A uniform square plate and a disc having same mass per unit area are kept in contact as shown in Fig. The side of square and diameter of circle are both equal to `L`. Locate the position of centre of mass of the system w.r.t. the centre of the square.

Explanation

Let, `sigma` be the mass per unit area
Square is treated as a point mass `m_(1)=a^(2)sigma` at its geometric centre.
Disc is treated as another point mass `m_(2)=(pia^(2)sigma)/(4)` at its centre.
If `m_(1)` is taken at the origin and line joining `m_(1)` and `m_(2)` as `X`-axis the position of centre of mass.
`X=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=(a^(2)sigmaxx0+(pia^(2))/(4)sigmaa)/(a^(2)sigma+(pia^(2))/(4)sigma)`
`=((pi)/(4)a)/(1+(pi)/(4))=(pia)/(pi+4)`
The centre of mass is at a distance `((pia)/(pi+4))` from the centre of square on the line joining the centres of the two objects.