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Motion Confined to a Cone

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    yDI8qJI.png

    2. Relevant equations
    [itex]\frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/itex]


    3. The attempt at a solution
    [itex]z = r\cos\alpha[/itex]
    [itex]s = r\sin\alpha[/itex]

    [itex]v^2 = \dot{r}^2 + r^2 \dot{\phi}^2 sin^2\alpha[/itex]

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

    [itex]\ell_z = I_z\omega_z = ms^2\dot{\phi} [/itex]
    [itex]\ell_z = mr^2\dot{\phi}\sin^2\alpha[/itex]

    The [itex]\phi[/itex] Euler Lagrange equation is just conservation of angular momentum.
    [itex]\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}}[/itex]

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

    the [itex]r[/itex] Euler Lagrange equation
    [itex]\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}} [/itex]

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

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  2. jcsd
  3. Apr 10, 2013 #2

    TSny

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    You just want to eliminate [itex]\dot{\phi}[/itex], not both [itex]\dot{\phi}[/itex] and [itex]r[/itex].
     
  4. Apr 11, 2013 #3
    Oops, I should have read that better. Thanks for answering both my questions this week, TSny!

    I got as far as

    [itex]\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[/itex] ????


    But I think I'm going to leave this incomplete. It's late and I'm recovering from a cold.
     
    Last edited: Apr 11, 2013
  5. Apr 11, 2013 #4

    TSny

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    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
     
  6. Apr 11, 2013 #5
    Yes, here is what I got.

    [itex]\ddot{\epsilon} = -\frac{3gcos(\alpha)}{r_0}\epsilon[/itex]
     
  7. Apr 11, 2013 #6

    TSny

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    I think that's right.
     
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