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Coriolis Effect for a level surface on Earth

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    A puck slides with speed v on a frictionless ice that is level in the sense that the surface is perpendicular to geff at all points. Show that the puck moves in a circle as viewed in the earth's rotating frame. Determine the radius of the circle and the angular frequency of the motion. Assume that the puck's circle is small compared to the radius of the earth.


    2. Relevant equations

    [itex]\vec{F}_{cor}=2m\vec{v}\times\Omega[/itex]

    3. The attempt at a solution

    I have a solution, I just wanna make sure that my logic is correct. I'm worried that I'm missing something subtle.

    Assuming that geff the normal force and the z component of the Coriolis force all balance (it's sitting on top of the ice at all times). Then the only thing I have to worry about is the Coriolis force.

    [itex]\vec{F}_{cor}=2m\vec{v}\times\Omega[/itex] dot product by v on both sides
    [itex]\vec{F}_{cor}\cdot\vec{v} = 2m(\vec{v}\times\Omega)\cdot\vec{v} [/itex]
    [itex](\vec{v}\times\Omega)\cdot\vec{v} = 0 [/itex] therefore:
    [itex]\vec{F}_{cor}\cdot\vec{v}=0[/itex] therefore, [itex]\vec{F}_{cor}\perp\vec{v}, \forall\vec{v}[/itex]

    Since the force is perpendicular to the velocity, it must go in a circle.
     
  2. jcsd
  3. Nov 14, 2013 #2

    D H

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    You didn't have to go through all those contortions to arrive at the fact that force is perpendicular to velocity. The cross product of two vectors that are not parallel or antiparallel to one another is a third vector that is perpendicular to both vectors involved in the cross product.

    That said, that force is perpendicular to velocity alone does not make for circular motion. You need some other condition. Hint: You were told to "assume that the puck's circle is small compared to the radius of the earth."
     
  4. Nov 14, 2013 #3
    The puck's circle is small implies v stays in a plane (in its reference frame). So I need to say that the motion is confined to a plane for a consistently perpendicular force to make a circle, right?
     
  5. Nov 14, 2013 #4

    D H

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    That's part of it. Do you get a circle if the force isn't constant?
     
  6. Nov 14, 2013 #5
    Ohhhhhh. I'm pretty embarrassed I missed that part. Thanks!
     
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