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_____
Let $\sigma$ be the mass density of circular disc.
Original mass of the disc, $m_{0}=\pi a^{2} \sigma$
Removed mass, $m=\frac{a^{2}}{4} \sigma$
Remaining, mass, $m^{\prime}=\left(\pi a^{2}-\frac{a^{2}}{4}\right) \sigma$
$=a^{2}\left(\frac{4 \pi-1}{4}\right) \sigma$
New position of centre of mass
$X_{C M}=\frac{m_{0} x_{0}-m x}{m_{0}-m}=\frac{\pi a^{2} \times 0-\frac{a^{2}}{4} \times \frac{a}{2}}{\pi a^{2}-\frac{a^{2}}{4}}$
$=\frac{-a^{3} / 8}{\left(\pi-\frac{1}{4}\right) a^{2}}=\frac{-a}{2(4 \pi-1)}=\frac{-a}{8 \pi-2}=-\frac{a}{23}$
$\therefore x=23$
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