Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 10 - Exercises and Problems - Page 185: 49


a) $\frac{2ML^2}{3} $ b) $\frac{2ML^2}{3} $ c) $\frac{4ML^2}{3} $

Work Step by Step

a) The rotational inertia depends on whether the rod is perpendicular or parallel to the axis of rotation. Thus, using the equations for moment of inertia, we find: $I=(\frac{1}{12}+\frac{1}{4})mL^2 =\frac{ML^2}{3} $ Since there are two of each rod, the total inertia is: $\frac{2ML^2}{3} $. b) While this is oriented differently, the mass of the uniform square is the same distance away on average, so the inertia is the same. Thus: $I= \frac{2ML^2}{3} $. c) We find the inertia of all of the four rods: $I=4(\frac{1}{12}+\frac{1}{4})mL^2 =\frac{4ML^2}{3} $
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