Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems - Page 224: 40

Answer

(a) $0.068~m\cdot N$ (b) Because of the friction from the potter's hands, the wheel would stop in 16 seconds.

Work Step by Step

(a) $\tau = r\cdot F = (0.045~m)(1.5~N) = 0.068~m\cdot N$ (b) We can find the magnitude of angular deceleration if the only torque is from friction. $I\alpha = \tau$ $\alpha = \frac{\tau}{I} = \frac{0.068~m\cdot N}{0.11~kg\cdot m^2}$ $\alpha = 0.618~rad/s^2$ We can use $\alpha$ to find the time to stop: $t = \frac{\Delta \omega}{\alpha} = \frac{0-(1.6~rev/s)(2\pi~rad/rev)}{-0.618~rad/s^2}$ $t = 16~s$ Because of the friction from the potter's hands, the wheel would stop in 16 seconds.
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