And now we're going to go back to the parallel axis theorem, but there's another one. We're going to do the parallel axis theorem for the angular momentum. So let's think about a problem like this. You've got a coordinate system with an origin here, and you've got a disk, and the center mass of the disk, they're just moving up in the plane, something like that, V, C, M, and the disk is also rotating in this sense like this. So let's see, the angular velocity vector is into the board, since it's spinning like that. And you just ask yourself, what is the total L of that disk? It's got some L because it's moving with respect to the origin. I accidentally drew it almost online with the origin, but I didn't quite. So we'll have some L due to its translation, and it'll also, clearly, have angular momentum due to its rotation. So how do we get the total L? Oops, I forgot about the origin. There we go. All right, so what this says is very similar to the parallel axis theorem for the moment of inertia, is that the total you can get by adding two terms. One has to do with the center of mass velocity, so the cross product of the position vector from the origin to the object, like that, R, C, M, and the momentum vector, which, of course, is just mass times the V, C, M. So that cross product would give you the thing acting as a point particle, and you just have to add I omega for the rotation about the center of mass, IC, M, M, and like that. And it's a little bit weird to say how is it parallel axis? Because this is about a point, and this has an axis. Well, you've got to have your R and your P in such a way that their cross product makes, is around, their cross product makes an L that is parallel to this L, which is in the direction of a make. So that's what makes it a parallel axis theorem, is that those two things need to, in the sense that these two are parallel. We can test it by seeing how useful it might have been. Not too long ago, I recall we were working on a disc bar, a problem, disc bar recently, and we had a bar like this, right? It had some length L. And we wanted to think about what happens when it gets struck in a way that it starts to rotate like that. So it has, again, it has omega into the board, but it also started to move forward. Remember the disc's struck it, and it moved like that. V, C, M like this. And let's see, the question is, and we did a lot of worrying and thinking about, if we consider the angular momentum about a certain point, it should be zero, but then it doesn't look like it's zero, because it translates forward. What I wanted to know is that really zero, what is it? We did this really long integral. Remember, it's really difficult to do that integral. But now we're going to show we could have solved it with parallel axis theorem. We could have said the angular momentum of the bar, due to this translation and this rotation, first, let's look at the R cross P part. P is just MV going this way. R, let's see, we're doing this with respect to the top, because it had zero angular momentum about that axis. Then let's see, it's R is down. R, C, M, and P is that way. Right? R cross P, so it would be out of the board, and it would be R is 1 half L. So that magnitude, that's MV, C, M. And then, sine of the angle between them, sine of the angle between them is 1, because theta is 90, and it's out of the board, so I'll just put out 100 like that. Okay, so that axis is this way. So now we just have to add the rotating part. So it's rotating like that. I, the moment of inertia, of the bar around its center of mass, is 1 12th ML squared. I do believe, I recall, 1 12th ML squared, times omega. And which way is this? Same direction as omega, this is in. So there's a chance those two are going to cancel. One is out and one is in. But if we want to have to be zero, we have to remember, in this specific problem, between the center of mass velocity and omega. That's what we got from all the elastic collision stuff. The solution was that the VCM, the velocity of the center of mass, was 2 fifths, the initial velocity. And the omega, the angular velocity was 12 over 5L. Initial velocity like that. All right, so now we can say, what's the total then? We'll plug these in. This is one half. So that becomes 2 1 1 1 5th. Right, the 2 is canceled. 1 5 L m. And then it's just vi instead of V. CM. And that's out of the board. And then this one is plus, let's see, 112 m L squared times omega, which is 12 fifths L vi. And that's in. So then we start canceling and stuff. And we see that goes with that. And we have 1 fifth m L L just like before vi. It's the exact same positive number, but 1 is not and 1 is not. Therefore, it equals 0. So that was a lot easier than the integral, is to use parallel axis theorem, to get the momentum about a certain point.