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: 3

Answer

(a) $v = 1.5~m/s$ (b) The fan turns through 12.5 revolutions while stopping.

Work Step by Step

(a) Initially, the fan rotates at 60 rpm which is 1 rev/s which $2\pi~rad/s$. Therefore $\omega_0 = 2\pi~rad/s$. We can find the rate of angular deceleration as the fan comes to a stop; $\alpha = \frac{\omega_f-\omega_0}{t}$ $\alpha = \frac{0-2\pi~rad/s}{25~s}$ $\alpha = -0.251~rad/s^2$ We can find the angular velocity after 10 seconds; $\omega = \omega_0+at$ $\omega = 2\pi~rad/s-(0.251~rad/s^2)(10~s)$ $\omega = 3.77~rad/s$ We can find the speed of the tip of the blade; $v = \omega~r$ $v = (3.77~rad/s)(0.40~m)$ $v = 1.5~m/s$ (b) We can find $\theta$ while the fan is stopping. $\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$ $\theta = (2\pi~rad/s)(25~s)-\frac{1}{2}(0.251~rad/s^2)(25~s)^2$ $\theta = 78.6~rad$ We can use $\theta$ to find the number of revolutions $N$. $N = (78.6~rad)(\frac{1~rev}{2\pi~rad})$ $N = 12.5~rev$ The fan turns through 12.5 revolutions while stopping.
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