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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Hard2 minPYQ_2020

PhysicsHardsingle choice

A thin, uniform metal rod of mass 'M' and length 'L' is swinging about a horizontal axis passing through its end. Its maximum angular velocity is ' $ω^{'}$ . Its centre of mass rises to a maximum height of $(g =$ acceleration due to gravity)

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Answer:

Solution:

$mgh = \frac{1}{2} I ω^{2} I = \frac{mL ^{2}}{3} ∴ mgh = \frac{1}{2} (\frac{mL ^{2}}{3}) ω^{2} h = \frac{1}{6 g} L^{2} ω^{2}$

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

⏱ 2mℹ️ Source: PYQ_2020

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