Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 51

Answer

$I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$

Work Step by Step

To find the moment of inertia of the object, we can add the moment of inertia of the rod, the moment of inertia of $M_1$, and the moment of inertia of $M_2$. Therefore: $I =\frac{1}{12}ML^2+M_1(\frac{L}{2})^2+M_2(\frac{L}{4})^2$ $I = \frac{1}{12}ML^2+\frac{1}{4}M_1~L^2+\frac{1}{16}M_2~L^2$
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.