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PhysicsRotationCalculation of M.I.Medium2 minQB

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A semicircular lamina of mass m has radius r and centre C . Its moment of inertia about an axis through its centre of mass and perpendicular to its plane is-

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Answer:

Solution:

$I_{C} = I_{CM} + m d^{2} \Rightarrow I_{CM} = \frac{1}{2} m r^{2} - m (\frac{4 r}{3 π})^{2} = (\frac{1}{2} - \frac{16}{9 π ^{2}}) m r^{2}$

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

⏱ 2mℹ️ Source: QB

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