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Free vortex

  1. Nov 30, 2003 #1
    Iam having some trouble understanding the nature of free vortexs, with the internet and libary being extremely scarce in answers. My understanding is that the angular momentum of the water causes a centripetal force to push the water outwards leaving a cavity in the middle. I understand that at any point the angular momentum times the radius is a constant. Can anyone give me a more indepth explanation on how vortices work? I need equations to work out circulation, vorticity and possibly the height and radius of a vortex.
  2. jcsd
  3. Dec 1, 2003 #2
    can anyone answer this instead? why is there greater centripetal force at the bottom of the beaker than the top
  4. Dec 1, 2003 #3
    It's rather a centrifugal force. From latin fugare=to flee. Because the water flees from the center.
    No. The velocity times the radius is a constant. In other words, the angular momentum is a constant.
    The idea of a water surface is, that at any point on the surface the total force on a particle is normal to the surface. With no motion, there's just gravity (vertical), so the surface is horizontal. With rotation, you get an extra centrifugal force [tex]F_{cf}=\frac{mv^2}{r}[/tex]acting horizontally. The surface will be normal to the vector sum of these 2 forces. Now, since [tex]vr = c[/tex] you get [tex]v=\frac{c}{r}[/tex] and thus [tex]F_{cf}=\frac{mc^2}{r^3}[/tex]. Thus, the centrifugal force will increase strongly towards the center, making the surface very steep.

    Note: This holds only for values of r above a certain limit. Below that, angular momentum and energy will be consumed by turbulence, the water starts behaving erratic, and <poetry on> your unlucky ship is destroyed in the great Maelstrom <poetry off>.

    The next step calls for differential equations to find the shape of the surface. Is that what you want to do?
    Last edited: Dec 1, 2003
  5. Dec 1, 2003 #4
    if you could guide me on the differentiation needed to work out the shape of the surface that would be helpful
  6. Dec 2, 2003 #5
    Well if a point on the surface has coordinates (r,h) then the tangent vector is (dr,dh). The force vector is (Fcf,-mg).

    Force being normal to tangent means
    dr \cdot F_{cf} - dh \cdot mg = 0.
    dr \frac{mc^2}{r^3} - dh \cdot mg = 0
    Can you integrate that to get h(r)?
    Last edited: Dec 2, 2003
  7. Dec 3, 2003 #6
    Thanks very much i can integrate from there. Just wondering in a vortex is the speed of the water rotating the same at any depth in the vortex?
  8. Dec 4, 2003 #7
    No. The idea is that it speeds up as it closes in. Did you read my post? I said vr = c. Yielding
    v = \frac{c}{r}
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