Question

A disc of radius R is cut out from a larger uniform disc of radius $2R$ in such a way that the edge of the hole touches the edge of the disc. Locate the center of mass of the remaining part.

Answer 1

Let $\sigma$ is mass per unit area.

Mass of disc of radius $2R$ is $M_1=\sigma \times 4\pi R^2$

Mass of disc of radius $R$ is $M_2=\sigma \times \pi R^2$

Centre of mass, $xcm=\frac{M_1{x}_1-M_2{x}_2}{M_1-M_2}$

$\Rightarrow x_{cm}=\frac{4\sigma \pi R^2\times 0-\sigma \pi R^2\times R}{4\sigma \pi R^2-\sigma \pi R^2}$ $\Rightarrow x_{cm}=-\frac{R}{3}$

Hence, $x_{cm}=\frac{R}{3}$ left to the centre of large disc as shown in figure.