# What is the meaning of change of angular momentum of a ballistic object during its flight?

In a 2D world, three stones, whose magnitude of initial velocities are 5000m/s, are thrown from the North pole towards the Equator with horizontal initial angles of 15o, 30o, 45o and 60o angles. Their trajectories are as below:

Their respective angular momentums untill they fall to the ground are as below:

I'm not a physicist, I'm a grad student in Electronics Engineer and I'm very unfamiliar to the term "momentum". This homework is about controlling flight of aerial objects. I only know angular momentum as a quantity that is product of mass and speed. I did my assignment up to this point. And now, I have to make some comments on it. But those curves of angular momentum doesn't mean anything to me.

What should I write in my report? Why is angular starting from a different positive value for each stone, and then decreasing in time, finally rising up to its initial value when the stone falls on the ground? What is the meaning of this physical phenomenon?

EDIT:

I iterated these two differential equations:

$\ddot{r} - r\dot{\theta}^2 + \frac{GM}{r^2} = 0 \\ \frac{d}{dt}(r^2\dot{\theta}) = 2r\dot{r}\dot{\theta} + r^2\ddot{\theta} = 0$

With initial conditions:

$r = R_0 \\ \theta = \frac{\pi}{2} \\ \dot{r} = 5000 sin(\gamma) \\ \dot{\theta} = -\frac{5000 cos(\gamma)}{R_0} \\$

Where;
$R_0 = 6378000 m$ (radius of the Earth)
$G = 6.6742 \times 10^{-11} N (\frac{m}{kg})^2$ (the gravitation constant)
$M = 5.9722 \times 10^{24} kg$ (mass of the Earth)
$\lambda = 15^o, 30^o, 45^o, 60^o$ (launch angles for each stone)

At each iteration step, I calculated the angular momentum per unit mass by the following formula:

$\mbox{Angular Momentum per Unit Mass} = r \dot{\theta}^2$

EDIT 2:

I realized that I was wrong with the formula of angular momentum per unit mass. I must have been:

$\mbox{Angular Momentum per Unit Mass} = r^2 \dot{\theta}$

When I modified the formula, I obtained these curves below in which angular momentum stays constant:

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The people who are most familiar with the term "momentum" are not physicians (doctors who cure other people) but physicists. An ex-student leader of the 1989 Velvet Revolution made the same mistake in a talk he gave to a conference of mathematicians and physicists in Prague – he admired the mathematicians and physicians over there. –  Luboš Motl Nov 20 '12 at 8:06
The angular momentum is constant in any system with a central force, so I don't understand how you get a variation in the angular momentum. How did you calculate the lines in the second graph? –  John Rennie Nov 20 '12 at 10:05
@Prathyush Thank you for the correction. But I realized a few minutes before you. :) –  hkBattousai Nov 20 '12 at 18:38
correct if im wrong but $L=p\times r=mr^2 \dot θ$. Which is infact your second equation –  Prathyush Nov 20 '12 at 18:41

Momentum, being it angular or not, is as you mentioned a product of mass and velocity. Intuitively it is a measurement of how much force it takes to change the way that object is moving (and the time you must apply that force for). For linear momentum, a flying cannonball is harder to stop than a ping pong ball, even if they're moving at the same speed. This is intuitive indeed, since you realize that the ping pong ball will be stopped quickly just by air resistance, whereas the cannonball will travel much further.

Angular momentum is the same concept but for rotation - how much Torque (angular force) does it take to change the spin on an object. I'm not totally clear from your post in which direction the object is rotating, but I hope the intuition helps. If the angular momenta remain constant throughout the flight of the object, that means that they are spinning constantly as they move through the air. Imagine throwing a ball and putting "backspin" on it.

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You said "it is a measurement of how much force it takes to change the way that object is moving". When the stones are thrown with 5000 m/s initial speed, they are difficult to stop. But when they reach their maximum height, they travel in a slower speed; I intuitively feel that they are "easier to stop" at that time. But why do they have the same momentum when they are easier to stop? I don't understand it. –  hkBattousai Nov 22 '12 at 8:28
Throughout the stones' flight, both gravity and air resistance are acting on them to change their momentum. Gravity exerts a force on an object pulling it towards the center of the earth - changing it's momentum. Air resistance applies the same way, though from the trajectories graphed it looks like air resistance is being ignored. So as soon as the rock leaves the hand of the person throwing it, gravity begins changing it's momentum. –  Harris M Snyder Nov 22 '12 at 21:42
Gravity, however does not exert Torque on the stone, so the stone's ANGULAR momentum shouldn't change until it hits the ground. There is nothing (in the absence of an atmosphere) that opposes the ball's spinning. Apologies for double post - character limit. –  Harris M Snyder Nov 22 '12 at 21:44
I think I understand it. What changes is "momentum", what stays fixed is "angular momentum". Gravity only changes its momentum, but it has no effect on its angular momentum. –  hkBattousai Nov 23 '12 at 5:01
You are correct. –  Harris M Snyder Nov 24 '12 at 6:05