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Physics - Rotation Question with Solution | TestHub

PhysicsRotationRotational KinematicsHard2 minPYQ_2003

PhysicsHardsingle choice

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$ . If radius of the ball be $R$ , then the fraction of total energy associated with its rotational energy will be:NEET 2003

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Solution:

Total energy $= \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$ $= \frac{1}{2} m v^{2} (1 + \frac{K ^{2}}{R ^{2}})$ $Rotational energy = \frac{1}{2} I ω^{2} = \frac{K ^{2} + R ^{2}}{1 + \frac{K ^{2}}{R ^{2}}}$ $Required fraction = \frac{K ^{2} / R ^{2}}{1 + K ^{2} / R ^{2}} = \frac{K ^{2}}{R ^{2} + K ^{2}}$

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Rotational Kinematics

⏱ 2mℹ️ Source: PYQ_2003

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