For a rotating fluid in a beaker, why does the depth of the parabola increase as the angular velocity increases (in terms of forces)? [closed]

Question

I am writing a highschool paper on this and wanted to know the relationship between the parabolic depth and the angular velocity of the rotating fluid in terms of the forces.

Answer 1

This is called Newton's bucket, and is most commonly solved by transforming into a reference frame which rotates at the same angular velocity as the water, such that the water appears stationary. When we do this, we must introduce a fictitious centrifugal force term of magnitude $mr\omega^{2}$ acting away from the axis of rotation!

Consider a small mass element $dm$ at the surface of the water, $x$ units to the right of the vertex of the parabola and $y$ units above the vertex. It's weight, $gdm$ acts downward, and the centrifugal force, $mx\omega^{2}$, acts to the right. If we add these two vectors, we obtain an effective weight.

The surface of the water is necessarily perpendicular to the effective weight, so if the "gradient of the weight vector" is $\frac{-mg}{mx\omega^{2}} = -\frac{g}{x\omega^{2}}$, then using standard geometry the gradient of the water at this point has to be $\frac{x\omega^{2}}{g}$!

So finally we write down the relationship $\frac{dy}{dx} = \frac{x\omega^{2}}{g}$, we can separate variables

$\int dy = \int \frac{x\omega^{2}}{g} dx \implies y = \frac{\omega^{2}x^{2}}{2g}+c$

As you should now be able to see, increasing $\omega$ will result in the parabola being stretched vertically (the arbitrary constant $c$ can just be taken to be the height of the vertex above the bottom of the bucket). This will make the well deeper relative to the top of the parabola at the edge of the bucket!