From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Explanation

[Surface density = `("mass")/("area")`]

Disc of radius R is a system of disc `(R )/(2)` and system made by remaining portion.
`therefore M=m_(1)+m_(2)`
Suppose the mass of disc of radius R is `M=piR^(2)rho` and coordinate of its centre O is `vec(r_(cm))=(0,0)` unit mass of disc of radius `(R )/(2),m_(1)=(piR^(2)rho)/(4)`
`therefore m_(1)=(M)/(4) [because piR^(2)rho=M]`
and it coordinate of it centre A is `vec(r_(1))=((R)/(2),0)` unit Now the mass of remaining portion `m_(2)` after the cut of disc of radius `(R)/(2)` from the the disc of radius R
`m_(2)=M-m_(1) [because M=m_(1)+m_(2)]`
`therefore m_(2)=M-(M)/(4)`
`therefore m_(2)=(3M)/(4)`
Suppose coordinate of `Bvec(r_(2))=(x,0)` unit.
From the defination of centre of mass
`Mvec(r_(cm))=m_(1)vec(r_(1))+m_(2)vec(r_(2))`
`therefore M(0,0)=(M)/(4)((R)/(2),0)+(3M)/(4)(x,0)`
Comparing X-coordinate
`therefore 0=(M)/(4)xx(R)/(2)+(3Mx)/(4)`
`therefore 0=(R)/(2)+3x [because" Dividing "(M)/(4)]`
`therefore 3x=-(R)/(2)`
`therefore x=-(R)/(6)`
Coordinate of `B,vec(r_(2))=(x,0)`
`vec(r_(2))=(-(R)/(6),0)` unit