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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2020

PhysicsEasysingle choice

The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is 'I'. Now, the rod is bent in a ring and its moment of inertia about diameter is ' $I_{1}^{'}$ . Then $\frac{I}{I _{1}}$ is

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Solution:

If $M$ is the mass of the rod and $L$ is its length, then $I = \frac{M L ^{2}}{3}$ Radius of the circular ring is given by $2 π r = L ∴ I_{1} ∴ \frac{1}{I _{1}} = \frac{L}{2 π} = \frac{M r ^{2}}{2} = \frac{M}{2} \cdot \frac{L ^{2}}{4 π ^{2}} = \frac{8 π ^{2}}{3}$

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

⏱ 2mℹ️ Source: PYQ_2020

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