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Sections
Angular Momentum & Newton's Second Law
Opening/Closing Arms on Rotating Stool
Spinning on String of Variable Length
Angular Momentum of Objects in Linear Motion
Intro to Angular Collisions
Conservation of Angular Momentum
Angular Collisions with Linear Motion
Intro to Angular Momentum
Jumping Into/Out of Moving Disc
Angular Momentum of a Point Mass
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Next SectionAngular Momentum of a Point Mass

Example #1: Jumping into a moving disc

Transcript

Hey guys. So another common example of the angular collision problem has you jumping into a disk with initial velocity. So let's check out a few possibilities, here we have a disk of mass 200 kilograms and 4 meters in radius. So, I'm going to say Big M 200 and the Big R 4 meters and the moment of inertia of a disk is 1/2 MR squared same with a solid cylinder I can actually go ahead and already plug this in here because we're going to be using this a lot and if I do this I get 400 or, actually I'm sorry, I get 1,600 this thing is squared, right? So 1600 is the moment of inertia of the disk, it spins around a perpendicular axis through its center. So perpendicular axis through its center of this just means that the disk spins like this like you would expect a normal, usual disk rotation should look like with the velocity of 2. So the Omega initial of the disk is 2 counter clockwise, counter clockwise is in the direction of the unit circle, it's a positive 2, okay? So the disk has I of 1,600 and a Omega initial of positive 2, find the new speed that the disk could have if an 80, if an 80 kilogram person, which we're going to treat as a point mass, jumps into the disk in a bunch of different situations, okay? So the person is going to come into the picture, little m equals 80, we're going to treat the person as a point mass which means we're going to use moments of inertia equation as mr squared, okay? And I want to know what is the new angular, the new angular speed that 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 disk spinning and you like laying on top of the disk you don't have a horizontal velocity, you're not doing anything that would cause this 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, you're 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 the disk is at rest and then you'll come running this way, right? And then you jump on the disk, what's going to happen is when you jump on the disk, the disk is going to spin. So you're contributing some initial momentum, here you're not contributing any initial momentum. Alright, and all of these questions were going to 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 going to use L as mvr instead but here the person doesn't have anything, doesn't have velocity. So in the beginning all we have is the angular momentum of the disk, okay? So this is going to be I, let's call, let's say the disk is 1 and the person is 2. So Big M is 1 person is 2, okay? So I1 Omega1 initial plus. Now the person is linear even though the person doesn't really move with any velocity at. So this would be m2, v2, r2 but this is 0 because the person enters with negligible speed v equals 0 equals I1, Omega1 final plus I2, Omega2 final, okay? The initial I of the disk the I of the disk is 1,600 the initial speed of the disk is positive 2. Now once the person enters the disk they will have the same angular momentum, right? If you step into a disk you will spin with the disk. So I can do this, Omega final can combine and then all I have is I of the disk and I of the person, I of the disk is 1600 but I of the person I have to calculate that real quick. So, let's go over here and say the I2 is going to be the moment of inertia of a point mass because that's how we're treating the person mr squared, little m is the mass of the person which is 80, R is distance from the center. So the person is entering the disk, you're entering the disk. So if you enter the disk without any speed is because you enter 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 disk, the only thing that matters is that you enter at the edge which means that your r is the entire radius of the disk, okay? So here r will be the radius of the disk because we enter the disk at its very edge but this has radius 4. So it's going to be like this and this, there's a square here and this is 1280, okay? So 1280 is the number that goes here, 1280 and when I move stuff around Omega final is 3200 divided by these two guys, you end up with one point one one radians per second, this should make sense because it's slower. So we started with an Omega of 2 positive and now a person entered it has more mass whenever you add mass to a rotating disk it will move at a slower rate. So that's it for part eight pretty straightforward. Let's do Part B, the Part B the person will jump into the disk at its edge with 90 meters per second directed towards the center. So, let me draw this here real quick, you jump directed in a direction, you jump at the edge directed towards the center. So the first one you did this and 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 I have a disk and you come running this way and you jump, you're going to add a momentum this way because you jump like this, think of you as pushing a force if you go the other direction that caused to go the other direction as well. Now if you come running to the disk. if you come running to the disk in a direction that is towards the center like a radial direction you actually don't contribute any momentum to the disk, it's just like if you are trying to push on a disk towards the radius, right? You can push but it's not going to 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 step in, you're not contributing any momentum 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 0 for you and the answer is the same as Part A. Omega final is just 1.11 radians per second because you don't contribute any momentum. Now for Part C and D you will contribute to momentum because you're jumping as I explained here you're jumping in a direction that will cause the disk to spin, okay? Alright. So for Part C you're jumping directed tangentially up. So this is Part C and this is Part D, in both cases you have the same velocity just different directions so let's do this real quick. Alright, so same setup, I'm going to have L initial equals L final, let's make some space here for Part C, L initial equals L final which means I can write this equation here I1, Omega1 initial plus m2, v2, r2 equals I1, Omega1 final plus I2 Omega2 final and all of these questions you will rotate with the disk once you land on it. So you are making, Omega1 and Omega2 at the end will be the same which means we can do this Omega final I1 plus I2, this doesn't change, okay? The initial Omega of the disk is, the initial I of the disk is 1600, the I doesn't really change. So the, not be initial but just the I of the disk is 1,600, the initial Omega that's what I meant is plus 2, okay? Now here the person which has mass of 80 is jumping in with a velocity of 9, I want to leave a little bit of space here. So we can put positive and negative, we'll talk about that in just a second and you are jumping tangentially at the edge of the disk which means your r is the entire radius here, okay? Your r is the radius and the radius is 4. So, I'm going to put a 4 here. Now we got to talk about this velocity, that's going to be the difference between Part C and D, it's just the sign of that velocity, okay? So initially the disk initially the disk is rotating like this with a positive 2 radians per second, okay? Look at the at the 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 to against the direction of the disk, right? So you're, the disk spin like this and then you're jumping in your jump, the disk spin like this, I got my disk is broken here, and you're 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 that's going to be clockwise opposite to the disk 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 velocity has to be negative even though it's going up because it's trying to cause negative rotation, okay? D is going to 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, this disk is rotating a particular direction and D is jumping in the same direction as 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 C 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. So we remember to just copy and paste that stuff obviously with a different number. So this is 30, this is 3,200, you 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 320 on this side and over here you have Omega final 1600 plus 12 plus 1280 and when you do all of this to get that Omega final is, Omega final is 0.11 radians per second. So the idea is that the disk was spinning with, in a particular direction with two 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 consequence of you jumping in, okay? For Part D we're just going to the same exact thing that we're going to plug a positive in here. So, let's copy this real quick 1600 plus 2 plus 80 times positive 9 times 4 equals Omega final and I have these two guys here, 1600 plus 1280 and if you do all of this, skip some steps, little bunch of algebra you just do all of this and you get, I have it here 2.11 radians per second. So let me talk about these two answers real quick, here you went from 2 all the way to 0.11. 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 more. So why was it that if I was going with 9 one way I lost almost all of my velocity, right? The change here is like 1., you lost like 1.9 of Omega and here you only gain 0.11. Well, it's because here not only are you going against the rotation so you caused it to slow down but you're also adding mass. So there's two things you are 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? I just want to talk about that try to make some sense as to why the the gaps between 2 and the final velocity here and 2 and the final velocity here were so significant when it seemed like maybe would be, the gaps might have been closer and it has to do with the fact that you're adding mass in both of these. Alright, so hopefully that makes sense, let me know if you have any questions, let's keep going.

Practice: 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?

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