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Homework Help: Motion on a paraboloid

  1. Dec 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A body of mass M moves (in a gravitational field g) on the inner surface of given by equation:

    [tex]z=\frac{1}{2a}(x^{2}+y^{2})[/tex]

    (a is positive)

    Reduce the question of finding the motion to quadratures.

    2. Relevant equations



    3. The attempt at a solution

    I used Lagrange equations (1st kind) to find relevant equations for x, y and z, and after separating variables, transformation to polar coordinates [tex](r, \phi, z)[/tex] etc. I came up with the following equation (C is a constant dependent on initial conditions):

    [tex]\ddot{r}-\frac{C}{r^{3}}=-\frac{1}{a^{2}}(r{\dot{r}}^{2}+r^{2}\ddot{r})-\frac{g}{a}r[/tex]

    I don't have any idea how to integrate this equation, but maybe I've done things in an unnecessarily complicated way...
     
  2. jcsd
  3. Dec 9, 2007 #2

    siddharth

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    I get a slightly different equation in r.

    Since

    [tex] L = \frac{m}{2}\left(\dot{r}^2 + r^2 \dot{\theta}^2 + \dot{z}^2 \right) - mgz[/tex]

    and

    [tex] z=\frac{r^2}{a^2}[/tex]

    [tex]\dot{z} = \frac{r \dot{r}}{a}[/tex]

    and using

    [tex] \frac{\partial L}{\partial \theta} = 0 = \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}[/tex]

    [tex]\frac{\partial L}{\partial r} = \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}[/tex]

    I get

    [tex]\frac{r \dot{r}}{a^2} -\frac{gr}{a} = \ddot{r} + \frac{2\dot{r}^2 + r^2 \ddot{r}}{a^2}[/tex]

    How does the constant depending on the initial condition (which should be the component of the angular momentum, I guess) come into the final equation for r?
     
  4. Dec 9, 2007 #3
    The way I actually obtained that equation is the following: because the body moves on the surface given by:
    [tex]f=z-\frac{1}{2a}(x^{2}+y^{2})=0[/tex]

    ,the reaction force must be proportional to the constraint function's gradient, i.e.:

    [tex]F_{r}=\lambda\nabla{f}[/tex] (Lagrange multipliers). So I get 3 equations:

    [tex]m\ddot{x}=-\lambda\frac{x}{a}[/tex]

    [tex]m\ddot{y}=-\lambda\frac{y}{a}[/tex]

    [tex]m\ddot{z}=-mg+\lambda[/tex]

    Using polar transformation leads to:

    [tex]\ddot{x}=\ddot{r}cos\phi-2\dot{r}\dot{\phi}sin\phi-r\ddot{\phi}sin\phi-r{\dot{\phi}}^{2}cos\phi[/tex]

    [tex]\ddot{y}=\ddot{r}sin\phi+2\dot{r}\dot{\phi}cos\phi+r\ddot{\phi}cos\phi-r{\dot{\phi}}^{2}sin\phi[/tex]
    [tex]z=\frac{1}{2a}r^{2}[/tex]

    [tex]\ddot{z}=\frac{1}{a}({\dot{r}}^{2}+r\ddot{r})[/tex]

    After plugging these expressions into the original equations (plus eliminating [tex]\lambda[/tex] using [tex]z[/tex]) I get two new equations (1) and (2) for [tex]r[/tex] and [tex]\phi[/tex]. Our aim is to separate variables, so I do the following trick:

    [tex](1)cos\phi+(2)sin\phi[/tex]
    [tex](1)sin\phi-(2)cos\phi[/tex]

    and I get another two equations:

    [tex](1') \ddot{r}-r{\dot{\phi}}^{2}=-\frac{1}{a^{2}}(r{\dot{r}}^{2}+r\ddot{r})-\frac{g}{a}r[/tex]
    [tex](2') 2\dot{r}\dot{\phi}+r\ddot{\phi}[/tex]

    (2') is easily integrable and yields:
    [tex]\dot{\phi}=\frac{C}{r^{2}}[/tex].

    Plugging this into (1') results in the equation I've written in my first post.

    Is there a mistake somewhere in these calculations? I know that this can be done quicker by writing apropriate lagrangians etc., but the problem's formulation was to do everything using this method (i.e. Lagrange multipliers).
     
    Last edited: Dec 9, 2007
  5. Dec 10, 2007 #4

    siddharth

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    I've not checked the calculations yet, but I think the final equation which describes the motion of r should be independent of the method. Since using the appropriate Lagrangian is much less time consuming, check if you get the same equation. That way, you'll know if there's a calculation error.
     
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