Question

From a uniform circular disc of radius $20\text{cm}$, two small circular discs of radius $10\text{cm}$ are removed in such a way that both have common tangent as shown in the figure given below. The distance of centre of mass of the remaining part from the centre of original disc is:

• A. $10\text{cm}$
• B. $0\text{cm}$
• C. $5\text{cm}$
• D. $20\text{cm}$

Answer 1

The correct option is B $0\text{cm}$

Given radius of original disc, $R=20\text{cm}$
Let, mass of original disc be $M$
Let the mass density of the disc be $\sigma$
$\Rightarrow \sigma \times \left(\pi R^2\right)=M$
$\Rightarrow \sigma =\frac{M}{\pi R^2}$

Mass of removed disc is $\sigma \times \frac{\pi R^2}{4}=\frac{M}{4}$
Both removed discs have mass $\frac{M}{4}$ each.
Let COM of the original disc be at the origin.


$x$-Coordinate of $COM$ of the remaining part is given by :-
$x_{COM}=\frac{M\left(0\right)-\frac{M}{4}\left(\frac{R}{2}\right)-\frac{M}{4}(-\frac{R}{2})}{M-\frac{M}{4}-\frac{M}{4}}$
$=\frac{0-\frac{MR}{8}+\frac{MR}{8}}{M-\frac{M}{2}}=0$
$\Rightarrow x_{COM}=0$
Similarly,
$y$-coordinate of $COM$ of remaining portion will be the same that of original disc.
$\Rightarrow y_{COM}=0$

i.e center of mass of the remaining part lies at the center of original disc.
So, the distance of the centre of mass of the remaining part from the center of mass of the original disc is $0$.