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: 32

Answer

.0086 percent

Work Step by Step

The translational Kinetic energy is: $K_t=\frac{1}{2}(.15)(33^2)=81.675 \ J$ The rotational Kinetic energy is: $K_r=\frac{1}{2}I\omega^2=\frac{1}{2}\frac{2}{5}(.15)(.037^2)(42^2)=.0724$ Thus, we see that .0086 percent of the energy is rotational.
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