Angular momentum of an extended object

- [Voiceover] So we saw in previous videos that a ball of mass m rotating in a circle of radius r at a speed v has what we call angular momentum, and the symbol we use for angular momentum is a capital L, and the amount of angular momentum that it would have would be the mass of the ball times the speed of the ball, so that means this is basically just the magnitude of the momentum, but then we multiply by the radius of the circle it's traveling in, and that gives us the angular momentum of this ball going in a circle, which is great and good to know, but sometimes you don't have a ball going in a circle and you wanna know the angular momentum. So, for instance, instead of this case, let's say you have this case, where instead of a ball going around in a circle, you've got a rod of mass m and radius R and the whole rod rotates around in a circle. Let's say the outside edge travels at a speed v just like the ball did. So the question is, will this rod also have an angular momentum that's equal to mvR and it won't. You can probably convince yourself of that because for the ball all the mass was traveling at a speed v and all the mass was at the outside edge of the circle that it traces out. In other words, all this mass is traveling at a radius of R. But for this rod, some of the mass, in fact, only part of the mass, only this outside edge of the mass, is actually traveling at a radius R. That's the part that travels at the full radius R. The rest of these pieces of mass, like this one in here, traces out a circle. It definitely traces out a circle, but the circle it traces out is not equal to the radius R. It's got a diminished R value. So how do we determine the angular momentum of an object whose mass is distributed in a way that some of the mass is close to the axis and some of the mass is far away from the axis. That's what we're gonna do in this video. That's the goal and the approach physicists take to this is almost always the same. We say, well, I've got the formula for the angular momentum of a single particle traveling at a single radius, so let's just imagine our continuous object being composed of a bunch of single masses all traveling at a single radius. So if I break this continuous mass up into individual pieces, right? So if I imagine it being broken up into all these little pieces, then if I found the angular momentum of each piece and added it up, I'd get the total angular momentum for the whole object. So let's try this. So the angular momentum of some piece of the object, let's say that little piece of mass, is gonna be, well the mass of that little piece, I'm gonna write m. It's not the entire mass of this entire rod, it'd just be the mass of that small piece times the speed of that piece times the radius that it is at. So to make this clear, let me just write this as like piece one, so this would be m1, v1 and R1 and this would be the angular momentum of that small piece, and you could do this for mass two over here, and you would get that the angular momentum of mass two would be m2, v2 and R2. Now keep in mind that these v's are all gonna be different so the speed out here at the outside edge is gonna be fastest. This speed's not gonna be as great, and this speed closer to the middle is even smaller because they're tracing out smaller circles in the same amount of time as these outside pieces trace out larger circles in the same amount of time. So at this point, you might be worried. You might be like, "This is gonna be really hard. We're gonna have to add all these up. They've all got different speeds. They're all at different radii. How are we gonna do this?" Well, you gotta have faith and something magical is about to happen. So let me show you what happens if we imagine adding all these up. I only draw two. You gotta imagine there's an infinite amount of these so that makes it seem even harder, but imagine breaking this up into an infinite amount of these little discrete masses and considering each individual angular momentum, they'd be very small because this m1 would be an infinitesimal very small piece of mass, and let's add them all up and see what we get. So if we add up all of the mvR's of every piece of mass on this rod, that would be the total angular momentum of the rod. So in other words, this is really just m1, v1, R1 m2, v2, R2 and so on. You'd have an infinite amount of them, right? I can't write them all out 'cause there's an infinite amount. But just imagine that. So what can we possibly do with this? How do we clean this up? When you're doing a physics problem, you don't want to solve an infinite series by writing each term out infinitely. We want a clever way to deal with this, and there's a really clever way to deal with this. Watch this. So if we write this as L equals the sum of mvR. One problem we have is that each mass has a different v. If I can pull things out of this summation, it would help me out 'cause it would simplify things. I could just factor them out, but right now I can't factor out the R, 'cause these all are different radii from the axis. You always measure your radius from the axis here, and they're all at different radii from the axis, and they all have different speeds. But, remember, we like writing quantities in terms of angular variables because the angular variables are the same for every point on this mass. So every point on this rotating rod has a different speed v but they all have the same angular speed omega, so that's a key. That's often what we do and we're gonna do that here. I'm gonna write this as summation of m, but instead of writing v, I'm gonna write this as R times omega. So, remember, for something rotating in the circle, the speed v is gonna be equal to R times omega and that's what I'm gonna substitute down here, so the speed at any point here is the radii of that point times the angular speed of this rod rotating in a circle, and I still have to multiply by the last R here, so this was v. We substituted in what v was, but we have to multiply by R and what do we get? We get that L is gonna be the summation of mR squared omega. And this is great. The omega is the same for every single mass in here. Every single mass travels at the same angular speed, so we could factor this out of the summation. Imagine all these terms would have an omega. We can factor that out, and I could just bring that outside of the summation. So I'll write this as the summation of mR squared, and to make this clear, I'm gonna put parentheses here. It's that summation and then that whole thing times omega, 'cause we're just factoring out omega. And you might not be impressed. You might be like, "All right. Big deal. We've still got an infinite sum in here. What the heck am I gonna do with that?" You don't have to do anything with that. This is where the magic happens. Look at what sum you've got. You've got the sum of all the mR squareds. Remember what mR squared was? Mr squared was the moment of inertia of a point mass, and if I add up all the mR squareds, I get the moment of inertia of the entire mass, this entire object. I get its total moment of inertia. So what we found was a really handy way to write the angular momentum of an object. It's just the moment of inertia of an object, I, times the angular speed of that object. So this is a great formula, and it totally makes sense for this reason. Think about regular momentum, right. Regular momentum, p, was just equal to mv. Well, if you then told me, "Determine the angular momentum," and I didn't wanna go through this derivation, I mighta just been like, "All right, angular momentum, shoot." Well, I'm just gonna replace mass with angular mass, and angular mass, the angular inertia, is just the moment of inertia, and I'll just replace the speed with the angular speed, and look, I just get this formula. So it makes sense because if you replace all the linear quantities with their angular counterpart, you indeed just get the angular momentum of a rotating object. So this is how you do it. If you've got an object, an extended object where the mass is distributed about the entire object, if you just take the moment of inertia of that object and multiply by its angular speed, you get its angular momentum. So, for instance, if this rod has a mass of let's say three kilograms, and that mass is evenly distributed. Let's say the radius of this object is two meters. So that's the distance from the axis to the outside edge. And let's say the angular speed of this object was, let's say 10 radians per second, we can figure out the angular momentum of this rod by saying that the angular momentum is gonna equal the moment of inertia. Well the moment of inertia of a rod about one end is equal to 1/3 mL squared. That's the moment of inertia of a rod about the end, and I then multiply by the angular speed of the object. So if I plug in numbers, I get the angular momentum of this rod is gonna be, I'll use purple here, 1/3 times three kilograms times the length of the object was two meters, and we square that, and then we multiply by the angular speed and that was 10 radians per second, which give us an angular momentum of 40 kilogram meters squared per second. So recapping, if you've got a point mass where all the mass rotates at the same radius and you wanna find the angular momentum, the easiest way to get it is probably with the formula mvR. However, if you have a mass whose mass is distributed throughout the object so that different points on the object are at different radii, the easiest way to get the angular momentum of that object is most likely with the formula I omega, where I is the moment of inertia of the object and omega is the angular velocity of the object.