Find the position of center of mass of the uniform lamina shown in figure, if small disc of radius `alpha/2` is cut from disc of radius a.

Explanation

Here, `A_1` = area of complete circular disc = pia^2 A2 = area of small circular disc =`pi(a/2)^2` = `pia^2 (x_1, y_1)` = coordinates of centre of mass of large circular disc = (0, 0) and `(x_2, y_2)` = coordinates of centre of mass of small circular disc = `(a/2,0)` Using `x_(cm) =(A_1x_1-A_2x_2)/(A_1-A_2)` We get `x_(cm)=(-(pia^2)/4(a/2))/(pia^2-(pia^2)/4)=(-1/8)/(3/4)a=-a/6 and y_(cm) = 0 as y_1and y-2` both are zero. Therefor, coordinatesof CM of the circular disc shown in figure are (-a/6,0)