From a uniform circular disc of diameter `d,a` circular disc of diameter `d//6` and having centre at a distance `d//4` from the centre of the disc is scooped out. Determine the centre of mass of remaining portion.

Explanation

Let `m` be the mass per unit area of the disc.

`:.` Mass of disc `M = m xx (piD^(2))/(4)`
It is concentrated at the centre `O` of the disc, Fig.
Mass of cut off portion,
`m_(1) = m xx (pi(D//6)^(2))/(4) = (mpi D^(2))/(144) = (M)/(36)`
It is concentrated at `O_(1)` , where `OO_(1) = D//4`.
Let the centre of mass of remaining portion of disc be at `O_(2)` where `OO_(2) = x`
Mass of remaining portion of disc
`m_(2) = m xx(piD^(2))/(4) - m(piD^(2))/(144) = (35M)/(36)`
Clearly, `m_(1) xx (D)/(4) = m_(2)x`
`x = (m_(1))/(m_(2))(D)/(4)`
`x = (M//36)/(35M//36) xx (D)/(4) = (D)/(35xx4) = (D)/(140)`
`:.` Centre of mass of remaining portion of disc is at `D//140` from centre `O` of the disc, to the left side.