# Thermodynamics of a floating cylinder

by private_donkey
Tags: cylinder, floating, thermodynamics
 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 \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} where $\rho$ is the density of the liquid. P is pressue, given by $P = \rho g z$, A is area, and thus A dz is volume of the displace liquid. $A$ can be given by $A = \pi f(z)^2$, where $f(z)$ 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 Thumbnails