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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Easy2 minPYQ_2019

PhysicsEasysingle choice

A thin metal wire of length ‘ $L$ ’ and uniform linear mass density ‘ $Q$ ’ is bent into a circular coil with ‘ $O$ ’ as centre. The moment of inertia of a coil about the axis $X X'$ is

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Answer:

Solution:

$I_{x x'} = \frac{m R^{2}}{2} + m R^{2}$
$= \frac{3}{2} . m R^{2}$
$∴ 2 π R = L ∴ R = \frac{L}{2 π}$
So $I_{x x'} = \frac{3}{2} (Q) . 2 π R . R^{2}$
$= 3 Q π (R^{3})$
$= \frac{3 Q π . L^{3}}{8 π^{3}}$

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

⏱ 2mℹ️ Source: PYQ_2019

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