Question

An isosceles right angled triangle of side $5\text{cm}$ is cut from a circular disc of radius $5\text{cm}$ as shown in the figure. What will be the coordinates of the centre of mass of the remaining disc? Origin is taken at the centre of the circular disc. Consider the material to be of uniform density.

• A. $-0.31\text{cm},-0.31\text{cm}$
• B. $-0.50\text{cm},-0.50\text{cm}$
• C. $-0.75\text{cm},-0.31\text{cm}$
• D. $-0.75\text{cm},-0.75\text{cm}$

Answer 1

The correct option is A $-0.31\text{cm},-0.31\text{cm}$

Considering the isosceles right angled triangle as the negative mass superimposed on the whole circular disc, will give the new shape.


Area of circular disc $A_1=\pi \times 5^2$
Area of triangular shape $A_2=\frac{1}{2}\times 5\times 5$

COM of the circular disc $(x_1,y_1)=(0,0)$

COM of the triangular shape is $(x_2,y_2)=(\frac{b}{3},\frac{h}{3})=(\frac{5}{3},\frac{5}{3})$


Thus, COM of new shape is:
$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$=\frac{(\pi \times 5^2\times 0)-(\frac{1}{2}\times 5\times 5\times \frac{5}{3})}{(\pi \times 5^2)-(\frac{1}{2}\times 5\times 5)}$

$\therefore x_{\text{CM}}=\frac{-125}{6\times 66}=-0.31\text{cm}$

Similarly,
$y_{\text{CM}}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}=-0.31\text{cm}$