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Accelerated mathemathical pendulum

  1. Apr 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Mathemathical pendulum with mass ##m## on a line ##L## is moving with constant acceleration in xz plane, ##\vec{a}=(a_x,a_z)##. Let's consider only the oscillations in xz plane. Find stationary value of ##\theta ## and find the frequency of small movements around the stationary angle if ##a_x=0##.

    wfferer.JPG
    2. Relevant equations



    3. The attempt at a solution

    Ok, let's put one coordinate system somewhere :D and another one on the top of line ##L##. Let's say that these two coordinate systems are connected with vector R.

    Let's express everything in accelerated system:

    ##\vec{r}=(Lsin\theta , Lcos\theta )## and ##\vec{R}=(-\frac{1}{2}a_xt^2,-\frac{1}{2}a_yt^2)##

    Both together give me ##v=(\dot{\theta }Lcos\theta -a_xt,-\dot{\theta }Lsin\theta -a_yt)## and ##v^2=\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x##

    So ##L=T-V=\frac{1}{2}m(\dot{\theta }^2L^2+t^2(a_x^2+a_y^2)+2\dot{\theta }Lt(sin\theta a_y-cos\theta a_x)+mgLcos\theta ##.

    I won't even write the equations from here one because this brings me to wrong simple solution:

    ##\ddot{\theta }+\frac{g}{l}\theta =0## which can not be ok. It looks like I am missing something when calculating the velocity? But what?
     
  2. jcsd
  3. Apr 25, 2014 #2

    Simon Bridge

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    I'd have tried:
    In the accelerated frame, rotate the axis to reflect the apparent direction of gravity - you should end up with the same relations as you are used to right?

    To recover the problem-statement x and y directions, transform back.

    Note: I originally interpreted the diagram as indicating that the pivot was subject to an applied force.
     
  4. Apr 25, 2014 #3


    That's the problem. If I would write everything one of you would for sure notice that I was not careful.

    Anyway, calculating everything ok brings me to

    ##\ddot{\theta }+\frac{1}{L}(a_xcos\theta +a_ysin\theta +g sin\theta )=0##

    And therefore stationary angle ##\theta _0=arctg(\frac{a_x}{a_y+g})##

    And for ##a_x=0## also

    ##\ddot{\theta }+\frac{1}{L}(a_y+g)\theta =0##
     
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