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: 9.53


$I = \frac{1}{2}MR^2$

Work Step by Step

Equation (9.20) states: $I = \int~r^2~dm$ Let $\rho$ be the area density of the disk. $I = \int_{0}^{R}~r^2~\rho~(2\pi~r~dr)$ $I = \rho\int_{0}^{R}2\pi~r^3~dr$ $I = \rho~(\frac{\pi~r^4}{2})\vert_{0}^{R}$ $I = \rho~(\frac{\pi~R^4}{2})$ $I = \frac{1}{2}~\rho~(\pi~R^2)~R^2$ $I = \frac{1}{2}~\rho~(Area)~R^2$ $I = \frac{1}{2}MR^2$
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