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 - Page 223: 30

Answer

$1.81 kg \cdot m^2$.

Work Step by Step

Assume this is a solid sphere. It is rotating around an axis passing through its center. $$I=\frac{2}{5}MR^2=\frac{2}{5}(10.8 kg)(0.648m)^2=1.81 kg \cdot m^2$$
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.