From a uniform disc of radius R, a small disc of radius `(R)/(2)` is cut and removed as shown in the diagram. Find the center of mass of the remaining portion of the disc.

Explanation

Let us consider the mass of the uncut full disc be M. Its center of mass would be at the geometric center of the disc on which the origin coincides. Let the mass of the small disc cut and removed be mn and its center of mass is at a position `(R)/(2)` to the right of the origin as shown in the figurr.

Hence, the remaining portion of the disc should have its center of mass to the left of the origin, say, at a distance x. We can write from the principle of moments,
`(M-m)x=(R)/(2)`
`x=((m)/(M-m))(R)/(2)`
If `sigma` is the surface mass density (i.e., mass per unit surface area). `sigma (M)/(piR^(2))` of small disc is,surface mass density surface area
m= surface mass density `xx` surface area
`m= sigma xx pi ((R)/(2))^(2)`
`m=((M)/(pi R^(2)))pi((R)/(2))^(2)=(M)/(pi R^(2)) pi(R^(2))/(4)=(M)/(4)`
substituting m in the expression for x `x=((M)/(4))/((M-(M)/(4)))xx(R)/(2)=((M)/(4))/(((3M)/(4)))xx(R)/(2)`
`x=(R)/(6)`
he center of mass of the remaining portion is at a distance `(R)/(6)` to the left from the center of the disc.