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Liquid oscillating in a U-tube

  1. Aug 19, 2013 #1
    1. The problem statement, all variables and given/known data
    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].



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



    3. 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. jcsd
  3. Aug 19, 2013 #2

    Simon Bridge

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    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.)
     
  4. Aug 19, 2013 #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.
     
  5. Aug 19, 2013 #4

    Simon Bridge

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    Hint: center of mass
     
  6. Aug 19, 2013 #5
    Ahh a terrific hint! But why is the potential energy of the right tube not negative?
     
  7. Aug 20, 2013 #6

    Simon Bridge

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    You were the one making the definitions. It can be negative if you want.
     
  8. Aug 20, 2013 #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.
     
  9. Aug 20, 2013 #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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  10. Aug 20, 2013 #9

    Simon Bridge

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    You've used the wrong speed.
    v should be a function of x.
     
  11. Aug 20, 2013 #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].
     
  12. Aug 20, 2013 #11

    Simon Bridge

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    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?
     
  13. Aug 22, 2013 #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]
     
  14. Aug 22, 2013 #13

    Simon Bridge

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    ... well done :)
     
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