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Coriolis acceleartion

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    If you project a particle vertically upward, and you neglect everything except gravity and the Coriolis acceleration, my book says that it will not land on the point where you projected it. Why does this make any sense since the Coriolis force should be antisymmetric on the ascent and on the descent and should thus cancel?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 7, 2008 #2
    Why not? - (w cross v) = w cross -v, right?
    Last edited: Feb 7, 2008
  4. Feb 7, 2008 #3


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    I think you might actually have to look at a solution to get the full story. That v -> -v reverses the coriolis acceleration is a valid observation IF you are treating the acceleration as an infinitesimal perturbation to the w=0 trajectory. But that just means that the displacement of the landing position is a higher order effect.
  5. Feb 7, 2008 #4


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    It's easiest to understand using angular momentum.
    To keep angular momentum constant, the angular velocity has to decrease as the altitude
  6. Feb 7, 2008 #5
    I don't see how that explains the lack of symmetry in the ascent and the descent.
  7. Feb 7, 2008 #6
    This is very sensible if you can manage to try it. Do you live somewhere where you can get on a rotating platform (a Merry-go-round) or one of those kid's spinners at the playground (if it's not covered by snow). In a pinch find a rotating stool which is fastened to the floor. Get a tennis ball. While spinning, toss it in the air. Watch what happens going up. Watch what happens coming down.
  8. Feb 8, 2008 #7


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    That is a different effect.
  9. Feb 8, 2008 #8
    Well, you're partly right. The movement outward is centrifugal force, but the movement CW or CCW is the coriolis force.
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