Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 75

Answer

The angular velocity of the tooth at t = 20 seconds is 2210 rpm.

Work Step by Step

We can find the angular velocity of the tooth at t = 20 seconds as; $\omega = \omega_0+\int_{0}^{20}\alpha(t)~dt$ $\omega = \omega_0+\int_{0}^{20}(20-t)~dt$ $\omega = \omega_0+(20t-\frac{1}{2}t^2)\vert_{0}^{20}$ $\omega = \omega_0+[(20)(20)-\frac{1}{2}(20)^2]$ $\omega = 300~rpm+(200~rad/s)$ $\omega = 300~rpm+(200~rad/s)(\frac{60~s}{1~min})(\frac{1~rev}{2\pi~rad})$ $\omega = 300~rpm+1910~rpm$ $\omega = 2210~rpm$ The angular velocity of the tooth at t = 20 seconds is 2210 rpm.
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