Hey guys. So here we have another classic problem in conservation of Angular Momentum, which is the problem where you have some sort of disc that spins and then there's either a person or an object on that disk. The person or object will either move closer to the center of the disc or away from the center of the disc. In doing this, you are going to change how fast the disc spins. Okay. So it's conservation of angular momentum. Let's check it out. You're moving on a rotating disc. So we have a disc with a mass equal to 200 kg. Its radius is 4 meters. Now this is a disc which means that the moment of inertia we're going to use for the disc is the moment of inertia of a solid cylinder, which is 12mr2. It spins about a perpendicular axis through its center. If you imagine a disc, a perpendicular axis means that the axis of rotation, an imaginary line, is perpendicular and makes a 90 degrees angle with the face of the disc so that the disc spins around this bottom axis. And there's going to be a person walking around on it. Right? So through its center at 2 radians per second. The disc is initially rotating with an angular velocity ωinitial=2 radians per second. And then you have a person. So this is big M. We're going to say that the person has mass little m equal to 80 kg, that falls onto the disc with no horizontal speed. Here is the disc, and the person will just like parachute into the disc, land on top of the disc. You're probably imagining that this will cause the disc to rotate at a lower rate because he added mass to it; it's heavier now. That's actually what's going to happen. So I'm going to draw a top view of the disc. Let's make that a little rounder. Alright. So here is the disc. The radius of the disk R is 4. The person is going to land somewhere over here at a distance of 3. So remember, radius is big R, distance from the center is little r, little r is 3. So the person lands there. Initially, there's no person. And then after the person lands, we will have a person, which means our omega will change. And what we want to know is what is the disc's new angular speed? So omega initial for the disc is 2. Omega final for the disc is what we're looking for, so question mark. Okay. So conservation of angular momentum. So I'm going to write that angular momentum initial equals angular momentum final. I'm going to expand this. This is Iinitialω and then this is Ifinalω. However, initially, there's only a disc. So I'm going to say that this is IbigM and ωbigM. But at the end, there are two things. There is big M and there's also little m, so we're going to do this, Ifinalωfinal of little m little m. Okay? So we added mass to the system. The system now has 2 I's instead of just 1. But the whole thing still has to be equal. Okay. So let's see. We're looking for omega final of the disc, which is this. Now let me point out to you that the final omega of the disc is the same as the final omega of the person because if you land on a disc that's spinning, right, if the disc is spinning, you land on it, you're going to rotate with the disc. Right? So you're rotating with the disc so you have the same omega. So these two guys here are actually the same which means you can do something like this. Instead of saying omega final big M and omega final little m, you can just call this omega final. And instead of having 2 variables, you have one variable, which is simpler. So you can do omega final in both of these. You can factor it out and then have IfinalbigM+Ifinallittlem. Okay?
Then you have IinitialbigMωinitialbigM, which we have. So we're looking for this. We have this. So all we got to do is calculate all the I's. Okay. So let's do that. The moment of inertia of the disc, in the beginning, the disc is by itself. So you have 12mr2, and I have all these numbers, so this is easy to calculate. 12 m is 200 and r is 4 squared. And when you do all of this, you get 1600. So this is going to be 1600 right here. Okay. So this number is 1600. The initial speed of the disc is 2 equals omega final, and then these two numbers here. Now at the end, after you land on it, well, the disc still has the same moment of inertia. Right? The mass of the disc didn't change. The radius of the disc didn't change. What changed was the mass of the whole system. So this is still 1600, but the difference is now that I have something else. And that's what we have to find. We have to find the final moments of inertia of this person. We're treating the person as a point mass. So we're going to use the moment of inertia of a point mass, which is mr squared where r is the distance from the center. The person is 80. It is a distance of 3, not 4 but 3 squared. So if you multiply, all of this, you get that this is 720. This is 720, and that's the number that goes right here, 720. Cool. So if you solve here, you have this is 3200 on the left. The stuff on the right side adds up to 2320. And if you divide, you get an omega of 1.38 radians per second. Now this should make sense. You started off at 2. You added mass. The disk became hea
16. Angular Momentum
Conservation of Angular Momentum
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