Question

Figure shows a uniform disc of radius $R$ from which a hole of radius $R/2$ has been cut out from left of the centre and is placed on the right of the centre of the disc. Find the $CM$ of the resulting disc.

Answer 1

Mass of the cut-out disc is
$m=\frac{M}{\pi R^2}\times \pi {\left(\frac{R}{2}\right)}^2=\frac{M}{4}$
Let centre of the disc is at the origin of the coordinates. Then we can write the $CM$ of the system as
$x_{CM}=\frac{M\overrightarrow{R}-m\overrightarrow{r}+m\overrightarrow{r}}{M-m+m}=\frac{M\times 0-\frac{M}{4}\left(\frac{-R}{2}\right)+\frac{M}{4}\left(\frac{R}{2}\right)}{M-\frac{M}{4}+\frac{M}{4}}=\frac{R}{4}$
$y_{CM}=0$