The ratio of rotational and translatory kinet | vedclass

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(C) Rotational kinetic energy is given by $K E_{R} = \frac{1}{2} I \omega^{2}$.Since the sphere is rolling without slipping,$\omega = \frac{v}{R}$.Sub

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$2/5$

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(C) Rotational kinetic energy is given by $K E_{R} = \frac{1}{2} I \omega^{2}$.Since the sphere is rolling without slipping,$\omega = \frac{v}{R}$.Substituting this,$K E_{R} = \frac{1}{2} I \left(\frac{v}{R}\right)^{2}$.The moment of inertia of a solid sphere about its center is $I = \frac{2}{5} m R^{2}$.Substituting $I$ into the equation: $K E_{R} = \frac{1}{2} \left(\frac{2}{5} m R^{2}\right) \frac{v^{2}}{R^{2}} = \frac{1}{5} m v^{2}$.Translational kinetic energy is $K E_{T} = \frac{1}{2} m v^{2}$.The ratio is $\frac{K E_{R}}{K E_{T}} = \frac{\frac{1}{5} m v^{2}}{\frac{1}{2} m v^{2}} = \frac{2}{5}$.

The ratio of rotational and translatory kinetic energies of a sphere is

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