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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Medium2 minPYQ_2021

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A solid sphere of mass $M$ , radius $R$ has moment of inertia ' $I$ ' about its diameter. It is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains ' $I$ '. Radius of the disc will be

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Answer:

Solution:

For solid sphere, $I = \frac{2}{5} MR^{2}$ For disc, $I^{'} = \frac{MR ^{'2}}{2} + MR^{'2} = \frac{3}{2} MR^{'2}$ $∵ I^{'} = I ∴ \frac{3}{2} M R^{'2} = \frac{2}{5} M^{2} R^{'} = \frac{2}{15} R$

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

⏱ 2mℹ️ Source: PYQ_2021

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