| MadSci Network: Physics |
>What governs the radius of curvature of a liquid meniscus? > >If a vertical cylindrical rod is immersed in a liquid, the liquid >will form a meniscus at the contact point between the rod and liquid. > >What governs the radius of curvature of this meniscus? Specifically, >is it a function of the rod diameter.This question was above my memory and library resources, but it did prompt me to add a gem to my library: Cyril Isenberg's "The Science of Soap Films and Soap Bubbles", Dover, 1978 and 1992. The factors governing the radii of curvature are the surface tension, rho, the gravitational constant, g, the densities of the two liquids, and the geometry of the surfaces that constrain the film. The situation you describe is not analyzed in the book above, but I believe that by looking at the Laplace-Young equation that we can conclude the varying the radius of the rod will have an effect on the radii of curvature of the meniscus. The Laplace-Young eqn. states that the pressure across a surface at a point will be:
p=rho(1/r1+r2)
where r1 would be the radius of the rod and r2 would be the radius of curvature of
the "meniscal" surface along a radius from the rod's axis. The pressure
(exterted by the force either upward or downward between the rod and the
contact ring) would equal the downward force, rho*g*y, where y is the height
above the main surface at a point on the meniscus. Solving for r2 gives a
formula for r2 that involves r1.
In thinking about this, I tried to compare the height of the meniscal rise at various points around an inside corner and concluded the there was more force per area of liquid near the apex of the corner than away from the corner and predicted that the rise is higher and the radius of curvature is lower around an inside corner than along the flat portions. Experiment confirmed this qualitatively.
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