## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 187: 45

#### Answer

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 $4.0~rad/s^2$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.