Just like kinematics, let's start in 1D because it's easiest. So let's calculate the moment of a rod. And moment, by the way, is how cool people say, moment of inertia, just say moment. The moment of a rod, length L mass m. Well, from that formula, we're going to need the mass density, right? So we're going to need the linear mass density, lambda. That's just the mass-prenant length. If it's uniform, often, it'll come like this. It'll say a uniform rod like that. And all they're saying is that it's this constant mass density. The mass is evenly distributed across the whole length, so we can just get the constant mass density as mass-prenant length. All right. So now, are we ready to calculate the moment? No. We're not ready to calculate the moment. We've got to know where the axis is. That's the whole. You can't just an object. It doesn't have a moment of inertia. You also have to have where the axis is. So now we're going to say about its center. So here's your axis as a right and middle. Now can we calculate the moment of inertia? No. Because that's not an axis. Now, if they tell you that, it's probably implied that the axis is perpendicular to the rod. So you can probably guess that's what they mean. But in principle, that could have been this axis down the road. So let's say yes. We're going about the center, about an axis perpendicular to the rod. Now, we have everything we need to know. We can say in a moment is the integral of lambda r squared. In this case, this is our axis, but we integrate along the geometry of the object. So we're going to put an x-axis like this. That. Yes. And if we have an axis like that, I would strongly recommend that you put the origin on the axis. So the origin of the x-axis is right in the middle of the rod, which means we're about to set up an integral. So you better realize, oh, this is an L over 2, and this is a minus L over 2. Yeah, we're probably going to need all that. So let's see, lambda, we know r squared, and then dx, like that. All right, so let's start plugging in things we know. So we have big m over L, m over L, and then r in this case, what is r? r is the distance of any little dm to the axis. So truly, just the x-position. As we add up all these other little dms to do the integral, each one is at a distance x. Some of them are at a distance negative x, but it's going to get squared. So it doesn't matter. So we'll just call that x squared. Because in this case, its x-position basically equals r. Once you square it to not deal with worry about the negative. And then it's just dx. And the limits of the integral are from minus L over 2 to L over 2, minus L over 2 to L over 2. All right, so now we just integrate. So let's see, let's pull out the constants. The integral of x squared is 1 third x cubed. And that's evaluated from minus L over 2 to L over 2. So now we remember how to evaluate limits M over 3L. And you plug in L over 2. That's L cubed over 8. L over 2 cubed minus. And then negative L over 2 cubed is negative L cubed over 8. I think it all is minus sign right for this to work out. You don't want to get zero or something worse. Let's see, so those are going to add. So that's going to make this a 4. So it's equal to M over 3L times L cubed over 4. And then that goes to that. And you get 112 mL squared. 112 mL squared. Oh, good. Is the moment of the run. And a quick check you can do on these is remember a moment of inertia. If you go back to your original definition, it's always a mass times, it's always kilograms times meter squared. So you should make sure that all the canceling you did ended up with a mass in a distance squared. So that's a run around that axis. Let's check it from another angle as well. So we did an axis here. But why don't we do the axis here just to see what happens? Just so you see a difference. Let's see. Whoops. We didn't clean up. Let's see. So now we're going to do real quick a run like this. Same rod, mass, length, all that's the same. It's just we're going to rotate around that axis. Let's see what that would do. The moment would be the integral of lambda r squared dx just like before. But now let's think about what's going to happen. Lambda is the same, m over l. And now you'd want to put the origin here. Because you always want the origin on your axis that way, wherever you are in x, that's the same as r. If you follow all that. So if the origin is here, then we're talking about taking x out to the point l. So the limits of our integral for adding up all the little mass elements from 0 to l. So 0 to l, each mass elements distance from the axis is just its x value, where it is an x. So r again just becomes x squared dx. So you might notice it's the same integral. It's just different limits. So let's see if that affects the answer. Let's see. So this is m over l. x squared is 1 third x cubed evaluated from 0 to l. So the limits are going to affect the answer. Because we're going to get i equals m over 3 l times l cubed minus 0. And what do we get? 1 third m l squared. So a much larger moment. 1 third is supposed to 1 12. And that makes sense because when you hold it this way, the mass is a lot more mass farther from the axis than if you hold it in the middle. And if you spin a rod, think about it, manipulating one around, it'll be less work to hold it in the middle, less torque than it will be to hold it on the end. But it's interesting. It's the same integral, the same setup, just different limits.