## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 296: 9.34

#### Answer

(a) $K = 1.33\times 10^6~J$ (b) The new angular speed is 2770 rpm.

#### Work Step by Step

(a) We can find the angular speed $\omega$ $\omega = (2400~rpm)(2~\pi~rad)(\frac{1~min}{60~s})$ $\omega = 80~\pi~rad/s$ We can find the rotational kinetic energy. $K = \frac{1}{2}I\omega^2$ $K = \frac{1}{2}(\frac{1}{12}mL^2)\omega^2$ $K = \frac{1}{2}[\frac{1}{12}(117~kg)(2.08~m)^2](80~\pi~rad/s)^2$ $K = 1.33\times 10^6~J$ (b) We can find the new angular speed $\omega_2$. $\frac{1}{2}(\frac{1}{12}(0.75~m)L^2)\omega_2^2 = \frac{1}{2}(\frac{1}{12}mL^2)\omega_1^2$ $0.75~\omega_2^2 = \omega_1^2$ $\omega_2 = \frac{\omega_1}{\sqrt{0.75}}$ $\omega_2 = \frac{2400~rpm}{\sqrt{0.75}}$ $\omega_2 = 2770~rpm$ The new angular speed is 2770 rpm.

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