## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 9 - Exercises and Problems: 43

#### Answer

$(0,\frac{h}{4})$

#### Work Step by Step

We follow the book's advice to integrate the mass element over dy. Using the volume equation as well as the density, $\rho$, we find: $M = \int_0^h \frac{\pi r^2 \rho}{h^2} y^2dy$ $M = \frac{\pi r^2 \rho h^3}{3h^2}$ $M = \frac{\pi r^2 \rho h}{3}$ Using the integral equation for the center of mass and plugging in the above value for M, we find that the y center of mass is: $y_{cm} = \frac{h}{4}$ We know that the shape is symmetrical around the center in terms of x, so we find that the x center of mass is 0.

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