Chapter 11: Rotational Dynamics and Static EquilibriumAnswers to Even-Numbered Conceptual Questions2. As a car brakes, the forces responsible for braking are applied at ground level. The center of mass of the caris well above the ground, however. Therefore, the braking forces exert a torque about the center of mass thattends to rotate the front of the car downward. This, in turn, causes an increased upward force to be exertedby the front springs, until the net torque acting on the car returns to zero.4. The force that accelerates a motorcycle is a forward force applied at ground level. The center of mass of themotorcycle, however, is above the ground. Therefore, the accelerating force exerts a torque on the cycle thattends to rotate the front wheel upward.6. Consider an airplane propeller or a ceiling fan that is just starting to rotate. In these cases, the net force iszero because the center of mass is not accelerating. However, the net torque is nonzero and the angularacceleration is nonzero.8. A car accelerating from rest is not in static equilibrium because its center of mass is accelerating. Similarly, anairplane propeller that is just starting up is not in static equilibrium because it has an angular acceleration.10. Yes. When an airplane’s engine starts up from rest the propeller has a nonzero rotational acceleration,though its translational acceleration is zero.12. The tail rotor on a helicopter has a horizontal axis of rotation, as opposed to the vertical axis of the mainrotor. Therefore, the tail rotor produces a horizontal thrust that tends to rotate the helicopter about a verticalaxis. As a result, if the angular speed of the main rotor is increased or decreased, the tail rotor can exert anopposing torque that prevents the entire helicopter from rotating in the opposite direction.14. No. If the diver’s initial angular momentum is zero, it must stay zero unless an external torque acts on her. Adiver needs to start off with at least a small angular speed, which can then be increased by folding into atucked position.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The force is applied in a direction perpendicular to the handle of the wrench and at the end of the handle.Strategy: Use equation 11-1 to find the force from the known torque and the length of the wrench.Solution: Solve equation 11-1 for F: τ =r F sinθ( )F =τr sinθ=15 N ⋅m0.25 m( )sin90°= 60 NInsight: A longer wrench can exert a larger torque for the same amount of force.2. Picture the Problem: The weed is pulled by exerting a downward force on the end of the tool handle.Strategy: Set the torque on the tool equal to the force exerted by the weed times themoment arm and solve for the force.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.11 – 1Chapter 11: Rotational Dynamics and Static Equilibrium James S. Walker, Physics, 4th EditionSolution: Solve equation 11-1 for F: τ =FweedrweedFweed=τrweed=1.23 N ⋅m0.040 m= 31 NInsight: The torque must be the same everywhere on the tool. Therefore, the hand must exert a 1.23 N ⋅m 0.22 m=5.6 Nforce to produce a 31-N force at the weed. The force is multiplied by a factor of 22 4 =5.5.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.11 – 2Chapter 11: Rotational Dynamics and Static Equilibrium James S. Walker, Physics, 4th Edition3. Picture the Problem: The arm extends out either horizontally or at some angle below horizontal, and the weight of the trophy is exerted straight downward on the hand.Strategy: The torque equals the moment arm times the force according to equation 11-3. In this case the moment arm is the horizontal distance between the shoulder and the hand, and the force is the downward weight ofthe trophy. Find the horizontal distance in each case and multiply it by the weight of the trophy to find the torque. In part (b) the horizontal distance is r⊥=r cosθ = 0.605 m( )cos22.5°=0.559 m.Solution: 1. (a) Multiply the moment arm by the weight: τ =r⊥mg = 0.605 m( )1.61 kg( )9.81 m/s2( )= 9.56 N ⋅m2. (b) Multiply the moment arm by the weight: τ =r⊥mg = 0.559 m( )1.61 kg( )9.81 m/s2( )= 8.83 N ⋅mInsight: The torque on the arm is reduced as the arm is lowered. The torque is exactly zero when the arm is vertical.4. Picture the Problem: The arm extends out either horizontally and the weight of the crab trap is exerted straight downward on the hand.Strategy: The torque equals the moment arm times the force according to equation 11-3. In this case the moment arm isthe horizontal distance between the shoulder and the hand, and the force is the downward weight of the crab trap.Solution: Multiply the moment arm by the weight: τ =r⊥mg = 0.70 m( )3.6 kg( )9.81 m/s2( )= 25 N ⋅mInsight: If the man bent his elbow and brought his hand up next to his shoulder, the torque on the shoulder would be zero but the force on his hand would remain 35 N or 7.9 lb.5. Picture the Problem: The biceps muscle, the weight of the arm, and the weight of the ball all exert torques on the forearm as depicted at right.Strategy: Use equation 10-3 to determine the torques produced by the biceps muscle, the weight of the forearm, and the weight of the ball. Sum the torques together to find the net torque. According tothe sign convention, torques in the counterclockwise direction are positive, and those in the clockwise direction are negative.Solution: 1. (a) Compute the individual torques using equation 10-3 and sum them: τbiceps=r⊥F = 0.0275 m( )12.6 N( )=0.347 N ⋅mτforearm=r⊥mg = 0.170 m( )1.20 kg( )9.81 m/s2( )=−2.00 N ⋅mτball=r⊥Wball= 0.340 m( )1.42 N( )=−0.483 N ⋅mτ∑=τbiceps+τforearm+τball=+0.347−2.00−0.483 N ⋅m= −2.14 N ⋅m2. (b) Negative net torque means the clockwise direction; the forearm and hand will rotate downward.3. (c) Attaching the biceps farther from the elbow would increase the moment arm and increase the net torque.Insight: The biceps would need to exert a force of at least 90.3 N in order to prevent the arm from rotating downward Copyright © 2010