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 $
Essential University Physics: Volume 1 (4th Edition)
by Wolfson, Richard
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
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