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 - Page 348: 9


$KE_{rot} = 2.57\times 10^{29}~J$

Work Step by Step

We can use $M_e = 5.97\times 10^{24}~kg$ as the mass of the earth. We can use $r = 6.38\times 10^6~m$ as the radius of the earth. We can find the rotational inertia of the earth as: $I = \frac{2}{5}M_e~r^2$ $I = \frac{2}{5}(5.97\times 10^{24}~kg)(6.38\times 10^6~m)^2$ $I = 9.72\times 10^{37}~kg~m^2$ We can find the angular velocity of the earth as it rotates as: $\omega = \frac{2\pi~rad}{(24~hr)(3600~s/hr)}$ $\omega = 7.27\times 10^{-5}~rad/s$ We can find the rotational kinetic energy of the earth; $KE_{rot} = \frac{1}{2}I\omega^2$ $KE_{rot} = \frac{1}{2}(9.72\times 10^{37}~kg~m^2)(7.27\times 10^{-5}~rad/s)^2$ $KE_{rot} = 2.57\times 10^{29}~J$
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