A uniform disc of radius R has a round disc of radius R/3 cut as shown in Fig. The mass of the remaining (shaded) portion of the disc equals M. Find the moment of inertia of such a disc relative to the axis passing through geometrical centre of original disc and perpendicular to the plane of the disc.

Explanation

Let the mass per unit area of the material of disc be `sigma`. Now the empty space can be considered as having density – `sigma` and `sigma`.
Now `I_0 = I_(sigma)+ I_(- sigma)`
`I_(sigma) = ( sigma pi R^2)R^2 //2 = M.I` of `sigma` about O m
`I_(-sigma) = (-sigma pi (R//3)^2(R//3)^3)/2 + [- sigma(R//3)^2](2R//3)^2`
= M.I or `- sigma` about 0
`:." "I_(0) = 4/9 sigma pi R^4`