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Find the circulation using equation of motion

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    v = ui + v j
    u = cos(x)sin(y)
    v = -sin(x)cos(y)
    Find the circulation around randa to the square defined by: x = y = [-0.5*pi,0.5*pi]



    2. Relevant equations

    Is there a rule that says which sides on the square that get i and whics gets -i when you draw the square?
     
  2. jcsd
  3. Mar 5, 2009 #2

    HallsofIvy

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    Re: Circulation

    The direction of flow is given by the equation of motion itself, not by any "rule".

    Assuming that v is the velocity vector of a flow, taking x=0.5pi gives v = cos(0.5pi)cos(y)i- sin(0.5pi)sin(y)j= -i. Taking x= -05pi gives v= i.
     
  4. Mar 5, 2009 #3
    Re: Circulation

    Thanks, so it is like this?

    right vertical line -j, x = 0.5*pi
    left vertival line j, x = -0.5*pi
    top horizontal line i, y = 0.5*pi
    bottom horizontal line -i , y = -0.5*pi
     
  5. Mar 5, 2009 #4

    HallsofIvy

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    Re: Circulation

    I realized I had miscopied your formula and rewrote what I had.
    At x= 0.5pi, cos(x)= 0 and sin(x)= 1 so v= -cos(y)j.
    At x= -04pi, cos(x)= 0 and sin(x)= -1 so v= cos(y)j.
    At y= 0.5pi, cos(y)= 0 and sin(y)= 1 so v= -cos(x)i.
    At y= -0.5pi, cos(y)= 0 and sin(y)= -1 so v= cos(x)i.

    In particular, notice that the flow is across the boundary of the square- at each point on the boundary, the flow is at right angles to the boundary so the total flow around the square is 0.
     
  6. Mar 5, 2009 #5
    Re: Circulation

    Thank you.
     
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