How Do Shell and Liquid Forces Achieve Equilibrium in a Hemispherical Container?

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SUMMARY

This discussion focuses on the equilibrium of liquid and shell forces in a rotating hemispherical container. The force balance equations derived include the downward force from the upper hemispherical shell, calculated as ##\frac{1}{3}\pi R^3ρg##, and the upward force from the liquid, which is equal to the same value. The participants explore the implications of the liquid's rotation with the shell, concluding that if the liquid rotates, the pressure distribution can be described using the equation ##P(x) = \frac{1}{2}ρ ω^2 R^2 + ρgR##, leading to a pressure of ##\frac{3}{2}ρgR## at the bottommost point. The discussion emphasizes the importance of understanding pressure gradients in rotating fluids.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly hydrostatics and dynamics.
  • Familiarity with rotational motion and its effects on fluid behavior.
  • Knowledge of pressure distribution in rotating fluids, including the concept of equipotential surfaces.
  • Ability to apply calculus, specifically integration, to solve fluid dynamics problems.
NEXT STEPS
  • Study the effects of rotation on fluid pressure in containers, focusing on the derivation of pressure equations.
  • Learn about the parabolic surface of rotating fluids and its implications in fluid dynamics.
  • Explore the concept of equipotential surfaces in rotating systems and their mathematical representation.
  • Investigate the conditions under which fluids inside rotating shells may or may not rotate with the shell.
USEFUL FOR

Students and professionals in physics, particularly those specializing in fluid dynamics, mechanical engineering, and applied mathematics, will benefit from this discussion.

  • #31
Ok . Thanks .
 
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  • #32
haruspex said:
The horizontal axis case for an arbitrary rotation rate is quite interesting.
I'm stumped on one aspect of this case. In order to know the pressure throughout, I first need to know the pressure at one point. In the rotation about a vertical axis we just assumed the pressure is zero at the top.

What if the pressure is zero at the top before any rotation, and then we start rotating about a horizontal axis? Is there a simple way to see what the pressure will be at some one point so I can determine the pressure everywhere else?
 
  • #33
TSny said:
I'm stumped on one aspect of this case. In order to know the pressure throughout, I first need to know the pressure at one point. In the rotation about a vertical axis we just assumed the pressure is zero at the top.

What if the pressure is zero at the top before any rotation, and then we start rotating about a horizontal axis? Is there a simple way to see what the pressure will be at some one point so I can determine the pressure everywhere else?
Yes, that's what makes it tricky.
I believe one is justified in saying that the minimum pressure is always zero. Below a threshold rotation that will always be at the top. In the limit it is at the centre.
 
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  • #34
haruspex said:
Yes, that's what makes it tricky.
I believe one is justified in saying that the minimum pressure is always zero. Below a threshold rotation that will always be at the top. In the limit is at at the centre.
OK. That makes sense to me. Then I think the threshold is for ω = √(g/R). For larger ω the point of zero pressure will be located a distance r above the center of the sphere where ω2r = g. Is that what you find?
 
  • #35
TSny said:
OK. That makes sense to me. Then I think the threshold is for ω = √(g/R). For larger ω the point of zero pressure will be located a distance r above the center of the sphere where ω2r = g. Is that what you find?
Yes.
 
  • #36
OK, thanks. That clears it up.
 

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