# Fluid in a rotating U-tube

1. Aug 4, 2008

### kidsmoker

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?

Thanks.

2. Aug 4, 2008

### tiny-tim

Hi kidsmoker!

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.

3. Aug 5, 2008

### kidsmoker

Ah yeah I get it now thanks. :)

4. Nov 7, 2009

### fluidistic

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 $$h=\frac{aL}{g}$$ 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 $$h=\frac{L^2 \omega ^2}{2g}$$ but I'm unsure of the integral I used.
The difference of pressure within a $$dr$$ element is $$\omega ^2 r \rho dr$$. I integrated this expression with respect to $$r$$, from $$0$$ to $$L$$.