Question

A circular plate of uniform thickness has a diameter of $28\text{cm}$. A circular portion of diameter $21\text{cm}$ is removed from the plate as shown. Take the origin at the center of mass of the complete plate. The position of the center of mass of the remaining portion will shift towards left from the origin by

• A. $5\text{cm}$
• B. $9\text{cm}$
• C. $4.5\text{cm}$
• D. $5.5\text{cm}$

Answer 1

The correct option is C $4.5\text{cm}$


Let $x_1$ be the position of the COM of the removed portion and $x_2$ be the position of the center of mass of the remaining portion. (taking origin at point $\text{O}$)
Due to symmetry, $Y_{COM}$ will be zero.

Area of circular portion which is removed $A_1=\pi (10.5{)}^2$
Area of portion remaining $A_2=\pi ({14}^2-{10.5}^2)$

Hence, $X_{COM}=\frac{A_1{x}_1+A_2{x}_2}{A_1+A_2}$
Here, $x_1=14-10.5=3.5\text{cm}$

Since centre of mass of the complete plate is at the origin
$0=\frac{\pi (10.5{)}^2\times 3.5+\pi ({14}^2-{10.5}^2)x_2}{A_1+A_2}$
$x_2=-\frac{(10.5{)}^2\times 3.5}{{14}^2-{10.5}^2}=4.5\text{cm}$
$x_2=-4.5\text{cm}$ {-ve sign is showing it is towards left of origin }