# A loop-the-loop track question

1. Mar 24, 2009

### ghostcloak

1. The problem statement, all variables and given/known data

A small solid marble of mass m and radius r will roll without slipping along the loop-the-loop track, shown below, if it is released from rest somewhere on the straight section of track. For the following answers use m for the mass, r for the radius of the marble, R for the radius of the loop-the-loop and g for the acceleration due to gravity.

Here's the picture:

From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop?

2. Relevant equations
Centripetal Acceleration = mg

Ki+mgh=Kf+mg(2R)

3. The attempt at a solution
Centripetal Acceleration = mg
Centripetal Acceleration = (mv^2)/R
(mv^2)/R = mg
v^2 = gR

Ki+Ui=Kf+Uf

Ki = 0
Ui = mgh
Kf = (mv^2)/2
Uf = mg(2R)

mgh = (mv^2)/2 + 2mgR
gh = (v^2)/2 + 2gR
gh = gR/2 + 2gR
h = R/2 + 2R
h = (5/2)R

In my attempt at the solution, I calculated (incorrectly) that the initial height needs to be 2.5*R. However, I did visit this old thread here: https://www.physicsforums.com/showthread.php?t=6357
The answer is actually 2.7*R, which is indeed the correct answer. Can someone help me out in identifying what I did wrong?

Thank you. :)

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Last edited: Mar 24, 2009
2. Mar 24, 2009

### alphysicist

HI ghostcloak,

This would be true if the ball was just sliding, but here it is rolling, so you need another kinetic energy term in your equation. Do you see what to do now?

3. Mar 24, 2009

### Delphi51

I got h = 2.5R as well.
Being picky about the marble radius, the height of its center at the top of the loop is a bit less (2R-r) and the radius of the loop a bit less so the answer comes out
h = 2.5R - 2r. That's further from the 2.7 R answer!

Taking rotational motion into account a slightly greater initial height will be needed to provide the rotational energy of the marble moving at linear speed v. However, this would also depend on the radius r of the marble, so you would get an even more complicated answer with r's in it.

I'm stumped. Looks like the 2.7 R is wrong.

4. Mar 24, 2009

### alphysicist

Hi Delphi51,

Since it is a small marble, we can assume that

$$R-r\approx R$$

I don't think that's right (as long as you use the above small r approximation). Remember that we can eliminate the angular speed in favor of the linear speed.

I got 2.7R, and ghostcloak said it was the correct answer.

5. Mar 24, 2009

### Delphi51

It would be most interesting and instructive if you could indicate where ghostcloak's solution is in error!

6. Mar 24, 2009

### alphysicist

It does not include the rotational kinetic energy of the sphere.

7. Mar 24, 2009

### Delphi51

Both ghostcloak and I got 2.5R; you got 2.7R. We did not include rotational energy, either. Where did you find the factor of 2.7?

8. Mar 25, 2009

### alphysicist

The error that you and ghostcloak are both making is that you are not including the rotational kinetic energy. If you include the rotational kinetic energy you will get the correct answer of 2.7R. (Near the end of his post he said that the correct answer is 2.7R.)

If an sphere is rolling you have to include the rotational kinetic energy if you want to find the speed. When a sphere is rolling without slipping (any sphere, with any radius, at any speed) close to 30% of the total kinetic energy is in the form of rotational kinetic energy. Ignoring that will give a bad answer.

9. Mar 25, 2009

### Delphi51

If we include rotational energy, won't the radius of the marble appear in the answer? Also a g, I think. I don't see how it could work to 2.7R that way. Have you actually worked it out or just noting, as I did, that the answer would be somewhat larger than 2.5R if rotational energy is taken into account?

10. Mar 25, 2009

### alphysicist

I have worked it out and got 2.7R. When you eliminate angular velocity (using v = r w), you will eliminate r. Later on in the calculation, each term will have a g so it will cancel also.

11. Mar 25, 2009

### Delphi51

Thanks for your patience, alphysicist. I see it now. I guess my physical intuition is failing in my old age. It still seems amazing that the rotational energy doesn't depend on the radius. The rotational energy term 0.5*I*(v/r)^2 doesn't really have an r in it because I = 0.4mr^2 and the r's cancel. You don't know until you try it mathematically.

12. Mar 25, 2009

### alphysicist

Exactly, and actually the whole business of being able to ignore the r in the potential energy but still having to take into account the rotational energy is the most interesting part of the problem to me. From my point of view, so much of physics is about knowing what you can safely ignore, and what you have to treat as carefully as possible. So here, the potential energy correction due to the finite radius gets smaller as the radius decreases, so it vanishes as r vanishes.

But the neat thing is that if you the look at the fraction

(rotational kinetic energy)/(total kinetic energy)

for any solid rolling sphere (that's not slipping), it's always 2/7, so the effect of including rotational kinetic energy does not scale with the radius (for the question that's asked here).

13. Mar 25, 2009

### ghostcloak

alphysicist, I just wanted to say thanks. I see what I did wrong after your first response. Makes sense now! :)

14. Mar 25, 2009

### alphysicist

Glad to help, and welcome to PF!

15. Mar 25, 2009

### Dr.D

Comment to ghostcloak: