Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    Staff: Mentor

    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

    User Avatar

    Staff: Mentor

    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

    Tide

    User Avatar
    Science Advisor
    Homework Helper

    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]\frac{ds}{dt}=v=x\frac{d\theta}{dt}=x\omega[/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

    User Avatar

    Staff: Mentor

    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

    Danger

    User Avatar
    Gold Member

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Centrifugal Force and shape of water
  1. Centrifugal Force (Replies: 12)

Loading...