College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 5 - Problems - Page 187: 44


The angular acceleration is $0.39~rad/s^2$

Work Step by Step

We can express the angular speed in units of $rad/s$: $\omega = 0.50~rev/s \times \frac{2\pi~rad}{1~rev} = \pi~rad/s$ We can express the angular displacement in units of radians: $\Delta \theta = 2.0~rev \times \frac{2\pi~rad}{1~rev} = 4\pi~rad$ We can find the angular acceleration $\alpha$: $\omega_f^2 = \omega_0^2+2\alpha~\Delta \theta$ $\alpha = \frac{\omega_f^2 - \omega_0^2}{2~\Delta \theta}$ $\alpha = \frac{(\pi~rad/s)^2 - 0}{(2)(4\pi~rad)}$ $\alpha = 0.39~rad/s^2$ The angular acceleration is $0.39~rad/s^2$
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