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Physics - Rotation Question with Solution | TestHub

PhysicsRotationMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A circular disc rolls down an inclined plane. The ratio of rotational kinetic energy to total kinetic energy is

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Answer:

Solution:

The ratio of rotational kinetic energy to total kinetic energy is

$\frac{K E_{R}}{K E_{T}} = \frac{\frac{1}{2} I ω^{2}}{\frac{1}{2} I ω^{2} + \frac{1}{2} M v^{2}}$

The moment of inertia of a ring
= $\frac{1}{2} M R^{2}$

$∴$ $\frac{K E_{R}}{K E_{T}} = \frac{\frac{1}{2} (\frac{1}{2} M R^{2}) {[\frac{v}{R}]}^{2}}{\frac{1}{2} \times \frac{1}{2} M R^{2} {[\frac{v}{R}]}^{2} + \frac{1}{2} M v^{2}}$

$= \frac{\frac{1}{2} (\frac{1}{2} M v^{2})}{\frac{1}{2} (\frac{1}{2} M v^{2}) + \frac{1}{2} M v^{2}}$

$\frac{K E_{R}}{K E_{T}} = \frac{1}{3}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:RotationSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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