Question

Figure given below shows a uniform disc of radius $8$ units. A hole of radius $4$ units has been cut out from left of its centre and is placed on the right of the centre of the disc. Find the $COM$ of the resulting shape.

• A. $(4,0)$
• B. $(2,0)$
• C. $(6,0)$
• D. $(3,0)$

Answer 1

The correct option is B $(2,0)$

Let mass of complete disc be $M$
Mass of cut out disc is $M_1=\frac{M}{\pi \times 8^2}\times \pi \times 4^2$
$\Rightarrow M_1=\frac{M}{4}$
Hence mass of the hole is $-M/4$ (negative mass for removed part)

Coordinate of $COM$ of complete disc $\left(M\right)$ are $(0,0)$
Coordinates of $COM$ of hole $\left(M_1\right)$ are $(-4,0)$
Coordinates of $COM$ of disc $\left(M_2\right)$ which is placed on the right are $(4,0)$

Then, $COM$ of the system is given by
$X_{COM}=\frac{M{X}_1+M_2{X}_2-M_1{X}_1}{M+M_2-M_1}$
$=\frac{M\times 0+\left(M/4\right)\left(4\right)-\left(M/4\right)(-4)}{M+M/4-M/4}=\frac{2M}{M}$
$\Rightarrow X_{COM}=2$
and $Y_{COM}=0$ [as figure is symmetric about $x$-axis]

Therefore, the coordinates of $COM$ of the system are $(2,0)$