Solving Physics Puzzle: Prove Guy Slides off Ice Mount at 2/3R Height

[FrostScYthe]

Hi, I encountered this problem while I was studying for a Physics test, and this is how far I got, hopefully someone can point me in the RIIGHT direction.
The proof goes like this, a guy is sitting on the top of an ice hemisphere, and someone pushes him off, and so he slides down the frictionless ice mount and falls out at the height of 2/3R... I'm actually supposed to prove that he falls off at that height.
http://img131.imageshack.us/img131/4333/superdrawing7mk.gif
Here's what I've done, in a few words, I've basically considered the initial potential energy at the top to be mgR and before the guy takes off, he'd have a potential energy of mgh plus a kinetic energy of 1/2Iw^2, then I substitute for the I and try to get rid of that w, and I can't, the book gives a little clue with the normal force, but I don't know how it applies in all of this.
Thanks in advance
Ok I've considered the energies like this: (trying to prove h = 2/3R btw)
mgr = (0.5)Iw^2 +mgh
mgr = (0.5)(1/2mr^2)V^2/r^2 + mgh
mgr = (mr^2)V^2/r^2 + mgh
mgr = mv^2 + mgh
gr = v^2 + gh
I don't know how to get rid of that V^2 for one thing, maybe I'm not doing it right, hopefully someone can help

 

[aboutguava]

Fn-mgcosx=-mv^2/R
gravitational potential is zero at the top then potential energy at the time shown is
u = -mgR(1-cosx)
he starts at rest conservation of energy gives
0 = .5mv^2-mgR(1-cosx)
substitute above equation into one for second law to obtain
gcosx = 2g(1-cosx)
cosx = 2/3
h = Rcosx = 2R/3

 

[amcavoy]

Looks ok. The angle (with the vertical) is about 48.3^o, so (2/3)R looks fine.