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Instantaneous velocity of point on a trochoid

  1. Jun 7, 2008 #1
    Hi
    I Googled trochoid, epitrochoid and hypotrochoid but wasn't able to find formulas for the instantaneous velocity of a point on trochoid curves. Does anyone know of an online reference (or know how to find the velocity)

    I can get rough estimates by calculating the distance between two "ticks" on a trochoid curve but I want to get exact values if possible.

    Thanks for any help
     
  2. jcsd
  3. Jun 7, 2008 #2

    HallsofIvy

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    Then you learned that parametric equations for a trochoid are
    [itex]x= a\phi- bsin(\phi)[/itex] and [itex]y= a- bcos(\phi)[/itex] where [itex]\phi[/itex] is the angle the line from the point to the center of the "rolling circle".

    I assume you also understand that a "point" on a tronchoid (or any other geometric figure) doesn't have a "velocity"- that's why just look up those words didn't give a "velocity". "Velocity" is a physics concept while "tronchoid" and "points" are purely geometric. Assuming you mean "the instantaneous velocity of a point moving with given speed on a tronchoid", then you get the instantaneous velocity, as Sir Isaac Netwons said, by differentiating the position vector: the x and y components of the velocity vector are given by
    [tex]\frac{dx}{dt}= \left[a -b cos(\phi)\right]\frac{d\phi}{dt}[/tex]
    and
    [tex]\frac{dy}{dt}= sin(\phi)\frac{d\phi}{dt}[/tex]

    In the simple case of "constant angular speed", [itex]\phi(t)= \omega t[/itex], that is
    [tex]\frac{dx}{dt}= \omega\left[a -b cos(\phi)\right][/tex]
    and
    [tex]\frac{dy}{dt}= \omega sin(\phi)[/tex]
     
  4. Jun 7, 2008 #3
    Thanks Halls. I'm guessing you're using [tex]\phi[/tex] to signify the angle but what are you using [tex]\omega[/tex] for? I tried Googling "omega + trig" and several variations on that, but omega seems to be used many different ways depending on the branch of mathematics. Could you give me a concrete example with actual numbers plugged in rather than [tex]\omega[/tex]?

    Thanks in advance.
     
  5. Jun 7, 2008 #4

    HallsofIvy

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    Once again, you will not find your answer in mathematics because your question is about physics. I don't know why you are "guessing" anything since I said

    1) 'where [itex]\phi[/itex] is the angle the line from the point to the center of the "rolling circle".'

    2) (as you quoted) "In the simple case of "constant angular speed", [itex]\phi(t)= \omega t[/itex]".
    [itex]\omega= \phi/t[/itex] is standard (physics) notation for the angular velocity.
     
  6. Jun 7, 2008 #5
    Thanks Halls. Much appreciated.
     
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