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.