JEE MainPhysicsModerateRigid Body Dynamics

An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. The line of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational and translational kinetic energies of the sphere just after
the impulse are equal, then h is equal to $\sqrt{\frac{2}{n}}R$ . Find n.

Show answer and solution

Transitional k.e.= rotational k.e.

\frac{1}{2}{mv}^{2} = \frac{1}{2}I\omega^{2}

V^{2} = \frac{2}{5}\omega^{2}R^{2}

....(i)
Impulse equation
Fdt = mV....(ii)
Fhdt

= I\omega = \frac{2}{5}mR^{2}\omega

....(iii)
dividing (2) and (3)

h = \frac{2}{5}\frac{R^{2}\omega}{v}

...(iv)
Solving (1) and (4)
h

= \sqrt{\frac{2}{5}}R

Source: Grass-G / P-10 Physics paper 1 30 October Jee Main.docx

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