Essential University Physics: Volume 1 (4th Edition)

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


$\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 $
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