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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Medium2 minPYQ_2020

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A solid sphere of mass 'M' and radius 'R' has moment of inertia 'l' 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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Solution:

Moment of inertia of solid sphere is $I_{s} = \frac{2}{5} MR^{2}$ Moment of inertia of the disc is given by about the given axis is $I_{d} ∴ r^{2} ∴ r = \frac{3}{2} Mr^{2} = \frac{4}{15} R^{2} ∴ \frac{3}{2} Mr^{2} = \frac{2}{5} MR^{2} = \frac{2 R}{15}$

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

⏱ 2mℹ️ Source: PYQ_2020

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