# Liquid oscillating in a U-tube problem

## 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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Redbelly98
Staff Emeritus
Homework Helper

## 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.

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?

Redbelly98
Staff Emeritus
Homework Helper
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.

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 ∫?