Determining Minimum Height for Marble to Complete Loop-the-Loop

[kamikaze1]

Homework Statement

A marble rolls down a track and around a loop-the-loop of radius R. The marble has mass m and radius r. What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?

Homework Equations

What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?

The Attempt at a Solution

Find initial energy: mgh
Final energy tabulation: mg(2R) + 0.5mrg + 0.5(2/5)mr^(2)(v/r)^2
Set equal to initial energy, I obtain 2R+7/10 =h, which isn't the right answer, what did I do wrong?

I found the velocity at top by using Newton's 2nd law, i get v = squareroot(rg). I find w using constraint of wr=v. I get 2/5(mr^2) using moment of inertia for a solid sphere.

Any help would be appreciated. I'm so confused why this doesn't work out.

 

[SchruteBucks]

I'm still new to this, but it looks like you've got everything right except for the last part of your final energy calculation (the 0.5(2/5)mr^(2)(v/r)^2 part). I'm not quite sure why it's there, but I'm pretty sure that's the part that's throwing you off.

 

[kamikaze1]

I'm still new to this, but it looks like you've got everything right except for the last part of your final energy calculation (the 0.5(2/5)mr^(2)(v/r)^2 part). I'm not quite sure why it's there, but I'm pretty sure that's the part that's throwing you off.

The last part is the kinetic rotational energy using I = (2/5)(M)(R^2)
Using kinetic rotational energy = 0.5Iw^2, I have
0.5(2/5)(MR^2)w.
using constraint wr=v, knowing that v=(rg)^0.5, I get
0.5(2/5)mr^(2)(v/r)^2

The answer isn't right.

 

[JHamm]

The energy at the top of the loop is the potential (which you have) plus it's kinetic, since it is moving in a circle there must be an acceleration at the top equal to $\frac{V^2}{R}$, for this to be a minimum (IE only just making it around the loop) this acceleration must be set equal to $g$ so we have
$$V^2 = gR \Rightarrow KE = \frac{1}{2}mgR$$
$$\displaystyle{\dot{. .}} mgh = mg(2R) + \frac{1}{2}mgR$$
Which you can then solve for the minimum height :)

EDIT: Just reread your question, the "effective" radius of the loop is changed because of the radius of the marble (the center of mass of the marble moves around a slightly smaller loops) and there is indeed rotational energy equal to $\frac{1}{2}\frac{2}{5}\frac{gR}{r^2}mr^2$ (again this has to be changed because the $R$ isn't really the radius of the loop that the center of mass moves over, but I'll leave that to you)

:)

 

[kamikaze1]

Never mind. Thanks for all your suggestions. I've just solved them. I left out the little tiny details which messed up the problems. I didn't take into account of the center of mass, which I paid dearly.
Here is how I solved it.
Initial energy: mg(h+r)
At the loop, the radius is R-r because we're taking into account of the center of mass. Hence using Newton's 2nd law.
m(v^2)/(R-r) = mg, solve for v = (g(R-r))^0.5 (1)

Final energy at top of the loop is:
mg(2R-r) + 0.5(m)(g(R-r)) + 0.5(2/5)(mr^2)(w^2).
Using velocity constraint, w=v/r=(1/r)(g(R-r))^0.5
Plug that in (1), and set (1) equal to initial energy, then solve for h.