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Question

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.
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Solution

I=Iremain+IR/3
Iremian=IIR/3
Iremian=MR22⎢ ⎢ ⎢ ⎢ ⎢M4(R3)22+M4(R3)2⎥ ⎥ ⎥ ⎥ ⎥
=MR22[MR272+3MR236]
=MR22[MR2+2MR272]
=MR223MR272
=36MR23MR272
=1124MR2
Hence, the answer is 1124MR2.


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