# Lagrangian for point mass on a hoop

1. Sep 23, 2012

### aaaa202

1. The problem statement, all variables and given/known data
See attached picture for problem.

2. Relevant equations
Lagranges equation.

3. The attempt at a solution
So I have found the lagrangian to be:

L = ½ma2(θ'2+$\omega$2sinθ) - mgacosθ

I think this is correct but I have some questions on the further solution of the problem. First of all - what is meant by the constants of motion? Is that the total energy? I can see in a sense that it should be conserved since the lagrangian doesn't depend on time. On the other hand - why would the energy be conserved from an intuitive point of view- i.e. if the hoop is moving at constant angular velocity regardless of the azimuthal angle then that must mean someone is compensating for the extra kinetic energy needed to maintain constant angular speed at larger angles (0;1/4$\pi$) - I don't see how you could phyiscally have the gain in potential energy to go into this work.
Overall I am a bit unsure of how to picture the situation. At first I thought of a ball lying on top of a ring that is rotating about the z-axis (because that's what it is rotating about right???)
but then that situation doesn't really constrain the mass to move on the hoop. Then I thought of the hoop as a torus in which the ball is free to move. Is this a better picture?
And furthermore - since there exists a solution where the ball remains stationary at another point than the bottom that must mean some force is opposing the force of gravity on the mass. What is that force. Is that the force upwards from the exterior of the torus?
And lastly, I don't understand why you don't have to take into account the initial conditions when finding the angular velocity for which that mass can remain stationary. Surely there must be a difference between the situation where the mass starts at the top and the bottom for instance?

#### Attached Files:

• ###### pointmasshoop.png
File size:
10.9 KB
Views:
247
2. Sep 24, 2012

### Hypersphere

It looks OK, yes. You have spherical symmetry, and the potential term behaves properly.

A constant of motion is really just some quantity that is conserved under the motion. Using Noether's theorem we know that it corresponds to a symmetry of the Lagrangian, and here you have rightly pointed out that the energy is a constant of motion.

Maybe it's more intuitive if you combine the two θ-dependent terms into an effective potential? Then compare this potential term with the kinetic term. I don't think it's impossible to imagine a constant total energy in this case, do you?

Also, I guess you should view the constant ω as being the result of a motor or something. That motor, however, isn't part of your system. You just consider this point mass, that for some reason is constrained to this hoop. That's what your Lagrangian describes. This mass has a total energy, which is conserved.

A torus with a ball is one way to look at it. I guess I would just see the mass as a thin ring encircling another thin ring. And yeah, it rotates about the vertical axis.

I find that when things rotate, it's hard to think in terms of contact forces. But yes, there is a force due to the rotating hoop structure. Generally though, I think it's easier to say that a force is just (minus) the gradient of the potential (you should be able to find the anwer to your question using this method), or just look at potential minima instead.

As for your question about initial conditions, well what is a stationary point? It's a point which, for some values of the parameters, the mass does not leave. It's not necessarily an attractive fixpoint, which pulls the mass in if it has initial values within some neighbourhood. The point is, if the mass is in a stationary point, then it's in a stationary point regardless of how it did get there.

3. Sep 25, 2012

### aaaa202

Thanks for the awesome answer. Cleared up some things. I just don't understand the energy part still. Are you sure that energy is conserved and it is not just some other constant of motion which is not the total energy? - I mean the hamiltonian can be conserved even though it's not the total energy.
Let me redo my argument that energy shouldn't be conserved. Imagine the mass starts at the bottom and rolls op due to some kinetic energy leftover. When the mass is at the bottom the rotation from the motor means nothing since the bottom is standing still - if the rotation is around the axis as drawn on the picture.
Now the ball moves up. To maintain the same angular velocity the motor must simply be doing some work on the ball for now the ball is a part of the rotating part of the hoop.

#### Attached Files:

• ###### hoop.png
File size:
1.9 KB
Views:
116
4. Sep 25, 2012

### Hypersphere

I think I was a bit sloppy last night, and yeah, you're quite right. When L is independent, then you have a conserved quantity. Goldstein calls this the energy function h, and when you go to the Hamiltonian formulation it's just the Hamiltonian H. Like you say, however, this is not necessarily the total energy of the mass on the hoop. This time it isn't, and energy enters or leaves the system to keep the constant angular velcity.