Question

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B $\frac{1}{3}$

Let the circular disc be placed in the $x-y$ plane.

Let the center of the complete disc of radius $2R$ be at origin and the center of the removed disc be on $x=R$. Let the mass of complete disc be $M$.

Then, position vector of center of mass of smaller disc is ${\overrightarrow{x}}_1=R\hat{i}$.

Let the center of mass of the new disc be at $(x,y)$. Then, its position vector is given by ${\overrightarrow{x}}_2=x\hat{i}+y\hat{j}$

Mass of disc is proportional to its area.

$\frac{M_1}{M}=\frac{A_1}{A}=\frac{\pi R^2}{\pi 4R^2}=\frac{1}{4}$

Mass of new disc is given by $M_2=M-M_1=\frac{3}{4}M$

Now, position vector of center of mass of complete disc is given by the formula,

${\overrightarrow{x}}_{cm}=\frac{M_1{\overrightarrow{x}}_1+M_2{\overrightarrow{x}}_2}{M_1+M_2}$

$\therefore 0=\frac{\frac{M}{4}R\hat{i}+\frac{3M}{4}(x\hat{i}+y\hat{j})}{M}$

Equating the co-efficients of $\hat{i}$ and $\hat{j}$ from both sides, we get:

$x=-\frac{R}{3},y=0$

Distance from center of big disc, $d=\frac{R}{3}=\alpha R$, given.

So, $\alpha =\frac{1}{3}$