Question

A circular hole of radius $3\, cm$ is cut out from a uniform circular disc of radius $6\, cm$. The centre of the hole is at $3 \,cm$, 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. $0.5 \,cm$
• B. $1\, cm$
• C. $1.5\, cm$
• D. $0.75 \,cm$

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, $M'=\pi r^{2} m$
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 $M'$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively.
The mass $M'$ 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_{C M}=\frac{M x_{1}-M' x_{2}}{M-M'}=\frac{M \times 0-M' \times 3}{\pi R^{2} m-\pi r^{2} m}=\frac{-3 M'}{\pi m\left(R^{2}-r^{2}\right)} $
$=\frac{-3 \pi r^{2} m}{\pi m\left(R^{2}-r^{2}\right)}=-\frac{-3 r^{2}}{R^{2}-r^{2}}=\frac{-3 \times 3^{2}}{6^{2}-3^{2}}=-1\, cm $
$\Rightarrow x_{C M}=-l \,cm$
Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be $1 \,cm$ left side.