Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 248: 65



Work Step by Step

The total moment of inertia $I=I_{station}+I_{people}$ First, the people stay on the outer surface of the cylinder, so $$I_{people}=\sum MR^2=(70kg)\times500\times(82.5m)^2$$ $$I_{people}=2.38\times10^8kg.m^2$$ So, $I_0=3\times10^9+2.38\times10^8=3.24\times10^9kg.m^2$ After people move to the axis, the distance from people to the axis $R=0$, so $I_{people}=0$ Therefore, $I_f=I_{station}=3\times10^9kg.m^2$ The angular momentum is conserved, so $$I_0\omega_0=I_f\omega_f$$ $$\frac{\omega_f}{\omega_0}\times100\%=\frac{I_0}{I_f}\times100\%=108\%$$ So the station's angular speed can have a maximum increase of $8\%$
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