## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 17 - Planar Kinetics of a Rigid Body: Force and Acceleration - Section 17.1 - Mass Moment of Inertia - Problems - Page 419: 5

#### Answer

$I_{y}=\frac{2}{5}mr^{5}$

#### Work Step by Step

$m=\int \rho dV=\rho \int dV= \rho \int \pi x^2 dy=\rho \pi \int (r^2-y^2)dy=\rho \pi(r^3-\frac{1}{3}r^3)=\frac{2}{3} \rho \pi r^3$ $I_{y}=\int dm(\frac{1}{2} x^2 )=\frac{\rho}{2} \int x^4 \pi dy=\frac{\rho}{2} \int (r^2-y^2)^2 \pi dy=\frac{\rho \pi}{2}(r^4y-\frac{2}{3}r^2y^3+\frac{y^5}{5})$ Now fill in $r$ for $y$ (integral). $I_{y}=\frac{2}{5} \rho \pi r^5$ $I_{y}=\frac{2}{5}mr^2$

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