Question

A uniform circular disc of radius $a$ is taken. A circular portion of radius $b$ has been removed from it as shown in the figure. If the centre of hole is at a distance $c$ from the centre of the disc, the distance $x_2$ of the centre of mass of the remaining part from the initial centre of mass $O$ is given by

• A. $\frac{\pi b^2}{(a^2-c^2)}$
• B. $\frac{c{b}^2}{(a^2-b^2)}$
• C. $\frac{\pi c^2}{(a^2-b^2)}$
• D. $\frac{c{a}^2}{(a^2-b^2)}$

Answer 1

The correct option is B $\frac{c{b}^2}{(a^2-b^2)}$

Consider the mass of the removed portion as negative.
Mass of remaining portion $=m_{rem}$
Mass of removed portion $=m_r$
$X_{com}=\frac{m_{rem}\times x_2-m_r\times c}{m_{rem}-m_r}$
Centre of mass of whole system was at point O and since the disc was uniform,
$x_2\text{(area of remaining portion)}=c\text{(area of removed disc)}$
$x_2(\pi a^2-\pi b^2)=c\left(\pi b^2\right)$
$x_2=\frac{c{b}^2}{a^2-b^2}$