Essential University Physics: Volume 1 (4th Edition)

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

Chapter 10 - Exercises and Problems - Page 192: 32


.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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.