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Spiral motion and the centripetal force

  1. May 24, 2012 #1
    Major brainfart here.

    Consider the spiral motion of a particle such that the distance from the origin to the particle is

    r(t) = e^t

    with a constant angular velocity.

    Now since the particle rotating with a constant angular velocity, I would think that the net force on the particle would have to be an inward centripetal force in the radial direction.

    However, there must also be a net outward force, in the radial direction, to account for the accelerating distance between the origin and the particle.

    Obviously this can't be right, but I can't pin down where I'm going wrong.

    Cheers!
     
  2. jcsd
  3. May 24, 2012 #2

    tiny-tim

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    hi mathman44! :smile:

    (try using the X2 button just above the Reply box :wink:)

    the radial acceleration is r'' - θ'2r :wink:

    (and can you prove that? o:))
     
  4. May 24, 2012 #3
    Isn't r'' the radial acceleration? Did you mean that r'' = -w^2 * r ?

    And isn't this just equivalent to saying that the net force is equal to the centripetal force?

    ps for a polar system I'm getting

    mrw^2 - dV/dr = mr''

    as the force equation using a simple lagrangian.
     
    Last edited: May 24, 2012
  5. May 25, 2012 #4

    haruspex

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    No. Consider a particle on a straight course missing the origin. It has an angular velocity around the origin, and if its speed changes appropriately it may even have a constant angular velocity for while (but will have to shoot off to infinity in a finite time).
    It's not at all clear that it would be a net outward or net inward. That will depend on the details of r(t).
    Radial acceleration means the component of the acceleration which is directed towards/away from the origin (not, acceleration in the value of the radius). As Tiny Tim says, this is given by r'' - θ'2r. E.g. in circular motion about the origin, r'' = 0, but the particle will have acceleration towards the origin.
    In your scenario, θ' is constant, and r'' = r. Note that if instead you had
    r(t) = ewt, where θ' = w
    then the radial acceleration would have been 0.
     
  6. May 25, 2012 #5

    tiny-tim

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    acceleration in polar coordinates :wink:

    $$ \boldsymbol{a}\ =\ (\ddot{r}-r\dot{\theta}^2)\hat{\boldsymbol{r}}\ +\ (r\ddot{\theta}+2\dot{r}\dot{\theta})
    \hat{\boldsymbol{\theta}} $$

    (the "hats" are unit vectors)
     
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