## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 11 - Exercises and Problems - Page 203: 61

#### Answer

$\omega = \frac{mgr}{L }$

#### Work Step by Step

Examining the values of omega and the angular displacement, we find the following equation: $\omega = \frac{\Delta L}{Lsin\theta \Delta t}$ We know that the value of torque is $\tau=\frac{\Delta L}{\Delta t}$. Thus: $\omega = \frac{\tau}{Lsin\theta }$ Using the definition of torque as well as the fact that the force acting is gravity, which equals mg, it follows: $\omega = \frac{mgrsin\theta}{Lsin\theta }$ $\omega = \frac{mgr}{L }$

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