How to solve loop the loop with rotational kinematics?

Question

A small spherical ball of radius r = 1.7 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 3.1 m. The ball has mass M = 345 g.

How high above the top of the loop must it be released in order that the ball just makes it around the loop?

Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases.


I used conservation of energy and set mgh = KE(rot) + KE(trans) + mg2R. Using V=sqrt(Rg), I found the minimum speed the ball must have at the top of the loop and found it to be 5.512 m/s. For the right side of the equation, I used KE(rot) = 1/2Iw(omega)^2. For a spherical ball, I =2/5MR^2.
Now I have
mgh = 1/2mv^2 +1/2(2/5)MR^2(w^2) + 2mgR
M's cancel
w = v/R
So I was left with
h = (1/2(v^2)+1/5(R^2)(v/R)^2 + 2gR)
----------------------------------
g
h = 8.37m
h-2R = H = 2.17m

Is that right? and how would I do it for a disk.

Answer 1

I didn't check the v=5.512, using it,
I got the same answer.
I agree with the expression, which I simplified to
mg(h-2R) = 1/2mv^2 +1/2(2/5)MR^2(w^2)
gh = 1/2v^2 +(2/10)(v^2)

h-2R=7/(10*g)v^2
=2.17 m

For a disk, the moment of inertia becomes
I=.5*m*r^2

mg(h-2R) = 1/2mv^2 +1/2(1/2)MR^2(w^2)
h-2R=3v^2/(4*g)
=2.322 m

j