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: 40


$c=\frac{2}{3}$ So, the ball is hollow.

Work Step by Step

We know that the translational kinetic energy is given as $K_t=\frac{1}{2}mv^2$ The rotational kinetic energy is given as $K_r=\frac{1}{2}I\omega^2=\frac{1}{2}(cmr^2)(\frac{v}{r})^2=\frac{1}{2}cmv^2$ Now the total kinetic energy is: $K_{total}=K_t+K_r$ $K_{total}=\frac{1}{2}mv^2+\frac{1}{2}cmv^2=\frac{1}{2}(c+1)mv^2$ Thus, the required ratio is $\frac{K_r}{K_{total}}=\frac{40}{100}=0.4=\frac{\frac{1}{2}cmv^2}{\frac{1}{2}(c+1)mv^2}$ This simplifies to: $c=\frac{0.4}{0.6}$ $c=\frac{2}{3}$ So, the ball is hollow.
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