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: 64

Answer

(a) $\Delta \theta = (\frac{\omega_0~T}{2})~rad$ (b) The turbine rotates through 19000 revolutions as it coasts to a stop.

Work Step by Step

(a) The turbine coasts to a stop from $\omega_0~rad/s$. During this time, the average angular speed is $\frac{\omega_0~rad/s+0}{2}$ which is $\frac{\omega_0~rad/s}{2}$. We can find the angle through which the turbine rotates as it stops: $\Delta \theta = (\frac{\omega_0~rad/s}{2})(T~s)$ $\Delta \theta = (\frac{\omega_0~T}{2})~rad$ (b) The turbine coasts to a stop from 3800 rpm. During this time, the average angular speed is $\frac{3800~rpm+0}{2}$ which is 1900 rpm. We can find the number of revolutions the turbine rotates through as it stops: $\theta = (1900~rpm)(10~min) = 19000~rev$ The turbine rotates through 19000 revolutions as it coasts to a stop.
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