newtonian mechanics - Shape of a rotating fluid surface

I was going through fluid dynamics and I came upon this section on the shape of a liquid surface in a rotating cylindrical vessel. The working in the book is something like this:

It considers a mass $m$ of water on the surface of the rotating liquid at a distance $x$ from its axis of rotation .It takes into account a pseudo force ($m \cdot x \cdot \omega^2$) radially outwards and its weight force ($mg$).It then proceeds to calculate the net force and the angle it makes with the vertical.From there through basic integration it gets the equation of the surface of liquid i.e. $y= \dfrac{(x^2 \cdot \omega^2)}{2g}$ .

I get this working.

But if I were to view the same rotation through an inertial frame,how would I proceed? I mean, there wouldn't be any pseudo force, rather a centripetal acceleration of same magnitude radially inwards.This is where I am facing certain doubts.

The forces on the liquid would be the centripetal force inwards and the weight force downwards.The net force would therefore be in a different direction,and consequently the slope of the surface at that point would be in a different direction.How is it possible for the same surface to have two different shapes in two different frames?

I would be obliged if someone could help me out by pointing out the flaw in my understanding.

Answer

Imagine a little piece of water on the top surface at radius $x$. The height of the water is $y(x)$.

By moving inward $\mathrm{d}x$, the water would reduce its energy by $m g \mathrm{d}y = m g y' \mathrm{d}x$. It therefore feels a force of magnitude $m g y'$ towards the center.

This force causes the water to accelerate towards the center. We know this to be the centripetal acceleration, so

$$m \omega^2 x = m g y'$$

from which we recover your original expression.

Your question says:

"The forces on the liquid would be the centripetal force inwards and the weight force downwards."

The forces on a bit of water are gravity downwards and a force from a pressure gradient. Without knowing the exact pressure at every point, we can still calculate the net force from energy, as I did above.

"THe net force would therefore be in a different direction"

The direction of the net force on any bit of water is horizontally towards the center. If you wanted to calculate all the force exactly, you would need the pressure everywhere in the fluid. Note that the pressure is not the usual $\rho g h$ that you may have learned in an introductory class.