Hey, everyone. So let's check out this practice problem. We've got 2 small masses at the end of a rod like this, and you might recognize this from a previous problem that we've already worked out. However, there's one key difference here: the rod itself actually has some mass. It's 5 kilograms and 2 meters long. I'm going to call the mass of the rod \( m_{\text{rod}} = 5 \) and the length \( l = 2 \). We also have the 2 small masses. I'm going to call this one \( m_{l} \) and this one \( m_{r} \). We're going to calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left.

The other key difference in this problem is that, previously, we were working with a perpendicular axis that spun around the center, but now, that's not what we're dealing with here. Here, the axis is going to go through the mass on the left and it's going to spin around like this. You can imagine these 2 balls on the end of the rod actually spinning around this one over here, kind of like a door hinge. So let's go ahead and get started.

If we want to calculate the moment of inertia of the system, it is made up of all of the moments of inertia of the objects that make it up. There are actually 3 objects here: there is the mass on the left, so I'm going to call this \( I_{l} \), plus the moment of inertia on the right, that's \( I_{r} \), and then plus the moment of inertia of the rod itself.

We just have to figure out which equations we're going to use for those three terms. For the mass on the left, hopefully, you realize that it's a point mass and so we're just going to use this equation over here: \( I_{l} = m_{l} \times r_{l}^{2} \). Same thing for the mass on the right: \( I_{r} = m_{r} \times r_{r}^{2} \). Now for the rod itself, the equation depends on where the rod is spinning. For example, if you're spinning around the end of the rod like this, it's a different moment of inertia than if you were spinning around the center. So, since we're dealing with spinning at the end of a rod, we're going to use \( I_{\text{rod}} = \frac{1}{3} m_{\text{rod}} l^{2} \).

All you have to do is figure out where are the distances, what are the \( r \) distances for each one of those masses here. Remember that \( r \) means the distance to the axis of rotation. The distance here for the left mass to the axis of rotation is actually 0, as the axis of rotation goes through the mass itself. So \( r_{l} = 0 \). The other distance, \( r_{r} \), is going to be 2, as the length of the rod itself is 2 and if the axis is all the way on the left, then that means it's also just going to be 2.

So, we're going to have \( 3 \times 0^{2} \) and that will go away. Even though this mass here has mass, there's no distance from the axis of rotation because that is the pivot point. So, there's no moment of inertia there. Now we have the mass on the right, which is \( 4 \times 2^{2} \), and the moment of inertia for the rod, which is \( \frac{1}{3} \times 5 \times 2^{2} \). Working this out, you get \( 0 + 16 + 6.67 = 22.67 \) kilogram meters squared. That is the moment of inertia.

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