Rotational Pressure & Paraboloidal Liquid Surface

[chouZ]

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 don't know how to bring g (the gravitational acceleration)..anybody has any idea?

 

[Fredrik]

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.

 

[m2003]

I would like to clarify that the equation provided in the prompt, dP/dR = (density)*w^2*R, is actually the Euler's equation for a rotating fluid, which describes the variation of pressure in the radial direction for a fluid mass rotating at a constant angular velocity w. This equation is derived from the Navier-Stokes equations and is widely used in fluid dynamics.

To show that the liquid surface is a paraboloidal form, we can start by considering the forces acting on the fluid mass. Since the fluid is rotating at a constant angular velocity, there is a centrifugal force acting outward. This force is given by Fc = m*w^2*R, where m is the mass of the fluid element and R is the distance from the central vertical axis. This force is balanced by the pressure gradient force, given by dP/dR.

Now, let's assume that the fluid surface is at a constant height, h, from the bottom of the cylindrical container. This means that the pressure at the surface is also constant, P0. Using this information, we can rewrite the pressure gradient equation as dP/dR = (P0-P)/h, where P is the pressure at a certain depth, h-R, from the surface.

Substituting this into the Euler's equation, we get (P0-P)/h = (density)*w^2*R. Solving for P and substituting it back into the pressure gradient equation, we get dP/dR = (density)*w^2*R - (density)*w^2*R^2/h. This can be simplified to dP/dR = (density)*w^2*(R-h/2).

Now, we can integrate this equation to get the pressure as a function of R: P = (density)*w^2*(R^2/2 - h*R + C). Since the pressure at the surface is constant, we can set C = P0. Therefore, the equation for pressure becomes P = (density)*w^2*(R^2/2 - h*R + P0).

We can now use this equation to find the shape of the liquid surface. The surface of the fluid is at constant pressure, P0, so we can set the equation equal to P0 and solve for R. This gives us R = h/2, which is a constant value. This means that the surface of the fluid is