Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 330: 16


(a) $I = 3.8\times 10^{-5}~kg~m^2$ (b) $I = 1.1\times 10^{-4}~kg~m^2$

Work Step by Step

(a) We can find the moment of inertia of the CD for a perpendicular axis through the center. $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(0.021~kg)(0.060~m)^2$ $I = 3.8\times 10^{-5}~kg~m^2$ (b) We can use the parallel axis theorem to find the moment of inertia about a perpendicular axis that is through the edge of the disk. Note that the distance $d$ from the center of mass to the edge of the disk is 6.0 cm. $I = I_{cm}+Md^2$ $I = 3.8\times 10^{-5}~kg~m^2+(0.021~kg)(0.060~m)^2$ $I = 1.1\times 10^{-4}~kg~m^2$
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