A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is

Moment of inertia of complete disc about O is I=12MR2. Mass of the cut-out part is m=M4. The moment of inertia of the cut-out portion about its own centre I0=12mr2=12M4R22=132MR2 because r = R/2. From the parallel axes theorem, the moment of inertia of the cut out portion about O is Ic=I0+mr2=132MR2+M4R22=332MR2∴ Moment of inertia of the shaded portion about O is Is=I−Ic=12MR2−332MR2=1332MR2.