# Formation process of parabola in rotating water cup

I have been looking at the well known problem of the parabola forming when e.g. a cup of water is rotating. I get how the equation for the parabola is derived, that is not the problem!

My problem is, that I do not understand how the parabola is formed in the first place, i.e. in the angular acceleration phase from omega=0 to the specific final angular frequency.

To illustrate my problem let's take of the deepest point of the parabola which coincides with the rotational axis. It is obviously lower than the original water level. That means a small portion of water (let's call it a droplet) must have experienced a net downward force during the transition phase. Where did this come from? When I think of all the droplets aligned with the rotational axis at t=0, they won't even experience a centrifugal force when the cup starts rotating, i.e. the net force is still zero.

Same thing with droplets on the far sides. They end up higher than the original water level, hence must have experienced a net upward force. I cannot see how this can happen.

• Gravity is always there, a downward force. Dec 18, 2023 at 11:42
• Not sure this explains anything. Sure, gravity is there but it does not produce any net force causing water to move. Dec 18, 2023 at 11:53
• Are you familiar with the Navier Stokes equations for fluid flow? If so, have you written these down for the case of this transient change for zero angular velocity everywhere to the final angularlar velocity? Dec 18, 2023 at 12:06
• @ChetMiller I don't think OP knows anything about these.. Dec 18, 2023 at 12:15
• You can't use Bernoulli principle. Here a link to a .pdf of notes and exercises about Fluid Mechanics (only Italian). You find this exercise solved at pag. 145, ex. 6.8 gitlab.com/davideMontagnani/fluidmechanics-ita/-/blob/master/… Dec 18, 2023 at 14:14

I'll assume you understand why the surface of the water in a rotating up forms a parabola in the steady state. This comes from considering the forces on a parcel of fluid at the surface in the rotating frame. There is the weight acting downwards, centrifugal forces acting away from the axis and a force normal to the surface, caused by the pressure of the surrounding fluid.

So how does the water transition from being flat in a non-rotating cup to achieving a curved surface, lower than its resting state in the middle and higher towards the edge? When a torque is applied to the cup to get it rotating, it will initially be uncoupled from the fluid. Friction with the fluid in contact with the cup will then begin to apply a torque to the fluid. Zooming into a fluid parcel at the edge of the fluid, it will experience a frictional force tangential to the rotation. In the rotating frame this means there is a growing (centrifugal) force acting away from the axis of rotation. This outward force squeezes the fluid parcel. But water is almost incompressible, so the only way it can go is upwards. The movement of water towards the outside and upwards means that there is less water nearer the centre and so its level falls (since there is a fixed volume of water in total).

The water that has been accelerated by the walls then exerts a torque on the water closer to the centre. This torque is a consequence of the friction between fluids with a non-zero relative velocity between them. This has the same consequence - as water is spun up there is a growing centrifugal force outwards and as the water moves in that direction it gets pushed upwards due to the incompressibility of the fluid around it.

Eventually the fluid reaches an equilibrium where that component of the fluid pressure that acts towards the centre provides sufficient centripetal force to accelerate the fluid in circular motion, which is satisfied when the surface assumes a parabolic shape.

Assume water rotating in a cylindrical bucket: I will take a random 'cylinder' of water from this(and treat it as separate from the rest of the liquid) Then drawing FBD of that cylindrical 'block' of water (with lateral area dA and length x): ($$P_o$$ is atmospheric pressure, $$\rho$$ is density of liquid) Since it is rotating with $$\omega$$, the acceleration of CM is $$\frac{\omega^2x}{2}$$
Then solve the forces along horizontal for this acceleration. You'll get: $$y=\frac{\omega^2x^2}{2}$$ Which takes the shape of a parabola. Since a bucket is 3D, there will be such parabolas in each vertical plane passing through it, creating a paraboloid.

• thanks, could you elaborate a bit more? Are you using bernouilli? Not entirely sure what is happening, sorry. Dec 18, 2023 at 12:58
• Not using Bernoulli. I'm using concepts of Newtonian mechanics (FBD, centripetal acceleration, center of mass) and the pressure at height h below surface of liquid. Dec 18, 2023 at 13:55
• "Assume water rotating in a cylindrical bucket:" This doesn't address the question. Dec 18, 2023 at 14:04
• @ProfRob in the limit of negligible viscosity, the vessel could have any shape. I guess Stuti Gupta used cylindrical coordinates to get the simplest form of an equation describing the physical process Dec 18, 2023 at 14:17
• This standard proof is for a volume of water that is already rotating and has already assumed a parabolic shape. That isn't what the question is asking - as the OP says, "I get how the equation for the parabola is derived, that is not the problem!" And in a world where friction and viscous forces don't exist, the parabola would never form and the water wouldn't be rotating. Dec 18, 2023 at 14:30

# Yes, it is friction

If the cylinder-water interface was frictionless, the cylinder could spin, and none of the water would move.

But there is friction, because there is drag between the cylinder and the water.

And there is friction between the water itself, no layer of water can move in respect to another layer without there being drag between them.

Hence, when the cylinder starts rotating, it will drag water along with it, which in turn will drag more water with it, and so on. And from then, you have a centrifugal force.

As the cylinder keeps spinning, it imparts more and more rotational energy on the water, increasing the centrifugal force.

The net downwards force comes from water moving out of the way, letting gravity do its thing.

My problem is, that I do not understand how the parabola is formed in the first place, i.e. in the angular acceleration phase from omega=0 to the specific final angular frequency.

When the container starts to rotate, each portion of water close to the walls get a tangential pull due to shear stresses between the container and water. Shear stresses between water particles communicate that movement until the centre. If the liquid has no viscosity, that process is not possible. The container would rotate leaving the water undisturbed.

Once in movement, the particles tends to keep in straight line (following the local tangent) with constant speed. Of course there are already particles where they go, which makes the movement complicated. But the basis for the start of the water rotation is shear stresses and inertia.

Same thing with droplets on the far sides. They end up higher than the original water level, hence must have experienced a net upward force. I cannot see how this can happen.

Note that force is proportional to acceleration not to velocity. For example, a ball moving on a horizontal surface can climb a small ramp without any upward force. On the contrary, the component of gravity force opposes its movement, and that is the reason for a decrease of its velocity.

It is similar for the water droplets. They tend to follow the tangent, but there are other water particles in the bigger radius, until the biggest one, and the result is a pile up of water, raising the level from the centre to the periphery. Of course the result of the increase of the level there is a decrease in the centre. The water is simply getting out of there.