Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 10 - Exercises and Problems - Page 192: 27

Answer

$2.6\times10^{19} \ Nm$

Work Step by Step

a) Using the radius of the earth and the equation for the moment of inertia of a solid sphere, we obtain: $ I =\frac{2}{5}mr^2 = \frac{2}{5}(5.97\times10^{24})(6.37\times10^6)^2=9.69\times10^{27}$ b) Using unit conversions, we know that this rotation corresponds to an angular acceleration of $2.68\times10^{-19} \ rads/s^2$ Thus, we find the torque using the rotational version of Newton's second law: $\tau = I\alpha=( 9.69\times10^{27})(2.68\times10^{-19} \ rads/s^2)=2.6\times10^{19} \ Nm$
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