Liquid oscillating in a U-tube problem

In summary: Ok so v= πr2 hthe h would be the length, I think.the r2 has to do with the raduis, but that's changing, doesn't chaning things like that have something to do with the ∫?Yes, changing the radius would affect the area and the potential energy.
  • #1
RockenNS42
52
0

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. I am 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
RockenNS42 said:

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. I am 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
Redbelly98 said:
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 isn't 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
RockenNS42 said:
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 isn't 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
Redbelly98 said:
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 that's changing, doesn't chaning things like that have something to do with the ∫?
 

What is the liquid oscillating in a U-tube problem?

The liquid oscillating in a U-tube problem is a physics problem that involves a U-shaped tube filled with a liquid, such as water or mercury. The liquid is allowed to flow through the tube and is subjected to an external force, causing it to oscillate back and forth.

What causes the liquid to oscillate in a U-tube?

The liquid oscillates in a U-tube due to the principle of conservation of energy. When the liquid is pushed up one side of the tube, it gains potential energy. As it flows back down the other side, this potential energy is converted into kinetic energy, causing the liquid to rise on the opposite side. This process repeats, resulting in the oscillation of the liquid.

What factors affect the oscillation of the liquid in a U-tube?

The oscillation of the liquid in a U-tube is affected by several factors, including the density of the liquid, the diameter of the tube, and the magnitude of the external force. The length of the tube and the shape of the U also play a role in the oscillation.

How can the liquid oscillating in a U-tube problem be applied in real life?

The liquid oscillating in a U-tube problem is often used to explain the movement of fluids in pipes and tubes, which has practical applications in fields such as engineering and fluid mechanics. It can also be used to demonstrate the principles of resonance and harmonic motion.

Is there a mathematical equation to describe the liquid oscillating in a U-tube problem?

Yes, there is a mathematical equation known as the Bernoulli's equation that can be used to describe the liquid oscillating in a U-tube problem. This equation takes into account the various factors that affect the oscillation and can be used to predict the behavior of the liquid in different scenarios.

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