A thin homogeneous lamina is in the form of a circular disc of radius R. From it a circular hole exactly hall the radius of the lamina and touching the lamina's circumference is cut off. Find the centre of mass of the remaining part.

Explanation

Let the centre of the lamina be the origin. Due to symmetry, the C.M. will lie on the x - axis. Let m be the mass of the circular lamina. Then, mass m. of the removed circular hole is
`m.=(M)/(pi R^(2))(pi r^(2))=(M)/(R^(2))((R )/(2))^(2)=(M)/(4)`
`therefore x_(cm)=(M.0-m..r)/(M-m.)`
The negative sign of m. denotes that it has been removed.
`x_(cm)=(-(M)/(4)(R )/(2))/(M-(M)/(4))=-(R )/(6)`
Thus, the centre of mass of the remaining part lies at a distance R/6 towards left of the origin, i.e., the initial centre of mass of the disc.