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Homework Help: Lagrangian and Jacobi Integral

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves on the surface of a paraboloidal bowl with position given by r=rcosθi+rsinθj+[itex]\frac{r^{2}}{a}[/itex]k
    with a>0 constant. The particle is subject to a gravitational force F=-mgk but no other external forces.
    Show that a suitable Lagrangian for the system is

    Find two constants of the motion

    If J[itex]^{2}[/itex]>2gp[itex]^{2}[/itex]/a where J and p are the initial values of the Jacobi function and the momentum conjugate to θ, show that in the subsequent motion the hight of the particle above to xy-plane varies between

    2. Relevant equations
    Lagrangian= kinetic energy-potential energy
    Kinetic energy=[itex]\frac{1}{2}[/itex]m||[itex]\dot{r}[/itex]||[itex]^{2}[/itex]

    3. The attempt at a solution
    I found the Lagrangian and I found p[itex]_{θ}[/itex] to be a constant of the motion. I also found the Jacobi to be a constant of the motion because the Lagrangian has no explicit time dependence. Using the definition of the Jacobi (and also because in this case it is equal to the total energy) I found it to be [itex]\frac{1}{2}[/itex]m[itex](\dot{r^{2}}[/itex][itex]([/itex]1+4[itex]\frac{r^{2}}{a^{2}})[/itex][itex][/itex]+r[itex]^{2}[/itex][itex]\dot{θ^{2}}[/itex])+[itex]\frac{mgr^{2}}{a}[/itex]

    I found p[itex]_{θ}[/itex] to be mr[itex]^{2}[/itex][itex]\dot{θ}[/itex]

    I tried substituting [itex]\dot{θ}[/itex]=[itex]\frac{p_{θ}}{mr^{2}}[/itex] into the expression for the Jacobi and rearranging for [itex]\dot{r}[/itex] and then integrating to find r but I ended up with a complicated function which I couldn't integrate. I also noticed that the solution of h is in the form of the quadratic equation so I don't know if that's anything to do with it. Am I completely on the wrong track or have I done something stupid that complicates everything? Any help will be greatly appreciated.
  2. jcsd
  3. Mar 15, 2013 #2


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    You don't have to do an integral. What can you say about [itex]\dot{r}[/itex] at the maximum and minimum heights?
  4. Mar 15, 2013 #3
    Thank you so much. I can't believe I missed that, I can console myself with the fact that so has everyone else I have spoken to. So I used [itex]\dot{r}[/itex]=0 and ended up with
    I'm guessing [itex]\frac{r^{2}}{a}[/itex]=h
    Is this because in the original r the [itex]\frac{r^{2}}{a}[/itex] is with k which is the vector in the z direction. Does that even make sense outside my head? Sorry if it doesn't
  5. Mar 15, 2013 #4


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    Yes, I believe that's correct. I didn't pay much attention to the parameterization of the radial vector at first and was confused for a bit too.
  6. Mar 15, 2013 #5
    Thank you, you have saved me several hours of unnecessary work and frustration :D
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