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: 39

Answer

$\omega = 98~rpm$

Work Step by Step

The change in angular velocity is equal to the area under the angular acceleration versus time graph from t = 0 to t = 3.0 s. $\Delta \omega = \frac{1}{2}(4~rad/s^2)(2~s)$ $\Delta \omega = 4~rad/s$ We can convert this to units of rpm. $\Delta \omega = (4~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$ $\Delta \omega = 38~rpm$ We can find the angular velocity at t = 3.0 s. $\omega = \omega_0+\Delta \omega$ $\omega = 60~rpm+38~rpm$ $\omega = 98~rpm$
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