>> Hello class, Professor Anderson here. We talked about the ball on the string problem. And we calculated what the angular momentum is. It has a magnitude of m times v times r. Ok as this thing goes around it has a particular direction and in this case it's actually coming out of the screen towards you but the magnitude of the angular momentum is mvr. But can we relate this to something else that we know? One of the things that we know for rotational motion is we like to use omega instead of v. The other thing that we talked about is instead of mass in rotational motion we should really use moment of inertia, I. So how do these two things tie together? Well if we have L equals mvr I remember that v equals omega r. And so I can substitute that in to L and I get m times v which is now omega r and I still have one more r hanging out there, ok, and so we can rewrite this as mr squared times omega. But we also know that a single mass on the end of a string we can write moment of inertia of that thing as what? Remember moment of inertia is sum of m sub i times r sub i squared. You have to sum over all the particles. And in this case we only have one particle and it is r away from the axis of rotation and so we just get mr squared. And so now look what happens, l, which was mr squared omega, now becomes i times omega. This is angular momentum in terms of angular velocity and moment of inertia. And this applies not only to a ball on the end of a string but any object that is rotating, ok, any object that is rotating, if it has a moment of inertia, i, it has as an angle of velocity omega, this is the angular momentum. How do you figure out the direction? You follow the right hand rule that we talked about before, ok? Let's see how this works in a couple examples.