An analysis in the non-inertial frame: ball in a rotating tube and the Taylor column

We use the axes and variable definitions as shown in figure 1 with the origin at the base centre of the tube. In the rotating tube, the ball experiences a spatially varying acceleration field, the effective gravity, which is a vector sum of the downward gravitational acceleration and centrifugal acceleration of the tube (g_eff = g + a_c). The equilibrium is achieved when ${{\bf{g}}}_{{\rm{e}}{\rm{ff}}}\cdot \hat{z}=0$, i.e. no acceleration component along the tube. A light ball floats to the top surface of the tube while a a heavy ball sinks to the bottom as the normal force is required to maintain the equilibrium of forces. Rolling friction is neglected in this and subsequent analyses since it is significantly smaller in magnitude than weight and hydrodynamic forces acting on the ball as the ball used was solid with little deformations.

Zoom In Zoom Out Reset image size

To qualitatively analyse the stability of the equilibrium, we resolve the forces acting on the light and heavy balls of the same dimensions with masses m and M respectively as shown in figure 2 where the normal contact force vector N is not depicted. The effective upthrust (−ρ_w V g_eff) acting on the ball is in the opposite direction to g_eff, which is equivalent in magnitude to the effective weight of fluid displaced. When displaced slightly from the equilibrium, the net force points towards the equilibrium for the light ball since effective upthrust is greater than its effective weight. The reverse is true for heavy balls. An implication of this will be that when the heavy ball is released below its equilibrium point, it will fall to the bottom of the tube and when it is released above the equilibrium, it will shoot to the top of the tube. The light ball will instead tend toward the equilibrium position regardless of its point of release in the tube.

Zoom In Zoom Out Reset image size

A quick derivation accounting only for centrifugal forces will show that the oscillation frequency ω of the light ball about the equilibrium is $\omega \approx {\rm{\Omega }}\sin \theta \sqrt{\tfrac{{\rho }_{w}}{{\rho }_{b}}-1}$ where ρ_w and ρ_b are the densities of water and the ball respectively. However, such oscillations are not experimentally observed due to the expectionally high damping.

It is not immediately clear what the exact expression of the buoyant force on the ball should be, given that the ball has a finite dimension in a spatially varying effective gravitational field, since centrifugal acceleration increases with distance to the axis.

The effective gravity at each point in the rotating frame of the tube is given by

$$\begin{eqnarray}\begin{array}{rcl}{{\bf{g}}}_{{\rm{e}}{\rm{ff}}} & = & {\bf{g}}-{\boldsymbol{\Omega }}\times ({\boldsymbol{\Omega }}\times {\bf{r}})\\ & = & \left[\begin{array}{c}g\sin \theta +{{\rm{\Omega }}}^{2}(x{\cos }^{2}\theta +z\sin \theta \cos \theta )\\ {{\rm{\Omega }}}^{2}y\\ -g\cos \theta +{{\rm{\Omega }}}^{2}(x\cos \theta \sin \theta +z{\sin }^{2}\theta )\end{array}\right].\end{array}\end{eqnarray} \tag{ 1 }$$

The condition for static equilibrium in the water

$$\begin{eqnarray}&&{\rm{\nabla }}p={\rho }_{w}{{\bf{g}}}_{{\rm{e}}{\rm{ff}}}\end{eqnarray} \tag{ 2 }$$

can then be directly integrated with respect to z to obtain pressure P.

Integrating again, this time across the surface of the ball where dS is a surface element on the ball, the expression for effective upthrust on the ball in the z-direction is obtained as follows:

$$\begin{eqnarray}&&{F}_{{\rm{up}},z}=\unicode{x0222f}_{S}p\,{{\rm{d}}{S}}_{z}=\displaystyle \frac{4}{3}{\rho }_{w}\pi {a}^{3}{g}_{{\rm{eff}},z}(z).\end{eqnarray} \tag{ 3 }$$

Note that this reduces to the expression for the force on the CM of the ball.

With this we return to derive the expression for equilibrium position. Accounting for the finite radius of the ball and the tube and the fact that the ball is in contact with the top inner surface of the tube we write

$$\begin{eqnarray}&&{z}_{0}=\displaystyle \frac{1}{\sin \theta }\left[\displaystyle \frac{g}{{{\rm{\Omega }}}^{2}\tan \theta }-d+(2R-a)\cos \theta \right],\end{eqnarray} \tag{ 4 }$$

where d is the distance from the bottom centre of the tube to the axis of rotation, illustrated in figure 1. z₀ is the z coordinate of the equilibrium position of the ball, according to the axes in figure 1. The first term in the brackets in equation (4) is the result of the balance of gravitational and centrifugal accelerations along the tube, while the second and third term are obtained by considering the geometry in figure 1.

Experimental verification of equation (4) is conducted in section 5.1.

While the Stokes' drag is usually used in flow with low Reynolds number, it should nevertheless give a resultant drag force of around the same order of magnitude in our system. With the following expression for Stokes' drag [8, 9]:

$$\begin{eqnarray}&&{{\bf{F}}}_{{\bf{D}}}=-6\pi \mu a{{\bf{v}}}_{{\bf{b}}},\end{eqnarray} \tag{ 5 }$$

where v_b = velocity of ball and μ = dynamic viscosity of water, we write Newton's second law as such, constraining the ball to move only in the z-axis:

$$\begin{eqnarray}&&\displaystyle \frac{7}{5}V{\rho }_{b}\ddot{z}=V({\rho }_{w}-{\rho }_{b})(g\cos \theta -z{{\rm{\Omega }}}^{2}{\sin }^{2}\theta )-6\pi \mu a\dot{z}\end{eqnarray} \tag{ 6 }$$

when the rolling condition

$$\begin{eqnarray}&&{\mu }_{s}\geqslant \displaystyle \frac{2}{7}\displaystyle \frac{{F}_{z}}{N},\end{eqnarray} \tag{ 7 }$$

where μ_s = static coefficient of friction is satisfied. Then the motion of the light ball with amplitude A will be described by

$$\begin{eqnarray}&&z={z}_{0}+{{A}{\rm{e}}}^{-\gamma t}\cos (\omega t)\end{eqnarray} \tag{ 8 }$$

with $\omega \approx {\rm{\Omega }}\sin \theta \sqrt{\tfrac{5}{7}(\tfrac{{\rho }_{w}}{{\rho }_{b}}-1)}$ and $\gamma =\tfrac{45}{28}\tfrac{\mu }{{a}^{2}{\rho }_{b}}.$

In this section we write the governing equations for the liquid flow in the rotating frame of the tube coupled with the motion of the ball. For the fluid, we write the incompressible Navier–Stokes equation as

$$\begin{eqnarray}&&\,\displaystyle \frac{\partial {\bf{u}}}{\partial t}+({\bf{u}}\cdot {\boldsymbol{\nabla }}){\bf{u}}=-\displaystyle \frac{1}{{\rho }_{w}}{\boldsymbol{\nabla }}p+\nu {{\boldsymbol{\nabla }}}^{2}{\bf{u}}-{\boldsymbol{\Omega }}\times ({\boldsymbol{\Omega }}\times r)-2{\boldsymbol{\Omega }}\times {\bf{u}}+{\bf{g}}\end{eqnarray} \tag{ 9 }$$

and the continuity equation

$$\begin{eqnarray}&&{\boldsymbol{\nabla }}\cdot {\bf{u}}=0.\end{eqnarray} \tag{ 10 }$$

With that, we impose the no-slip boundary conditions on the surface of the ball:

$$\begin{eqnarray}&&{\bf{u}}={{\bf{v}}}_{{\bf{b}}}+{{\boldsymbol{\Omega }}}_{{\bf{b}}}\times ({\bf{r}}-{{\bf{r}}}_{{\bf{b}}})\forall | {\bf{r}}-{{\bf{r}}}_{{\bf{b}}}| =a,\end{eqnarray} \tag{ 11 }$$

where ${{\boldsymbol{\Omega }}}_{{\bf{b}}}\,=$ angular velocity of ball and r_b = position of ball in tube and on the inner surface of the tube:

$$\begin{eqnarray}&&{\bf{u}}=0\,{\rm{on}}\,\hspace{0.1cm}{x}^{2}+{y}^{2}={R}^{2},z=0,z=l,\end{eqnarray} \tag{ 12 }$$

where l = length of tube

With boundary conditions equations (11) and (12), we solve for u and p in equations (9) and (10). To obtain the force and torque on the ball, we first express the stress tensor as

$$\begin{eqnarray}&&{\boldsymbol{\sigma }}=\mu ({\boldsymbol{\nabla }}{\bf{u}}+{\boldsymbol{\nabla }}{{\bf{u}}}^{{\bf{T}}})+p{{\bf{I}}}_{{\bf{3}}},\end{eqnarray} \tag{ 13 }$$

where ${{\bf{I}}}_{{\bf{3}}}=\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right].$

With that we integrate for the hydrodynamic force and torque on the ball

$$\begin{eqnarray}&&{{\bf{F}}}_{{\bf{H}}}=\unicode{x0222f}_{{\bf{S}}}{\boldsymbol{\sigma }}\cdot \,{\rm{d}}{\bf{S}},\end{eqnarray} \tag{ 14 }$$

$$\begin{eqnarray}&&{{\boldsymbol{\tau }}}_{{\bf{H}}}=\unicode{x0222f}_{S}({\bf{r}}-{r}_{{\bf{b}}})\times ({\boldsymbol{\sigma }}\cdot \,{\rm{d}}{\bf{S}}).\end{eqnarray} \tag{ 15 }$$

Including the effects of the fictitious forces and torques in the rotating frame of the tube, we write the net force and torque on the ball as

$$\begin{eqnarray}&&{\bf{F}}=m\left[{\bf{g}}-2{\boldsymbol{\Omega }}\times {{\bf{v}}}_{{\bf{b}}}-{\boldsymbol{\Omega }}\times ({\boldsymbol{\Omega }}\times {{\bf{r}}}_{{\bf{b}})}\right]+\unicode{x0222f}_{S}{\boldsymbol{\sigma }}\cdot \,P{\rm{d}}{\bf{S}},\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&{\boldsymbol{\tau }}=-I({\boldsymbol{\Omega }}\times {{\boldsymbol{\Omega }}}_{{\bf{b}}})+\unicode{x0222f}_{S}({\bf{r}}-{r}_{b})\times ({\boldsymbol{\sigma }}\cdot \,{\rm{d}}{\bf{S}}),\end{eqnarray} \tag{ 17 }$$

where $I=\tfrac{2}{5}{{ma}}^{2}$ for a homogeneous ball.

The general equations of motion for the ball are

$$\begin{eqnarray}&&m\ddot{{{\bf{r}}}_{{\bf{b}}}}={\bf{F}}+{\bf{f}}+{\bf{N}},\end{eqnarray} \tag{ 18 }$$

$$\begin{eqnarray}&&I\dot{{{\boldsymbol{\Omega }}}_{{\bf{b}}}}={\boldsymbol{\tau }}+{{\boldsymbol{\tau }}}_{{\bf{f}}},\end{eqnarray} \tag{ 19 }$$

where N = normal contact force between ball and tube, ${{\boldsymbol{\tau }}}_{f}\,=$ torque from static friction to prevent slipping of ball. Equations (18) and (19) together with (9) and (10) are solved using the finite element method [10], where implicit time-stepping is implemented using the second-order backward difference formula [11]. Detailed procedures on the numerical evaluation of N and the numerical methods used are included in the appendices.

Sample plots of the velocity magnitude is shown in figure 3, which shows the Taylor column being formed and trailing the ball. Quantitative results will be discussed in section 5.2.

Zoom In Zoom Out Reset image size

The equilibrium position z₀ of the ball is plotted against Ω and θ in figures 6(a) and (b) respectively.

Zoom In Zoom Out Reset image size

Increasing either Ω or θ will cause a resultant increase in the magnitude of the centrifugal acceleration a_c and the equilibrium shifts closer to the bottom of the tube to balance the component of a_c and g along the tube. The x-intercept present in figure 6(b) is resultant of the finite distance from the bottom of the tube to the rotation axis in our experimental set-up (figure 4).

A ball of dimensions a = 3.5 mm is released from the top of tube of dimensions R = 18 mm, l = 0.2 m (i.e. ${{\bf{r}}}_{{\bf{b}}}{| }_{t=0}=-(R-a)\hat{{\bf{x}}}+0\hat{{\bf{y}}}+(l-a)\hat{{\bf{z}}}$). The trajectory of the ball for Ω = 9.7 rad s⁻¹, θ = 63.2° is shown in figure 7 a plot of z against t is shown in figure 8.

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

From figure 7, when the ball is released from the top of the tube, it is initially deflected due to the Coriolis force acting on the ball. The ball then performs small oscillations on the surface of the tube which are quickly damped out. The ball tends towards its equilibrium position as described in section 5.1.

We can clearly see from figure 8 that solely accounting for Stokes drag is insufficient to model the phenomenon accurately since a theoretical low drag on the ball will lead to the ball exceeding the equilibrium position and even shooting below the bottom of the tube. Both the wall effect and the presence of the Taylor column significantly increases drag on the ball.

The fact that the ball stays in contact with the wall increases the shear stress damping the ball's motion since no-slip condition is held at both the stationary wall surface and the moving ball surface. However, since the ball is undergoing pure rolling most of the time, this wall effect is not extremely significant. The non-laminar flow in this set-up may also explain the deviation of Stokes' law, but both these reasons do not account for the drag force being more than 2 orders of magnitude off when calculated from figure 8.

Rather, it is more likely to be the formation of a Taylor column which significantly impedes the ball's motion. As illustrated in figure 9, in the rotating frame of the tube, the Coriolis force provides for centripetal acceleration of moving fluid elements, causing them to circulate in a column parallel to the tube's rotation axis. For a slow moving object in the fluid medium, this circulating column of fluid follows the object closely. The presence of this Taylor column as numerically predicted in figure 3 is verified by the experimental image in figure 10.

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

Note that the Taylor column will only be parallel to the rotation axis in the quasistatic regime (when the ball is moving slowly). A sufficiently high velocity of the ball will cause the Taylor column to no longer be parallel to Ω, but trail the ball as shown in our numerical predictions of figure 3.

The accuracy of our numerical model can be verified by extracting a parameter which characterises the ball's motion. In the following graphs figures 11(a) and (b), v_av, the average velocity of the ball in the first 1s of its motion, is plotted against Ω and θ.

Zoom In Zoom Out Reset image size

As the ball initially moves down the tube with v_b in the −z direction, the Coriolis force acting on the ball serves to deflect the ball to the +y-direction with ${{\bf{F}}}_{{\rm{c}}{\rm{oriolis}}}=-2m({\boldsymbol{\Omega }}\times {\boldsymbol{u}})$. The ball then arrives at a maximum deflection y_max before upthrust restores the ball back to the equilibrium of y = 0. The dependence of y_max on Ω and θ is shown in figures 12(a) and (b).

Zoom In Zoom Out Reset image size

In a rotating frame, the Coriolis force typically results in oscillations of an object. However, such oscillations are not significantly observed in our set-up due to the high damping in the system, attributed to the formation of the Taylor column.