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Thermodynamics of a floating cylinder 
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#1
Aug1314, 01:44 PM

P: 2

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 [tex] \begin{align} 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 \end{align} [/tex] 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? 


#2
Aug1914, 09:08 PM

Admin
P: 9,700

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?



#3
Aug2114, 11:47 AM

P: 2

Not sure...Would a depressed liquid surface have a buoyancy? Because you are displacing liquid against gravity right...?



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