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

Question

Infinity Learn · Accepted Answer

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 $$Is=I$$−$$Ic=12MR2$$−$$332MR2=1332MR2$$.

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

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