Liquid oscillating in a U-tube problem

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Homework Statement


A U-tube has vertical arms of radii 'r' and '2r', connected by a horizontal tube of length 'l' whose radius increases linearly from r to 2r. The U-tube contains liquid up to a height 'h' in each arm. The liquid is set oscillating, and at a given instant the liquid in the narrower arm is a distance 'y' above the equilibrium level.

Show that the potential energy of the liquid is given by...

U = (5/8)gρπ(r2)(y2)

'y' is the change in height.

Homework Equations


So i have the potential energy of system like his is given by
U = gρπ(r2)(y2)

The Attempt at a Solution



Im not sure where this 5/8's comes from. Im pretty sure its from the fact that the raduis of the tube is changing from r to 2r. Any hints?
 

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  • #2
Redbelly98
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Homework Equations


So i have the potential energy of system like his is given by
U = gρπ(r2)(y2)
For "Relevant equations", you are supposed to list any equation(s) that you'd use as a starting point for solving the problem. What you wrote above would be part of your "attempt at a solution".

So, what equation (or equations) have you used as a starting point for this problem?

The Attempt at a Solution



Im not sure where this 5/8's comes from. Im pretty sure its from the fact that the raduis of the tube is changing from r to 2r. Any hints?
Yes, it does have to do with the changing radius.

Another way to think about it: the scenario described is equivalent to taking some water from the "2r" side of the tube and putting it in the "r" side. I think you have figured out the volume and mass of that amount of water (if not -- do it). What is needed is to calculate a change in height, or something along those lines, for moving the water.
 
  • #3
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Another way to think about it: the scenario described is equivalent to taking some water from the "2r" side of the tube and putting it in the "r" side. I think you have figured out the volume and mass of that amount of water (if not -- do it). What is needed is to calculate a change in height, or something along those lines, for moving the water.
Ok In class we did a tube that was all the radius
We found that the m was the density x the Area
Would that be the same here? But isnt the area different b/c of the different r values?
And if it pluged that value into p=m/v, is that how I'd find the volume?
 
  • #4
Redbelly98
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Ok In class we did a tube that was all the radius
We found that the m was the density x the Area
Almost, not quite though. Mass is density x Volume.
Would that be the same here? But isnt the area different b/c of the different r values?
And if it pluged that value into p=m/v, is that how I'd find the volume?
It would be, if you knew both p and m. But we don't know m, so it won't help here.

Instead, you first need to use geometry to figure out the volume.
 
  • #5
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Almost, not quite though. Mass is density x Volume.

It would be, if you knew both p and m. But we don't know m, so it won't help here.

Instead, you first need to use geometry to figure out the volume.
Ok so v= πr2 h
the h would be the length, I think.
the r2 has to do with the raduis, but thats changing, doesnt chaning things like that have something to do with the ∫?
 

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