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Rotating circular pan of liquid

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A circular pan of liquid with density [itex]\rho[/itex] is centered on a horizontal turntable rotating with angular speed [itex]\omega[/itex]. Atmospheric pressure is [itex]P_{a}[/itex]. Find expressions for (a) the pressure at the bottom of the pan and (b) the height of the liquid surface as functions of the distance [itex]r[/itex] from the axis, given that the height at the center is [itex]h_{0}[/itex].

    2. Relevant equations
    In a flow tube
    [tex]P+\frac{1}{2}\rho v^2 + \rho g h = C[/tex]
    where C is a constant


    3. The attempt at a solution
    I boiled up a few equations using the above equation
    [itex]E_{d}[/itex] is the energy density
    at the center and at height [itex]h_0[/itex]:
    [tex]E_{d} = P_{a} + \rho g h_{0}[/tex]

    at the center and at the bottom:
    [tex]E_{d} = P_{b}(0)=P_{a}+\rho g h_{0}[/tex]
    where [itex]P_{b}(r)[/itex] is the pressure at the bottom measured r from the center

    r from the center and at the surface of the liquid (height h)
    [tex]E_{d}=P_{a} +\frac{1}{2}\rho r^2 \omega ^2 + \rho g h[/tex]

    r from the center and at the bottom
    [tex]P_{b}(r) + \frac{1}{2}\rho r^2 \omega ^2 = P_{b}(0)[/tex]

    Mixing all these together out pops
    [tex]P_{b}(r) = P_{a} + \rho g h_{0}-\frac{1}{2}\rho r^2 \omega ^2[/tex]
    and
    [tex]h=h_{0} - \frac{1}{2g}r^2 \omega ^2[/tex]
    but the answer has a plus sign where I have my negative signs. Pretty sure I am wrong but would have to fudge the equations to make it work.

    I made a couple of assumptions:
    For a given vertical segment of water, it is all travelling at the same angular speed [itex]\omega[/itex] and the atmopsheric pressure at the different liquid heights at the surface is the same.
     
  2. jcsd
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