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Anyone familiar with centrifugal potential and brachistochrone in polar coords?

  1. Jan 16, 2008 #1
    Anyone familiar with "centrifugal potential" and "brachistochrone" in polar coords?

    Hi there,

    The issue appears within this problem:

    a bead is sliding without friction on a straight wire; the wire rotates in a plane:
    \omega= const
    No external fields, no gravity.



    *******************Which is true:

    1) bead's velocity **IS** v = \omega r, because in a rotating frame the kinetic energy = centrifugal energy:
    (1/2) m v^2 = (1/2) m (r^2) (\omega^2)

    2) bead's velocity **IS NOT** v = \omega r, because expressing v in polar coordinates r,phi:

    v = \sqrt(\dot{r}^2 + (r^2)*(\dot{phi}^2))
    The bead turns with the wire, so \dot{phi} = \omega
    Then v = \sqrt(\dot{r}^2 + (r^2)*(\omega^2)) > \omega r,
    because thoughout the motion \dot{r} is never zero.

    Any ideas?
    ---------------------------

    Once we know the answer, let's try to build the functional
    T = \int (ds/v) for the above problem. Try to express it in polar coordinates. Any luck?
    It is really useful if the integral can be Euler-Lagrange minimized for any constraint (not just a straight line). For instance, how do I write T = \int (ds/v) when the bead is constrained to slide on a string shaped as an Archimedean spiral r = \phi \prime?

    When this one is minimized with Euler-Lagrange, it should give a straight-line trajectory seen from the inertial frame of reference.

    Any input is appreciated.
     
  2. jcsd
  3. Jan 17, 2008 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    z_offer09, it's kind of hard to follow the math in this thread. Since you seem to be familiar with LaTeX, take a look at this thread to see how it's implemented in this forum.
     
  4. Jan 21, 2008 #3
    Here is the question posted neatly:

    Hi there,

    The issue appears within this problem:

    a bead is sliding without friction on a straight wire; the wire rotates in a plane:
    [tex] \omega= const [/tex]
    No external fields, no gravity.



    *******************Which is true:

    1) bead's velocity **IS** [tex]v = \omega \ \ r[/tex], (this should read v = \omega * r )because in a rotating frame the kinetic energy = centrifugal energy:
    [tex]\frac{1}{2} m v^2 = \frac{1}{2} m r^2 \omega^2[/tex]

    2) bead's velocity **IS NOT** [tex]v = \omega \ \ r[/tex], because expressing [tex]v[/tex] in polar coordinates [tex]r,\phi[/tex] in the inertial frame:

    [tex]v = \sqrt{\dot{r}^2 + r^2\dot{\phi}^2}[/tex]
    The bead turns with the wire, so [tex]\dot{\phi} = \omega[/tex]
    Then [tex]v = \sqrt{\dot{r}^2 + r^2\omega^2} > \omega r [/tex]
    because thoughout the motion [tex]\dot{r}[/tex] is never zero.

    Any ideas?
    ---------------------------

    Once we know the answer, let's try to build the functional
    [tex]T = \int \frac{ds}{v}[/tex] for the above problem. Try to express it in polar coordinates. Any luck?
    It is really useful if the integral can be Euler-Lagrange minimized for any constraint (not just a straight line). For instance, how do I write [tex]T = \int \frac{ds}{v}[/tex] when the bead is constrained to slide on a string shaped as an Archimedean spiral [tex]r = \phi '[/tex]?

    When this one is minimized with Euler-Lagrange, it should give a straight-line trajectory seen from the inertial frame of reference.

    Any input is appreciated.
     
    Last edited: Jan 21, 2008
  5. Apr 6, 2008 #4
    Just wanted to bump this thread seeing as I have the same question.

    Any input from the higher powers?
     
  6. Apr 6, 2008 #5
    They both look very odd, the first more so. Stick to one frame and one frame only. Either do it in the non-rotating frame, in which case the latter is true, or do it in a rotating frame, in which case there is just an effective potential that pushes things out from the origin.
     
  7. Apr 6, 2008 #6
    You never stated what the actual question/problem was, I mean what are you really looking for? I will assume it's the trajectory.

    Then do this
    (1) Transform to frame co-rotating with the wire, use standard result from Marion and Thornton for either the effective force or potential from doing that non-inertial uniformly rotating reference frame transformation.
    (2) Construct the Lagrangian in that co-rotating frame, and then write down and solve the Euler-Lagrange equation, it's at least only 1d since you transformed away the 2d motion.
    (3) Now that you have the solution transform back to the inertial reference frame you started with.
     
  8. Apr 7, 2008 #7
    Well, if you set it up looking at the problem in the 1d, non-rotating frame of the wire, what would be the constraints used for the Lagrangian Multiplier?

    Also how would you set up the equation of motion in this same frame?
     
  9. Apr 8, 2008 #8
    I have to say that "non-rotating frame of the wire" is confusing. The frame attached to the wire is rotating, the inertial frame is the one in which the wire appears to be rotating.

    In the frame that co-rotates with the wire you do not need a Lagrangian multiplier, simply do everything in 1d.

    In the inertial frame the constraint that forces the bead to move with the wire would be (in polar coordinates) [tex]\theta - \theta_0 = \omega t[/tex]. So you can add to your Lagrangian the term [tex]\lambda (\theta - (\theta_0 + \omega t))[/tex], where [tex]\lambda[/tex] is your Lagrangian multiplier.

    The reason that is what the constraint should be is due to the fact that the bead must rotate at the same rate as the wire so that it doesn't leave the line that the wire lies on, but the radial coordinate should be independent of that constraint, and it's time evolution given by Newton's 2nd Law/Euler-Lagrange equation.
     
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