Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 247: 46



Work Step by Step

The parallel axis theorem states that $$I=I_{cm}+Mh^2$$ For a solid cylinder of radius $R$, we have $I_{cm}=\frac{1}{2}MR^2$ The axis in this exercise lies on the surface of the cylinder, so the distance between that axis and the axis passing through the center of mass $h=R$ $$I=\frac{1}{2}MR^2+MR^2=\frac{3}{2}MR^2$$
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.