Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 8 - Rotational Motion - Problems: 52

Answer

48.77 J

Work Step by Step

The sum of the translational and rotational kinetic energy makes the total kinetic energy. The angular velocity is given by $\omega=\frac{v}{R}$. The rotational inertia of a sphere about an axis through its center is $I=\frac{2}{5}mR^2$. $$KE_{total}=KE_{transl}+KE_{rot}=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=\frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{5}mR^2)\frac{v^2}{R^2}=\frac{7}{10}mv^2=(0.7)(7.25kg)(3.1m/s)^2=48.77J$$
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