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Physics - Rotation Question with Solution | TestHub

PhysicsRotationMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

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Answer:

Solution:

Loss in PE = Gain in KE

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} T ω^{2} (v = R ω)$

$m g h = \frac{1}{2} m {(R ω)}^{2} + \frac{1}{2} (\frac{m R^{2}}{2}) ω^{2}$

$g h = \frac{3}{4} R^{2} ω^{2}$

$\Rightarrow ω^{2} = \frac{h g h}{3 R^{2}} \Rightarrow ω = \frac{2}{R} \sqrt{\frac{g h}{3}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:RotationSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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