How High Does a Rolling Cylinder Reach on a Smooth Incline?

[nathangrand]

1. Homework Statement

A cylindrical hoop rests on a rough uniform incline. It is released and rolls
without slipping through a vertical distance h0. It then continues up a perfectly
smooth incline. What height does it reach?

2. Homework Equations

GPE=mgh
Rotational Ke =0.5Iw²
Linear Ke = 0.5mv²
Moment of inertia of the hoop, I = mr² where r is the radius, and m the mass
Friction Force = UN where U is the coefficient of friction, N the normal reaction force

3. The Attempt at a Solution :
Ok..I'm sort of thinking now that at bottom of the slope, mgh0=Ke[Rot] +Ke[Lin]
mgh0={{0.5Iw²}}+0.5mv²
mgh0= {{0.5mv²}} + 0.5mv² by using v=wr and equation for I

So half the energy is as rotational kinetic energy?
This rotational energy will have to be dissipated as the hoop rolls up the other slope and comes to rest
meaning that only half the energy -- the translational half -- is available for climbing the slope
and hence why the hoop only reaches half the height it started at (I have the answers!)

Can't help but thinking this is a bit of a rubbish explanation and very fudged..can someone tell me if my thinking is right and explain?


Any help appreciated as always guys :)

 

[nathangrand]

sorted :) would delete the post but not sure how