Vibhor
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Ok . Thanks .
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.
PREREQUISITESStudents and professionals in physics, particularly those specializing in fluid dynamics, mechanical engineering, and applied mathematics, will benefit from this discussion.
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.haruspex said:The horizontal axis case for an arbitrary rotation rate is quite interesting.
Yes, that's what makes it tricky.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?
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?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.
Yes.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?