Liquid oscillating in a U-tube

In summary: I'm really confused about this part)In summary, the potential energy of the liquid is given by U = \frac{5}{8}g\rho\pi r^{2}y^{2}.
  • #1
mbigras
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2

Homework Statement


A U-tube has vertical arms of radii [itex]r[/itex] and [itex]2r[/itex], connected by a horizontal tube of length [itex]l[/itex] whose radius increases linearly from [itex]r[/itex] to [itex]2r[/itex]. The U-tube contains liquid up to a height [itex]h[/itex] in each arm. The liquid is set oscillating, and at a given instant the liquid in the narrower arm is a distance [itex]y[/itex] above the equilibrium level.
(a) Show that the potential energy of the liquid is given by
[tex]U = \frac{5}{8}g\rho\pi r^{2}y^{2}[/tex].
(b) Show that the kinetic energy of a small slice of liquid in the horizontal arm (see the diagram) is given by
[tex]dK = \frac{1}{2}\rho\frac{\pi r^{2}dx}{\left(1+x/l\right)^{2}}\left(\frac{dy}{dt}\right)^{2}[/tex]
(Note that, if liquid is not to pile up anywhere, the product velocity Xcross section must have the same value everywhere along the tube.)
(c) Using the result of part (b), show that the total kinetic energy of all the moving liquid is given by
[tex]K = \frac{1}{4}\rho \pi r^{2}\left(l + \frac{5}{2} h\right) \left(\frac{dy}{dt}\right)^{2}[/tex]
(ignore any nastiness at the corners.)
(d) From (a) and (c), calculate period of oscillations if [itex]l = 5h/2[/itex].



Homework Equations


[tex]U = mgh[/tex]
[tex]m = V \rho[/tex]
[tex]V = \pi r^{2} * height[/tex]



The Attempt at a Solution


Right now I'm working with part (a). From reading in the book:
"The increase of gravitational potential energy in the situation shown in Fig. 3-8 corresponds to taking a column of liquid of length [itex]y[/itex] from the left-hand tube, raising it through the distance [itex]y[/itex], and placing it on the top of the right-hand column. Thus we can put [itex]U = g \rho A y^{2}[/itex]"

The book gives a hint:
"The side arms need not, in fact, be vertical, as long as they are straight; the cross sections need not be equal, as long as they are constant; and the connecting tubing may be of different cross section again, provided the appropriate geometrical scaling factors are using to express the displacement and speed of any part of the liquid in terms of those in either of the side arms."

When I read this hint, it feels like, alright all these words make sense to me by themselves. The horizontal tube scales up in a linear way, the radius of the vertical right-hand tube scales up by a factor of 2, so the Volume scales up by a factor of 4. There's some displacement in the left-hand tube [itex]y[/itex] which if it's scaled properly will equal some displacement in the right-hand tube. But putting them together I feel myself asking:

How do I calculate the appropriate geometrical scaling factors? and How do I use these factors once I've calculated them?

I calculated the factor that the horizontal tube scales up to be:
[tex]\pi r^{2}l / \left( \pi \int _{0} ^{l} \left(\frac{r}{l} x + r\right)^{2} dx \right) = \frac{7}{3}[/tex]
but I'm not really sure if this is the idea that the hint is pointing to.
 

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  • #2
I would ignore the hint myself.
Just use that the liquid in the narrow arm is a distance y higher than equilibrium and follow your nose.
It should just be straight geometry: if the narrow tube increases height by y, then the wide tube must decrease height by how much? What does than mean for potential energy?

(Assume oscillations are small - so there is always liquid above the U-bend and none spills out the open tops.)
 
  • #3
If the left tube increases its height by [itex]y[/itex] then the right tube decreases its height [itex]\frac{y}{4}[/itex]. Choosing the [itex]y = 0[/itex] to be the equilibrium point then the left tube has some positive potential energy and the right tube has some negative potential energy:
[tex]V = \pi r^{2} y = 4 \pi r^{2} y^{\prime}[/tex]
[tex]y^{\prime} = \frac{y}{4}[/tex]
[tex]U = U_{left} + U_{right}[/tex]
[tex]= mg(y - y/4) = \frac{3}{4} \pi r^{2} y g y = \frac{3}{4}g \rho \pi r^{2} y^{2}[/tex]
so something seems to be missing from this answer.
 
  • #4
Hint: center of mass
 
  • #5
Ahh a terrific hint! But why is the potential energy of the right tube not negative?
 
  • #6
You were the one making the definitions. It can be negative if you want.
 
  • #7
The center of mass hint was really the way to go. I found the volume that the liquid takes up in the left tube and use that to find how that same volume of liquid sits in the right tube. Then you can see the change in height of the center of mass and use mgh.
 
  • #8
For part (b):

[tex]R = \frac{r}{l} x + r \\
dK = \frac{1}{2}dm v^{2} \\
dm = \pi R^{2} dx \rho \\
v^{2} = \left(\frac{dy}{dt}\right)^{2} \\
dK = \frac{1}{2} \rho \pi dx \left(\frac{r}{l} x +r \right)^{2} \left(\frac{dy}{dt}\right)^{2} \\
dK = \frac{1}{2} \rho \pi r^{2} dx \left(\frac{x}{l}+1\right)^{2} \left(\frac{dy}{dt}\right)^{2}[/tex]

It looks like the book divides by [itex]\left(\frac{x}{l} + 1\right)^{2}[/itex] instead of multiplying. Could it be because I set up my function for the radius in an unrealistic way? (see the attached image)
 

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  • #9
You've used the wrong speed.
v should be a function of x.
 
  • #10
The question wants it in terms of [itex]\frac{dy}{dt}[/itex] but because all the liquid is moving the same speed I think that [itex]\frac{dy}{dt} = \frac{dx}{dt}[/itex].
 
  • #11
If dx/dt were the same for all parts of the horizontal tube, then what happens as the liquid moves between narrow and wide sections?
 
  • #12
Yes! Thank you Simon. It's the velocity X cross sectional Area that is equal everywhere so:
[tex]
\frac{dy}{dt} \pi r^{2} = \frac{dx}{dt} \pi R^{2}
[/tex]
 
  • #13
... well done :)
 
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1. What is a liquid oscillating in a U-tube?

A liquid oscillating in a U-tube refers to a phenomenon where a liquid inside a U-shaped tube moves back and forth in a rhythmic manner due to the forces of gravity and surface tension.

2. Why does liquid oscillate in a U-tube?

Liquid oscillates in a U-tube due to the difference in air pressure on either side of the liquid column. When the liquid level on one side is higher than the other, the heavier column of liquid will move down, causing the liquid to oscillate back and forth.

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

The oscillation of liquid in a U-tube is affected by the density and viscosity of the liquid, the diameter and length of the tube, and the difference in air pressure between the two sides of the U-tube.

4. How is liquid oscillation in a U-tube used in science?

Liquid oscillation in a U-tube is used in various scientific experiments to demonstrate the principles of fluid mechanics, such as the effects of gravity, surface tension, and air pressure. It is also used in devices like barometers and manometers to measure pressure.

5. Can the oscillation of liquid in a U-tube be controlled?

Yes, the oscillation of liquid in a U-tube can be controlled by adjusting the factors that affect it, such as the diameter and length of the tube, the density and viscosity of the liquid, and the air pressure difference. However, it will always occur to some extent due to the forces of gravity and surface tension.

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