Question

A circular plate of uniform thickness has a diameter of $56\text{cm}$. A circular portion of diameter $42\text{cm}$ is removed from the edge of the plate. The position of centre of mass of the remaining portion is

• A. $5\text{cm}$ from centre of bigger circle to the left.
• B. $8\text{cm}$ from centre of bigger circle to the right.
• C. $10\text{cm}$ from centre of bigger circle to the left.
• D. $9\text{cm}$ from centre of bigger circle to the left.

Answer 1

The correct option is D $9\text{cm}$ from centre of bigger circle to the left.

Taking origin at $O$ as shown in figure. Remaining portion can be obtained by superimposing a disc ( diameter $42\text{cm}$) of negative mass and a disc of positive mass ( diameter $56\text{cm}$).


For bigger disc, $x_{\text{com}}=-28\text{cm}$
For smaller disc, $x_{\text{com}}=-21\text{cm}$
$\Rightarrow$Since the obtained shape body is symmetrical about $x-$axis, the $\text{COM}$ of system will lie along $x-$axis only, i.e $y_{\text{com}}=0$
Assuming the uniform surface density for the circular plate.
The $x-$coordinate of centre of mass is given by -
$x_{\text{com}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$A_1=\text{area of bigger disc}$
$A_2=\text{area of smaller disc}$
$\Rightarrow x_{\text{com}}=\frac{\left[\right(\pi \times {28}^2)\times (-28\left)\right]-\left[\right(\pi \times {21}^2)\times (-21\left)\right]}{(\pi \times {28}^2)-(\pi \times {21}^2)}$
$\Rightarrow x_{\text{com}}=-37\text{cm}$
Position of $\text{COM}$ of system from centre of bigger disc is $\Rightarrow -37-(-28)=-9\text{cm}$
Here, $-9\text{cm}$ shows that $\text{COM}$ of the remaining portion is at a distance of $9\text{cm}$ on left side from centre of bigger circular plate.