Question

A circular hole of radius $3cm$ is cut out from a uniform circular disc of radius $6cm$. The centre of the hole is at $3cm$, from the centre of the original disc. The distance of centre of gravity of the resulting flat body from the centre of the original disc is:

• A. $0.5cm$
• B. $1cm$
• C. $1.5cm$
• D. $0.75cm$

Answer 1

Answer: (B)

Let mass per unit area of the disc $=m$
Total mass of disc, $M=\pi R^{2}m$
Mass of the scooped out portion of the disc, $M^{\prime }=\pi r^{2}m$
For small objects, centre of gravity and centre of mass are at same position.
We take the centre $O$ of the disc as the origin.
The masses $M$ and $M^{\prime }$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively.
The mass $M^{\prime }$ of the removed portion is taken negative.

The $x$-coordinate of the centre of mass of the remaining portion of the disc will be
$x_{CM}=\frac{M{x}_1-M^{\prime }{x}_2}{M-M^{\prime }}=\frac{M\times 0-M^{\prime }\times 3}{\pi R^{2}m-\pi r^{2}m}=\frac{-3M^{\prime }}{\pi m\left(R^2-r^2\right)}$
$=\frac{-3\pi r^{2}m}{\pi m\left(R^2-r^2\right)}=-\frac{-3r^2}{R^2-r^2}=\frac{-3\times 3^2}{6^2-3^2}=-1cm$
$\Rightarrow x_{CM}=-lcm$
Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be $1cm$ left side.