Lagrangian of a particle moving on a cone

In summary: But I think I might have figured it out. I'm going to partial-differentiate the Lagrangian with respect to the time derivative of r, and then use the resulting equation to get the z-component of angular momentum.?In summary, the particle is confined to move on the surface of a circular cone with its axis on the vertical z axis, vertex at the origin (pointing down), and half-angle a. The Lagrangian L in terms of the spherical polar coordinates r and ø is v^{2} = \dot{r}^{2} + r^{2}sin^{2}(\phi)\dot{\theta}^{2} + r^{2}\
  • #1
Oijl
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0

Homework Statement


A particle is confined to move on the surface of a circular cone with its axis on the vertical z axis, vertex at the origin (pointing down), and half-angle a.

(a) Write the Lagrangian L in terms of the spherical polar coordinates r and ø.

(b) Find the two equations of motion. Interpret the ø equation in terms of the angular momentum l[tex]_{z}[/tex], and use it to eliminate ø-dot from the r equation in favor of the constant l[tex]_{z}[/tex]. Does your r equation make sense in the case that l[tex]_{z}[/tex] = 0? Find the value r[tex]_{o}[/tex] of r at which the particle can remain in a horizontal circular path.

(c) Suppose that the particle is given a small radial kick, so that r(t) = r[tex]_{o}[/tex] + ε(t), where ε(t) is small. Use the r equation to decide whether the circular path is stable. If so, with what frequency does r oscillate about r[tex]_{o}[/tex]?

Homework Equations


v[tex]^{2}[/tex] = [tex]\dot{r}[/tex][tex]^{2}[/tex] + r[tex]^{2}[/tex]sin[tex]^{2}[/tex]([tex]\phi[/tex])[tex]\dot{\theta}[/tex][tex]^{2}[/tex] + r[tex]^{2}[/tex][tex]\dot{\phi}[/tex][tex]^{2}[/tex]
l = r X mv


The Attempt at a Solution



Okay, so the langrangian L = T - U.

U is easy enough, saying the only potential energy is gravitational energy, so U = mgrcos[tex]\phi[/tex].

But T = (1/2)mv[tex]^{2}[/tex], and v[tex]^{2}[/tex] = [tex]\dot{r}[/tex][tex]^{2}[/tex] + r[tex]^{2}[/tex]sin[tex]^{2}[/tex]([tex]\phi[/tex])[tex]\dot{\theta}[/tex][tex]^{2}[/tex] + r[tex]^{2}[/tex][tex]\dot{\phi}[/tex][tex]^{2}[/tex]
Now, I'm told that the cone on which this particle moves has a half-angle of [tex]\alpha[/tex]. Then, I know, [tex]\phi[/tex] = [tex]\alpha[/tex] = const., so [tex]\dot{\phi}[/tex] = 0. Right?

With [tex]\dot{\phi}[/tex] being zero, v[tex]^{2}[/tex] reduces to [tex]\dot{r}[/tex][tex]^{2}[/tex] + r[tex]^{2}[/tex]sin[tex]^{2}[/tex]([tex]\phi[/tex])[tex]\dot{\theta}[/tex][tex]^{2}[/tex].

But this still has a theta coordinate in it. How can I express the Legrangian in just r and [tex]\phi[/tex]?

And then, after that, how do I relate l and [tex]\dot{\phi}[/tex] so as to eliminate the latter from the r equation of motion?

But first: How can I express the Legrangian in just r and [tex]\phi[/tex]?
 
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  • #2
Here the angle [tex]\theta[/tex] is fixed and equal to the angle of the cone. Theta is the angle between the position vector and the z axis.
[tex]\phi[/tex] is the angle around the axis z and is the one that contributes to the velocity.
 
  • #3
That's great news! I had always seen spherical coordinates written with the theta and phi's meaning opposite of this.

Now, then, I need to find out how to get the z-component of the angular momentum. How can I do this?
 
  • #4
What are the canonical momenta you get from this Lagrangian?
 
  • #5
That sounds familiar, and helpful! But I don't remember... what's "canonical momenta"?
 
  • #6
Hmm, maybe "canonical" wasn't the right term. Anyway, you partial-differentiate the Lagrangian with respect to the time derivative of a coordinate. What you get is a momentum.
 
  • #7
Oh, well, my equations of motion were, for r and phi, respectively,

m(r-doubledot) = rm(phi-doubledot)^2
and
(d/dt)mr(phi-dot)=0

So the r-component of momentum is m(r-dot),
and the phi-component of momentum is (phi-dot)mr
?

The form of the r-component is familiar to me, thinking of p=mv and all.

But then how do I move from here to writing the z-component of angular momentum?
 
  • #8
I think you dropped the exponent on the r. When you differentiate wrt [tex]\dot{\phi}[/tex], you get [tex]mr^2\dot{\phi}[/tex]. Does that look familiar?
 
  • #9
Oh... yes, it looks familiar, but I truly never learned angular momentum in elementary physics, so could you remind me what it is?
 
  • #10
Actually, I think that expression is still wrong. I was using the Lagrangian in your initial post, but you have [itex]\theta[/itex] and [itex]\phi[/itex] switched in there. You may want to try recalculating the momentum more carefully.

You can obtain the rotational formulas by using the corresponding linear formulas and replacing the variables with their rotational analogues. So for linear momentum, you have p=mv. To get angular momentum, you replace m by the rotational mass I and v by the angular velocity [itex]\omega[/itex] to get [itex]L=I\omega[/itex]. You just need to identify what I and [itex]\omega[/itex] are in this situation.
 
  • #11
This is what I'm trying to do:

(b) Find the two equations of motion. [Got it.] Interpret the ø equation [[tex]\dot{\phi}[/tex]mr^2 = 0] in terms of the angular momentum l[tex]_{z}[/tex], and use it to eliminate [tex]\dot{\phi}[/tex] from the r equation in favor of the constant l[tex]_{z}[/tex].

First off, what is l[tex]_{z}[/tex]?
That's the z-component of the angular momentum.
Angular momentum is the cross product of the position vector and the momentum vector.
The momentum vector is the velocity vector times the scalar quantity of mass.
But how do I get the z-component? I don't know how to relate the time derivatives of Cartesian coordinates to spherical, or spherical to Cartesian.
 

1. What is the Lagrangian of a particle moving on a cone?

The Lagrangian of a particle moving on a cone is a mathematical function that describes the energy of the particle in terms of its position and velocity on the surface of a cone. It takes into account the potential and kinetic energy of the particle, as well as the constraints imposed by the cone's shape.

2. How is the Lagrangian derived for a particle on a cone?

The Lagrangian is derived using the principle of least action, which states that the path taken by a particle between two points is the one that minimizes the total action (the integral of the Lagrangian) along the path. This results in a set of equations that describe the particle's motion on the cone.

3. What does the Lagrangian tell us about the particle's motion on the cone?

The Lagrangian provides information about the dynamics of the particle, such as its position, velocity, and acceleration as it moves on the cone. It also allows us to calculate the forces acting on the particle and predict its future motion.

4. How does the Lagrangian change if the cone is tilted?

If the cone is tilted, the Lagrangian will change to take into account the new potential and kinetic energy terms caused by the tilt. This will result in a different set of equations of motion for the particle.

5. Can the Lagrangian be applied to other curved surfaces?

Yes, the Lagrangian can be applied to any curved surface or manifold, not just a cone. It is a powerful mathematical tool for studying the dynamics of particles and systems in various geometries and is widely used in physics and engineering.

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