What is the Release Height for a Ball and Disk in a Loop-the-Loop Problem?

[ruffrunnr]

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

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
-----------------------------
g

Help anyone?

 

[member 553083]

For the disk, you can use the same equations as for the ball. The only difference is that for the disk, I = 1/2MR^2. So the equation becomes:h = (1/2(v^2)+1/4(R^2)(v/R)^2 + 2gR) ---------------------------------- gh = 7.25m h-2R = H = 1.05mThe ratio of heights h for the two cases is 8.37/7.25 = 1.15