Difference in liquid column heights in a rotating U-tube

AI Thread Summary
The discussion centers on calculating the height difference between liquid columns in a rotating U-tube. The key equation derived for linear acceleration is h = aL/g, where 'a' is the acceleration. The participant considers substituting a = w^2L for the rotating case but questions whether to use average acceleration due to its variation with distance from the axis. The pressure difference is analyzed using a pressure gradient approach, leading to the conclusion that the height difference in the rotating scenario is h = (L^2ω^2)/(2g). The discussion highlights the complexities of integrating pressure differences in a rotating system.
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Homework Statement



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?

Homework Equations



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

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.
 
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kidsmoker said:
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?

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:
 
Ah yeah I get it now thanks. :)
 
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.
 
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