## 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 - Page 112: 30

#### Answer

(a) $\omega = 3.75~rad/s^2$ (b) $\omega = 5.0~rad/s^2$ (c) $\omega = 5.0~rad/s^2$

#### Work Step by Step

The change in angular velocity is equal to the area under the angular acceleration versus time graph. Note that the initial angular velocity is zero. From t = 0 to t = 2 s: $\alpha(t) = (\frac{-5}{2}~t+5)~rad/s^2$ $\omega(t) = \omega_0+\int_{0}^{t}\alpha(t)~dt$ $\omega(t) = 0+\int_{0}^{t}(\frac{-5}{2}~t+5)~rad/s^2~dt$ $\omega(t) = (\frac{-5}{4}~t^2+5t)~rad/s$ (a) At t = 1 s: $\omega = [\frac{-5}{4}~(1)^2+(5)(1)]~rad/s$ $\omega = 3.75~rad/s^2$ (b) At t = 2 s: $\omega = [\frac{-5}{4}~(2)^2+(5)(2)]~rad/s$ $\omega = 5.0~rad/s^2$ (c) From t = 2 s to t = 3 s, the area under the angular acceleration versus time graph is zero. Therefore, the angular velocity does not change from t = 2 s to t = 3 s. At t = 3 s: $\omega = 5.0~rad/s^2$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.