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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Hard2 minPYQ_2020

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

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Solution:

$I_{T otal disc} = \frac{MR ^{2}}{2}$ As mass is proportional to area, $M_{Removed} = \frac{M}{4}$ Now, about the same perpendicular axis: $I_{Removed} = \frac{M}{4} \frac{( R /2 ) ^{2}}{2} + \frac{M}{4} (\frac{R}{2})^{2} = \frac{3 MR ^{2}}{32}$ $\Rightarrow I_{Remaining Disc} = I_{Total} - I_{Removed}$ $= \frac{MR ^{2}}{2} - \frac{3 MR ^{2}}{32}$ $= \frac{13 MR ^{2}}{32}$

Stream:MHT_CETSubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2020

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