Consider two cylinders that start down identical inclines from rest except that one is frictionless. Thus one cylinder rolls without slipping, while the other slides frictionlessly without rolling. They both travel a short distance at the bottom and then start up another incline. (a) Show that they both reach the same height on the other incline, and that this height is equal to their original height. (b) Find the ratio of the time the rolling cylinder takes to reach the height on the second incline to the time the sliding cylinder takes to reach the height on the second incline. (c) Explain why the time for the rolling motion is greater than that for the sliding motion.

a)  m*g*hi = m*g*hf

         h = h

b)  ratio = 1.22

c) the potential energy is converted to both rotational kinetic energy and translational kinetic energy

Explanation:

For a frictionless system : initial energy = final energy

i.e. Ei = Ef

a) Prove

Total Initial energy for First cylinder ( Ei ) = m*g*hi      

The final energy of the first cylinder ( Ef ) = m*g*hf

and this energies are purely potential energies

According to conservation of energy :

m*g*hi = m*g*hf

         hi = hf   ( and this holds for both cylinders because they where both kept at the same height initially  )

i.e. initial height = final height ( hence it is proved that both cylinders will reach their initial heights )

b) Determine the ratio of the time taken by both cylinders

lets assume height of incline ( h ) = Lsin

where L = distance covered by cylinder

Final velocity of first cylinder ( V1 )  =  

Final velocity of  second cylinder ( V2 ) =

( following the application of the law of conservation of energy )

Given that ; Time taken = displacement / velocity  

hence the ratio = ( h / v1 ) / ( h / v2 )

where h = Lsin

the ratio = 1.22

c) The time for the rolling motion is greater than the time for the sliding motion because the potential energy is converted to both rotational kinetic energy and translational kinetic energy

the answer is b i think