Question

A uniform disc of mass $m$ and radius $R$ is cut from the right side of the larger disc of mass $M$ and radius $2R$ and kept on the left side as shown in the figure (shaded portion). Find the $COM$ of the system w.r.t the origin $\text{O}$.

• A. $(\frac{R}{3},0)$
• B. $(\frac{R}{2},0)$
• C. $(\frac{-R}{3},0)$
• D. $(\frac{-R}{2},0)$

Answer 1

The correct option is D $(\frac{-R}{2},0)$

From the data given in the question,
$\begin{array}{ccc}& \text{Mass}& \text{COM}\\ \text{Larger disc}& \text{M}& (0,0)\\ \text{Removed disc}& \text{m}& (R,0)\\ \text{Inserted disc}& \text{m}& (-R,0)\end{array}$

From symmetry, we can say that $y_{COM}$ will not change. Mass density is uniform.
So, $\frac{M}{\pi (2R{)}^2}=\frac{m}{\pi R^2}$
$\Rightarrow m=\frac{M}{4}$

Now,
$x_{COM}=\frac{M_1{x}_1+M_2{x}_2+M_3{x}_3}{M_1+M_2+M_3}$
Here, $x_1=$ $x$ coordinate of $COM$ of the original disc
$M_1=$ Mass of original disc
$x_2=$ $x$ coordinate of $COM$ of the removed disc
$M_2=$ Mass of removed disc
$x_3=$ $x$ coordinate of $COM$ of the inserted disc
$M_3=$ Mass of inserted disc

So,
$x_{COM}=\frac{M\left(0\right)+(-m)\left(R\right)+\left(m\right)(-R)}{M+(-m)+\left(m\right)}$
$=\frac{-2mR}{M}=\frac{-2\times \frac{M}{4}\times R}{M}$
$x_{COM}=\frac{-R}{2}$

Hence, $COM$ of the system shifts towards left from the origin by $\frac{R}{2}$