Lagrangian for point mass on a hoop

In summary, the conversation discusses a problem involving Lagrangian and a point mass constrained to a rotating hoop. The Lagrangian is correctly derived and the concept of constants of motion and their relation to energy conservation is explained. Different ways of visualizing the situation are also discussed. It is noted that there is a force opposing gravity due to the rotating hoop structure and that the concept of stationary points does not depend on initial conditions.
  • #1
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Homework Statement


See attached picture for problem.


Homework Equations


Lagranges equation.


The Attempt at a Solution


So I have found the lagrangian to be:

L = ½ma2(θ'2+[itex]\omega[/itex]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[itex]\pi[/itex]) - 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?
 

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  • #2
aaaa202 said:

The Attempt at a Solution


So I have found the lagrangian to be:

L = ½ma2(θ'2+[itex]\omega[/itex]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.

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.

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[itex]\pi[/itex]) - I don't see how you could phyiscally have the gain in potential energy to go into this work.

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.

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?

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.

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?

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
Hypersphere said:
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.

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.
 

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  • #4
aaaa202 said:
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.

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.
 
  • #5


I would suggest breaking down the problem into smaller parts to better understand the situation. The Lagrangian you have found is correct, and it describes the energy of the system as it moves along the hoop. The constants of motion in this case refer to the quantities that remain constant during the motion, such as energy or angular momentum. These quantities are conserved because the Lagrangian does not depend on time, meaning the system is in equilibrium and there are no external forces acting on it.

To better visualize the situation, you can think of the hoop as a circular track with the mass constrained to move only along the track. The hoop is rotating about the z-axis, so the mass is moving in a circular motion on the track. The force that keeps the mass on the hoop is the normal force from the hoop, which is directed towards the center of the circle.

When finding the angular velocity for which the mass can remain stationary at a certain point, you do not need to take into account the initial conditions because the system is in equilibrium. This means that the forces acting on the mass are balanced, and the mass can remain stationary at any point on the hoop as long as the forces remain balanced.

I hope this helps clarify the situation and how to approach the problem. Remember to always break down complex problems into smaller parts and use physical intuition to understand the situation.
 

What is the Lagrangian for a point mass on a hoop?

The Lagrangian for a point mass on a hoop is a mathematical expression that describes the energy of the system. It takes into account the kinetic energy and potential energy of the mass as it moves along the hoop.

How is the Lagrangian derived for a point mass on a hoop?

The Lagrangian is derived using the principles of Lagrangian mechanics, which involves finding the difference between the kinetic and potential energies of the system. In the case of a point mass on a hoop, the mass is constrained to move along a circular path and the Lagrangian takes into account the forces acting on the mass.

What is the significance of the Lagrangian for a point mass on a hoop?

The Lagrangian is significant because it allows for the equations of motion of the system to be derived using the principle of least action. It also provides a more elegant and efficient way of describing the dynamics of the system compared to traditional methods such as Newton's laws of motion.

What are the variables in the Lagrangian for a point mass on a hoop?

The variables in the Lagrangian for a point mass on a hoop include the mass of the point mass, the radius of the hoop, the position and velocity of the mass along the hoop, and any external forces acting on the mass.

How is the Lagrangian used to solve problems involving a point mass on a hoop?

The Lagrangian is used to set up the equations of motion for the system, which can then be solved using mathematical techniques such as calculus and differential equations. It allows for a more systematic and elegant approach to solving problems involving a point mass on a hoop compared to traditional methods.

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