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

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 63

Answer

(a) $v = 12~m/s$ (b) The disk rotates through 36 revolutions.

Work Step by Step

(a) We can find the angular speed after the $\frac{1}{2}~second$ acceleration period. $\omega = \alpha ~t$ $\omega = (600~rad/s^2)(\frac{1}{2}~s)$ $\omega = 300~rad/s$ We can find the speed of the dot on the edge of the disk. $v = \omega ~r$ $v = (300~rad/s)(0.040~m)$ $v = 12~m/s$ (b) We can find the angle $\theta_1$ through which the disk rotates in the first $\frac{1}{2}~second.$ $\theta_1 = \frac{1}{2}\alpha~t^2$ $\theta_1 = \frac{1}{2}(600~rad/s^2)(0.5~s)^2$ $\theta_1 = 75~rad$ We can find the angle $\theta_2$ through which the disk rotates in the next $\frac{1}{2}~second.$ $\theta_2 = \omega_f~t$ $\theta_2 = (300~rad/s)(0.5~s)$ $\theta_2 = 150~rad$ The disk rotates through a total angle of $75~rad+150~rad$ which is $225~rad$. We can express this in revolutions. $\theta = (225~rad)(\frac{1~rev}{2\pi~rad})$ $\theta = 36~rev$ The disk rotates through 36 revolutions.
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