Does air flow direction affect the speed of a spinning table tennis ball?

AI Thread Summary
The discussion centers on the effects of air flow direction on the behavior of spinning table tennis balls, particularly regarding topspin and backspin. It questions whether a ball spinning against the air flow will float more, while spinning with the air flow will cause it to dive faster. The Magnus Effect is referenced to explain how air pressure differentials created by the ball's spin influence its trajectory. The conversation also critiques the use of the Bernoulli effect in open systems, emphasizing the role of air viscosity and friction. Overall, the interaction between spin, air flow, and pressure differentials is crucial in understanding ball dynamics in table tennis.
beatenbob
Messages
6
Reaction score
0
Well, I have read a thread regarding a table tennis paradox and I wondered why a top spin ball will dive faster. I have a question here: Does the direction of air flow will affect the ball whether it will float more or dive faster? I mean if the ball spins in the opposite direction of air flow(counter air resistance), will it float more?? And vice versa, if the ball spins in the direction of air flow,will it be diving faster?

If so,by referring the uploaded figure below, can someone tell me, in between which region the air pressure will be lower and where the ball would curve...Thanks a lot
 

Attachments

Physics news on Phys.org
The diagram at this website explains the separation of air flow component of Magnus Effect with a backspin example: the air is accelerated downwards, and the ball is accelerated upwards.

Magnus Effect .htm

Wikipedia has a similar diagram, but refers to Bernoulli effect, which I personally don't like when describing open systems (the air isn't contained). It also refers to the air stream detachment as described above. This diagram shows topspin, air is accelerated upwards and the ball is accelerated downwards.

http://en.wikipedia.org/wiki/Magnus_effect

What's common to all of these explanations is that a thin layer of air spins with the ball because of surface friction and viscosity (friction within the air itself).

In addition to air stream detachment, the air that is spinning "forwards" (in the same direction as the ball lateral movment), causes more forwards acceleration of the nearby air than the air that is spinning backwards, also due to viscosity. This differential in acceleration of air also contributes to a pressure differential, which accelerates the air in one direction and the ball in the other direction.
 
Last edited:
Did you read the previous thread on exactly the same subject, beatenbob? It has all the answers in it...
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top