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Surface between 2 rotating fluids.

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Oil and water are put in a cylindrical container. They can rotate at various frequency around the rotational symmetry axis.
    Consider gravitational and radial forces to find the form of the surface between the fluids.
    no friction, no mixing
    the surface has the form of a parabloid.
    surfaces are not touching each other or the bottom of the container.

    2. Relevant equations
    [itex]F_r=mr\omega ^2[/itex]
    [itex]F_g=mg[/itex]


    3. The attempt at a solution
    I have found a solution, but it only applies when both (oil-air ,water-oil) parabloids have the form of [itex]y=r^2[/itex], but observation suggests, that if the water is still and the oil rotates, that the oil-water surface is of the form [itex]y=-r^2[/itex]

    Edit:
    fluid3.png
    i found the upper surface by saying:
    the slope(derivative of the function y(r)) is given by the fact, that the resulting force is applied orthogonal to the surface.
    this gives:
    [itex]\frac{dy}{dr}=\frac{F_r}{F_g}=\frac{mr\omega ^2}{mg}=\frac{f\omega ^2}{g}[/itex]
    Integrating:
    [itex]y=\frac{r^2 \omega ^2}{2g}+C[/itex]

    Edit 2:
    Now I thought I could simply pull of this principle like in this picture to find the surface between the two fluids, simply adding the respective forces, but this stops making sense as soon as you see the surface between water and oil having the form [itex]y=-r^2[/itex]
    fluid.png
    Once the surface turns this way, the vectors stop making sense.
    fluid 2.png
     
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 30, 2012 #2

    haruspex

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    Try working with the internal pressure at a point at radius r and height y.
    Can you figure out the two partial derivatives?
    Start with the oil.
     
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