Ignoring the nasty shear field around a sphere, this experiment is roughly like a creep test. The stress is the weight of the ball (less the Archimedean upthrust) divided by its cross section and the strain is the penetration depth. In all cases, if the stress is less than the yield stress then the ball does not penetrate. If the stress is greater than the yield stress, then it descends indefinitely with the speed calculated from total stress minus yield stress and the viscosity. As far as I can see, for a yield stress fluid, in no case does the ball stop at some finite penetration depth. PS Fun fact: I live in Annecy. I was in Lausanne two days ago.

Cher Sylvain,

For getting of the evaluation that is interest to you I can to suggest to consider the mechanical action, which will be performed in the considered system. Herewith, for analysis simplification, it is better to use not the usual interpretations of this notion (Lagrange or Hamilton's interpretations), but to represent this quantity as the product of the work  dA  that was done in the system on the time  dt , during of which this work was commited. If you want, I can additionally to prove mathematically the validity of such interpretation. Then for a non-conservative system, you will get that

                                              dAdt = dAdisp dt + dAr dt,                                 (1)

where  dAdispdt  is the action, directly connecting with the displacement; dAr dt  -  the action to overcoming of resistance. 

For this case equality (1) will take the form:

                                             dAdt = d(mv)ds + dEdspdt,                                (2)  

where  mv  is the momentum of a ball;  ds  is the path way, passed by the falling ball;  dEdsp  - the amount of energy, dissipated during of the period of time  dt.

Equation (2) is a full analog of the mathematical expression that describes the mechanical action in Hamilton's interpretation. Doing this way, you will be able to reduce your task to the case, the simplest form of that is considered in our and, I think, in your secondary schools in the course of physics (the energy conservation law at the free fall of a body and at its falling in some medium). It goes without saying that the amount of work  dAr  and, consequently, the amount of energy  dEdsp  are determined by the viscous properties of the medium.

Two quick thoughts: 1) I assumed that the ball was placed on the sample surface, i.e. zero initial velocity. The results are obviously different if it is dropped onto the sample, see Eugene's post; 2) I hope that we are not doing your homework.

It does not matter whether newtonian or non hieve a terminal velocitynewtonian fluid the falling ball will reach a terminal velocity when the weight of the  ball equal the drag forces plus the bouyancy force. the ball will continue to fall at constant speed. similar effect when parachutist drop from aircraft.

Dear all,

Thanks for all of your contribution. But I don't really agree with you.

I'm here considering a fluid with a yield stress, which means that it is non-newtonian. For this kind a fluid, a flow is possible only if a sufficient shear stress is applied. Thus, if we consider a ball which is light enough, the gravity force is always overcame by the friction force, even for a non-moving ball. Thus, if this kind of ball penetrates a fluid with a yield stress, the ball slows down until it stops. For a fluid with a very high yield stress, the gravity can even be neglected. 

What I'm looking for is a method (analytic formulae as it is available for newtonian fluid or numerical simulation, I don't care) which is able to determine the friction force applied on a ball moving in a fluid exhibiting a yield stress (let's start with a Bingham plastic, which the simplest model describing this kind of fluid). So, I'm looking for someone which has suffisant knowledge to solve this dynamic fluid problem.

Thank you very much!