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Centrifugal Force and shape of water

  1. Sep 30, 2005 #1
    "Centrifugal" Force

    Say you have a cylinder of radius r with πr2h liters of water in it. If you put it on some sort of turntable and spin it with an angular velocity ω, is there a way to tell what the shape of the water will be? I mean, if you look at the cross section will you be able to find out if it forms for example, a parabola?

    This isn't a homework problem or anything, I'm just curious. Thanks for your help.
  2. jcsd
  3. Sep 30, 2005 #2

    Doc Al

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    Actually, the rotating water surface will assume the form of a parabola. The way to figure that out is to examine the forces on a element of the water surface. The net force is centripetal.
  4. Sep 30, 2005 #3
    Relative to some origin (let's say the vertex), is there a way to find the equation of that parabola or is the shape the most that can be determined?
  5. Sep 30, 2005 #4

    Doc Al

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    According to my calculation, the equation (using the vertex as the origin) should be:
    [tex]y = \frac{\omega^2}{2 g} x^2[/tex]
  6. Sep 30, 2005 #5


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    Said another way,the surface of the water will be an equipotential surface in a corotating frame of reference.
  7. Oct 1, 2005 #6
    Are you rotating just the turntable or the cylinder as well? In the former case I think some water will fall out due to its inertia of rest.Tell me if I am wrong,and also if water would fall out in the latter case too.
  8. Oct 1, 2005 #7
    Is this shown by the following?:

    [tex]E_k=E_p\implies v=\sqrt{2gh}[/tex]

    [tex]y=h;\quad x=r[/tex]


    [tex]\sqrt{2gy}=x\omega\implies y=\frac{{\omega}^2}{2g}x^2[/tex]

    This was the only way I could derive the equation you posted above, and I figured that the kinetic energy would be the same as the potential energy because if you look at a single molecule of water, it is doesn't experience any friction because it is rotating with every other molecule. Also, I just said that the distance from the center of the cylinder r was x (since the origin is the center) and the height above that center is y. Am I correct with what I posted above?

    Thanks again.
  9. Oct 2, 2005 #8

    Doc Al

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    I can't say I understand your derivation or reasoning, but maybe I'm just dense today.

    Here's how I would derive it (from an inertial frame). Take an element of mass on the surface. The force due to the water/air pressure must be perpendicular to the surface. Call that force F. The only other force is the weight, mg. Applying Newton's 2nd law to vertical and horizontal components gives:
    [tex]F \cos \theta = mg[/tex]
    [tex]F \sin \theta = m\omega^2 x[/tex]
    where [itex]\theta[/itex] is the angle the tangent to the curve makes with the horizontal axis, thus:
    [tex]\tan \theta = dy/dx = \frac{\omega^2}{g} x[/tex]
    [tex]y = \frac{\omega^2}{2g} x^2[/tex]

    Viewed from the rotating frame, the derivation is even simpler. Just realize that the surface of the fluid will be perpendicular to the direction of the apparent "gravity": the sum of real gravity plus the outward centrifugal force.
  10. Oct 2, 2005 #9


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    For an excellent practical application of this idea, check this out.
  11. Oct 2, 2005 #10
    Ahh, ok that makes sense. I don't know what I was thinking when I derived it the other way :uhh:

    Thanks for the help :smile:
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