but not on top of m 2

(Assume the positive x-direction is to the right and the y-direction is upward.) 1.5 m T 45 0 F 1 F2 1.5 m F R 3m mg You can choose where to place the pivot point.

5 Question IV Free Response (25 pts) Show all work in sufficient detail that the grader (Prof. Buck) understands exactly what you are calculating. Homework Problem and setup in class A large gate weighing 175 N is supported by hinges at the top and bottom of the wooden frame, and is further supported by a wire. (Assume the positive x-direction is to the right and the y-direction is upward.) 1.5 m T 45 0 F 1 F2 1.5 m F R 3m mg You can choose where to place the pivot point. Consider the pivot point to be at top hinge. The conditions for equilibrium: F = 0 F 1 + F 2 + Tsin45 = mg AND F R = Tcos45 (2) = 0 F R (1.5m) + Tsin45 (1.5m) = mg (1.5m) (a) (10 pts) What must the tension in the wire be for the force on the upper hinge to have no horizontal component? Since F R = Tcos45, then substitute into torgue equation gives Tcos45 + Tsin45 = mg (the lengths divide out). Thus T = mg/(cos45 + sin45) = 124N (b) (7 pts) What is the horizontal force on the lower hinge? (Take the direction of pushing the gate away from the hinges to be positive and the direction of pulling the gate towards

6 the hinges to be negative.) F R = Tcos45= 87.5 N (c) (8 pts) What is the sum of the vertical forces on the hinges? From the first of the force equations: F 1 + F 2 + Tsin45 = mg Thus, F 1 + F 2 = mg - Tsin45 = 175N 87.5N = 87.5 N Note: you will get the same answer even if you chose only one vertical force on just one of the hinges. It also would not matter if one hinge force was up and the other down or even both were chosen to be down. Finally, it matters not where you pick your pivot point.

7 Name Student ID last first V [20 points] Tutorial. In case 1, a puck of mass m, radius r, and uniform mass density is moving with velocity toward a disk of mass M and radius R Case 1 as shown in the top-view diagram at right. The disk is at rest, and neither the puck nor the disk is rotating. m, r In this problem, you will consider 6 other cases that differ from case 1 in one or two ways. All differences will be made explicit (i.e., if any quantities are not explicitly mentioned, they have the same value in the two cases being considered). In this problem, take angular momentum to mean angular momentum of the system of all objects in that case with respect to the disc s center at the instant shown. 1 [4 pts] The velocity of the puck in case 2 is in a different direction than that in case 1 as shown at right (but the speed is the same). momentum in case 2 compare to that in case 1? 2 2 C. L 2 Case 1 r Case 2 L X = p r. The two pucks have the same mass, speed, and thus the same p. However, r is smaller for case 2. r 2. [3 pts] The speed of the puck in case 3 is twice that in case 1. momentum in case 3 compare to that in case 1? 3 3 C. L 3 3. [3 pts] The puck in case 4 is closer to the disk than in case 1 as shown at right. momentum in case 4 compare to that in case 1? 4 4 C. L 4 Case 1 Case 3 2 L X = p r. The two pucks have the same mass but the speed is twice as large in case 3, thus p is twice as large. The pucks have the same r. Case 1 Case 4 L X = p r. The two pucks have the same mass, and speed, and thus the same p. They also have the same r. B PHYS 121A Exam 3 ME-UWB149A094T-EF(CNL)_SOL.doc

8 Name Student ID last first 4. [3 pts] The radius of the puck in case 5 is twice that in case 1 (but the masses are the same). momentum in case 5 compare to that in case 1? 5 5 C. L 5 5. [4 pts] The mass of the puck in case 6 is twice that in case 1, and its location is different as shown at right. momentum in case 6 compare to that in case 1? 6 6 C. L 6 6. [3 pts] In case 7, the disk is removed. momentum of the puck in case 7 compare to that of the puck-disk system in case 1? 7 7 C. L 7 Case 1 Case 5 m, r m, 2r L X = p r. The two pucks have the same mass, and speed, and thus the same p. They also have the same r. Since the puck is not rotating, how its mass is distributed about its center of mass does not affect its angular momentum. Case 1 Case 6 m 2m L X = p r. The two pucks have the same speed, but the mass is twice as large in case 6, thus p is twice as large in that case. However, r is half as large in that case. Case 1 Case 7 Disk is removed The disk is at rest, thus the magnitude of its angular momentum is zero. Therefore its presence (or absence) does not affect the angular momentum of the system. B PHYS 121A Exam 3 ME-UWB149A094T-EF(CNL)_SOL.doc

1 MR SAMPLE EXAM 3 FALL 2013

1 MR SAMPLE EXAM 3 FALL 2013 SAMPLE EXAM 3 FALL 013 1. A merry-go-round rotates from rest with an angular acceleration of 1.56 rad/s. How long does it take to rotate through the first rev? A) s B) 4 s C) 6 s D) 8 s E) 10 s. A wheel,