## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 7 - Problems - Page 269: 98

#### Answer

The center of mass of the disk now that the hole has been drilled is $(0.5~cm,0)$

#### Work Step by Step

Let $M$ be the mass of an entire disk with radius 3 cm. We can find an expression for the area of the missing piece which has a radius of 1.5 cm. $\pi(\frac{r}{2})^2 = \frac{\pi~r^2}{4}$ Since this area is $\frac{1}{4}$ of the area of the entire disk of radius 3 cm, the mass of the missing piece is $\frac{M}{4}$. Then the mass of the remaining part of the disk is $\frac{3M}{4}$ The center of mass of the entire disk would be the origin. Let $x$ be the x-coordinate of the center of mass of the remaining part of the disk with a mass of $\frac{3M}{4}$. We can find $x$: $\frac{(\frac{3M}{4})(x)+(\frac{M}{4})(-1.5~cm)}{\frac{3M}{4}+\frac{M}{4}} = 0$ $\frac{3x}{4} = \frac{1.5~cm}{4}$ $x = 0.5~cm$ By symmetry, the y-coordinate of the center of mass is zero. The center of mass of the disk now that the hole has been drilled is $(0.5~cm, 0)$.

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