A small ball of radius r rolls down without sliding in a big hemispherical bowl. of radius R. What would be the ratio of the translational and rotational kinetic energies at the bottom of the bowl?

Options

A. `2:1`
B. `3:2`
C. `4:3`
D. `5:2`

Answer

D

Explanation

`(K.E._(T))/(K.E._(R))=(1/2 Mv^(2))/(1/2Iomega^(2))=(1/2Mv^(2))/(1/2 2/5 MR^(2)(v/R)^(2))=5:2`