AP Physics 1: Rotational Motion & Dynamics: Problem Set

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4 51. A student of mass 42 kg is standing at the center of a merry-go-round of radius 3.4 m and a moment of inertia of 840 kg-m 2 that is rotating at ω = 1.8 rad/s. The student walks to the outer edge of the merry-go-round. What is the angular velocity of the merry-go-round when he reaches the edge? VI. General Problems 83. A very light cotton tape is wrapped around the outside surface of a uniform cylinder of mass M and radius R. The free end of the tape is attached to the ceiling. The cylinder is released from rest and as it descends it unravels from the tape without slipping. The moment of inertia of the cylinder about its center is I = MR2. a. On the circle above, show all the forces applied on the cylinder. b. Find the acceleration of the center of the cylinder when it moves down. c. Find the tension force in the tape. 84. A uniform cylinder of mass M and radius R is fixed on a frictionless axle at point C. A block of mass m is suspended from a light cord wrapped around the cylinder and released from rest at time t = 0. The moment of inertia of the cylinder is I = MR2. a. On the circle and the square above, show all the applied forces on the cylinder and the block. b. Find the acceleration of the block as it moves down. c. Find the tension in the cord. d. Express the angular momentum of the cylinder as a function of time t.

5 85. A uniform cylinder of mass M and radius R is initially at rest on a rough horizontal surface. A light string is wrapped around the cylinder and is pulled straight up with a force T whose magnitude is 0.80 Mg. As a result, the cylinder slips and accelerates horizontally. The moment of inertia of the cylinder is I = MR2 and the coefficient of kinetic friction is a. On the circle above, show all the forces applied on the cylinder. b. In terms of g, determine the linear acceleration, a of the center of the cylinder. c. Determine the angular acceleration, α of the cylinder. d. Explain the difference in results of linear acceleration a, and Rα. 86. A billiard ball of mass M and radius R is struck by a cue stick along a horizontal line though the center of mass of the ball. The ball initially slides with a velocity v 0. As the ball moves across the rough billiard table its motion gradually changes from pure translational through rolling with slipping to rolling without slipping. The moment of inertia of the ball is I = MR2 and the coefficient of kinetic friction is µ. a. Express the linear velocity, v of the center of mass of the ball as a function of time t while it is rolling with slipping. b. Express the angular velocity, ω of the ball as a function of time while it is rolling with slipping. c. Find the time at which the ball begins to roll without slipping. d. When the ball is struck it acquires an angular momentum about the fixed point A on the surface of the table. During the motion the angular momentum about point A remains constant despite the friction force. Explain why this occurs.

6 87. A block, A of mass, M is suspended from a light string that passes over a pulley and is connected to block B of mass 2M. Block B sits on the surface of a smooth table. Block C, of mass 3M, sits on the top of block B. The surface between block C and block B is not frictionless. When the system of three blocks is released from rest, block A accelerates downward with a constant acceleration, a, and the two blocks on the table move relative to each other. The moment of inertia of the pulley is I = 1.5 MR 2. Present all results in terms of M, g, and a. a. Find the tension force in the vertical section of the string. b. Find the tension force in the horizontal section of the string. The acceleration of block A was determined from a series of experiments: a = 2 m/s. c. Find the coefficient of kinetic friction between the two blocks on the table. d. Find the acceleration of block C. 88. A pulley of radius, R and moment of inertia, I = 2 MR 2 is mounted on an axle with negligible friction. Block A, with a mass M, and Block B, with a mass 3M, are attached to a light string that passes over the pulley. Assuming that the string doesn t slip on the pulley, answer the following questions in terms of M, R, and fundamental constants. a. at is the acceleration of the two blocks? b. What is the tension force in the left section of the string? c. What is the tension force in the right section of the string? d. What is the angular acceleration of the pulley?

7 89. A solid uniform sphere of mass, M and radius, R is placed on an inclined plane at a distance, h from the base of the incline. The inclined plane makes an angle, θ with the horizontal. The sphere is released from rest and rolls down the incline without slipping. The moment of inertia of the sphere is I = MR2. a. Determine the translational kinetic energy of the sphere when it reaches the bottom of the inclined plane. b. Determine the rotational kinetic energy of the sphere when it reaches the bottom of the inclined plane. c. Determine the linear acceleration of the sphere when it is on the inclined plane. d. Determine the friction force acting on the sphere when it is rolling down the inclined plane. e. If the solid sphere is replaced with a hollow sphere of identical mass and radius, how would that change the answer to question (b)? Explain. 90. A solid uniform cylinder of mass M and radius R is placed on an inclined plane at a distance h from the base of the incline. The inclined plane makes an angle θ with the horizontal. The cylinder is released from rest and rolls down the incline without slipping. The moment of inertia of the cylinder is I = MR2. a. Determine the translational kinetic energy of the cylinder when it reaches the bottom of the inclined plane. b. Determine the rotational kinetic energy of the cylinder when it reaches the bottom of the inclined plane. c. Determine the linear acceleration of the cylinder when it is on the inclined plane. zzxzx d. Determine the friction force acting on the cylinder when it is rolling down the inclined plane. x e. If the solid cylinder is replaced with a hollow hoop of identical mass and radius, how would that change the answer to question (b)? Explain.

8 91. In a physics experiment, students made a lab cart of a wooden block of mass 5m and four wheels each of mass, m and radius r. The moment of inertia of each wheel is I = mr2. The cart is released from rest and rolls without slipping from the top of an inclined plane of height h. After the cart reaches the bottom of the inclined plane, it collides with an elastic spring with negligible mass and a spring constant k. a. Determine the moment of inertia of all four wheels. b. Determine the speed of the cart at the bottom of the inclined plane. c. Determine the maximum compression of the spring after the collision. 92. A bicycle wheel of mass M and radius R is placed on a vertical axle and is able to rotate without friction. A dart of mass, m is fired toward the wheel with an initial velocity V o. The dart strikes the wheel and sticks in the tire. The moment of inertia of the wheel is I = MR 2. a. What is the initial linear momentum of the dart? b. Is the linear momentum of the dart conserved? Explain. c. Is the angular momentum of the dart conserved? Explain. d. What is the initial angular momentum of the dart with respect to the center of the wheel? e. What is the angular velocity of the wheel after the dart strikes? f. Find the ratio of the final kinetic energy of the system, after the dart strikes, and the initial kinetic energy of the dart.

9 Answers Chapter Questions 1. Real matter is not just a point it has a structure. It doesn t just move in a linear fashion it can rotate about itself. 2. The axis of rotation is a line about which the object rotates. For a tire, the axis goes through the empty space within the tire if the tire is on a car, the axle would be the axis of rotation, so, no, the axis doesn t touch the rubber in the tire. 3. The degree was an angle of measurement chosen by humans based on culture or their system of mathematics. The radian is a property of the circle, as it relates the angle subtended by a length on the circumference of the circle (arc length) to the radius of the circle. 4. Both bugs move through the same angular displacement, but the bug near the edge of the record moves through a greater linear displacement (arc length). 5. Tangential acceleration and centripetal acceleration. Both increase as an object is further from the axis of rotation. 6. Both rings are moving with the same angular acceleration and velocity. However, the further you are away from the center of the merry-go-round, the greater the tangential and centripetal acceleration and the velocity. This is what you feel, so if you are prone to motion sickness, you want slower magnitudes of the linear quantities. You should choose the inner ring. 7. Angular acceleration is assumed to be constant. If you know the linear kinematics equations, you just replace the linear variables, x, v and a, with their angular equivalents, θ, ω and α. No the reason why objects move is covered in Dynamics. Kinematics just deals with how objects move, not why. 8. Torque. By increasing the distance of where the force is applied to the object to be rotated. Also, the force should be applied perpendicular to the line connecting the force application with the rotated object. 9. Torque is dependent upon the distance between the force application and the object. It also depends on the angle that it makes with a line connecting the force to the object. Both are maximized by pushing at the far end of the door in a perpendicular fashion, resulting in a greater angular acceleration. The door stop is positioned as far away from the door hinges as possible to increase its torque which opposes any torque applied to/by the door to shut it. 10. Torque is a vector it has a direction and a magnitude. Torque is maximized when the force is applied perpendicular to the line connecting it and the rotated object. Work and Energy are scalars and they are maximized when a force is applied parallel to their motion. 11. A longer wrench will maximize the torque due to a given force (your strength), and will result in a greater angular acceleration of the wheel nuts. The goal is to provide a greater torque then what currently is holding the nuts in place friction between the nut and the wheel rim and bolt. 12. The fulcrum should be closer to the heavy rock. The rock is providing a torque at the fulcrum point in one direction (let s say, clockwise), and you are providing a torque in the opposite direction (counter clockwise). To lift the rock, the torque that you provide has to be greater than the rock s torque. As you have a limited amount of strength (force), by having a longer section of the board on your side of the fulcrum will maximize your applied torque. 13. The moment of inertia of the hollow cylinder is greater than the sphere (assuming equal masses and radii). The moment of inertia depends on the mass distribution of the object the more mass that is located further from the rotation axis, the greater the moment of inertia. More of the mass of the hollow cylinder is further from the rotation axis so it has a greater moment of inertia than the sphere. 14. The hollow cylinder has a greater moment of inertial. Therefore, its rotational kinetic energy is greater than the rotational kinetic energy of the solid cylinder. Since the total energy of each rotating object is the same that leaves less translational kinetic energy for the hollow cylinder. Therefore, its velocity is less, and it will reach the bottom of the inclined plane after the solid cylinder. 15. The linear distance (translational) covered by an object in translational motion equals the arc length distance (rotational) traveled by a point on the rotating object. If this condition holds, then the equation v = rω can be used. 16. She should throw her arms out to the side, increasing her moment of inertia (more of her mass is distributed further from her axis of rotation). Since there is no external torque applied (because she did

11 rad rad/s 2 ; 195 revolutions s N-m N-m. The lug nut will not move it requires a torque of 135 N-m before it will have an angular acceleration m N N-m; 1.3 N-m N-m x10 7 rad/s; 2.1x10 5 m/s. These numbers are way too high. But, what s missing is the friction force provided by the threads of the lug nut, which opposes the applied force and reduces the total torque. The wrench would start applying the force, the lug nut would move and the wrench would lose contact, reducing the applied torque and the angular acceleration. This is where a simple basic physics model fails in explaining a fairly simple project N-m 48. a) 6.2 kg-m 2 b) 11 kg-m 2 c) 10 kg-m kg-m a) N-m b) N-m N-m N-m m x10 30 m. Archimedes might have been wrong here. The limit of the observable universe (that which we have received light from is 2.6x10 26 m. Archimedes would need more room (or more strength) N-m; 1.7 N-m N-m N-m 58. a) 14 kg-m 2 b) 22 kg-m 2 c) 24 kg-m N-m 60. a) N-m b) N-m m/s. No, it only depends on g and the height of the incline. 62. They both get to the bottom at the same time; 3.2 m/s J J 65. Zero work. At the point of contact between the cylinder and the ground, the cylinder is moving first down then up. The friction exerted by the incline on the cylinder is parallel to its linear movement. Hence the frictional force is perpendicular to the displacement of the cylinder at the point of contact zero work x10 4 J; 3.10 kw m/s; No, it is mass and radius independent. 68. Both arrive at the same time; 3.3 m/s J J 71. Yes. The frictional force is along the incline, and the displacement of the sphere is along the incline (it is sliding, not rotating). W = Fdcosθ and θ = x10 4 J; 2.05kW kgm 2 /s 74. I=0.73I ω=ω 0/ kgm 2 /s rad/s kgm 2 /s 79. ω=0.64ω m/s kgm 2 /s rad/s

12 General Problems: 83. a. F T Mg b. c. 84. a. F Support F T b. Mg F T mg c. d. 85. a. F T F N Mg F FR b. 0.08g c.. d. Because the cylinder slips, αr is greater than a. 86. a. v = v μgt b. ω = c. d. Point A, with the addition of the friction force, does not rotate, as the rotational forces cancel out. 87. a. M(g- a) b. Mg 2.5Ma

13 c d. 0.3 m/s 88. a. b. c. 2mg d. 89. a. b. c. d. e. The moment of inertia, I, would increase. This would result in a lower tangential velocity, as more of the total Kinetic Energy is transformed into rotational Kinetic Energy. 90. a. b. c. d. e. It would lower the moment of inertia I, increasing the value of the answer to (b). 91. a. 2mr b. c. 92. a. mv

14 b. No. There is an unbalanced force at the center of the wheel. c. Yes. Torque at the center of the wheel is equal to 0. No external torque acts on the system. d. mrv sin θ e. () f. 93. a. b. c. d. No. Initially, linear momentum was equal to. The final linear momentum is equal to, showing that they are not conserved. This is because there is an unbalanced force at point C. e. Yes. The torque at point C is equal to zero. No external torque acts on the system. 94. a. b. c. M v

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