A circular plate of unifrom thickness has a diameter of `56 cm`. A circular portion of diameter `42 cm` is removed from the edge of the plate as shown in Fig. Find the position of centre of mass of the remaining portion.

Explanation

Suppose mass per unit area of circular plate `= m`.
`:.` Total mass of circular plate,
`M = m xx pi(d//2)^(2) = pi m((56)/(2))^(2) = 784 pi m`
This is supposed to be concentrated at the centre `O` of the disc.
Mass of cut portion, `M_(1) = m xx pi ((d_(1))/(2))^(2) = pi m(42//2)^(2) = 441 pi m`
This is supposed to be concentrated at `O_(1)`, where `O_(1) O = AO - AO_(1) = 28 - 21 = 7 cm`.
Mass of remaining portion of disc, `M_(2) = M - M_(1) = 784 pi m - 441pi m = 343 pi m`
Let it be concentrated at `O_(2)`, where `OO_(2) = x`.
As `x_(cm) = (M_1)x_(1) + (M_(2)x_(2))/(M_(1) + M_(2))` and `x_(cm) = 0 (at O) :. M_(1)x_(1) + M_(2)x_(2) = 0`
`441pi m xx 7+ 343 pi m x = 0` or `x = (-441pi m xx 7)/(343 pi m) =- 9cm`
Hence, centre of mass of remaining portion of disc is at `9 cm` from centre of disc on the right side of `O`.