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Incompressible fluid in horizontal test tube

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    An incompressible fluid with density ρ is in a horizontal test tube of inner cross-sectional area A. The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed ω. Gravitational forces are negligible. Consider a volume element of the fluid of area A and thickness dr' at a distance r' from the rotation axis. The pressure on its inner surface is p and outer surface is p+dp.
    (a) Show that dp=ρω2r'dr'.
    (b) If the surface of the fluid is at a radius ro where the pressure is po, show that the pressure p at a distance r≥ro is p=po+ρω2(r2-ro2)/2.
    (c) An object of volume V and density ρob has its centre of mass at a distance Rcmob from the axis. Show that the net horizontal force on the object is ρVω2Rcm, where Rcm is the distance from the axis to the center of mass of the displaced fluid.
    (d) Explain why the object will move inward if ρRcm>ρobRcmob and outward if ρRcm<ρobRcmob.
    (e) For small objects of uniform density, Rcm = Rcmob. What happens to a mixture of small objects of this kind with different densities in an unltracentrifuge?

    2. Relevant equations

    ω=v/r
    A= π*r^2
    F=ma
    p=F/A

    3. The attempt at a solution

    I have really tried everything I could but i simply dont understand the way how to tackle these problems. I would really like to know how these answers come about. Can anybody please help me with discussing this??
    I would really appreciate it!
     
  2. jcsd
  3. Dec 4, 2013 #2
    That small volume element is in circular motion. What is its acceleration? What are the forces acting on it?
     
  4. Dec 4, 2013 #3
    a=ω^2r is the acceleration. The force acting on it comes from the pressure on its inner surface i guess? p=F/A with F=ma so substituing the above equation for acceleration would give: p= m*ω^2*r/A
    I dont know how to deal with the area A. I need to show what dp is, so I don't need the entire area A. But how can I tackle this?
     
  5. Dec 4, 2013 #4

    haruspex

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    Consider a disk element thickness dr. What is its mass?
     
  6. Dec 4, 2013 #5
    Hmm.. Its volume times its density? The volume should be A*dr I guess?
     
  7. Dec 4, 2013 #6
    But there's no A in the equation which I should get. I'm so lost!!
     
  8. Dec 4, 2013 #7
    I solved a!! Finally. But what about b? Should i integrate the answer from a?
     
  9. Dec 4, 2013 #8
    Integrate the left hand side of (a) from ##p_0## to ##p##, and the right hand side from ##r_0## to ##r##.
     
  10. Dec 11, 2013 #9
    How can I show that the net horizontal force is ρVω2R?
    F=ma= and a = ω^2r but how can I show that it is 2R? And can anybody help me with the explanations for questions d and e ?
     
  11. Dec 11, 2013 #10
    In part c, you have to recognize that you've created artificial gravity (in the horizontal direction). The acceleration you calculated is the artificial gravitational acceleration. You need to use Archimedes principle. The displaced mass of fluid is ρV. Parts d and e can be explained also by artificial gravity arguments.

    Chet
     
  12. Dec 11, 2013 #11

    haruspex

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    Consider the situation without the object in the fluid. The region that the object occupies would instead be occupied by the same volume of the fluid, in exactly the same location.
    Would that body of fluid be subject to the same external forces as the object?
    Would that body of fluid move along the tube as a result of those forces?
     
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