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Kinetic energy in a u0tube

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    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 kinetic energy of a small slice of liquide in the horzional arm is given by

    dk= (1/2)ρv2 * (∏r2)dx/ (1+x/l)2 (1)

    where v is the velocity
    ρ is denisty
    l is the lengh from the centre of one arm to the centre of the other
    x is the distance the slice is from the middle if the smaller arm

    2. Relevant equations

    k = 1/2mv2=1/2ρAlv2 (2)

    3. The attempt at a solution

    So at first I seen the dK and dx and thought to intergrate, but I got a really messy problem so I figured that was it. SO i thought Id prove it bit by bit
    i got that the l in (2) is the 1/(1+x/l)2 from (1) but Im not sure how to go about proving this is the case...
     
  2. jcsd
  3. Oct 10, 2012 #2
    The kinetic energy of the small slice is mv^2/2 where m = ρdV

    dV = ∏R^2dx

    And R = R(x)

    Your expression,

    dk= (1/2)ρv^2 * (∏r^2)dx/ (1+x/l)^2

    Should read,

    dk= (1/2)ρv^2 * (∏r^2)dx (1+x/l)^2

    With no division symbol?
     
  4. Oct 10, 2012 #3
    it the text it has one

    heres a pic of the page
     

    Attached Files:

  5. Oct 11, 2012 #4
    Got it now I think. The velocity v is the velocity in the smaller vertical tube. You know that the product of the velocity of the fluid and the cross-sectional area at a point is constant?

    Got to go to work, can help afterwards, hopefully someone else can fill in before then.
     
  6. Oct 11, 2012 #5


    That. Little was enough! I got it thanks :)
     
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