Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 10 - Section 10.3 - Rotational Inertia and the Analog of Newton’s Law - Example - Page 182: 10.7

Answer

$\frac{1}{2}MR^2 $

Work Step by Step

We know the following value for $dm$: $ dm = \frac{2Mrdr}{R^2}$ Thus, we find: $I = \int_{0}^{R}r^2dm=I = \int_{0}^{R}r^2 \frac{2Mr}{R^2}dr = \frac{1}{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.