What is the Conservation of Angular Momentum for Motion Confined to a Cone?

[MisterX]

Homework Statement

Homework Equations

\frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}

The Attempt at a Solution

z = r\cos\alpha
s = r\sin\alpha

v^2 = \dot{r}^2 + r^2 \dot{\phi}^2 sin^2\alpha

\mathcal{L} = \frac{1}{2}m\dot{r}^2 + \frac{1}{2}mr^2 \dot{\phi}^2 sin^2\alpha- mgr\cos\alpha

\ell_z = I_z\omega_z = ms^2\dot{\phi}
\ell_z = mr^2\dot{\phi}\sin^2\alpha

The \phi Euler Lagrange equation is just conservation of angular momentum.
\frac{\partial\mathcal{L} }{\partial \phi} = 0 = \frac{d}{dt}\left(mr^2\dot{\phi}sin^2\alpha \right) = \dot{\ell_z} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}

\dot{\ell_z} = m\sin^2(\alpha)(2r\dot{r}\dot{\phi} + mr^2\ddot{\phi}) = 0

the r Euler Lagrange equation
\frac{\partial\mathcal{L} }{\partial r} = mr\dot{\phi}^2\sin^2\alpha - mgcos\alpha = m\ddot{r} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{r}}

I'm not sure how to put this in terms of \ell_z in a way that gets rid of all the \dot{\phi} and r. There's that extra power of \dot{\phi}.

 

[TSny]

I'm not sure how to put this in terms of \ell_z in a way that gets rid of all the \dot{\phi} and r. There's that extra power of \dot{\phi}.

You just want to eliminate \dot{\phi}, not both \dot{\phi} and r.

 

[MisterX]

Oops, I should have read that better. Thanks for answering both my questions this week, TSny!

I got as far as

\ddot{\epsilon} \approx \frac{\ell_z^2}{m^2 r_0^3 sin^2\alpha \cos\alpha}(1 - \frac{3\epsilon}{r_0}) - gcos\alpha ?


But I think I'm going to leave this incomplete. It's late and I'm recovering from a cold.

 

[TSny]

I got as far as

\ddot{\epsilon} \approx \frac{\ell_z^2}{m^2 r_0^3 sin^2\alpha \cos\alpha}(1 - \frac{3\epsilon}{r_0}) - gcos\alpha ?

But I think I'm going to leave this incomplete. It's late and I'm recovering from a cold.

Hey, that looks good. Can you argue that $\frac{\ell_z^2}{m^2 r_0^3 sin^2\alpha \cos\alpha}$ cancels $- gcos\alpha$ leaving a single term on the right proportional to $\epsilon$?

Take care

 

[MisterX]

Yes, here is what I got.

\ddot{\epsilon} = -\frac{3gcos(\alpha)}{r_0}\epsilon

 

[TSny]

I think that's right.