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

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 the centre of mass of the large circle
`=(0,0)`
`(x_(2),y_(2))=` coordinates of centre of mass of the small circle
`=(a/2,0)`
Using `x_(CM)=(A_(1)x_(1)-A_(2)x_(2))(A_(1)-A_(2))` we get
`x_(CM)=(pia^(2)xx0-(pia^(2))/4xxa/2)/(pia^(2)-(pia^(2))/3)=-a/6`
`y_(CM)=`(as `y_(1)` and `y_(2)` both are zero)
Therefore coordinates of CM of the lamina shown in figure are `(-a//6, 0)`