JEE MainPhysicsToughRotational motion

An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. Theline of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational andtranslational kinetic energies of the sphere just after
the impulse are equal, then h is equal to

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}

...(1)
Impulse eq. Fdt = mV...(2)

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

...(3)
Dividing (2) and (3)

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

...(4)
Solving (1) and (4)

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

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx

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