# How does angular momentum affect the trajectory of a projectile?

I am thinking about a football thrown in a very tight, very fast spinning spiral. If the football is thrown upwards at a high angle, is it possible for the football to not turn over at the top of its trajectory? If so, what type of calculations are needed to show the conditions of angle, velocity, angular velocity, etc. that will determine if the football turns over and lands nose down or doesn't turn over and lands tail down?

• I understand that you are not interested in the angular momentum with respect to some external axis, but in spin. And I understand that you paint the football so as to have head and tail. Is that true? Feb 15, 2015 at 23:35
• the law of this site is that we are forbidden to solve home-exercises. We are allowed only to give some main ideas, s.t. the user could solve the problem by himself. Feb 15, 2015 at 23:58

Key to understanding this question is the fact that a "football", in this case, is a hand-egg:

When it is given spin about its long axis, it will act a bit like a gyroscope: it tends to maintain its attitude.

So the question becomes - when is that desire of the ball to keep pointing in the same direction greater than the torque due to airflow that tries to get it to point in the direction it is going?

To answer this you need to know the torque on the ball due to it moving at an angle to the air (air flow not parallel to the main axis) - that is actually a hard problem that is usually solved by CFD - computational fluid dynamics. In other words - we get the computer to do the math. When it is done you will find that the torque tends to want to turn the ball at right angles to the airflow - so if it is pointing up a little, the air will try to tip it more.

Now we have to add the angular momentum of the ball- when the ball is spinning fast (assuming a right hand throw) the angular momentum vector points forwards, and the torque vector points to the right. This means the ball will start to veer to the right - it precesses. The more it veers tithe right, the more the torque starts to point sideways, etc - in other words, the ball will wobble about the direction of the velocity.

This spinning and wobbling results in a stable spiral, but I believe that it also means that the ball will always continue to point along the trajectory. Incidentally this spinning motion is the same motion that a rifle imparts to a bullet - it makes the bullet fly more true since any errors in the shape will average out as the ball completes on complete "wobble".

In summary - if my understanding of football dynamics is correct, the ball will continue to precess about the velocity vector. The faster it spins the more tightly it will spiral about that path. The higher the angular momentum the slower the precession. There might be a limit where the precession is so slow that the ball appears to continue pointing up - but without a good CFD a model it would be hard to calculate.

Lucky for us, somebody else has done the calculation which you can find (rather poor copy) here

• Thank you for answering my question. I took my last physics class 20 years ago so this was not a homework question. I've always wondered why some American football punts turn over and others don't. May 26, 2015 at 2:19

There is no problem to get the time $T_1$ to reach the peak, and the time $T_2$ to fall on the ground if you know $v_{y,0}$, the projection on the vertical of the total velocity under which you threw the ball. For this calculus you can consider only the center-of-mass movement of the ball.

About the spinning movement, I don't understand in football but I don't see any reason that the spinning axis and angular velocity be dependent on the center-of-mass movement. Iff this independence is true, the spinning axis and angular velocity should be invariant, $\vec {\omega_0}$. Thus, the total angle by which your ball rotates until the peak, or until the ground, depends on the initial conditions.