A loop rolls down on an inclined plane. The fraction of its kinetic energy that is associated with only the rotational motion is.

Options

A. `1:2`
B. `1:3`
C. `1:4`
D. `2:3`

Answer

A

Explanation

Total energy of rolling body = translational K.E. + Rotational K.E.
`=1//2Mv_(c.m.)^(2)+1//2I_(c.m.)omega^(2)`
Where I is the moment of inertia about an axis passing through its C.M. & perpendicular to the plane of body.
For hoop `I_(cm)=MR^(2),v_(c.m.)=Romega`
So `("Rotational K.E")/("total energy")=(1//2I_(c.m.)omega^(2))/(1//2Mv_(c.m.)^(2)+1//2I_(c.m.)omega^(2))`
`=(1//2MR^(2)omega^(2))/(1//2MR^(2)omega^(2)+1//2MR^(2)omega^(2))=1:2`