Question

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.

Answer 1

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`.