A system is made of two small masses (M LEFT = 3 kg, M RIGHT = 4 kg) attached to the ends of a 5 kg, 2-m long thin rod, as shown. Calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left.

Hey, everyone. So let's check out this practice problem. So 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. But there's one key difference here, which is now the rod itself actually has some mass. It's 5 kilograms and 2 meters long, so this whole entire rod here, I'm gonna call this m rod is 5 and the length is 2. And we also have the 2 small masses. I'm gonna call this 1 ml and this 1 mr. And what happens here is we're gonna calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left. So what happens is the other key difference in this problem is that in the other one we were working with a perpendicular axis that sort of spun around the center like this, but now that's not what we're dealing with here. Here the axis is gonna go through the mass on the left and it's gonna spin around like this. Right? So you imagine these 2 balls in the end of the rod and it's actually sort of spinning around this one over here. So so you can kind of think of it as like a like a door hinge almost. Right? So let's go ahead and get started. If we wanna calculate the moment of inertia of the system, this is gonna be made up of all of the moments of inertia of the objects that make it up. Right? There's actually 3 objects here. There is the mass on the left, so I'm gonna call this IL, plus the moment of inertia on the right, so that's IR and then plus the moment of inertia of the rod itself. Now, we just have to figure out which equations we're gonna use for those three terms. So for the eye on the left, right, this mass on the left here, we just have to figure out is it a point mass or is it a rigid body or a shape or something like that. Hopefully, you guys realize that it's a point mass and so we're just gonna use this equation over here. So this is gonna be ml times rl square. Same thing gonna be for the mass on the right. So it's gonna be m r times r r squared. And now for the rod itself, well, that actually depends on the axis or where the rod is spinning. What you might notice here is in this table we've got 2 different moments of inertia that have to do with spinning around different places of the rod. 1 is at the end and one is at the center. But the moments of inertia in rotation problems, where you're spinning is actually very very important here. So 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 like that. So hopefully, guys realize we're dealing with spinning at the end of a rod, so this is the right right moment of inertia to use. We're not gonna use the center one. Alright? So this is the equation that we're gonna pop into our moment of inertia. So this is gonna be 1 third and this is gonna be m rod times l squared. Okay? So all you have to do is we have some distances here for our l and our r. We have the l, right? We have the length of the rod itself. We got the mass of the rod and the mass of the two point masses. All you have to do is figure out where are the distances, what are the r distances for each one of those masses here. Okay. So remember that the distance, that little r means the distance to the axis of rotation. So what is the distance here for the left mass to the axis of rotation? Well, hopefully, you guys realize it's actually 0. If the perpendicular axis goes through the mass itself, then there is no distance to the center of rotation. That r l is actually just equal to 0. That is the axis of rotation. There's no distance. So, therefore, the other distance here, this r r is gonna be what? Well, we've got the length of the rod itself is 2 and if the axis is all the way on the left, then that means this is also just gonna be 2. Okay? So what that means here is that we're gonna have 3 times 0 squared and that will go away. So even though this mass here has mass, there's no distance from this from the axis of rotation because that is the pivot point. So there's no moment of inertia there. Right? So there's no resistance to changes in rotation. So now we have this one over here, which is 4, and this is gonna be 2 squared, and this is gonna be 1 third times the mass of the rod, which is 5 times the length, which is 2 squared. Okay? So if you work this out, what you're gonna get here is you're gonna get, 0 plus this is gonna be 16 plus, and this is gonna be 6.67 repeated. So this whole entire thing here is gonna be 22.67 kilogram meters squared. That is the moment of inertia. Alright. So that's it for this one. Let me know if you have any questions.

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A system is made of two small masses (MLEFT = 3 kg, MRIGHT = 4 kg) attached to the ends of a 5 kg, 2-m long thin rod, as shown. Calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left.

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A system is made of two small masses (M_LEFT = 3 kg, M_RIGHT = 4 kg) attached to the ends of a 5 kg, 2-m long thin rod, as shown. Calculate the moment of inertia of the system if it spins about a perpendicular axis through the mass on the left.