The ratio of rotational and translatory kinet | vedclass

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Question

Rotational $K E=\frac{1}{2} I \omega^{2}$

but $\omega=\frac{v}{R}$

$	herefore$ Rotational $K E=\frac{1}{2} I \frac{v^{2}}{R^{2}}$

but the mo

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

Rotational $K E=\frac{1}{2} I \omega^{2}$

but $\omega=\frac{v}{R}$

$	herefore$ Rotational $K E=\frac{1}{2} I \frac{v^{2}}{R^{2}}$

but the moment of inertia of a sphere is $I=\frac{2}{5} m R^{2}$

$	herefore$ Rotational $K E=\frac{1}{2}\left(\frac{2}{5} m R^{2}ight) \frac{v^{2}}{R^{2}}$

$\mathrm{Or}$

Rotational $K E_{R}=\frac{1}{5} m v^{2}$

Translational $K E_{T}=\frac{1}{2} m v^{2}$

Ratio $\frac{K E_{R}}{K E_{T}}=\frac{\frac{1}{5} m v^{2}}{\frac{1}{2} m v^{2}}=\frac{2}{5}$

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

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