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The moment of inertia of a solid cylinde...

The moment of inertia of a solid cylinder about its axis is given by `(1//2)MR^(2)`. If this cylinder rolls without slipping the ratio of its rotational kinetic energy to its translational kinetic energy is -

A

`1 : 1`

B

`2 : 2`

C

`1 : 2`

D

`1 : 3`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(R) = (1)/(2) I omega^(2) rArr K_(R) = (1)/(2) xx (1)/(2) mR^(2) (v^(2))/(R^(2))`
`K_(R) = (1)/(2).(1)/(2) mv^(2) rArr K_(R) = (1)/(2) K_(r) rArr (K_(R))/(K_(r)) = (1)/(2)`.
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