## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 10

#### Answer

$v = 2.4~m/s$

#### Work Step by Step

We can find the rotational inertia of the disk as: $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(0.10~kg)(0.040~m)^2$ $I = 8.0\times 10^{-5}~kg~m^2$ We can use the rotational kinetic energy to find the angular velocity. $KE_{rot} = 0.15~J$ $\frac{1}{2}I~\omega^2 = 0.15~J$ $\omega^2 = \frac{(2)(0.15~J)}{I}$ $\omega = \sqrt{\frac{(2)(0.15~J)}{I}}$ $\omega = \sqrt{\frac{(2)(0.15~J)}{8.0\times 10^{-5}~kg~m^2}}$ $\omega = 61.2~rad/s$ We can find the speed of a point on the rim. $v = \omega ~R$ $v = (61.2~rad/s)(0.040~m)$ $v = 2.4~m/s$

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