College Physics (4th Edition)

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

Chapter 5 - Problems - Page 187: 45


The angular acceleration is $4.0~rad/s^2$

Work Step by Step

We can express the angular displacement in units of radians: $\Delta \theta = 8.0~rev \times \frac{2\pi~rad}{1~rev} = 16\pi~rad$ We can find the angular acceleration $\alpha$: $\Delta \theta = \frac{1}{2}\alpha~t^2$ $\alpha = \frac{2\Delta \theta}{t^2}$ $\alpha = \frac{(2)(16\pi~rad)}{(5.0~s)^2}$ $\alpha = 4.0~rad/s^2$ The angular acceleration is $4.0~rad/s^2$
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