Magnus Effect Equations and Mechanics

[lebronJames24]

Homework Statement

Not necessarily homework question, but part of homework project.

Relevant Equations

In description

Ok I need to clear up a few things. First, in (Figure 1), a soccer ball (the circle in the photo), would have the magnus force towards the lower pressure because the universe favors high entropy? Also what are the forces are active in figure 2? Lastly what to the variables in Figure 3 mean?

(Figure 1)

(Figure 2)

(Figure 3)

 

[lebronJames24]

also why do some shapes display the magnus force while others do not?

 

[haruspex]

The equation in fig 3 matches that at https://en.wikipedia.org/wiki/Magnus_effect#Pressure_gradient_force, where all of the variables are defined.

 

[haruspex]

First, in (Figure 1), a soccer ball (the circle in the photo), would have the magnus force towards the lower pressure because the universe favors high entropy?

Not sure what your thinking is there, but the proximate cause is that the higher pressure acting over the same area implies the greater force, so the net force is towards the lower pressure.

 

[lebronJames24]

Not sure what your thinking is there, but the proximate cause is that the higher pressure acting over the same area implies the greater force, so the net force is towards the lower pressure.

im just trying to understand why the force point in the lower pressure direction

 

[haruspex]

im just trying to understand why the force point in the lower pressure direction

I don't know how to explain that any more clearly than I already have.

 

[lebronJames24]

The equation in fig 3 matches that at https://en.wikipedia.org/wiki/Magnus_effect#Pressure_gradient_force, where all of the variables are defined.

they do not define what p equals. and also if there are only two objects in the system, those being the ball and the air, then why do we need two u's? shouldnt there only be one surface?

 

[haruspex]

they do not define what p equals. and also if there are only two objects in the system, those being the ball and the air, then why do we need two u's? shouldnt there only be one surface?

The equation considers the side of the ball rotating forwards and the side rotating backwards separately. The pressure each side depends on the square of the speed relative to the air, $u_1, u_2$.
$\Delta p$ is the pressure difference.

 

[lebronJames24]

I don't know how to explain that any more clearly than I already have.

The equation considers the side of the ball rotating forwards and the side rotating backwards separately. The pressure each side depends on the square of the speed relative to the air, $u_1, u_2$.
$\Delta p$ is the pressure difference.

also what side is designated as u1 and which u2?

 

[Lnewqban]

also what side is designated as u1 and which u2?

That is the whole point of the rotation, to have two very different relative velocities on both sides.
As the rotating body deflects the airstream to one side, the airstream pushes it in the opposite direction.

 

[haruspex]

also what side is designated as u1 and which u2?

Since the expression is in the order $u_1^2-u_2^2$, $u_1$ must be the side with the greater relative speed to the air.

 

[lebronJames24]

I don't know how to explain that any more clearly than I already have.

let me see if get the idea. pressure equals force/area. since both of the areas of half of the ball is the same, that coefficient remains constant. however, since more drags force is necessary for the ball to be in equilibrium position (not spinning) for the side that goes against the air flow, that region has a higher pressure. is this statement correct?

 

[haruspex]

for the ball to be in equilibrium position (not spinning)

It is surely spinning. If it has radius r, velocity v and spins at rate ω, then on the forward spinning side (the underside in your diagram) the relative velocity is $v+\omega r$ and on the backward spinning side it is $v-\omega r$ (which could be negative).
Since the pressure rises as the square of the relative speed, the pressure differential is proportional to $(v+\omega r)^2-(v-\omega r)^2=4v\omega r$.