Landing and moving on a disc

by Patrick Ford
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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 in that disk, and the personal object will either move closer to the disk or closer to the center of the disc or away from the center of the disc. And in doing this, you are going to change how fast the disk spins. Okay, so it's conservation of angular momentum. Let's check it out. So you're moving on a rotating disc, so we have a disc of Mass. 200. Mass equals 200. Radius is 4 m. Now. This is a disk. This means that the moment of inertia we're going to use for the disk is the moment of inertia of a solid cylinder, which is half M R Square. Um, it spends about a perpendicular access through its center. Um, if you have a disc, here's a disk perpendicular axis means that your axis of rotation imaginary line perpendicular means it makes 90 degrees with the face of the disk. So it means that the disk spends like this about an axis, and there's gonna be like a person here walking around. All right, um, so through its center, at two ratings per second. So the disk is initially rotating with an omega initial of two radiance per second on, then you have a person. So this is Big M. We're going to say that the person has massed little M equals 80 that falls into the disk with no horizontal speed. So here is the disk, and the person will just, like parachute into the disc land on top of the disk. You're probably imagining that this will cause the disk to rotate at a lower rate lower speed because you added mass to its It's heavier. Now that's actually what's going to happen. So I'm going to draw a top view of the disk, so let's make it a little rounder. Alright, so here's disk. The radius of the disk R is for the person is going to land somewhere over here at a distance of three. So remember, radius is big. Our distance from the center is little. Our little our is three, so the person lands there. Initially, there is 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 disks? New angular speed. So Omega initial for the disk is to Omega Final for the disk is what we're looking for. So question Mark. Okay, So conservation of angular momentum. So I'm gonna write that angular momentum Initial equals angular momentum final. Uh, I'm gonna expand this. This is I initial Omega initial, and then this is I final Omega final. However, in the beginning, there's Onley a disk. So I'm gonna say that this is I of Big M in Omega of Big M But at the end, at the end, there are two things. There is big, um, and there's also little m So we're gonna do this. I f omega f off little m a little. Okay, so we added mass to the system. So the system now has two l's instead of just one, 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 make of the disk is the same as the final omega of the person. Because if you land on a disk that's spinning, right, if the disks spinning you land on it, you're going to rotate with the disc, right? So you rotate with the disk, so you have the same omega. So these two guys here are actually the same, Which means you could do something like this instead of saying Omega final Big M and Omega Final little M you could just call this Omega final. And instead of having two variables, you have one variable, which is simpler so you can do a mega final in both of these, you can factor it out and then have I final M plus I final little m. Okay, then you have I initial Big m then Omega Initial Big M, which we have. So we're looking for this. We have this, so all we gotta do is calculate all the eyes. Okay, so let's do that. The I'm gonna do this off to the side over here and the moment of inertia of the disc. In the beginning, the disk is by itself. So you have half m r squared and I have all these numbers, So this is easy to calculate. Half M is 200 and our is four squared. And when you do all of this, when you do all of this, you get 1600. So this is gonna be 1600 right here. Okay, so this number is 1600. The initial speed of the disk is too equals Omega Final. And then these two numbers here now at the end, after you land on it. Well, the disk the disc still has the same moment of inertia, right? The mass of the disk didn't change. The radius of the disk didn't change. What change 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 it. The final. The final moments of inertia off this person we're treating the person is a point mass. So we're gonna use the moment of inertia off a point mass, which is M R. Squared, where r is distance from the center. The person is 80. It is a distance not four, but three three square. So if you multiply all of this, you get that? This is 7. 20. This is 7. 20 and that's the number that goes right here. 7. 20. Cool. So if you solve here you have This is on the left. Uh, the stuff on the right side adds up to 23 to 0. And if you divide, you get in omega of 1.38 radiance per second. Now, this should make sense. You started off at two. You added mass. The disk became heavier. Therefore, the disk slows down a little bit, and now it has an Omega final off 1.38. Okay, so I went from 2 to 1 point 3/8. This is part A. I'm gonna make some space here and then solve part B. Part B is calculate the person's new tangential speed. Okay, If you are a point mass going around in a circular path, not only do you have in omega, but you have an equivalent tangential speed that is tied to your omega, and that's given by our Omega. So that's all we gotta do is plug this in. The person spins with the disk, so The person also has an omega of 1.38. Okay. And 1.38. Now, what are do you think I use? Well, R is the distance that the person has from the center you are. The radius of the disk is four. But you're spinning around the circular path of radius of three. So that you put it three here. All right. And if you multiply this answers 4.14. Very straightforward. That is your tangential velocity. As a result, part C part C. So you landed on the disk. Now you're going to walk towards the center of the disc, okay? And this is what we talked about in terms of moving inside of the disk. So here we're adding mass to a disk when you land and removing within the disk after you land. So it says the person starts walking towards the disc center, calculate the disc's speed once the person reaches it. So I want to draw here. Quick diagram. You got the person here. That's our equals three. Okay, Max, what's going do this, Um, and the person is initially at r equals three right there. So that's my are think of this as your our initial, okay? And then the person is going toe walk in this direction here until he gets to the center. So the person is going to have our final off zero. Okay, So you think the disk is gonna be spinning faster or slower? Faster or slower? Well, one of the ways you can think about this is similar to when you close your arms, you're gonna spin faster. The effective total radius of this system, um, is going to be smaller. They're more mass is closer to the axis of rotation. Therefore, it spins faster. Okay, so the disk will speed up. How do we do this? Well, conservation of angular momentum. Every time you have something that spins and something about that system changes like the radius of the earth mass, then we're gonna right. The conservation of angular momentum equation l initial equals l final. In the beginning, we have in the beginning here. We're talking about this to this. So in the beginning, we have disc. We have disc plus guy. When the guy is at R equals three. And at the end, we also have disc plus guy, but here. The guy is in a position of r equals zero. The guy sits on the axis of rotation. Okay, so let's expand this equation Here. I initial mega initial I Final Omega Final, and for both of these, we have guy and, um, we have guy and disk. So what you can do is you can write I as I disk. Plus I guy, it's the moment of inertia. The system. You can do it together like this Omega initial equals I disk plus I guy Omega Final. So this is obviously idisk initial I guy initial idisk Final I guy final. The moment of inertia of the disc doesn't really change because the disc didn't change its mass or its radius. What will change is the I put a little delta here for change. What will change is the final moment of inertia of the guy will be different than the initial. Okay, so here I have half M R square. Remember the moment of inertia of the guy since returning has a point. Mass is m little r squared. This is little our initial um, this is the number that's changed and this is a three in the beginning in the zero at the end, Omega initial. So the guy lands at this thing and hey automatically picks up a speed of 1.38. So that is the Omega initial off. The guy in the disk together. One point. I'm 38. It's not too, too. Was the speed of the disc before you landed on it? Okay, on then, here on the other side. You have same thing, half M r squared. That doesn't change. But this is going to be m our final square in this final here, our final is zero. Okay. And Omega Final is what we're looking for. We have all the numbers, so should be able to plug everything in and software made a final. So the rest here is just algebra. Most important step was getting, um, here. Okay, so let's do this. Half the mass of the disk is 200 actually. Already calculated this whole thing here before, This is just 1600 right? Yeah, this is 1600. Plus this thing here, which is Did we have this before? The mass of the guy is 80 distances, three squared. Feel like we have that before Yeah, that's 7 20 right? So we have 1600 plus 7. 20 and the whole thing is 1 38. We had these two guys from before from part a, the end of party. And then here we have the same 1600. But now check it out. This whole thing will be zero because the are is zero. He's sitting on the axis of rotation Ah, point mass. Sitting on your axis of rotation doesn't contribute towards your moment of inertia, right? It's as if it wasn't there. It's basically the same thing as if this guy just got, like, picked off from the disc. Alright, so I'm gonna put a plus. You're here just to be very clear and omega final, So Omega Final will be. Now, once you plug this in here, this entire left side will be 3200. I divide by 1600 I get the Omega Final is to radiance per second. Now, where have we seen to before? Two was the original speed that the disc had without the guy, which kind of goes with what I just mentioned that if you walk to the center, you have no contribution towards the moment of inertia. It's as if the guy got picked off. So you're back to the original part of the problem, which is disk without the person was spending with two. You add a person, you're slower. The person walks to the center, which means that the person's contribution contribution to moments of inertia disappears. So you're back to where you were before the person was on the disc. Cool. So too, is not a coincidence. It should make sense. Hope the whole question makes sense. Let me know if you have any questions and let's keep going.