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

Published by Pearson

# Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 298: 9.57

#### Answer

The piece travels a horizontal distance of 7.68 meters.

#### Work Step by Step

We can find the angular speed $\omega$ when the piece breaks loose. $\omega^2 = 2\theta \alpha$ $\omega = \sqrt{2\theta \alpha}$ $\omega = \sqrt{(2)(155~rev)(2.00~rev/s^2)}$ $\omega = 24.9~rev/s$ We can find the horizontal speed when the piece breaks off. $v = (24.9~rev/s)(2\pi)(0.120~m)$ $v = 18.77~m/s$ We can find the time it takes the piece to drop a vertical distance of 0.820 meters. $y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(0.820~m)}{9.80~m/s^2}}$ $t = 0.409~s$ We can find the horizontal distance $x$. $x = v~t = (18.77~m/s)(0.409~s)$ $x = 7.68~m$ The piece travels a horizontal distance of 7.68 meters.

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