Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 10 - Section 10.3 - Rotational Inertia and the Analog of Newton’s Law - Example - Page 174: 10.4


$.29 \ kgm^2$

Work Step by Step

Recall, the moment of inertia is equal to the sum of the moments of inertia of every aspect of the object, including the masses. Thus, we find: $I = m(\frac{1}{4}L)^2 + m(\frac{3}{4}L)^2 = \frac{5mL^2}{8}= \frac{5(.64)(.85)^2}{8}= .29 \ kgm^2$
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.