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What's the friction force over a rotating ball under the water (or any other fluid)? The center of the ball is at rest, so I'm only interested in the slowing down of the rotation of the ball.

My first intuition is that it should be proportional to $v^2$, as in the usual drag force, but I'm not sure. The standard drag force is given by

$$ F_d = \frac{1}{2} \rho v^2 S C_x $$

where $\rho$ is the density of the liquid, $S$ is the cross sectional area, and $C_x$ is the drag coefficient.

But as far as I know, this is for an object moving in the liquid, not just rotating. Do I need to change something to adapt this expression to my case? Should it be proportional to the full area of the ball instead of just the cross-sectional area?

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  • $\begingroup$ Please clarify: Is this about the friction against the linear motion of a ball that is also rotating? Or are you interested in the slowing down of the rotation of a ball whose center is at rest? In the first case, what is the angle between $\vec{v}$ and the axis of rotation? Are you aware of the Magnus effect? Do you maybe know the Reynolds number? $\endgroup$
    – kricheli
    Commented Jun 12, 2023 at 21:43
  • $\begingroup$ I'm interested in the slowing down of the rotation of a ball whose center is at rest. I'll edit the question. $\endgroup$
    – alexmolas
    Commented Jun 13, 2023 at 8:57

2 Answers 2

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tldr: Fluid dynamics is full of quick and dirty ways to estimate things that would be very hard to calculate exactly. It is based on hard science, but you often look for the easiest way to get an answer. In this case, the drag on a flat plate is very similar to your situation.


There are two sources of drag on a body moving under water.

First, viscosity. Water moving relatively slowing flows in layers. The layer right next to the sphere moves at the speed of the sphere. Successive layers farther away move relative to each other. The velocity of the water changes smoothly to the velocity of the undisturbed fluid far away. Each layer exerts a friction force on adjacent layers.

If water moves too fast, it becomes turbulent. The motion is chaotic and hard to calculate. But the idea is the same. Friction between layers causes viscosity.

Second, inertial forces. An object moving through water pushes water ahead of it out of the way. Water flows around behind and fills in the space behind the sphere. Water that was at rest is accelerated and then decelerated back to rest. This takes energy, which comes from the kinetic energy of the sphere. That is pushing on the water ahead of the sphere causes a reaction force that pushes the sphere back.

Usually one of these forces is much bigger than the other. You typically can ignore the smaller force. The ratio of these forces is called the Reynold's number. $Re = f_{inertial}/f_{viscous}$. There is a rule of thumb to estimate it. It is crude because you don't need finesse if one force really is much bigger.

A large Reynold's number leads to turbulent flow. You can imagine that a lot of kinetic energy tends to stir up the water as it swirls around obstacles, and large friction forces tend to damp out swirly motion. By and large, real flows are turbulent.

Here, the sphere is rotating without translating. The water is accelerated purely by friction forces. You would think the Reynold's number is $0$.

Your equation assumes the fluid is moving over the object. The Reynold's number $> 0$. Here are some examples. See Shape Effects on Drag

enter image description here

What you want is more like the flow over a flat plate. Here is a video that goes through how to calculate that. DRAG ON FLAT PLATE. They get about the same function you do,

$$C_{D_f} = \frac{F_D}{\frac{1}{2}\rho V^2 A}$$

except that $A$ is the surface area of the flat plate, not the cross section.

To get the drag coefficient , $C_{D_f}$, the video talks about how the flow changes from the leading edge toward the tail. The boundary layer thickness grows. At some distance downstream from the leading edge, a transition to turbulence can occur. A rough leading edge can trigger turbulence, even at lower velocities. A thin boundary layer thickness or turbulence leads to a high shear stress. You have to integrate the shear stress over the plate.

In some ways, you have a simpler situation. You expect the thickness of the boundary layer to be the same all the way around a latitude circle of the sphere, and to behave as if the leading edge was far upstream. This means the shear stress is uniform and small. It is up to you if you want to make the sphere smooth or rotate slowly to avoid turbulence.

In other ways, your case is more complex. In the video, a flat plate has a uniform shear stress across the plate. For a sphere, the velocity changes along longitude lines. You will have to figure that out.

They also talk about Reynold's number. They do not justify why it is not $0$ or why you use the surface area of the plate rather than the cross section that blocks the flow. Fluid dynamics is full of rough approximations and rules of thumb. In any event, a long plate triggers turbulent flow, and the crude approximation used for Reynold's number helps tell you when.

The video concludes that $C_{D_F}$ is a function of Reynold's number and roughness. Rather than calculate it, they look it up on a Moody diagram. You could do the same.

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If the sphere rotates without advancing overall, what counts is rather the moment of the actions of viscosity (moment calculated at the center of the sphere).

The answer in this link should give you a clue: viscous torque on a spinning sphere

I'm not sure the answer given is complete. I suspect that it corresponds to situations for which the Reynolds number $Re=\mu (R \omega) R/\eta$ is much lower than 1. We can expect regime changes when the Reynolds number increases.

Hope it can help and sorry for my poor english.

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