Question

From a uniform disk of radius $R$, a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc.Locate the centre of gravity of the resulting flat body.

Answer 1

If the mass per unit area of the disc is $m$,

then mass of the portion remove from the disc is $M^{\prime }=\pi (R/2{)}^2\times m=(\pi R^2/4)m=M/4$

In figure centre of mass of $M$ is at $O$ and for $M^{\prime }$ is at $O^{\prime }$

But $O{O}^{\prime }=R/2$.

After mass $M^{\prime }$ is removed the remaining portion can be taken as two masses $M$ at $O$ and $-M^{\prime }=-M/4$ at $O^{\prime }$ we taking $-M^{\prime }$ because we are removing this mass.

distance of centre of gravity(P) of remaining part is :

$X=\frac{M\times 0-M^{\prime }\times R/2}{M+M^{\prime }}$

$X=\frac{-M/4\times R/2}{M-M/4}$

$X=-R/6$

Minus sign indicates that $P$ is to the left of $O$.