Question

Find the position of centre of mass of the uniform lamina shown in Fig.

Answer 1

Here, $A_1=$ area of complete circle $=\pi a^2$
$A_2=$ area of small circle $=\pi \left(\frac{a}{2}\right)=\frac{\pi a^2}{4}$
$(x_1,y_1)=$ coordination of centre of mass of the large circle $=(0,0)$
$(x_2,y_2)=$ coordination of centre of mass of the small circle $=(\frac{a}{2},0)$
Using $x_{CM}=(A_1{x}_1-A_2{x}_2)/(A_1-A_2)$, we get
$x_{CM}=\frac{\pi a^2\times 0-\frac{\pi a^2}{4}\times \frac{a}{2}}{\pi a^2-\frac{\pi a^2}{4}}=-\frac{a}{6}$
$y_{CM}=0$ (as $y_1$ and $y_2$ both are zero)
Therefore, coordinates of $CM$ of the lamina shown in Fig. are $(-a/6,0)$.