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Hamiltonian of a pendulum constrained to move on a parabola

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data

    The point of suspension of a simple pendulum of length [itex] l [/itex] and mass [itex] m [/itex] is constrained to move on
    a parabola [itex] z = ax^2 [/itex] in the vertical plane. Derive a Hamiltonian governing the motion of the
    pendulum and its point of suspension. This is a two-dimensional problem.


    2. Relevant equations

    eq(1) [itex] H=T+U [/itex]

    3. The attempt at a solution

    [itex] T=\frac{m}{2}[\dot{x}^2+\dot{z}^2] [/itex]

    [itex] U=mgz [/itex]

    eq(2) [itex] H = \frac{m}{2}[\dot{x}^2+\dot{z}^2]+mgz [/itex]

    Applying the constraint [itex] z=ax^2 [/itex], [itex] \dot{z} = 2ax\dot{x} [/itex]

    [itex] H = \frac{m}{2}[\dot{x}^2+4a^2x^2\dot{x}^2]+mgax^2 [/itex]

    The resulting Hamilton's equations I got were:

    eq(3) [itex] \dot{p_x} = -(4ma^2x\dot{x}^2+2mgax) = -\frac{∂H}{∂x} [/itex]

    eq(4) [itex] \dot{x} = \frac{p_x}{m(1+4a^2x^2)} = \frac{∂H}{∂p_x} [/itex]

    and [itex] p_x = \frac{∂L}{∂\dot{x}} = m(1+4a^2x^2)\dot{x} [/itex]

    This problem seemed uncharacteristically easy, so I definitely appreciate anybody who checks it over (: Thank you in advance!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 3, 2014 #2

    Matterwave

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    The constraint you implemented is for the point of suspension for the pendulum. The pendulum bob need not move on that parabola.

    What you have there is the solution to a point mass constrained to move on a parabola, when instead you want to solve for a pendulum whose point of support is on a parabola.
     
  4. Apr 3, 2014 #3
    Okay how does it look now?
     
  5. Apr 3, 2014 #4

    Matterwave

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    You've still just applied the constraint to the bob itself, and not the point of support...all you've done is make a change of variables.

    You need a set of coordinates for the pendulum support, and then a set of coordinates for the bob. There should be two constraints, one for where the support may go, and one for where the bob may go since the pendulum is fixed to the support.

    For example, I should have the point of support at coordinates:

    [tex](x_s,z_s)[/tex]

    The bob would then have coordinates:

    [tex](x_b,z_b)[/tex]

    I should express the bob coordinates in terms of the support coordinates. And then implement the constraint on my support coordinates!
     
  6. Apr 3, 2014 #5
    then constraint I should apply should be [itex] (x_s-x_b)^2+(z_s-z_b)^2=l^2 [/itex] ?
     
  7. Apr 3, 2014 #6

    Matterwave

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    Yes, but there's a much simpler way of applying that constraint, which you sort of did from your previous attempt, by using lcos(theta) and lsin(theta)
     
  8. Apr 3, 2014 #7
    Okay I'm going to try to resolve it using this new information. Thanks for pointing that out by the way.(:
     
  9. Apr 3, 2014 #8
    Okay so far I think I've managed to express the bob coordinates in terms of the support coordinates

    [itex] x_b=x_s+x [/itex] and [itex] z_b = z_s+z [/itex] such that [itex] x^2+z^2 = l^2 [/itex] but I'm not exactly sure whether this is helpful.
     
  10. Apr 3, 2014 #9

    Matterwave

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    That would work except the algebra will get very annoying very quickly. Try this:

    [tex]x_b=x_s+l\sin (\theta)[/tex]

    and

    [tex]z_b=z_s-l\cos (\theta)[/tex]

    Draw a picture to see how these coordinates work.

    Notice that your constraint is automatically satisfied with these coordinates.
     
  11. Apr 3, 2014 #10
    I actually just had that thought and doing it that would also mean that I end up 2 canonical coordinates which is exactly what I want. Thank you for your help you are a life saver! (:
     
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