Spinning ball deviation from straight path | Magnus effect

[JD_PM]

Homework Statement

Please see below pic. Let's focus on obtaining how much will the ball deviate from a straight path and assume the given spin of ##11600## rpm and ##C_L = 0.4## to be the lift coefficient

Relevant Equations

N/A

Let's focus on obtaining how much will the ball deviate from a straight path and assume the spin of the ball to be ##11600## rpm and ##C_L = 0.4## to be the lift coefficient

A pitcher is able to make a baseball follow a curved path by impinging spin on the ball (which triggers nonsymmetric flows about the ball and a force being exerted on the ball perpendicular to its angular velocity. Such phenomena is known as Magnus effect).

My approach:

I assumed the ball to deviate in a circular path of radius ##R##. Then, by geometry

We obtain

\begin{equation}
x+R\cos \theta = R \Rightarrow x = R \Big[ 1 - \sqrt{1-\Big( \frac{L}{R}\Big)}\Big] \tag{1}
\end{equation}

There are two unknowns in this equation: the deviation ##x## and the radius ##R##. Hence, we need to find another equation.

The force of lift is perpendicular to the angular velocity of the baseball and points outwards. The centripetal force counters the lift i.e. their magnitudes are equal. This is why the ball follows a curved path.

Out of ##F_c = F_L## we obtain (the lift force is given to be ##F_L = \frac 1 2 \rho A v^2 C_L## where ##C_L = 0.4## is the given lift coefficient, ##\rho = 1.225## Kg/m##^3## is the ambient air density, ##A= \pi R^2 = \frac{\pi}{4} D^2## is the baseball area and ##v## is baseball's speed)

$$F_c = F_L \Rightarrow m\frac{v^2}{R} = \frac 1 2 \rho A v^2 C_L \Rightarrow R = \frac{8m}{\pi \rho D^2 C_L} = 137.64 \ \text{m}.$$

Hence we plug R into ##(1)## and obtain ##x=1.18 \ \text{m}##.

Despite getting the right answer, the TA told me that the approach is not correct. Let me quote his explanation

"Indeed if you want to be accurate, you have to consider that the ball moves on a curved path. However, the path doesn’t have to be a circular arc (to put it in your words, R and the center of the circle change while the ball moves). Also, the angular velocity isn’t that of the ball rotating around that assumed center, instead it is the rotation of the ball around its own axis. Hence, the centripetal force you stated is incorrect.In order to solve this problem you should think of the following

1- The drag is negligible, so the initial velocity of the ball in the x direction doesn't change.

2- The y velocity component is 0 at the beginning but it increases because the lift acts in this direction.

3- The second assumption is that the y-component of the velocity remains negligible comparing to the x-component, and so the x-component can be used in the lift equation, which makes the lift constant.

Here's a sketch"

I would like to discuss with you guys how do you see his approach and whether you would use an alternative one.

Thank you

 

[TSny]

I think your method is correct and is more accurate than the method outlined by the TA. The TA's approach works well because the deflection is small. Both methods give the same result to 3 significant figures.

I noticed a little typographical error:

\begin{equation}
x+R\cos \theta = R \Rightarrow x = R \Big[ 1 - \sqrt{1-\Big( \frac{L}{R}\Big)}\Big] \tag{1}
\end{equation}

##\frac{L}{R}## should be squared.