Question

From a uniform disc of radius $R$, a small disc of radius $\frac{R}{2}$ has been cut out from the left having centre at $(-\frac{R}{2},0)$ and is placed on the right with centre at $(\frac{R}{2},0)$ as shown in the figure. Find the centre of mass of the resulting system.

• A. $x_{\text{com}}=\frac{R}{2}$
• B. $x_{\text{com}}=\frac{R}{4}$
• C. $y_{\text{com}}=0$
• D. $y_{\text{com}}=\frac{-R}{2}$

Answer 1

Answer: (B), (C)

$y_{\text{com}}=0$


Let $M$ be the total mass of disc.
Mass of the cut out portion $\left(m\right)$,
$m=\frac{M}{\pi R^2}.\pi {\left(\frac{R}{2}\right)}^2$
$\Rightarrow m=\frac{M}{4}$

After the smaller disc has been cut from the original and is placed on the right of the centre of the disc, the remaining portion is considered to be a system of three masses.
The three masses are:
$M$ concentrated at $(0,0)$,
$-m(=M/4)$ concentrated at $(-\frac{R}{2},0)$ and
$m(=M/4)$ concentrated at $(\frac{R}{2},0)$

COM of the new system:
$x_{\text{com}}=\frac{M{x}_1-m{x}_2+m{x}_3}{M-m+m}[\because x_1=0,x_2=\frac{-R}{2},x_3=\frac{R}{2}]$
$x_{\text{com}}=\frac{M\left(0\right)-\frac{M}{4}(-\frac{R}{2})+\frac{M}{4}\left(\frac{R}{2}\right)}{M}=\frac{R}{4},y_{\text{com}}=0\text{[by symmetry]}$

Thus, option (b) and (c) are the correct answers.