Question

A square of side $2.5\text{cm}$ is cut out from a circular disc of radius $10\text{cm}$ as shown in the figure. Take the origin at the centre of the circular disc. What will be the coordinates of the centre of mass of this new shape with respect to the origin? Consider the material used to be of uniform density.

• A. $(-0.325\text{cm},-0.425\text{cm})$
• B. $(-0.025\text{cm},-0.025\text{cm})$
• C. $(-2.25\text{cm},-1.025\text{cm})$
• D. $(-2.5\text{cm},-2.5\text{cm})$

Answer 1

The correct option is B $(-0.025\text{cm},-0.025\text{cm})$

Considering the removed square shape as negative mass superimposed on the whole circular disc, will give the new shape.



Area of circular disc $A_1=\pi \times {10}^2{\text{cm}}^2$
Area of square $A_2=(2.5{)}^2{\text{cm}}^2$

Taking origin at $O$,
COM of circular disc is $(x_1,y_1)=(0,0)$
and COM of the cut square is $(x_2,y_2)=(\frac{2.5}{2}\text{cm},\frac{2.5}{2}\text{cm})$

Therefore, coordinates of COM of the whole system is given by:
$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}=\frac{(\pi \times {10}^2)\times 0-(2.5{)}^2\times \frac{2.5}{2}}{(\pi \times {10}^2)-(2.5{)}^2}$

$\therefore x_{\text{CM}}=-0.025\text{cm}$

Similarly,
$y_{\text{CM}}=-0.025\text{cm}$

$\therefore (x_{\text{CM}},y_{\text{CM}})=(-0.025\text{cm},-0.025\text{cm})$