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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)

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g

h = 8.37m

h-2R = H = 2.17m

That is the correct answer according the homework but then it asks for a disk with the same radius and here is where I don't get the right answer.

mgh = 1/2mv^2 +1/2Iw^2 + mg2R

mgh = 1/2mv^2 +1/2*1/2mR^2w^2 + mg2R

m's cancel

h = 1/2v^2 +1/4R^2(v/R)^2 +2gR

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g

Help anyone?