From a disc of radius Rand mass M a circular hole of diameterR whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Question

From a disc of radius Rand mass M a circular hole of diameterR whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

Infinity Learn · Accepted Answer

Mass per unit area of disc $$=M$$$$π$$$$R2Mass$$ of removed portion of disc,M'$$=M$$$$π$$$$R2$$×π$$R22=M4Moment$$ of inertia of removed portion about an axis passing through centre of disc O and perpendicular to the plane of disc,IO'$$=IO$$'$$+M$$'$$d2=12$$×$$M4$$×$$R22+M4$$×$$R22=MR232+MR216=3MR232When$$ portion of disc would not have been removed, the moment of inertia of complete disc about centre O $$isIO=12MR2$$ So, moment of inertia of the disc with removed portion is $$I=IO-IO$$'$$=12MR2-3MR232=13MR232$$

From a disc of radius $R$ and mass $M$ a circular hole of diameter $R$ whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

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