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Homework Help: Fluid in a rotating U-tube

  1. Aug 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A U-shaped tube with a horizontal segment of length L contains a liquid. What is the difference in height between the liquid columns in the vertical arms if the tube is mounted on a horizontal turntable, and is rotating with angular speed w, with one of the vertical arms on the axis of rotation?

    2. Relevant equations

    a=w^2r towards the centre
    p = p(o) + density*g*h

    3. The attempt at a solution

    A previous part to the question was finding the difference in height if the U-tube had acceleration a towards the right. I did this and found h=aL/g.

    I was thinking I could just substitute a=w^2L into this equation but then I had second thoughts since the acceleration decreases as you get closer to the axis of rotation. Do you take the average acceleration intead?

  2. jcsd
  3. Aug 4, 2008 #2


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    Science Advisor
    Homework Helper

    Hi kidsmoker! :smile:

    In the first case (linear acceleration),

    if the linear density (mass/length) is ρ, and the pressure gradient is Q, then a length dr in the middle will have mass ρ dr and acceleration a, and so mass x acceleration = aρ dr = pressure difference = Q dr,

    so Q = aρ, and so total pressure difference = aρL.

    Does that help you with the circular case? :smile:
  4. Aug 5, 2008 #3
    Ah yeah I get it now thanks. :)
  5. Nov 7, 2009 #4


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    Gold Member

    Although I took vector calculus, I didn't understand what you mean by "gradient pressure", tiny-tim. I think I should google for it.

    For the first question I get that the answer is [tex]h=\frac{aL}{g}[/tex] and the column of liquid that has a higher high is the right one.

    I tried the second part of the question, namely the first question that kidsmoker asked, and reached [tex]h=\frac{L^2 \omega ^2}{2g}[/tex] but I'm unsure of the integral I used.
    The difference of pressure within a [tex]dr[/tex] element is [tex]\omega ^2 r \rho dr[/tex]. I integrated this expression with respect to [tex]r[/tex], from [tex]0[/tex] to [tex]L[/tex].
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