## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 187: 47

#### Answer

(a) The magnitude of the angular acceleration is $73.3~rad/s^2$ (b) The washer makes 23.3 revolutions.

#### Work Step by Step

(a) We can express the angular velocity in units of $rad/s$: $\omega = 1400~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 146.6~rad/s$ We can find the angular acceleration $\alpha$: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{146.6~rad/s - 0}{2.0~s}$ $\alpha = 73.3~rad/s^2$ The magnitude of the angular acceleration is $73.3~rad/s^2$ (b) We can find the angular displacement in this time: $\Delta \theta = \frac{1}{2}\alpha~t^2$ $\Delta \theta = \frac{1}{2}(73.3~rad/s^2)(2.0~s)^2$ $\Delta \theta = 146.6~rad$ We can express the angular displacement in units of revolutions: $146.6~rad \times \frac{1~rev}{2\pi~rad} = 23.3~rev$ The washer makes 23.3 revolutions.

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