Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 349: 22


(a) $\tau_g = 34.3~N~m$ (b) $\tau_g = 24.3~N~m$

Work Step by Step

(a) The gravitational torque comes from the center of mass of the arm $m_a$ and the ball $m_b$ in the person's hand. We can find the magnitude of the gravitational torque; $\tau_g = r_a\times m_a~g+r_b\times m_b~g$ $\tau_g = (0.35~m)(4.0~kg)(9.80~m/s^2)+ (0.70~m)(3.0~kg)(9.80~m/s^2)$ $\tau_g = 34.3~N~m$ (b) We can find the magnitude of the gravitational torque when the arm is at an angle of $45^{\circ}$: $\tau_g = r_a\times m_a~g+r_b\times m_b~g$ $\tau_g = (0.35~m)(4.0~kg)(9.80~m/s^2)~sin(45^{\circ})+ (0.70~m)(3.0~kg)(9.80~m/s^2)~sin(45^{\circ})$ $\tau_g = 24.3~N~m$
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