Find the position of centre of mass of a uniform disc of radius R from which a hole of radius is cut out. The centre of the hole is at a distance R/2 from the centre of the disc.

Options

A. `(Rr^(2))/(2 (R^(2) - r^(2))` towards right of O
B. `(Rr^(2))/(2(R^(2) - r^(2))` towards left of O
C. `(2Rr^(2))/((R^(2) + r^(2))` towards right of O
D. `(2Rr^(2))/((R^(2) + r^(2))` towards left of O

Answer

B

Explanation

For a disc, mass is proportional to surface area. So
Mass of cut portion : `m_(1) = k pi r^(2)`
Mass of remaining portion : `m_(2) = kpi, (R^(2) - r^(2))`
Applying `m_(1) x_(1) = m_(2)x_(2)`, where `x_(1) = R//2`, and solving to get
`x_(2) = (Rr^(2))/(2 (R^(2) - r^(2)))`