What is the shape of a rotating mercury surface in an astronomical telescope?

[Greenhippo]

Been looking through my notes for a basis to start this question but can't seem to find a starting basis due to poor handouts (or more likely bad note taking by me)

Heres the question
A cocnave astronomical telescop mirror may be made by rotating a circular tank of mercury. Find an experssion for hte shape of the surface in terms of the density, distance from the center and rotation rate.

my initial thoughts are along the lines of when the 'internal resistance force' of the mercury is equal to the centripetal force required to keep it in place. But the problem lies in i have no clue how to calcuate this 'internal resistance force' of the mercury, any information you can offer would be appreciated

 

[barob1n]

I think it can be done by considering a differential bit of mass (dm) = differential bit of volume (dv) times the density (p): dm=p*dv. The forces acting on 'dm', for example at the surface, are the centrifugal, gravitational forces, and bouancy force. F_grav = -p*dv*g and F_centrigual=-P*dv*r^2*(angular velcocity). The liquid at the surface will be at rest when it has assumed it's final shape; F_grav + F_centri = F_bouy. The bouancy force will be perpendical to the surface and equal (but opposite) to the sum of the gravitational and centrifugal forces. I don't think you need to know anything specific about the bouy force, just that it's perpendiclar to the surface.

The trick is that you now know the tangent (dy/dx) at every point along the surface. It should be equal to the the gravitational force divided by the centrifugal acceleration. Integrate y w/respect to x.

Even better. Use energy arguments to show that it'll be a parabola. If a differential bit of mass is to stop moving, it's kinetic energy due to rotation will have to go into raising it to a certain height, or
(1/2)*m*v^2 = m*g*y ==> y = (1/2g)*w*r^2 where v^2 = w*r^2.

I don't see how the mass, or density, is not going to divide out.

 

[arildno]

A couple of hints:
1. The surface must be an isobar, that is a surface of constant pressure (equal to the ambient air pressure)

2. Pressure has mainly two roles to play in this problem:
For a given particle in the fluid, the pressure force acting upon it must counteract gravity to keep the particle from falling downwards; in addition, the pressure force must provide the particle's centripetal acceleration so that it moves about the axis of rotation in a circular fashion.