What Minimum Height Ensures a Marble Stays on a Loop-the-Loop Track?

[ghostcloak]

Homework Statement

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?

Homework Equations

Centripetal Acceleration = mg

Ki+mgh=Kf+mg(2R)

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. :)

 

[alphysicist]

HI ghostcloak,

Homework Statement

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?

Homework Equations

Centripetal Acceleration = mg

Ki+mgh=Kf+mg(2R)

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

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?

 

[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.

 

[alphysicist]

Hi 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!

Since it is a small marble, we can assume that

<br /> R-r\approx R<br />

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

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

 

[Delphi51]

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

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

 

[alphysicist]

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

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

 

[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?

 

[alphysicist]

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?

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.

 

[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?

 

[alphysicist]

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?

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.

 

[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.

 

[alphysicist]

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.

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).

 

[ghostcloak]

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

 

[alphysicist]

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

Glad to help, and welcome to PF!

 

[Dr.D]

Comment to ghostcloak:

In your original post, under relevant equations, you had
Centripetal Acceleration = mg
this is not true. The mg on the right represents a force, whereas the left side speaks of an acceleration. You might want to be alert to small things like that which will trip you up.