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Thermodynamics of a floating cylinder

  1. Aug 13, 2014 #1
    I am wondering if my reasoning is correct for determining the energy due to the buoyancy of a deformed water surface.

    Essentially, one has a floating cylinder that depresses the surface of a liquid in an infinite tank as seen in the figure. I want to compare the energy of a flat surface with the energy of the deformed surface due to the displacement of the liquid by air (I know there are other energies involved but for the time being I am only looking at the energy due to the buoyancy of the displace liquid due to the deformed surface), not including the liquid displaced by the cylinder.

    My reasoning is
    E_{deformed} - E_0 &= \int_0^{-h} -F_{buoyancy} dz \\
    &= \int_0^{-h} -P A dz \\
    &= \rho g \int_0^{h} z \pi f(z)^2 dz

    where [itex] \rho [/itex] is the density of the liquid. P is pressue, given by [itex] P = \rho g z [/itex], A is area, and thus A dz is volume of the displace liquid. [itex] A [/itex] can be given by [itex] A = \pi f(z)^2 [/itex], where [itex] f(z) [/itex] describes the radius from the centre of the cylinder to the edge of the meniscus to make infinitesimal disks as a function of z. g is gravity accel.

    Is this reasoning correct?

    Attached Files:

  2. jcsd
  3. Aug 19, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
  4. Aug 21, 2014 #3
    Not sure...Would a depressed liquid surface have a buoyancy? Because you are displacing liquid against gravity right...?
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