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: 25


A torque of $0.28~N~m$ will bring the balls to a halt.

Work Step by Step

We can express the initial angular velocity in units of rad/s. Note that since the rotation is clockwise, the angular velocity is negative. $\omega_0 = -(20~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega_0 = -2.094~rad/s$ We can find the angular acceleration if the object comes to rest in 5.0 seconds. $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{0-(-2.094~rad/s)}{5.0~s}$ $\alpha = 0.419~rad/s^2$ We can find the moment of inertia of the object. Note that the center of mass is 33 cm from the 2.0-kg ball. $I = m_1~R_1^2+m_2~R_2^2$ $I = (2.0~kg)(0.33~m)^2+(1.0~kg)(0.67~m)^2$ $I = 0.67~kg~m^2$ We can find the required torque. $\tau = I~\alpha$ $\tau = (0.67~kg~m^2)(0.419~rad/s^2)$ $\tau = 0.28~N~m$ A torque of $0.28~N~m$ will bring the balls to a halt.
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