U Tube: Find Height Shift Due to Rotation

[ritwik06]

Homework Statement

A U tube of uniform cross sectional area A and base length 'L' contains a liquid of density "p". It is rotated about one of its vertical limbs with constant angular velocity [tex]\omega[/tex]. If the diameter of the tube is small compared to the base length, find the difference in levels of liquids in the to limbs.

The Attempt at a Solution

the shift will be because of the force exerted by the rotation on the liquid contained in the base.
I integrate to find this force [I am not posting the integration, because I think its right, if your answer dos not match with mine, tell me I will post my working on this integration.]
The force on the liquid in th base=[tex]\frac{Ap\omega^{2}L^{2}}{2}[/tex]

Now the work done by this force = change in potential energy due to rise of liquid in one limb
[tex]\frac{Ap\omega^{2}L^{2}}{2} * h = \frac{AP h^{2}g}{2}[/tex]

Is this step right to find "h"?
I was also thinking what is the net work done on water. Is it zero, because the rise in one limb=fall in the other?

 

[Irid]

I'm not sure about the work, but you could most easily equate the force due to rotation with the force due to difference in liquid heights (Newton's 3rd law). Further, there's no need to integrate, because Bernoulli's law gives in this situation
[tex]P = \frac{\rho v^2}{2} = \frac{\rho (\omega L)^2}{2}[/tex]

 

[ritwik06]

I'm not sure about the work, but you could most easily equate the force due to rotation with the force due to difference in liquid heights (Newton's 3rd law). Further, there's no need to integrate, because Bernoulli's law gives in this situation
[tex]P = \frac{\rho v^2}{2} = \frac{\rho (\omega L)^2}{2}[/tex]

Sorry friend, but I can't make out what you said. I haven't studied Bernoulli yet. Please help m with my own method, thanks

 

[Irid]

What I said is

[tex]F_{\text{rotation}} = F_{\text{\hidrostatic}}[/tex]

or,

[tex]\frac{\rho \omega^2 L^2}{2} = \rho g \Delta h[/tex]

[tex]\Delta h = \frac{\omega^2 L^2}{2g}[/tex]

Your method is a little fishy, because the force you're talking about is actually horizontal, and you apply it for calculating work, which was done in vertical motion. If you want to do energy conservation, fine, I think it's possible, but I'm not sure how.