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16. Angular Momentum

Jumping Into/Out of Moving Disc


Jumping into a moving disc

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Hey, guys. So another common example of a angular collision problem has you jumping into a disc with initial velocity. So let's check out a few possibilities. Um, here we have a disc of mass, 200 kg and 4 m and radius. So I'm gonna say Big M 200 big are 4 m, and the moment of inertia of a disc is half m r squared Samos a solid cylinder. I can actually go ahead and already plugged this in here because we're gonna be using this a lot. And if I do this, um, I get 400 or actually, I'm sorry, I get 1600. Thing is squared, right? So 1600 is the moment of inertia of the disc. It spins around a perpendicular access through its center so perpendicular access through a center of disgust me that that this things like this, like you would expect a Nora usual disc rotation to look like with the velocity of two. So the omega initial of the disc is to counter clockwise. Counterclockwise is in the direction of the unit circle. It's a positive two. Okay, So the disk has I of 1600 Omega initial of positive to find a new speed at the disk would have if a 80 if 80 kg person which we're gonna treat is a point mass jumps into the desk and a bunch of different situations. Okay, so the person is gonna come into the picture. Little M equals 80. We're gonna treat the person as a point mass, which means we're gonna use moments of inertia equation as m r. Squared. Okay. And I want to know what is the new angular. The new angular speed at this will have. What is Omega final? Okay, so the first situation the person steps into the disk with negligible speed. Negligible speed is zero. It's the same thing as if the disks spinning. And you're like laying on top of the disk. You don't have a horizontal velocity. You're not doing anything that would cause a disk to move faster or slower. So before we start, I want to talk about that a little bit. Let's say the disk is here, and if you simply enter the disk, you don't affect your not adding velocity to the disk in any particular direction. The disk will become slower because there's more mass now, but you're not contributing to velocity as opposed to let's say if a disk is at rest and then you'll come running this way, right, and then you jump into this. What's gonna happen is when you jump into this, that this is gonna spin. So you're contributing some initial momentum here. You're not contributing any in their show momentum, All right? And all of these questions, we're gonna solve this using the conservation of angular momentum, so l initial equals l final. And remember, if you have an object which in this case, the person is moving in a linear motion, we're gonna use L as M V R instead. But here, the person doesn't have anything, doesn't have a velocity. So in the beginning, all we have is, um, the angular momentum of the disc. Okay, so this is gonna be I. Let's call. Let's say the disk is one, and the person is too. So big M is one person is too. Okay, so I won Omega one initial. Plus, Now the person is linear, even though the person doesn't really move with any velocity yet, So this would be m, um m two v to R two. But this is zero because the person enters with negligible speed. V equals zero equals, um one I one Omega one final, plus I to Omega two final. Okay, um, the initial I of the disc, the eye of the discus 1600. The initial speed of the disk is positive, too. Now, once the person enters the disk, they will have the same angular momentum, right? If you step into a disc, you will spend with the disk so I could do this or make a final can combine. And then all I have is I of the disk and I of the person I have the disk is 1600 but I have the person I have to calculate that real quick. So let's go over here and say the I two is going to be the moment of inertia of a point mass because that's how we're treating the person. M r squared little m is the mass of the person, which is 80 are is distance from the center. So the person is entering the disc. Um, you're entering the disk, so if you're entering the disk without any speed, it's because you're entering at the very edge of the disk, so you can assume that the person just steps in here because the person is negligible speed. It doesn't matter the direction with which the person comes into the disc. The only thing that matters is that you enter at the edge, which means that your are is the entire radius of the disk. Okay, so here are will be the radius of the disk because you enter the disk and it's very edge. The disk has radius four. So it's gonna be like this and this. There's a square here and this is 12 80. Okay, so 12 80 is the number that goes here 12 80. And when I move stuff around Omega Final is 3200, divided by thes two guys. You end up with 1.11 radiance per second. This should make sense because it's slower. So we started with an omega of two positive, and now a person enter. It has more mass. Whenever you add mass to a rotating disc, it will move at a slower rate. So that's it, for part a pretty straightforward let's do part B part B the person will jump into the disk at its edge with 9 m per second directed towards the center. So let me draw this here real quick. You jump directed in the direction you jump at the edge, direct towards the center. So the first one you did this on the second one. Let's say you jumped over here and you jumped with the direction that points towards the center. Okay, so let's think about what would happen here, right? So I was talking about how if you have a disk and you come running this way and you jump, you're gonna add a momentum this way because you jumped like this. Think of us pushing a force. Um, if you go the other direction that caused to go the other direction as well. Now, if you come running into the disc, if you come running into a disc in the direction that's towards the center, like a radio direction, you actually don't contribute any momentum to the disc. Um, it's just like if you are trying to push on a disk towards the radius, right, you can push, but it's not gonna cause rotation. So the idea is that jumping with the velocity if directed towards the center is actually the same thing as if you just stepped in. You're not contributing any mo mentum even though you have, um, even though you have linear velocity, you're not doing anything. So you just That's all you need for part B. You just have to realize that you're not adding any momentum, which means that this is also zero for you. And the answer is the same as part. A Omega final is just 1.11 radiance per second because you don't contribute any momentum now for part C and D, you will contribute momentum because you're jumping. As I explained here, you're jumping in a direction that will cause the disk to spin. Okay, All right. So for part c, you're jumping, directed tangentially up. So this is part C. And this is part D. Uh, in both cases, you have the same velocity, just different reactions. Um, so let's do this real quick. Alright? So same set up. I'm gonna have Ellen s show equals l final. Whoops. Let's make some space here for part C l initial equals l final, which means I can write this equation here, I won Omega one initial plus, um m two v to r two equals. I won Omega one final, plus I to Omega two final. In all of these questions, you will rotate with the disc once you land on it. So you're a mega niche Omega one and Omega to at the end will be the same. Which means we can do this or make a final. I won. Plus I to This doesn't change. Okay, um, the initial make of the disk is the initial I of the disc is 1600. The idea that doesn't really change. So the not the initial, but just the eye of the disc. 1600 thean initial Omega. That's what I meant is plus two. Okay, Now, here the person which has mass of 80 is jumping in with the velocity of nine. I wanna leave a little bit of space here so we can put positive or negative. We'll talk about that in just a second. And you are jumping tangentially at the edge of the disc. Which means your are is the entire radius here. Okay, your R is the radius, and the radius is four so I'm gonna put a four here. Now we gotta talk about this velocity. That's gonna be the difference between parts C and D is just a sign of that velocity. Okay, so initially the disk initially, the disk is rotating like this with a positive to radiance per second. Okay, look at the A C. Arrow here, right for part C. You're jumping that way, and I hope you can see how jumping that way would go against which caused the disk to spin. It's basically going against the direction of the disc, right? So you're the disks spinning like this, and then you're jumping in your jump, The disks spinning like this I got my disk is broken here. Ah, Ndure running that way. So if this is positive, this velocity has to be negative because that velocity is trying to produce a counter rotation. A rotation is going to be clockwise opposite to the disc. Um, therefore would be negative. So you can think of this almost as if this was a force that would produce a torque in this direction. But in this case, it's just a force that's trying to produce a rotation in that direction. This direction is clockwise. Therefore, this linear Veloz has to be negative even though it's going up because it's trying to cause negative rotation. Okay, um, D is gonna be the same thing, but it's going to be positive because it's trying to help the disk. Let me get my broken disk again. Um, this disk is rotating a particular direction, and D is jumping in the same direction of the rotation. So it's adding, So if this is positive, then this is positive as well, because they're both spinning in the same direction. Cool. So Parts C and D are exactly the same, except that for see, this is negative. And for d, that is positive. So it's gonna be the same thing except positive. Okay, so let me just put a little plus here. Um eso we remember to just copy and paste that stuff, Obviously not with a different number. So this is 30 thistles. 3200. Um, em todo let me make sure I got everything here, so actually, I just did this whole thing together. So when you combine all of this, you get a 3. 20 on the side and over here you have Omega Final 1600 plus 12 plus 12. 80. And when you do all of this, you get that Omega Final is Omega Final is 0 11 0. radiance per second. So the idea is that the disk was spinning with, um, in a particular direction with to you jumped against it, and you brought it all the way down to 0.11. So you had enough velocity that you almost caused the disk to spin. In fact, had you been going a little bit faster, the disk would be spinning in the opposite direction as a, um, as a consequence of you jumping in okay for party, we're just going to the same exact thing. But we're gonna plug a positive in here. So let's copy this real quick. 1600 plus two plus 80 times positive. Nine boom times four equals omega final. And then I have these two guys here 1600 plus 12 80. And if you do all of this, skip some steps, do a bunch of algebra, you do all of this and you get I have it here, um, 2.11 to radiance per second. So Let me talk about these two answers real quick here you went from to all the way to 20.11. So you change your velocity a lot. Here you went from 2 to 2.11. So you might be thinking. Actually, that's not that much mawr. So why was it that if I was going with 91 way I lost almost all of my velocity? Right? The change here is like one point you lost 1. of of of omega, and here you'll only gain 0.11. Well, it's because here not only are you going against the rotation, so you call it to slow down, but you're also adding mass. So there's two things you're doing to make this thing go slower. You're adding mass and you're going against rotation. Here you are going in the direction of the rotation, which helps, but you're also adding mass, which hurts rotation. And the net result of these two things ended up being a small gain in rotation. Cool. So I just want to talk about that. Try to make some sense as to why, um, the gaps between two and and the final velocity here and two. And the final velocity here were so significant when it seemed like, um, maybe it would be the gaps might have been closer. And it has to do with the fact that you're at a mass in both of these. All right, so hopeful. That makes sense. Let me know if you have any questions. Let's keep going.

A 200 kg disc 2 m in radius spins around a perpendicular axis through its center, with a person on it, at 3 rad/s counter-clockwise. The person has mass 70 kg, is at rest (relative to the disc, that is, spins with it) at the disc’s edge, and can be treated as a point mass. If the person jumps tangentially out of the disc with 10 m/s (relative to the floor), as shown by the red arrow, what new angular speed will the disc have as a result? If the person steps out into ice with negligible speed of his/her own, what speed would it have upon exiting?