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

[Vibhor]

Ok . Thanks .

 

[TSny]

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?

 

[haruspex]

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.

 

[TSny]

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?

 

[haruspex]

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.

 

[TSny]

OK, thanks. That clears it up.