Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hydrodynamic coincidence

  1. Dec 27, 2007 #1
    Given a comparatively tall cylinder full of water, is it possible to punch in its side three similar holes so that their streams all intersect at one point?
     
  2. jcsd
  3. Dec 27, 2007 #2
    I think it is not possible!

    In order to find if there is an intersection point we have to find the equations that describe the trajectories of its stream of water and demand that they have a common point.
    Thus let one hole to live at height [itex] h [/itex] from the ground while the surface of the water in the cylinder lives at height [itex] H[/itex] (at some instance of time). Furthermore we assume that the velocity of the surface is negligible with respect to the velocity of the water leaving from the hole. Then from Bernoulli's law we have

    [tex] \rho\,g\,H=\frac{1}{2}\,\rho\,v^2+\rho\,g\,h\Rightarrow v^2=2\,g\,(H-h)[/tex]​

    When the water leaves the hole it feels only its weight, so the equations of motion are


    [tex] \left\{\begin{array}{l}x=v\cdot t\\ y=h-\frac{1}{2}\,g\,t^2\end{array}\Rightarrow y=h-\frac{g}{2\,v^2}\,x^2\Rightarrow y=h-\frac{x^2}{4\,(H-h)} \quad (1)[/tex]

    I choosed my coordinate system having its origin at the bottom corner of the cylinder, right below the holes.
    Now take the first two holes living at heights [itex]h_1,\,h_2[/itex] respectively. In order the trajectories to have a common point they must both fulfill equation (1) for some coordinates [tex](x_o>0,y_o>0)[/tex], i.e.


    [tex]\left{\begin{array}{l} y_o=h_1-\frac{x_o^2}{4\,(H-h_1)}\\y_o=h_2-\frac{x_o^2}{4\,(H-h_2)}\end{array}\Rightarrow \left\{\begin{array}{l}x_o=2\,\sqrt{(H-h_1)(H-h_2)} \\ y_o=h_1+h_2-H \end{array}[/tex]

    Now for the 3rd hole in height [tex]h_3[/tex] equation (1) merely defines [tex] h_3 [/tex] if we plug in it the above values of [tex](x_o,y_o)[/tex]. Thus we arrive to


    [tex]y_o=h_3-\frac{x_o^2}{4\,(H-h_3)}\Rightarrow \frac{(h_3-h_1)(h_3-h_2)}{H-h_3}=0\Rightarrow h_3=h_1 \quad \text{or} \quad h_3=h_2 [/tex]

    Thus the 3rd hole must be identified with the 1st or the 2nd hole, which means that we can't have three holes with a common intersetion point for their streams.
     
  4. Dec 27, 2007 #3
    Rainbow Child,

    Solved like a true physicist!
     
  5. Dec 27, 2007 #4

    dst

    User Avatar

    Very nice :biggrin:

    Could you achieve the 'three-point intersection' with a conical container though?
     
  6. Dec 27, 2007 #5
    Yes, but only by nit-picking your question way beyond your intention.
     
  7. Dec 28, 2007 #6
    For example, he never said the cylinder had to be oriented vertically...
     
  8. Dec 28, 2007 #7
    Sorry, I messed up the quote feature again. I was replying to the OP, not dst.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Hydrodynamic coincidence
  1. Hydrodynamics Question (Replies: 4)

Loading...