Inclined plane physics problem

Question

A sphere with mass m is put on top of an inclined plane with height h. In the first case the sphere is moving downwards without friction, and in the second case with very little friction. Find the ratio of both cases of sphere's velocities at the bottom of the inclided plane

Sorry for my English

Answer 1

For the case with little (or no) friction, the sphere will just slide down so we only have translational KE

Therefore mgh = (0.5)mv^2

Therefore v =sqrt(2gh)

For the rolling sphere we must include rotational kinetic energy =(0.5)Iω^2

Since I =(2/5)mr^2 and ω=v/r

The rotational KE = (1/5)mr^2

The new energy at the bottom = KE + RKE = (7/10)mv^2 = mgh

Therefore v = sqrt(10gh/7)

Therefore the ratio of their velocities (without friction/with friction) = sqrt((2gh)/(10gh/7)) = sqrt(1.4) = 1.18