 Okay, so I said that I would show you the derivation of the moment of inertia of a disk. So for discrete masses, the moment of inertia is, which I like to call the rotational mass, because it does the same thing in rotational dynamics that mass does in linear dynamics. Some over all the masses, the mass of each mass, the distance from the axis of rotation squared. This formula assumes that there is a fixed axis of rotation. Okay, so you get a scalar value for the moment of inertia. It's really a little bit more complicated than that, but leave it like that. Okay, so I already said that, I already said that in class I derived the moment of inertia for a thin ring. If you rotate about this axis and it has a radius r and a mass m, then i ring equals mr squared. And you know, what if I stack rings up like this? It doesn't matter, right? They're all on the same axis that have the same thing, so it doesn't matter how thick that ring is this way. Okay, and I'm assuming it's pretty thin that way, because I'm going to use this to determine the moment of inertia for a disk. So here's a solid disk. It also has a mass m and a radius r. And so now what I need to do is say, okay, how would I go about and add up all these, break this into an infinite number of little masses like this and find dm and find r and integrate. So I would say i equals the integral over the whole volume of dm r squared. Now, and that's the way you would do it. So how do you integrate over the volume of a disk? And what's the mass element? Well, that clearly depends on the volume element. And in Cartesian coordinates, the volume element's pretty easy, but the integral would be tough. So you could do it in, you could do a two-dimensional surfacing integral because the height doesn't matter, and you could do it that way. But I'm not going to do that because, you know, we like to cheat, not cheat, do things smart. Okay, so here's, you know, I don't have to break it up into pieces like that. What if I take my disk and I break it into rings like that? Okay, so now I have, this disk is a sum of a whole bunch of rings. And I know the moment of inertia for a ring. I just, I just said what it was. So I could say this is a ring of radius lowercase r. It has a mass of dm. Then in that case, the little piece of the component of the moment of inertia due just to that ring, di would be dmr squared. So all I need to do is add up i equals, the integral from, I'm going to add them up from r equals zero to r equals r, r equals zero to r equals r of dmr squared. So couldn't you just integrate that? No, you can't. Because your integration variables are, and your, I mean your integration variables dm and you have an r, they don't really match. You want to get them both into the same terms. So let me, we place dm in terms of r. So what's the mass of this little ring right here? Well, let's say that it has a thickness of dr. Okay, then what I'm going to assume is that the area of this ring, the ratio of the area of this ring to the whole disk is the same as the ratio of the masses. So I could say dm over the total mass m is going to be the area of this little ring, which is going to be, I can write it as 2 pi r dr. Because 2 pi r is the circumference around times the thickness of dr. Okay, that worked just fine. Over the area of the whole thing was going to be pi big r squared. So if I solve this for dm, I get this. And the pi is canceled. So instead of dm, I'm going to put that in. So I have i equals the integral of dm, which is going to be 2r dr over r squared, big r squared, m r squared. Right? Okay. So now let's pull this stuff out that's not, that's a constant. I get 2m over r squared. And then I get the integral of r cubed dr from 0 to r. So the integral of r cubed dr is going to be running out of space up here. I'm going to put it right here. i equals 2m over r squared. And then I get r to the fourth over 4, 0 to r. So I get 1 half m r to the fourth over r squared, r squared. Boom. That's it. That's the moment of inertia of a disc, 1 half m r squared. It doesn't depend on the thickness of the disc.