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 12 - Rotation of a Rigid Body - Exercises and Problems: 2

Answer

(a) $\alpha = 420~rad/s^2$ (b) The drill turns through 8.4 revolutions in the first 0.50 seconds.

Work Step by Step

(a) We first convert the final angular velocity to units of rad/s as: $\omega_f = (2000~rpm)(\frac{1~min}{60~s})(\frac{2\pi~rad}{rev})$ $\omega_f = 209.4~rad/s$ We then find the angular acceleration.: $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{209.4~rad/s-0}{0.50~s}$ $\alpha = 420~rad/s^2$ (b) We first find $\theta$ during the first 0.50 seconds; $\theta = \frac{1}{2}\alpha~t^2$ $\theta = \frac{1}{2}(420~rad/s^2)(0.50~s)^2$ $\theta = 52.5~rad$ We then use $\theta$ to find the number of revolutions $N$: $N = (52.5~rad)(\frac{1~rev}{2\pi~rad})$ $N = 8.4~rev$ The drill turns through 8.4 revolutions in the first 0.50 seconds.
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