Question

A circular plate of uniform density has a diameter of $56\text{cm}$. A circular portion of diameter $42\text{cm}$ is removed from the original circular plate as shown in figure. Find the position of the $\text{COM}$ of the remaining portion, from the centre of the original plate.

• A. $7\text{cm}$ towards left
• B. $7\text{cm}$ towards right
• C. $9\text{cm}$ towards left
• D. $9\text{cm}$ towards right

Answer 1

The correct option is C $9\text{cm}$ towards left

Let $M$ be the mass of parent/original circular plate with radius$28\text{cm}$
$M_1$ be the mass of removed plate

Let mass per unit area be $m$, then
$M=m.\pi (28\times {10}^{-2}{)}^2\text{kg}$
$M_1=m\times \pi (21\times {10}^{-2}{)}^2\text{kg}$
$\therefore \frac{M}{M_1}={\left(\frac{28}{21}\right)}^2=\frac{7^2\times 4^2}{7^2\times 3^2}=\frac{16}{9}...\left(1\right)$

Replacing the removed mass by $-ve$ mass and assuming the masses to be at their respective centres of mass, $\text{COM}$ of remaining part is given by:
$x_{\text{com}}=\frac{M{x}_1-M_1{x}_2}{M-M_1}$

Here:
$\begin{array}{ccc}& \text{Mass}& \text{x-coordinate of COM from origin}O\\ \text{Parent}& M& 0\\ \text{Removed}& M_1& +7\text{cm}\end{array}$


$O\to$ origin taken at centre of original plate

From Eq. $\left(i\right)$ and $\left(ii\right)$:
$x_{\text{com}}=\frac{M\left(0\right)-\left(M_1\right)\left(7\right)}{M-M_1}$
$\Rightarrow x_{\text{com}}=\frac{(-M_1)\left(7\right)}{\frac{16}{9}{M}_1-M_1}$
$\Rightarrow x_{\text{com}}=\frac{-7M_1}{\frac{7M_1}{9}}$
$\therefore x_{\text{com}}=-9\text{cm}$

Due to symmetry along $x-$axis, $\text{COM}$ of remaining plate will lie along $x-$axis at distance $9\text{cm}$ towards left from the centre of original plate.