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Q.

A ring is rolling without slipping. Its energy of translation is $E$ . Its total kinetic energy will be :-

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a

$3 E$

b

$E$

c

$4 E$

d

$2 E$

answer is B.

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Detailed Solution

Translational $K E = \frac{1}{2} m v^{2} = E$

Total $K E = \frac{1}{2} m v^{2} (1 + \frac{K^{2}}{R^{2}})$

$= \frac{1}{2} m v^{2} (1 + 1) = 2 E$

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A ring is rolling without slipping. Its energy of translation is E. Its total kinetic energy will be :-