The pressure difference P across a surface of radius of curvature r with surface tension γis P = 2γ/r. The hydrostatic pressure of a fluid of density ρ with column height h in gravity g is P = ρgh. Suppose there is a circular hole of radius R in the bottom of the jar. A bead of fluid extends beneath the hole, and if R << h, we may neglect the variation of hydrostatic pressure within the bead, so that the bead is a spherical cap of radius of curvature r. Setting the pressures equal, rh = 2γ/ρg. For the bead to be stable, r >= R, so there is a maximum column height H with RH = 2γ/ρg. On Earth g = 9.8 m/s^2, and for water, ρ = 1000 kg/m^3, γ = 0.073 N/m. Thus RH = 1.5e-5 m^2 = 15 mm^2. For example, a 1 mm radius hole should support a 15 mm column height. Note that one must use a jar, not a tube, in order to avoid the attractive capillary force. -- Gene
________________________________ From: James Propp <jamespropp@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, August 11, 2014 9:14 PM Subject: [math-fun] Dumping water
Say I have a jar full of water covered by a thin plate. Then I turn the jar upside down, holding the plate firmly against the lip of the jar to prevent water from spilling. Then I whisk away the plate so that instead of the plate pushing against the water, only the air beneath the jar is pushing upward.
Of course, the water will leave the jar. But what will the geometry of the process be? The water can't leave as a cylindrical slug; intuitively, it seems that the process has instability, so that spontaneous fingering in the air-water interface will break the initial cylindrical symmetry.
Maybe Doc Edgerton made high-speed pictures of this?
Jim Propp