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 circle `= pia^(2)`
`A_(2)`=area of small circle.
=`pi(a/2)^(2)` = `(pia^(2))/4`
`(x_(1),y_(1))` = coordinates of center of mass of large circle = (0,0)
`(x^(2),y^(2))` = coordinates of center of mass of small circle =`(alpha/2,0)`
Using, O `x_(CM)` = `(A_(1)x_(1)-A_(2)x_(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_(1) and `y_(2)` both are zero. Therefore, coordinates of CM of the lamina shown in figure are `(-a/6,0)`.