Fluid pressure

  • Thread starter chouZ
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A fluid mass is rotating at constant angular velocity, w, about the central vertical axis of a cylindrical container. The variation of pressure in the radial direction is given by:
dP/dR= (density)*w^2*R

Show that the liquid surface is a paraboloidal form; that is a vertical cross section of the surface is the curve y = (w^2*R^2)/2g


MY ATTEMPT:
Since the form of the liquid surface is due to the pressure on the outside of the cylinder, making the liquid level high there. My idea is to find the pressure. So I integrated the variation of pressure given and found:

P = (<density>*w^2*R^2)/2 + C

I try all what possible for me to try to get the y given above but I cant..i dont know how to bring g (the gravitational acceleration)..anybody has any idea???:confused:
 

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  • #2
Fredrik
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Imagine a drop of liquid at the surface. What forces are acting on it? You know it's being pulled on by a gravitational force G and a centrifugal force C. Those two don't cancel each other, so there must be a third force that's equal in magnitude and oppositely directed to the vector sum G+C. This is the normal force N. It's perpendicular to the surface. This means that if you know G and C, you can calculate the direction of N, and the slope dy/dr of the surface. You should be able to figure out the rest.
 

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