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Trajectory of pendulum in frame of rotating disk under it

  1. Aug 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the pendulum depicted in the adjacent figure: a mass m
    is attached to non stretching chord of length `. Directly below the
    pendulum is a circular disc rotating with constant angular velocity
    w. We attach to the disk a frame whose x-axis is in the plane of the
    pendulum and the pendulum is in the extremal position to the right
    at t = 0.

    Compute the coordinates x and y of the pendulum in the rotating
    frame as a function of t when the maximal angle of deflection qmax
    is assumed small. And what curve does the motion of the pendulum correspond to in
    the rotating frame if w = √(g/l). (Which is angular freq. of pendulum)


    Example of the situation (Starting from 0:49):



    2. Relevant equations

    Same way as with Foucalt's pendulum, I've derived equation $$\ddot{u}+2ωi\dot{u}+Ω^{2}u=0$$
    Where u is complex variable $$x+iy$$
    ω is angular frequency of the disk and Ω of the pendulum
    And the coordinate axes (x,y) in rotating plane is considered as complex plane here.


    3. The attempt at a solution

    Solving characteristic equaton yields$$m^2 + 2ωim+Ω^2=0⇒m=-iω±√(i^2ω^2-Ω^2)$$

    Now if this is solved for u, I get$$u=Ae^{-i(ω+√(Ω^2+ω^2))t}Be^{-i(ω-√(Ω^2+ω^2))t}$$

    But then, if Ω = ω it seemingly should be that motion of pendulum in frame of rotating disk is circular. And if I plug that in the equation above, the trajectory seems to do same kind of "flower" as Foucalt's pendulum.

    Then if the characteristic equation would instead yield $$⇒m=-iω±√(ω^2-Ω^2)$$ then the latter term would cancel out, and we would get (I suppose) circular motion of form:$$Ccos(√(g/l)t)+iDsin(√(g/l)t)$$

    Which I have assumed what would be right answer. So what is problem with that 2. order characteristic equation, why doesnt imaginary unit's second power appear in the first terminside the square root? Or could that be right, and someting else wrong here?

    In example of Foucalt's pendulum, that equation was solved to form where there isnt 2. power of i in the first term inside sq. root, as in the term stayed positive, so that's why my first assumpton is that theres something I'm missing.
     
  2. jcsd
  3. Aug 14, 2016 #2

    Simon Bridge

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    ... but does it yield that?
    On what basis are you making this assumption?
    ... which 2nd order equation? (This is where it is useful to number your equations.)
    Do you mean: ##m = -i\omega \pm \sqrt{\omega^2-\Omega^2}## ?
    Then the answer is "I dunno" - you were the one who wrote it: why did you leave it out?
    It looks like you left it out because you thought it would be nice to do so.

    Your objection to just following the maths seems to be:
    ... since the pendulum should pass through the origin in both reference frames, circular motion is a problem, and I don't see how that somehow turns into the "flower" trajectory.
    I'm guessing you have covered Foucault's pendulum in class, with a derivation. The approach here should be just about identical with F's P at a pole.
    So what happens with that case that is different from this one?


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