University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 297: 9.52


(a) $I = \frac{1}{12}ML^2$ (b) $I = \frac{1}{12}ML^2$

Work Step by Step

(a) The moment of inertia for a thin rod rotating around the center is $I = \frac{1}{12}ML^2$. If the rod is bent into a $90^{\circ}$ angle, the mass is still spread out at the same distance from the axis. Therefore, the moment of inertia does not change. $I = \frac{1}{12}ML^2$ (b) When the rod is bent, let's draw one side on the +x-axis and one side on the +y-axis with the vertex at the origin. Then the center of mass is at the point (L/8, L/8). The midpoint of the line connecting the two ends is at the point (L/4, L/4). Note that the distance from this point to the center of mass is the same as the distance from the origin to the center of mass. Therefore, by the parallel axis theorem, the moment of inertia for this axis is the same as part (a): $I = \frac{1}{12}ML^2$.
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