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

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 65

#### Answer

(a) The angular acceleration is $-101~rad/s^2$ (b) The drill rotates through 50 revolutions as it stops.

#### Work Step by Step

(a) We can convert 2400 rpm to units of rad/s: $\omega = (2400~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 80\pi~rad/s$ We can find the angular acceleration. $\alpha = \frac{\omega-\omega_0}{t}$ $\alpha = \frac{0-80\pi~rad/s}{2.5~s}$ $\alpha = -101~rad/s^2$ The angular acceleration is $-101~rad/s^2$. (b) $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$ $\theta = (80\pi~rad/s)(2.5~s)+\frac{1}{2}(-101~rad/s^2)(2.5~s)^2$ $\theta = 313~rad$ We can convert the angle from radians to revolutions. $\theta = (313~rad)(\frac{1~rev}{2\pi~rad})$ $\theta = 50~rev$ The drill rotates through 50 revolutions as it stops.

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