A circular portion of radius $R_2$ has been removed from one edge of a circular disc of radius $R_1$. The correct expression for the centre of mass for the remaining portion of the disc is

• Correct answer
  A

  $-\frac{R_2^2}{R_1+R_2}$
• B

  $-\frac{R_2^2}{R_1-R_2}$
• C

  $\frac{R_2^2}{R_1+R_2}$
• D

  $-\frac{R_1^2}{R_1+R_2}$

Solution

Given situation is shown below.
Here we treat removed mass as negative mass. As solid is symmetrical about $X$ - axis, its CM lies along $X$ - axis. Now $x_{\mathrm{CM}}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}$ $=\frac{d \times \pi R_1^2 \times 0-d \times \pi R_2^2\left(R_1-R_2\right)}{d \times \pi R_1^2-d \times \pi R_2^2}$ Here, $d=$ Area wise mass density $\Rightarrow x_{\mathrm{CM}}=\frac{-R_2^2\left(R_1-R_2\right)}{R_1^2-R_2^2}$ $=\frac{-R_2^2\left(R_1-R_2\right)}{\left(R_1+R_2\right)\left(R_1-R_2\right)}=\frac{-R_2^2}{R_1+R_2^{\prime}}$

Hence, the correct answer is option: 'A'.