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: 67

Answer

$\alpha = 0.75~rad/s^2$

Work Step by Step

We can find the average angular speed during the period of acceleration: $\omega_{ave} = \frac{3.5~rev/s+6.0~rev/s}{2} = 4.75~rev/s$ We can find the average speed of the car during the acceleration period: $v_{ave} = \omega_{ave}~(2\pi~r)$ $v_{ave} = (4.75~rev/s)[(2\pi)(0.32)~m/rev]$ $v_{ave} = 9.55~m/s$ We can find the time to travel 200 meters. $t = \frac{x}{v_{ave}} = \frac{200~m}{9.55~m/s}$ $t = 20.9 ~s$ We can find the angular acceleration. $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{(6.0~rev/s-3.5~rev/s)(2\pi ~rad/rev)}{20.9~s}$ $\alpha = 0.75~rad/s^2$
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