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Classical Mechanics: Coriolis Effect Problem

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A bird of mass 2 kg is flying at 10 m/s in latitude of 60° N, heading due East. Find the horizontal and vertical components of the Coriolis force acting on it.

    2. Relevant equations
    The Coriolis Force, F = 2mwv. Where ∧ shows the cross product between angular frequency vector, w, and change in the position vector, v.

    Θ will be the co-latitude -- that is, 90°- 60° = 30°.

    3. The attempt at a solution
    I started by deciding that my coordinates would be oriented so that x points East, y points North, and z points straight up (away from the earth). Thus, I believe, w = {wcosΘ, 0, 0} since the bird flies only East.

    So taking the Cross product with v = {x' , y' , z'} (where ' indicates the change in position), I receive the following vector {0 , -z'cosΘ , y'cosθ}. Now, I've shown the product vector without the coefficients, because my confusion arises at the presence of the y' and z's. Exactly what am I to do about them?

    It's one of those problems where I can't tell if I'm missing something terribly basic, or having been working under a more general misapprehension. I'd very much appreciate any help!

    p.s. This is my first post in the forum, and so I'm sure I've broken a plethora of the rules/etiquettes for which you must forgive me.

    p.p.s. This is not a homework problem, just a kind of review (which makes the fact that I'm struggling with it so much more embarrassing), so don't feel ashamed at helping me cheat!
  2. jcsd
  3. Nov 25, 2014 #2


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    Homework Helper

    Welcome to PF!

    You miss a minus sign. The Coriolis force is F = -2mwv.

    The angular velocity is a vector parallel to the axis of rotation of Earth and pointing upward. In your coordinate system it has both y and z components, and zero x component.

    The velocity vector is (10, 0,0) as it has only East (x) component.
  4. Nov 25, 2014 #3
    Oh my god. Thank you so much!
  5. Nov 25, 2014 #4


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    Homework Helper

    You are welcome. :)
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