A ring rolls down an inclined plane. At the bottom its kinetic energy is E. Theratio of its rotational K.E. to the translational K.E. is :

Options

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

Answer

C

Explanation

Rotational K.E. `=(1)/(2)Iomega^(2)=(1)/(2)mr^(2)omega^(2)=(1)/(2)mv^(2)`
Translational K.E. = `(1)/(2)mv^(2)`
`therefore (R.K.E.)/(T.K.E.)=(1)/(1)`