Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 10 - Exercises and Problems - Page 185: 31


(a) $3.2\times 10^{38}Kgm^2$ (b) $1.8\times 10^{34}Nm$

Work Step by Step

(a) We know that Rotational inertial of solid sphere$=\frac{2}{5}MR^2$ We plug in the known values to obtain: Rotational inertial of solid sphere$=\frac{2}{5}(2.3\times 1.99\times 10^3)(13155)^2=3.2\times 10^{38}Kgm^2$ (b) We can find the torque as follows $\tau=I\alpha $ We plug in the known values to obtain: $\tau=3.2\times 10^{38}\times 5.6\times 10^{-5}=1.8\times 10^{34}Nm$
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