Physics - Rotation Question with Solution | TestHub

PhysicsRotationTranslational + RotationalMedium2 minPYQ_2022

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A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Options:

Answer:

Solution:

Under pure rolling condition: $v = ω R$

Translational $K . E = \frac{1}{2} m v^{2}$

Rotational $K . E = \frac{1}{2} \times \frac{2}{5} m R^{2} {(\frac{v}{R})}^{2} = \frac{1}{5} m v^{2}$

Total $K . E = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = \frac{7}{10} m v^{2}$

So, $\frac{Rotational K \cdot E}{Total K \cdot E} = \frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}} = \frac{2}{7}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Translational + Rotational

⏱ 2mℹ️ Source: PYQ_2022

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