How Does a Rotating Mercury Tank Form a Parabolic Mirror?

[rx_ke25]

A concave astronomical telescope mirror may be made by rotating a circular tank of mercury. Find an expression for the shape of the surface in terms of the density of mercury, the radius from the centre and the rotaion rate

so far I've come up with:
KE=1/2MV^2
In the case of circular motion the relation v = ωr holds, hence
KE=1/2MW^2R^2
The gravitational potential energy is given by
potential energy= mgh
where g is the acceleration of gravity and h is the height of the mercury's surface above some arbitrary elevation, for instance, we can set h = 0 to be the lowest mercury surface.

We set the potential energy equal to the kinetic energy to find the mirror's shape:
H=(1/2g)W^2R^2
which is by definition a parabola

im unsure how to include a density term into this formula any help appreciated
Thanks.

 

[anathema]

Thank you for your interesting question. To incorporate the density of mercury into the equation, we can use the formula for gravitational potential energy in terms of mass and density:

Potential energy = mgh = ρVgh

where ρ is the density of mercury, V is the volume of mercury, g is the acceleration of gravity, and h is the height of the mercury's surface above some arbitrary elevation.

Since we are dealing with a circular tank of mercury, the volume V can be expressed as the area of the circle (πr^2) multiplied by the height of the mercury (h):

V = πr^2h

Combining this with the previous equation, we get:

Potential energy = ρπr^2gh

Setting this equal to the kinetic energy (1/2MW^2R^2) and solving for h, we get:

h = (1/2ρπgr^2)W^2R^2

This gives us the shape of the surface of the mercury in terms of its density, the radius from the center, and the rotation rate.

I hope this helps. Let me know if you have any further questions.