A uniform circular ring is rolling on a horizontal surface without slipping. If its total kinetic energy is E, then its rotational and translational kinetic energies are respectively

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$\frac{E}{2}, \frac{E}{2}$

$\frac{2 E}{3}, \frac{E}{3}$

$\frac{3 E}{4}, \frac{E}{4}$

$\frac{E}{3}, \frac{2 E}{3}$

answer is A.

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Detailed Solution

$T r a n s l a t i o n a l k i n e t i c e n e r g y = \frac{1}{2} m v^{2} R o t a t i o n a l k i n e t i c e n e r g y = \frac{1}{2} I ω^{2} = \frac{1}{2} m r^{2} ω^{2} = \frac{1}{2} m v^{2} s o b o t h a r e e q u a l a n d \frac{E}{2} e a c h .$

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