When we immerse a long rod that is rotating about its axis in a liquid an upward or a downward motion along the rod can be seen.

When the rod is immersed in a Newtonian fluid, like water or alcohol, a downward motion can be seen. these fluids have a low viscosity and in this paragraph we say that the viscosity is zero. Let's say that the angular velocity of the rod is constant. We can use the Bernoulli-Equation with the points P_{1} and P_{2}

P_{1}, z_{1} = 0 and because r_{1} goes to infinity v_{1} is about zero. On each point of the surface we have the pressure P_{1} = P_{2} = P_{0}. So we have

Now we want to find out how v_{2} depends on r. Therefore we use the definition of the angular momentum I_{2}. For an element of the liquid with the mass m and the momentum p_{2}, I_{2} = r_{2} X p_{2}. We can use that because we have said that the viscosity is nearly zero, so there is no transmission of angular momentum. Its absolute value is | I_{2} | = r_{2}mv_{2}sin(r_{2}, p_{2}). Due to the fact that v_{2} is rectangular to r, the expression sin(r_{2}, p_{2}) is one. Therefore
| I_{2} | = r_{2}mv_{2} or v_{2} = I_{2}/mr_{2}.

The angular momentum and the mass are constant, so we can see that v is inversely proportional to r. That means that v equals to a constant C divided by r. If we put that in the equation above we have

The picture of the experiment does not really show a -1/r^{2} graph. That's because I have drawn what we could see if we rotate the rod in water, but water has a viscosity above zero. I think however that we can use it for describing a situation with fluids of small viscosity.

In a viscous flow around a rod an upward motion of the liquid can be seen. Tensile stress causes a force to the rod. You can do an easy experiment in order to show that tensile stresses in liquids are possible. When a siphon is slowly removed from a Newtonian fluid, it breaks very close to the surface, but some complex fluids can be pulled up many centimetres above the free surface. That's only possible if tensile stresses exist in these liquids. In order to understand how these forces come together we look at shear tensions that have an effect on elastic bodies. In a linear elastic body the tensile stresses sigma_{1} and sigma_{2} do not exist and the stress tau is linear to the strain gamma. In a non-linear body, like rubber, tau is not linear to gamma and in the case of simple shear the difference between sigma_{1} and sigma_{2} is always bigger than zero.
Now we can use Hoff's elastic-viscous analogy. When an elastic material (generally non-linear) is compared with a non-linear viscous fluid, it can be seen that the strain (elastic case) and the strain rates (viscous strain) are identical, under the condition that the two stresses (in viscous and in elastic case) are equal. That means that when an elastic material is described by epsilon_{ij} = f(sigma_{ij}) the strain rate of viscous material has to be
_{ij} = f(sigma_{ij}).

(Click to enlarge sketch)

When looking at an element of the volume in the form of a ring the shear stress causes a force to the rod and as a consequence the liquid makes an upward motion on it.