From a uniform disc of radius `R`, a circular section 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 mass of the resulting flat body.

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

Answer

B

Explanation

(b) For a disc, masss is proportional to surface area. So Mass of cut portion : `m_1 = k pi r^2`,
Mass of remaining portion : `m_2 = k pi(R^2 - r^2)`
Apply `m_1 x_1 = m_2 x_2`, where `x_1 = R//2`, and solve to get
`x_2 = (R r^2)/(2(R^2 - r^2))`.