## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 5 - Problems - Page 185: 5

#### Answer

(a) The wheel rotates through an angle of $157~rad$ in $1.0~s$ (b) The linear speed of a point on the rim is $47.1~m/s$ (c) The frequency of rotation is $25.0~s^{-1}$

#### Work Step by Step

(a) We can find the angular speed: $\omega = \frac{(2.0~rev)(2\pi~rad/rev)}{0.080~s} = 157~rad/s$ We can find the angle through which the wheel rotates in $1.0~s$: $\theta = \omega ~t = (157~rad/s)(1.0~s) = 157~rad$ The wheel rotates through an angle of $157~rad$ in $1.0~s$ (b) We can find the linear speed of a point on the rim: $v = \omega~r = (157~rad/s)(0.30~m) = 47.1~m/s$ The linear speed of a point on the rim is $47.1~m/s$ (c) We can find the frequency of rotation: $f = \frac{\omega}{2\pi} = \frac{157~rad/s}{2\pi} = 25.0~s^{-1}$ The frequency of rotation is $25.0~s^{-1}$

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.