Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 289: 37

Answer

$I = 0.097~kg~m^2$

Work Step by Step

We can use the parallel axis theorem to find the rotational inertia: $I = I_{com}+Mh^2$ $I = \frac{1}{12}ML^2+Mh^2$ $I = \frac{1}{12}(0.56~kg)(1.0~m)^2+(0.56~kg)(0.30~m)^2$ $I = 0.097~kg~m^2$
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