Question

Figure shows a uniform disc of radius R, from which a hole of radius $\frac{R}{2}$ has been cut out from left of the center and is placed on right of the center of disc. Find the centre of mass of the resulting disc.

• A. $(\frac{R}{4},0)$
• B. $(\frac{-R}{4},0)$
• C. $(\frac{R}{6},0)$
• D. $(\frac{-R}{6},0)$

Answer 1

The correct option is A

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

Mass of the cut out disc
$m=\frac{M}{\pi R^2}\times \pi {\left(\frac{R}{2}\right)}^2=\frac{M}{4}$
let center of the disc be at the origin of coordinates, then we can write the C.M. of the system as;
${\overrightarrow{x}}_{cm}=\frac{\overrightarrow{MR}-\overrightarrow{mr}+\overrightarrow{m{r}^{\prime }}}{M-m+m}=\frac{MO-\frac{M}{4}\left(\frac{-R}{2}\right)+\frac{M}{4}\left(\frac{R}{2}\right)}{M-\frac{M}{4}+\frac{M}{4}}=\frac{R}{4}and{y}_{cm}=0$