Question

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

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

Answer 1

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

Let the mass per unit area be $\sigma$.
Then the mass of the complete disc
$=\sigma \left[\pi \right(2R{)}^2]=4\pi \sigma R^2$


The mass of the removed disc $=\sigma \left(\pi R^2\right)=\pi \sigma R^2$
Let us consider the above situation to be a complete disc of radius $2R$ on which a disc of radius $R$ of negative mass is superimposed.
Let $O$ be the origin. Then the above figure can be redrawn keeping in mind the concept of mass as:


$X_{c.m}=\frac{\left(4\pi \sigma R^2\right)\times 0+(-\pi \sigma R^2)R}{4\pi \sigma R^2-\pi \sigma R^2}$
$\therefore X_{c.m}=\frac{-\pi \sigma R^2\times R}{3\pi \sigma R^2}$
$\therefore R_{c.m}=-\frac{R}{3}$
Hence distance of COM from centre of big disc is $\frac{R}{3}$
$\Rightarrow \alpha =\frac{1}{3}$

Why this question? "Shift in COM will be away from removed mass which is indicated by negative sign in here."