Consider a bucket of water rotating about a vertical axis passing through the ce

Question

Consider a bucket of water rotating about a vertical axis passing through the center of the bucket. Given that the only forces acting on an atom at the surface of the liquid are the gravitational attraction of the Earth and the normal force of the liquid underneath it, show that the liquid's surface becomes a paraboloid, or more specifically, the height of the liquid at a radial distance r away from the axis of rotation, relative to the liquid's height at the axis, given by h = (w^2/2g) r^2 where w is the angular speed of rotation.

Explanation / Answer

Let's work in a rotating reference frame and consider the forces on the fluid particles.

Centrifugal force = m omega^2 r (radially out)
Gravitational force = mg (down)

Integrate the forces to get potentials:
gravitational PE = mgh
centripetal potential = - 1/2 m omega^2 r^2

So at an equipotential surface (which the surface of the fluid must be), the two potentials' sum is constant

mgh - 1/2 m omega^2 r^2 = constant

h = (omega^2/2g) r^2 + constant/mg

This is, of course, the equation of a parabola.

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