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A body is rolling down an inclined plane...

A body is rolling down an inclined plane. If kinetic energy of rotation is `40 %` of kinetic energy in translatory start then the body is a.

A

ring

B

cylinder

C

hollow ball

D

solid ball

Text Solution

Verified by Experts

The correct Answer is:
d
Rotational kinetic energy is
`K_(R) = 1/2Iomega^(2) = 1/2Mk^(2) (v/r)^(2)` `(therefore I=Mk^(2)` and `v=Romega)`
`=1/2Mv^(2)(k^(2)/R^(2))`
Translation kinetic energy is
`K_(T) = 1/2Mv^(2)`
As per question, `K_(R) = 40% K_(T)`
`therefore 1/2Mv^(2)(k^(2)/R^(2)) = 40% " of " 1/2 Mv^(2)` or `k^(2)/R^(2) = 40/100 = 2/5`
For solid sphere, `k^(2)/R^(2) = 2/5`
Hence, the body is solid ball.
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