Register to reply 
Surface tension of a torus raindrop 
Share this thread: 
#1
Nov2512, 07:55 AM

P: 818

1. The problem statement, all variables and given/known data
When calculating the difference in pressure inside a spherical raindrop, the force exerted by the surface tension is calculated to be 2pi*R*gama, where R is the radius of the drop and gamma is dE/dS (dyne/cm). When the shape of the raindrop is said to be that of a torus, the force exerted by the surface tension is calculated to be 2pi*R*gamma + 2pi(R+2r)*gamma (please see attachment). My question is simply why does the force in the case of the torus have two components, whereas in the case of a sphere it has only one? 2. Relevant equations 3. The attempt at a solution 


#2
Nov2512, 08:29 AM

Mentor
P: 11,840

In both cases, the expression is the length of the boundary (as seen in a 2Dprojection) multiplied by gamma. A circle has one boundary, the torus has 2 (inner+outer).



#3
Nov2512, 09:06 AM

P: 818

But isn't the circle's circumference considered a boundary? Or shouldn't it be? Is it not a thin layer of fluid verging on air?



#4
Nov2512, 09:49 AM

Mentor
P: 11,840

Surface tension of a torus raindrop
You get a layer of fluid/air in contact there. 


#5
Nov2512, 10:49 AM

P: 818

Okay. Thank you.



Register to reply 
Related Discussions  
Surface Area of a Torus  Calculus & Beyond Homework  1  
Is Horntorus a valid genus1 Riemann Surface?  Differential Geometry  2  
Surface area of a beaker of water w/ surface tension?  Chemistry  1  
Surface Integrals and Average Surface Temperature of a Torus  Calculus & Beyond Homework  4  
Minimal Surface evolution from Torus  reversible?  Beyond the Standard Model  0 