# Lagrangian of a particle moving on a cone

1. Feb 18, 2010

### Oijl

1. The problem statement, all variables and given/known data
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$$_{z}$$, and use it to eliminate ø-dot from the r equation in favor of the constant l$$_{z}$$. Does your r equation make sense in the case that l$$_{z}$$ = 0? Find the value r$$_{o}$$ 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$$_{o}$$ + 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$$_{o}$$?

2. Relevant equations
v$$^{2}$$ = $$\dot{r}$$$$^{2}$$ + r$$^{2}$$sin$$^{2}$$($$\phi$$)$$\dot{\theta}$$$$^{2}$$ + r$$^{2}$$$$\dot{\phi}$$$$^{2}$$
l = r X mv

3. 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$$\phi$$.

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

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

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

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

But first: How can I express the Legrangian in just r and $$\phi$$?

2. Feb 18, 2010

### nasu

Here the angle $$\theta$$ is fixed and equal to the angle of the cone. Theta is the angle between the position vector and the z axis.
$$\phi$$ is the angle around the axis z and is the one that contributes to the velocity.

3. Feb 18, 2010

### Oijl

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. Feb 18, 2010

### vela

Staff Emeritus
What are the canonical momenta you get from this Lagrangian?

5. Feb 18, 2010

### Oijl

That sounds familiar, and helpful! But I don't remember... what's "canonical momenta"?

6. Feb 18, 2010

### vela

Staff Emeritus
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. Feb 18, 2010

### Oijl

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. Feb 18, 2010

### vela

Staff Emeritus
I think you dropped the exponent on the r. When you differentiate wrt $$\dot{\phi}$$, you get $$mr^2\dot{\phi}$$. Does that look familiar?

9. Feb 18, 2010

### Oijl

Oh.... yes, it looks familiar, but I truly never learned angular momentum in elementary physics, so could you remind me what it is?

10. Feb 18, 2010

### vela

Staff Emeritus
Actually, I think that expression is still wrong. I was using the Lagrangian in your initial post, but you have $\theta$ and $\phi$ 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 $\omega$ to get $L=I\omega$. You just need to identify what I and $\omega$ are in this situation.

11. Feb 18, 2010

### Oijl

This is what I'm trying to do:

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

First off, what is l$$_{z}$$?
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.