Question

Find the position of centre of mass of a uniform disc of radius $R$ from which a hole of radius $r$ is cut out. The centre of the hole is at a distance $R / 2$ from the centre of the disc.

• A. $\frac{R r^{2}}{2\left(R^{2}-r^{2}\right)}$ towards right of $O$
• B. $\frac{R r^{2}}{2\left(R^{2}-r^{2}\right)}$ towards left of $O$
• C. $\frac{2 R r^{2}}{\left(R^{2}+r^{2}\right)}$ towards right of $O$
• D. $\frac{2 R r^{2}}{\left(R^{2}+r^{2}\right)}$ towards left of $O$

Answer 1

Answer: (B)

For a disc, mass is proportional to surface area. So
Mass of cut portion: $m_{1}=k \pi r^{2}$,
Mass of remaining portion: $m_{2}=k \pi\left(R^{2}-r^{2}\right)$
Apply $m_{1} x_{1}=m_{2} x_{2}$, where $x_{1}=R / 2$, and solve to get
$x_{2}=\frac{R r^{2}}{2\left(R^{2}-r^{2}\right)}$