## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 350: 53

#### Answer

(a) $I = 0.010~kg~m^2$ (b) $I = 0.030~kg~m^2$

#### Work Step by Step

(a) We can find the moment of inertia for an axis through the center. $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(2.0~kg)(0.10~m)^2$ $I = 0.010~kg~m^2$ (b) We can use the parallel axis theorem to find the moment of inertia for an axis through the edge of the disk. Note that the distance $d$ from the center of mass to the edge is 10 cm. $I = I_{cm}+Md^2$ $I = 0.010~kg~m^2+(2.0~kg)(0.10~m)^2$ $I = 0.030~kg~m^2$

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