Consider a bucket holding some viscous, homogeneous and incompressible fluid (e.g. water). The bucket starts to rotate with constant angular frequency $\omega$ around a vertical axis passing through the center of its bottom surface as shown in figure 1. Due to friction with the walls, the fluid is influenced by the rotation of the bucket and at some point will also be rotating with angular frequency $\omega$. What is the shape of the fluid surface at this point?

The easiest way to solve this problem is by using a non-inertial frame of reference which rotates with the bucket. The vertical axis $z'$ of this rotating frame is chosen to be the axis of rotation of the bucket (see figure 1). At the point at which the fluid is rotating with the bucket with angular frequency $\omega$, the fluid is seen as being at rest in this rotating frame.

Fig. 1: | A rotating bucket holding some viscous, homogeneous and incompressible fluid. The reference frame $S$ sees the bucket rotating with angular frequency $\omega$ around the $z$ axis, while in the frame $S'$, the bucket is at rest since $S'$ is also rotating with respect to $S$ with angular frequency $\omega$. |

Let $S$ be a reference frame in which the bucket is rotating and $S'$ a frame which rotates with the bucket as shown in figure 1. As proven in a previous post, the following equation relates the force ${\bf F}_S$ experienced by a fluid element of mass $m$ in $S$ and the effective force ${\bf F}^{\textrm{eff}}_{S'}$ experienced by this same fluid element in $S'$ (terms which trivially vanish were omitted below): $$ {\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf F}_S + m\omega^2 \rho\hat{\pmb\rho} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_forces1} $$ where ${\bf a}_{S'}$ and ${\bf x}'$ are the acceleration and position of the fluid element as measured in $S'$ respectively, ${\pmb\omega} = \omega\hat{\bf z}$ is the angular velocity of the bucket as measured in $S$, $\rho$ is the distance between the fluid element and the $z$ (or $z'$) axis and $\hat{\pmb\rho}$ is the unit vector which points radially outwards from the $z'$ axis (see figure 2).

Fig. 2: | Forces acting on a fluid element of mass $m$ on the surface of the fluid (assumed to be already rotating with the bucket). The figure shows a cross section of the bucket which passes through the $z$ axis. The fluid height $h(\rho)$ at a given point on the surface depends only on the distance $\rho$ between this point and the $z$ axis. |

The term ${\bf F}_S$ is the sum of the only two forces experienced in the non-rotating frame $S$: the gravitational force and the force exerted on the fluid element by the surrounding fluid (up to this point, we have not explicitly assumed that the fluid element is on the surface, so the equation below applies to any fluid element): $$ {\bf F}_S = m{\bf g} + {\bf F}_{\textrm{fluid}} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_forces2} $$ Since in $S'$ the fluid is at rest, ${\bf a}_{S'} = {\bf 0}$. This fact, together with equations \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_forces1} and \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_forces2}, yields the following: $$ {\bf 0} = {\bf F}_{\textrm{fluid}} + m\left({\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}')\right) = {\bf F}_{\textrm{fluid}} + m\left({\bf g} + \omega^2 \rho\hat{\pmb\rho}\right) \label{post_195438f696c2c11e15340392ea2bf1cc_eq_acc} $$ Figure 2 shows these three forces. In equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_acc}, the centrifugal acceleration term $\omega^2 \rho\hat{\pmb\rho}$ can be interpreted as an additional component which, together with the gravitational acceleration ${\bf g}$, yields an effective gravitational acceleration ${\bf g}^{\textrm{eff}}$ which is no longer homogeneous in space: $$ {\bf g}^{\textrm{eff}} = \bf g - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = \bf g + \omega^2 \rho\hat{\pmb\rho} \label{post_195438f696c2c11e15340392ea2bf1cc_effec_gravity} $$ The term ${\bf F}_{\textrm{fluid}}$ is a very interesting one since it is always orthogonal to the surface of the fluid (see figure 2). To understand this point, imagine the case in which there is no rotation, i.e., ${\pmb\omega} = {\bf 0}$. In this case, the frames $S$ and $S'$ coincide and the fluid is at rest on both. From equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_acc}, we see that ${\bf F}_{\textrm{fluid}} = - m{\bf g}$, which means ${\bf F}_{\textrm{fluid}}$ is the force exerted by the surrounding fluid against the gravitational force: the fluid element is not moving, so the surrounding fluid must be providing the force ${\bf F}_{\textrm{fluid}} = -m{\bf g}$ to keep it in place. Since the fluid surface is horizontal in this case, then ${\bf F}_{\textrm{fluid}}$ is orthogonal to the fluid surface. If the fluid rotates, ${\bf F}_{\textrm{fluid}}$ plays exactly the same role, but now the effective gravitational force given in equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_effec_gravity} is no longer homogeneous in space but depends explicitly on the position ${\bf x}'$: ${\bf F}_{\textrm{fluid}}$ will still be orthogonal to the surface of the fluid for every fluid element on the surface; a tangential force component would set the fluid element in motion since it would not be able to resist this force (but a perpendicular force is resisted by a pressure gradient created inside the fluid).

As can be seen on figure 2, the height $h(\rho)$ of a point on the surface of the fluid depends only on the distance between this point and the $z$ axis due to the cylindrical symmetry of the bucket around the $z$ axis. Since the derivative $dh/d\rho$ is the tangent to the curve $h(\rho)$ at a point which is at a distance $\rho$ from $z$, then: $$ \displaystyle\tan\phi = \frac{dh}{d\rho} = \frac{m\omega^2\rho}{mg} = \frac{\omega^2\rho}{g} \Longrightarrow \frac{dh}{d\rho} = \frac{\omega^2\rho}{g} \label{post_195438f696c2c11e15340392ea2bf1cc_eq_deriv} $$ Integrating both sides of equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_deriv} gives us the explicit form of $h(\rho)$: $$ \boxed{ \displaystyle h(\rho) = h(0) + \frac{\omega^2\rho^2}{2g} } \label{post_195438f696c2c11e15340392ea2bf1cc_eq_parab} $$ where $h(0)$ is the height of the water at $\rho = 0$, i.e., at the surface fluid element which is right at the rotation axis. Equation \eqref{post_195438f696c2c11e15340392ea2bf1cc_eq_parab} gives us the shape of the fluid surface: it is parabolic!