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Lagrange equation particle on an inverted cone

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Derive the equations of motion and show that the equation of motion for [itex]\phi[/itex] implies that [itex]r^2\dot{\phi}=K[/itex] where [itex]K[/itex] is a constant
    2. Relevant equations
    Using cylindrical coordinates and [itex]z=\alpha r[/itex]
    The kinetic and potential energies are
    [tex]T=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right][/tex] and
    [tex]V=mg\alpha r[/tex]
    The Lagrangian is
    [tex]L\left(r,\dot{r},\dot{\phi}\right)=\dfrac{m}{2}\left[\left(1+\alpha^{2}\right)\dot{r}^{2}+r^{2}\dot{\phi}^{2}\right]-mg\alpha r[/tex]
    The equation of motion is
    [tex]\dfrac{d}{d t}\left(\dfrac{\partial L}{\partial\dot{\mathbf{q}_{k}}}\right)-\dfrac{\partial L}{\partial\mathbf{q}_{k}}=0[/tex]
    3. The attempt at a solution
    The equation of motion for [itex]r[/itex]
    [tex]\ddot{r}+\alpha^{2}\ddot{r}-r\dot{\phi}^{2}+g\alpha =0[/tex]
    The equation of motion for [itex]\phi[/itex]
    [tex]r^{2}\dot{\phi}-r^{2}\ddot{\phi} =0[/tex]
    Is this correct? If so how does it imply that [itex]r^2\dot{\phi}=K[/itex] is a constant?
    Thanks
     
  2. jcsd
  3. Apr 23, 2015 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your ##\phi##-equation is incorrect:
    $$\frac{\partial L}{\partial \dot{\phi}} = m r^2 \dot{\phi},$$
    so
    $$\frac{d}{dt} (m r^2 \dot{\phi}) = \frac{\partial L}{\partial \phi} = 0 .$$
     
  4. Apr 23, 2015 #3
    Hi, got it thanks. Integrating the left hand side gives
    [tex]r^2\dot{\phi}=K[/tex]
    Meaning I can eliminate [itex]\dot{\phi}[/itex] from the equation of motion for [itex]r[/itex]
     
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