A uniform disc of radius `R=20cm` has a round cut as shown in figure. The mass of the remaining (shaded) portion of the disc equals `m=7.3kg`. Find the moment of inertia of such a disc relative to the axis passing through its centre of inertia and perpendicular to the plane of the disc.

Explanation

For simplicity let us use a mathematical trick. We consider the portion of the given disc as the superposition of two complete discs (whithout holes), one of positive density and radius R and other of negative density but of same magnitude and radius `R//2`.
As (area) `alpha` (mass), the respective masses of the considered discs are `(4m//3)` and `(-m//3)` respectively, and these masses can be imagined to be situated at their respective centres (C.M). Let us take point O as origin and point x-axis towards right. Obviously the C.M. of the shaded position of given shape lies on the x-axis. Hence the C.M. (C) of the shaded portion is given by
`x_c=((-m//3)(-R//2)+(4m//3)0)/((-m//3)+4m//3)=R/6`
Thus C.M. of the shape is at a distance `R//6` from point O toward x-axis.
Using parallel axis theorem bearing in mind that the moment of inertia of a complete homogeneous disc of radius `m_0` and radius `r_0` equals `1/2m_0r_0^2`. The moment of inertia of the small disc of mass `(-m//3)` and radius `R//2` about the axis passing through point C and perpendicular to the plane of the disc
`I_(2C)=1/2(-m/3)(R/2)^2+(-m/3)(R/2+R/6)^2`
`=-(mR^2)/(24)-(4)/(27)mR^2`
Similarly `I_(1C)=1/2((4m)/(3))R^2+((4m)/(3))(R/6)^2`
`=2/3mR^2+(mR^2)/(27)`
Thus the sought moment of inertia,
`I_C=I_(1C)+I_(2C)=(15)/(24)mR^2-(3)/(27)mR^2=(37)/(72)mR^2`