Question

A square shaped hole of side $l=\frac{a}{2}$ is carved out at a distance $d=\frac{a}{2}$ from the centre ‘O’ of a uniform circular disk of radius $a$. If the distance of the centre of mass of the remaining portion from O is$-\frac{a}{X}$, value of $X$ (to the nearest integer) is

Answer 1

Let $\sigma$ be the mass density of circular disc.

Original mass of the disc, $m_0=\pi a^2\sigma$
Removed mass, $m^{\prime }=\left(\frac{a^2}{4}\right)\sigma$


New position of centre of mass
$X_{CM}=\frac{m_0{x}_0-mx}{m_0-m}=\frac{\pi a^2\times 0-\frac{a^2}{4}\times \frac{a}{2}}{\pi a^2-\frac{a^2}{4}}$
$X_{CM}=\frac{\frac{-a^3}{8}}{(\pi -\frac{1}{4}{a}^2)}=\frac{-a}{2(4\pi -1)}=\frac{-a}{8\pi -2}=-\frac{a}{23}$

$\therefore X=23$.