Normally (in air with a non-zero viscosity), when I give a ping-pong ball top-spin, the air above the ball tends to decrease in velocity while the air on the downside tends to decrease, resulting in a downward motion, because of the Bernoulli Effect.

This is what I think:

If the viscosity (inner friction of the medium, which is air in this case) of the air is zero (hypothetical), the air molecules will affect the ball only in a thin layer that has the thickness of the maximum variation in heigth on the surface of the ball (the fact that the air has zero viscosity doesn't mean there can't be friction between the air and the ball), because of the lack of viscosity doesn't let the air outside this thin layer take part in the rotation of the air. Of course, it takes longer for the ball stop rotating, as more energy is transferred to the air with viscosity.

This results in two very thin layers on the up- and down-side of the ball where there is gonna be a tiny pressure difference (less pressure on the downside, more pressure on the upside). So the effect is much less than in air with a viscosity (unless the surface of the ball is perfectly smooth, in which case there is no difference), and the ball will deflect down a little bit.

Am I right? Or does zero-viscosity air also means that there is no friction between the air and the rough surface of the ball?


1 Answer 1


The problem with your reasoning is the "little layer of air moving". There can be no such thing in a medium with zero viscosity. Even if you make the surface of your ping pong ball deliberately rough, it is mathematically impossible to introduce vorticity (a net rotation) in a liquid with zero viscosity that starts out without such rotation. And that means that if you could entrain a layer in one direction, another layer would have to spring to life traveling in the opposite direction.

And I don't think that makes sense. See for example this paper, section 8.

  • $\begingroup$ @Floris-But what happens if I attach equal flat pieces of a light metal, perpendicular to the ball's surface (uniformly distributed over the ball)? The air bumps into these pieces of metal, reducing the speed of air on the upside of the ball, and increasing it on the downside, resulting in a downward force (of course the ball will stop spinning very fast). In contrast to what you say (in a liquid with zero viscosity that starts out without such rotation), the ball starts with a top-spin rotation. So will it, in this case, be possible to make the topside air velocity smaller, and Vice -Versa? $\endgroup$ Jan 30, 2017 at 21:45
  • $\begingroup$ @Floris-Or is it for a spinning ball impossible to stop spinning, jut like it is impossible to impart rotation (via a rough surface) on a ball without rotation? After all, when the ball spins you impart negative rotation, which is the same as imparting positive rotation. $\endgroup$ Jan 30, 2017 at 21:49
  • $\begingroup$ @Floris-Does the ball experience a force that stops the movement of the ball? After all, the air momentum is imparted to the ball, eventually stopping the ball. This has nothing to do with a zero-viscosity air. $\endgroup$ Jan 30, 2017 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.