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 181: 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: $I = m(\frac{1}{4}L)^2 + m(\frac{3}{4}L)^2 = \frac{5mL^2}{8}= \frac{5(.64)(.85)^2}{8}= .29 \ kgm^2$
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