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# ch09 - CHAPTER 9 ROTATIONAL DYNAMICS ANSWERS TO FOCUS ON...

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Unformatted text preview: CHAPTER 9 ROTATIONAL DYNAMICS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ___________________________________________________________________________________________ 1. (d) A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. A body, such as a bicycle wheel, can be moving, but the translational and angular accelerations must be zero ( a = 0 m/s 2 and α = 0 rad/s 2 ). 2. (e) As discussed in Section 9.2, a body is in equilibrium if the sum of the externally applied forces is zero and the sum of the externally applied torques is zero. 3. (b) The torque τ 3 is greater than τ 2 , because the lever arm for the force F 3 is greater than that for F 2 . The lines of action for the forces F 1 and F 4 pass through the axis of rotation. Therefore, the lever arms for these forces are zero, and the forces produce no torque. 4. (b) Since the counterclockwise direction is the positive direction for torque, the torque produced by the force F 1 is τ 1 = - (20.0 N)(0.500 m) and that produced by F 2 is τ 2 = +(35.0 N)[(1.10 m)(cos 30.0 ° )]. The sum of these torques is the net torque. 5. (e) The clockwise torque produced by F 2 is balanced by the counterclockwise torque produced by F . The torque produced by F 2 is (remembering that the counterclockwise direction is positive) τ 2 = + F 2 [(80.0 cm - 20.0 cm)(sin 55.0 ° )], and the torque produced by F is τ = - (175 N)(20.0 cm). Setting the sum of these torques equal to zero and solving for F 2 gives the answer. 6. (d) The sum of the forces ( F- 2 F + F ) equals zero. Select an axis that passes through the center of the puck and is perpendicular to the screen. The sum of the torques [- FR + 2 F (0) + FR ] equals zero, where R is the radius of the puck. Thus, the puck is in equilibrium. 7. Magnitude of F 1 = 12.0 N, Magnitude of F 2 = 24.0 N 8. (c) The horizontal component of F 3 is balanced by F 1 , and the vertical component of F 3 is balanced by F 2 . Thus, the net force and, hence, the translational acceleration of the box, is zero. For an axis of rotation at the center of the box and perpendicular to the screen, the forces F 2 and F 3 produce no torque, because their lines of action pass through the axis. The force F 1 does produce a torque about the axis, so the net torque is not zero and the box will have an angular acceleration. 179 ROTATIONAL DYNAMICS 9. Distance of center of gravity from support = 0.60 m. 10. (b) The moment of inertia of each particle is given by Equation 9.6 as 2 I mr = , where m is its mass and r is the perpendicular distance of the particle from the axis. Using this equation, the moment of inertia of each particle is: A: 2 0 0 10 m r , B: 2 0 0 8 m r , C: 2 0 0 9 m r ....
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ch09 - CHAPTER 9 ROTATIONAL DYNAMICS ANSWERS TO FOCUS ON...

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