# Near a hard-wall, in which direction does a rigid spherical particle move when a positive torque is exerted upon it?

Consider a small rigid spherical particle of radius $a$ immersed in a viscous incompressible Newtonian fluid of shear viscosity $\eta$ in close vicinity to a hard-wall with stick (no-slip) boundary conditions, located at $z=0$. A constant (external) torque $T_x$ is applied on the particle directed along the $x$ axis in the positive direction. According to the low Reynolds number hydrodynamics, the particle translational velocity is computed as [1] $$V_y = \mu_{yx}^{tr} \, T_x \, ,$$ wherein $\mu_{yx}^{tr}$ is the translational-rotational (tr) coupling mobility function (bridging between the particle velocity in drection $y$ to the torque applied in direction $x$).

For $T_x > 0$ (oriented along the $x$ axis in the positive direction), does the particle velocity $V_y < 0$ or $V_y > 0$. My calculations lead to $V_y > 0$ but I find that counter intuitive (analogy with a sphere rolling on a hard-wall). In fact [2, Eq. (B2)] $$\mu_{yx}^{tr} = \frac{1}{6\pi\eta a^2} \frac{3}{32} \left( \frac{a}{h} \right)^4 \, ,$$ with $h>0$ being the distance between particle center and the wall.

Any help would be highly appreciated and rated

Thank you

Federiko

• What answer is given in the paper which you have referenced? The title is Simulation of hydrodynamically interacting particles near a no-slip plane wall so I think it ought to provide some results to this problem. Commented Mar 27, 2017 at 0:46
• @sammygerbil Yes. The paper provides the expression stated above. However, I find that the plus sign is counter-intuitive. Can you justify that by physical argument? Thanks Commented Mar 27, 2017 at 6:07
• I am not able to understand what the TR coupling function is, nor what sign convention is being used here. I would expect the particle to move in the same direction as when it is touching the wall with a no-slip contact. Commented Mar 27, 2017 at 15:51
• Thanks @sammygerbil for your feedback. Here we simply use the Cartesian system coordinates with the torque oriented along the $x$ axis in the positive direction. Accordingly, if the coupling mobility TR is positive (which is the case apparently) the particle undergoes translational motion along the $y$ direction in the positive direction. This is counter-intuitive and I was wondering whether the physics is here is in contrast to my and our intuition.. best federiko Commented Mar 27, 2017 at 16:29

## Scenario A

Your question didn't define that the direction of any motion of the particle towards the wall (may be none), or more importantly the zone that the wall occupies. Given your difficulty with the sign I'm reluctant to assume particle motion is in the positive x direction or that the wall occupies $x>x_{wall}$
Most consistent with your question would be a wall located either at $z>z_{wall}$ or $z<z_{wall}$ and any particle motion along z towards it... as then the vector torque in the x direction would create rotation that was asymmetric in the y direction (torque being cross product of distance and force) along the wall, i.e. rolling along y, which would cause $V_y$. Each of these choices would have a different sign for $V_y$.

## Scenario B (seems this is not the case - may be deleted)

If rather your $T_x$ is a Force along x that is causing a torque, then the direction of the y co-ordinate deviation is dependent on the direction of rotation, which is dependent
not only on the direction of the "force causing torque",
but also on the relative position of its point of application to the object centre of mass.
i.e. if the centre of mass is a above the torsional force will cause rotation one way, if below it will be the other way. This consequently impacts the sign of the y direction, and could be relevant.
Your question says the body is a sphere of diameter h, but it's not clear on the "point of application of the torsional force" relative to the sphere centre (c of m).

## Scenario C (seems this is not the case - may be deleted)

if your $T_x$ is a component of a vector torque, then the interesting component to produce rolling directed along the y axis would be the one perpendicular to both the wall surface and the concluded resultant $V_y$. if the wall is at x>0 then. This would by $T_z$ ($T_y$ rotates perpendicular to y). You haven't described $T_z$ in your question, but perhaps this can be ruled out as in your first "given" equation it relates $V_y$ to $T_x$. Unless it's in error, seems less likely.

I hope one of these will be your scenario.

OK - Seems like scenario A is established.

So the dilemma is that the direction of rotation is the opposite you would get from a frictional contact analogy?

I imagine this is because the motion is not due to friction but a pressure differential. Where the rotation direction is towards the wall the pressure will be higher, where it is away from the wall it will be lower. Thus the rotating particle will move in the opposite direction to the one it would if it were rolling on the surface, from high to low pressure. When/if it contacts the surface or friction becomes more significant (perhaps due to high viscosity) that could change - the direction of y motion could reverse.
I'm not sure on the precise interpretation of your $\mu$ thus far but having the term inversely proportional to $h^4$ supports this not being a friction effect, and doesn't seem glaringly inconsistent. Something probably to do with relative/apparent cross sectional areas of the object and the obstruction to flow around it.

• Thanks for your valuable comment. I apologize for the missing information. The particle is freely moving above a wall located at $z=0$ under the action of a torque directed along the $x$ direction. As a result, the particle undergoes translational motion along the $y$ direction as follows from the aforementioned pair-mobility component. Commented Mar 30, 2017 at 6:11
• I added to my answer. Commented Mar 30, 2017 at 13:18
• You mean that the physics here could be counterintuitive right? Thanks for your answer by the way Commented Mar 30, 2017 at 13:26
• If by counterintuitive you mean "not like the way friction works", then yes. (Intuition can be a bit subjective. I think the pressure explanation reasonably intuitive - assuming it continues to "hold water" [I should say "hold fluid" :-), possibly you'ld rather I didn't]) Commented Mar 30, 2017 at 13:27
• OK. Well understood Commented Mar 30, 2017 at 13:40

If your set-up is the same as in Wall forces on a sphere in a rotating liquid-filled cylinder, with the fluid rotating in a drum, then for low Reynolds numbers the force $$F_W$$ on the particle is repulsive, away from the wall.

Quoting from the middle of page 3 :

... it is clear that there are two mechanisms which contribute to $$F_W$$.

The first is the vorticity distribution in the wake behind the sphere. This diffuses outward, but this process is asymmetric due to the presence of the wall. It leads to a wall force away from the wall.

On the other hand the accelerated flow through the gap between the sphere and the wall produces an attractive force.

The first mechanism is dominant over a wide range of Reynolds numbers.

Edit

The direction of the velocity depends on the sign of $$\mu$$ (as well as that of $$T$$), which depends on the coupling mechanism. eg If torque on a propeller is in the +x direction, whether the propeller moves forward or backwards or doesn't move at all depends on how the propeller blades are oriented.

But I don't see how a sign comes out of (or into) the equation for $$\mu$$ which you have posted.

• Thanks for your comment. This system is actually different. Under the action of the torque, the particle undergoes translational motion parallel to the wall (along the y direction and not a drift force away from the wall as it is discussed in the paper you referred to.) The question is whether the direction of motion is along y+ or y- best federiko Commented Mar 28, 2017 at 13:29
• A bounty has just started Commented Mar 28, 2017 at 17:27
• Good idea. That might prevent the question from being closed! But how can you award a bounty of 100 points when you don't have sufficient rep? Commented Mar 28, 2017 at 23:05
• I had a score of 119 now it is only 19. The most important thing is to get the answer .. Commented Mar 29, 2017 at 10:33
• Ah, I see. The bounty is deducted immediately. Commented Mar 29, 2017 at 18:18