Physics: The Definition of Angular Momentum

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We defined an angular analog of force; we called it torque. Torque was x Force. So, suppose we want to define an angular analog of momentum? How about calling the angular momentum ? It seems like a reasonable definition. We need a symbol for angular momentum and the conventional symbol that I will use is L (I have no idea why L is used), but that's the symbol for angular momentum. You know, when I think about regular old linear momentum, I'm thinking about "oomph." If you've got some big old heavy object going this way, it's got oomph that way; and the more mass it has, the more velocity it has, the more oomph it has.
So, somehow, I am expecting that my angular momentum, L, should tell me about oomph. And I look at this equation; it's not clear yet, so we are going to have to do a little bit of thinking about this to see whether it is a good definition, a physical meaningful and useful definition. And it is, of course.
: it's defined with respect to an origin, just like torque. It's meaningless to talk about the torque, you have to talk about the torque of a force about a point; similarly, you have to talk about the angular momentum that a particle has that a particle has with respect to an origin. Because this r-vector needs to start at the origin and point to wherever the particle is moving with momentum p.
So, you need to pick an origin, and let's just take a look at a particular sample problem. It's just kind of a generic r and p just so that we can review the mathematics of cross products. So, it's a 3D picture. Angular momentum is kind of an intrinsically three-dimensional quantity. And, here's the XY plane, so think of this as a flat surface. And here's a particle with mass M, and it's moving this way-it's got momentum p, so it's got mass M and velocity V this way. Here's my origin, so here's the vector, which goes from the origin, pointing towards the object: that's the position vector r; and what's ?
You know the rule for cross products. You have to point your fingers along the direction of the first one, and curl them toward the direction of the second one. This is one of those horrible diagrams where you got to bend yourself all up. Here's the r-direction, and you curl your fingers toward p, and then, if your thumb is on your right hand is perpendicular; it points straight up, so the angular momentum of this particle around this origin is up, and that's what I've shown on the picture. And the magnitude of the angular momentum is just magnitude of r times magnitude of p times the sine of the angle between them, which in this case is five. So you can always just compute angular momentum.
You look at this formula, and it still not exactly clear why we're so interested in it. Let me give you an example. This is an example of a complex system. You know, when we dealt with momentum, we weren't usually so much interested in the momentum of a single particle, we were usually in momentum when we had problems that had a bunch of particles, at least two, doing something. It's the same story often with angular momentum. In any case, it's nice to think about rigid bodies, rotating around, and figuring out their angular momentums.
So the simplest complicated object that I could think of was a mass attached mass-less rod to a pivot point, swinging around in a circle. So, here's my pivot point, here's my mass, and it's swinging around in a circle with some angular velocity omega. I'm going to compute the angular momentum L.
Before I even do that, let's just think about what the answer should be. I've got a formal definition for L, . But come on, momentum is MV, mass times velocity. So, I'm thinking, "surely angular momentum should equal to the angular analog of mass, times the angular analog of velocity: L equals I." So, let me put a question mark over that equation, let's see, at least in this case, does it work out?
I've got a momentum in what direction? I'm swinging around in a circle, so instantaneously, my velocity is this way, my momentum is that way, and L is defined to be . Remember, I am finding the angular momentum about this point; I am picking a very natural point, the center of the circle. You can pick any point you want, but always pick the natural center if you can... I want the magnitude of r - that's presumably the radius of the circle - and I want the magnitude of p, that's just mv, and I want the sine of the angle between the radius vector and the momentum vector, and thus the sine of 90 is one. So there's the formula for the magnitude of L. And, let me rewrite this in a suggestive way by multiplying and dividing by r. So it's mR^2 . So, I stare at that formula, and I say, "mR^2, that's the moment of inertia of m about the center. So this is I and v over R, that's the formula for omega. So, sure enough, L is equal to I." So it worked.
And, in fact, it's better than that. Because the r-vector crossed with the momentum vector, points up out of the page, and if you have circular motion running around in a circle, you curl your fingers around, the omega is also pointing up out of the page, so not only is L equal to Iin magnitude, it's really equal as a vector. So, in this example, L is equal to I. This is not a vector, of course.
Is this always true? First of all, look at this formula; now we see finally, that, yes, angular momentum is angular oomph. If you have some big massive object, it's got a big moment of inertia, it's got lots of mass spread out and it's spinning around, it's got a big moment of inertia. And, imagine that it has a big angular velocity: omega, angular frequency. Then it's got lots of angular oomph. Does it have any momentum? No, its center of mass is sitting still. It's got no velocity of the center of mass. And yet, surely, this thing has lots of some kind of momentum. It's going in a circle, it's going to want to keep going in a circle. So, this formula is beginning to make more intuitive sense to me.
Alas, I wish I could just write this down as the fundamental definition of angular momentum rather than , but I can't; and I can't because, it's usually true, but there are some weird exceptions in physics where you have to fix this formula up a little bit. This formula is correct if you have some object which is symmetrical, like a sphere or a hoop or a wheel. And it's spinning around a fixed axis which is its symmetry axis. So, you know, you imagine, for the instance like the demo that I frequently like to show of a bicycle wheel spinning around a fixed axis through its hub, then this formula is great. It works just fine and so most of the problems we're going to work, we'll use L equals to I. If you have some weirdo shape and it's rotating around some funny axis, you have to do a little bit more fancy mathematics, which really belongs to an advanced physics course.
So, this is not the fundamental definition. This one is, but it's nice to know that for the kinds of systems we're going to working problems with, yes. You can think of angular momentum in this nice physical way.
Let me work another problem. It's a problem where I'm just going to calculate angular momentum. Just to show you how you go about doing it. Now, I have an object, consider this a point object. I want to find the angular momentum of this object when I drop it. That's a meaningless statement. I want to find the angular momentum of this object about an origin. So, let me pick my origin to be here (arbitrarily, I can pick any origin I want). And I drop the object, and while it's dropping, I want to figure out angular momentum with respect to my origin. So, I draw the picture. Here's is my origin, here's where the ball started its falling. At some moment in time, it's made it this far. And it's going downward. And I can easily figure out its momentum because this is just a free fall problem. So, as a function of time, a velocity final, it's just velocity-initial plus GT, constant acceleration. So, let's figure out what angular momentum is. L is equal to , that's always the fundamental definition.
You know, you look at this example and you say, "wait a minute, nothing's rotating; this is not a rigid body." But that's okay. This formula works fine always. In fact, this is the formula that you use for a single point and it's nice to see that no matter what, if you have a collection of points, if you can figure out the angular momentum for each little chunk, then you just add them up and you'll figure out the total angular momentum of the system. You can do that for any system. So, it's nice to be able to figure out angular momentum for any arbitrary little chunk of matter moving in some arbitrary direction with some momentum.
The cross product has a direction and a magnitude. The magnitude is the magnitude of p, that's m times the magnitude of velocity, and I already said velocity, just by simple kinematics is G times Time, and you want to multiply the magnitude of p. Well, there are two things you can do, you can multiply by the magnitude of r here, this length, times the sign of theta. I prefer to think of the cross product as the magnitude of p times r-perpendicular. What's r-perpendicular in this picture? You go from your origin to the line of the momentum vector, that's this dashed line. You just go straight across; perpendicular to it, and this distance here will be r-perpendicular. It's just B, so, bingo, I've got the angular momentum.
And what's the direction of angular momentum? It's r crossed with p, it's into the page. Kind of makes a little bit of sense, event though this isn't exactly rotation motion. It's not rotational motion at all. It's a straight down falling object. Still, at this moment in time, I can sort of picture that there's some sort of sense of rotation to the problem instantaneously, which is down and into the page. So the direction of L is into the page. Here's the magnitude. This formula as it stands is correct. It's not all that illuminating. But we'll come back to this example and see how we can make some use of this.
So, angular momentum: it's a vector; it's defined by for a single object; it's the sum of for all the little pieces if you have a system, whether that system is rigid or not-doesn't matter, you can define the total angular momentum.
Is it useful? It's very useful. Remember how useful momentum was. And when momentum was especially an interesting physics concept. Momentum was good when you had complicated internal forces: explosions and train crashes. But this system as a whole has no outside force. And we're going to see the same story with angular momentum. Angular momentum is especially interesting in problems when you've got lots of particulars, and their interacting in some complicated way that's hard to describe. But the whole system, you can talk about the external torque rather then the external forces. And especially if that's equal to zero and we get conservation of angular momentum. That's as useful, as powerful an idea, as conservation of linear momentum was.
The Physics of Extended Objects
Angular Momentum
The Definition of Angular Momentum Page [3 of 3]