Hey, guys. So let's check out another example of a moment of inertia problem. So here we have a solid disk. Remember, a solid disk has the same moment of inertia as a solid cylinder. And when I tell you the shape, I'm telling you which equation to use for I. So \( I \) of the disk is \( \frac{1}{2} m r^2 \). I'm also given that it has a radius of 4, so \( r = 4 \) and a mass of 10. Okay. And then we're going to add 2 small objects on top of it. The fact that it's saying 2 small objects and it's not giving me a shape for the object, is an indication that these are going to be treated as point masses. Okay? And these are the 2 objects here. So I'm going to call this 1. So this is \( m_1 \) and this is \( m_2 \). Okay.

So the object on the left has a mass of 2 kilograms and it's placed halfway between the disk's center and the edge. Now, the distance between the center and the edge is the radius, right, which is 4. If you are halfway between the center and the edge, your distance is half the radius. Okay. So this distance here is half the radius. So I'm going to call this \( r_1 \) because it's the distance for mass 1 and it is half of the radius, which is 2. And the other object is 3 kilograms in mass, and it's placed at the edge of the disk. So, if you're all the way at the edge, your distance, let's call this \( r_2 \), is the same as the radius, which is 4. Okay?

So I'm giving you all the information you need to calculate the system's moments of inertia. Now, the system is a combination of the disk with the masses. So \( I_{\text{system}} \) is \( I_1 + I_2 + I_{\text{disk}} \). And remember, for every one of these, you have to determine if this is a point mass or if this is a shape. Well, 1 and 2 are point masses. We talked about this here. So I'm going to write \( m_1 r_1^2 + m_2 r_2^2 \). And the disk is a shape, and it has a moment of inertia given by \( \frac{1}{2} M R^2 \). Now all we have to do is plug in the numbers, very straightforward. So I'm going to do the masses: 2, 3, and 10. So I'm going to calculate \( 2 \times 2^2 + 3 \times 4^2 + \frac{1}{2} \times 10 \times 4^2 \). Okay?

And then all we have to do is plug in the Rs. This \( R \) here is the radius; it's very straightforward. We know that. The radius is 4. Now these Rs, we have to slow down for a little bit. These are the distances between the center, the axis of rotation, which is in the center, and where the object is. Little Rs are the distance between the object and the center. And we already had these figured out here: 2 and 4. So let's just do this real quick. This is going to be \(2 \times 2^2 = 8\), this is going to be \(3 \times 4^2 = 48\), and this is going to be \(\frac{1}{2} \times 10 \times 4^2 = 80\). Okay. So we have \(8 + 48 + 80 = 136\) kilograms meter squared.

Cool. That's it for this one. Hope you got it. Let me know if you have any questions.