16. Angular Momentum

Intro to Angular Collisions

# Intro to Angular Collisions (Two discs)

Patrick Ford

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Hey, guys. So in this video, we're going to start talking about angular collision and angular. Collision happens when you have either, um, two discs that are rotating and if you like, push them against each other so that they now spend together. So that's an angular collision. Or if you have a knob checked, moving in linear motion, that hits another object that will rotate. So, for example, if you have a bar and let's say some object comes and hits the bar, the bars fixed here so the bar will rotate. Okay, that would be an angular collision problems. Well, let's check it out. So angular. Collision happens. Angular collisions will happen when one of the two objects at least one. So I should say one or more off. The two objects either is rotating or rotates as a result, the problem I mentioned here, this hits this object will rotate as a result. So this is an angular collision. Okay, so there are actually three different setups of collisions. You can have a collision where two objects have linear motion. Um, like, for example, two boxes going towards each other. In this case, this is a linear collision and we're gonna use this is linear collision. And we're gonna use conservation of linear momentum p to solve this problem. If you have two objects that are rotating, um, it should be obvious that this is an angular collision or rotational collision. So the example I gave two discs that are spinning and you push them together. So they spent together, and we're gonna use this is an angular collision. We're gonna use conservation of angular momentum to solve this. So the non obvious cases if you have one V and one omega, which is the example I gave with the bar. Right. So this object moves of the V hits the bar here cause the bar to spin the object has v the bar will, We'll get Omega as a result. Um, so we use this to solve this. This is actually angular collision. Even though there's one of each. It's not a linear collision. It's considered an angular collision, and we're gonna solve it using the conservation of angular momentum. L You can think of it as l basically supersedes as long as you have one Omega l will take over for P. Okay, so, um this already brave, Mentioned briefly, similar to linear collisions. We're gonna use the conservation of of momentum equation. But we're gonna use obviously the angular version. Okay, that's what this is. We're gonna use conservation of L instead of conservation of people. So instead of p nature, because p fine, I'm gonna write Ellen Show Echoes equals L finer. Remember the conservation of momentum equation for linear momentum? If you expanded the whole equation, you would have m one V one MTV to M one V one. MTV, too. It's the same thing here, but M v will be replaced by I Omega. So it's gonna be I one Omega one initial, plus I to Omega to initial equals. I won Omega One Final I to Omega to finally quote, that's the conservation of angular momentum equation. Now, if you have a point mass in linear motion, we're going to use the linear version of the angular momentum equation. What is this? So if you have the situation, I keep describing if you have an object that collides against the bar so you have a mass m here moving this way with the velocity V and he hits the bar at a distance hits the bar right there at a distance. Little are from the axis of rotation. We're gonna use the equation that l equals M V R. Let me put this over here as well. L equals M V R. Okay, I guess it might make more sense to put this over the here on the other side. So these guys were hanging out together. L equals M V. R. Is what we're going to use now. What that means is that for that object, instead of using Iomega instead of using Iomega, you're going to use this, so I'm gonna right here instead. Okay, We're gonna do this. Don't worry. We'll do an example. Um now, in this equation are, as I mentioned here, R is the distance between where the linear object collides right here, red dot in the the axis of the rotating object, the blue dot Right, So it's just the our vector between those two points. And the last point I wanna make here is before we do an example is if you have a situation where you have a rotating disc and you add mass to the disk that is technically angular collision problem that we could have solved that without talking about angular collisions just by using conservation of angular momentum. Okay. And the reason we could do that without worrying about, uh, you know, different implications of linear collision mixed with angular collision is that these questions were simpler if the mass was at rest. So if you just add a mass in there, it's a much simpler problem. Don't worry about it. We'll get there as well. All right, so here I have two disks. Um, the blue disk. If you read the whole thing here, the blue disk is spinning notice. However, disk of Radius six in the disk of radius three. So this is the six, obviously. So I'm gonna call this r one equals 6 m and has a mass of 100. So mass one equals 100 then this is I'm gonna call. This are, too. Is 3 m and he has a mass m r to M two of 50. Okay, it says here, that's the 100 kg. So the outer one, the bigger one, um, spins clockwise, clockwise. Looks like this, um, at 120 rpm. So I'm gonna say that the rpm is 1 21 20. I'm gonna call this negative 1. 20. because it's clockwise. Clockwise is negative, okay? And it spins around a perpendicular access to its center. What that means is, if you have a disc perpendicular axes just means that you're imaginary. Axis line, um, runs 90 degrees to the face of the disk. So it just means that the disks spinning like this standard rotation for a disk. Okay, a second solid disc, which is the darker one there is carefully placed on top of the first disk. Andi causes the disk to spin together, so imagine one disk is already rotating, right? The blue disk is already rotating, and then you add the, um, the great disk on top of it. And now the blue disk is essentially carrying the great disk, and they're spinning together. How? How do you think this is going to affect the speed, the English speed omega or the rpm of the blue disk? I hope you're thinking if you add some stuff on top, it's gonna spend slower, and that's what's gonna happen. Okay, we're gonna have a lower rpm. So this question is asking us to find the new rates in our PM that the disks who have in two different situations. So first we're gonna add a disk. We're gonna add the smaller disc here at rest, so we just lower it slowly and in another situation, we're gonna have it where the disk on top was actually rotating in the opposite direction. So now we're gonna have a disk spinning in one way and the other disks spinning the other way, and we're gonna land the one disk into the other. Okay, so let's do that on the first one. We're saying that the initial omega of disk to is zero, but this one has an initial rpm. This is our pm of one off negative 1. 20. There's two ways you can go about this. Um, you have omega and you have rpm. I'm giving you rpm. And I'm asking you for our PM The question here is what is the, um this is initial. What is our PM Final of the whole system. They will rotate together, right? So, what is our PM? Fine of the whole system I'm give you on our PM. I'm asking for an rpm. But remember the momentum equation. The angular momentum equation has omega and not rpm. So you have two choices you can convert from RPM. You can go from rpm to omega, do your calculations and then come back to our PM. Or you could just rewrite the theme angular momentum equation. The L equation, uh, in terms of rpm instead of omega. I'm gonna do that instead because I want to show you how that would look. Okay, So, conservation of angular momentum, you're doing something that changes the rotation of a system. So we're gonna start with l I equals L f. Okay. In the beginning, all you have is you have the blue disks spinning by itself. Um, and the great disc doesn't spend it. Also, all I have is I one Omega one initial at the end. They're going to spend together. They both have rotation. So I'm gonna have to, um if you want, you could have written it this way, right? I to Omega to initial and just say that this is zero, because that because that rest okay, and then this is gonna be I want Omega One final, plus I to Omega two final. I hope you realize that this is going to be the same. Okay, Omega one final equals Omega two final. So we're just gonna call it Omega Final? Because they are, um, they're going to spin together as a result. Okay, you may even remember that these situations where two objects collide and after the collision, they move with the same speed is called a completely in elastic collision. So this is technically a completely in elastic, angular collision. Cool. Fascinating. All right, so we're gonna be able to do this here, Omega Final I one plus I to, um And this here is just I won Omega one initial and then what I wanna do We don't have Omega's We have, um we have our PM's. So I wanna rewrite Omega's in terms of rpm. So I won. Instead of omega, I'm gonna have two pi r p m one initial divided by 60 equals, um, two pi rpm final divided by 60. And that is times I one plus I to Okay, I'm gonna cancel two pies. I'm gonna cancel the sixties and you end up with this. I'm giving you that. This is 1 20 I'm asking you for this. All you gotta do now is plug in I one and I to So let's do that's real quick. So I'm gonna go off to the side here and find I won. I won is half m one are one squared half M. One is 103 radius is six squared. And if you do this, you get 1800 for I to you have a half m two r two squared. I'm gonna calculate this off to the side because we're gonna use these lots. This is gonna half squared. So that z gonna be to 25. Okay. Yep. To 25. All right. So let's plug these numbers back in here. I one is gonna be 1800 times the initial rpm. The initial rpm. Is this one right here of the first disk? It's negative. 1 20 um, equals rpm final, which is our target variable. And then we're gonna add the two eyes. So 1800 plus 2 to 5. I combined this. I move it over to the other side, and I get that the final rpm of the joint system is going to be negative +107 Negative 107 Now, this should make sense that this was spinning with 1 20. Once you added something to the top of the disc, it now slows down a little bit. Goes from 1 20 to 107 still spins in the same direction, which is the negative direction. So the final rpm is 107 I guess we should stay here clockwise. Um, now, that's part A where the disk that we put the smaller, great disc had no initial speed. Now, for part B, that disk will have initial speed, and we're tight on space here. But I'm gonna cram it in here, and we're gonna be fine. So for part B, um, same equation. I won Omega one initial plus I to Omega two in their show equals I won Omega one final plus I to Omega two final. Remember, I can rewrites omega as two pi r p m over 60. Okay, that's what we did here. So I'm gonna rewrite all of these Omega's as two pi r p m over 60. So I'm gonna have every single one of these four terms will have a two pie in the 60 so I can cancel the two pi in 60 on all of them. Essentially, I'm replacing W just with our PM because two pi in 60 will cancel everywhere. I one is the same here. 18 hundreds. I'm gonna write 1800 rpm. One initial plus 2 rpm to initial equals. The Omega here is the same because they spent together, and then I just add up. And I'm also gonna rewrite this as our pm a swell. So, rpm final of both. This is our target variable. And I'm gonna add up the eyes, so it's gonna be 1800 plus 2. 25. Okay, the rpm of the first one in this problem, Um, the bigger disc rotates with 1 20 clockwise, so this is gonna be negative 1. 20. And it's saying here, that's the second disc for Part B would be spinning counterclockwise so positive within rpm of 3. 60. So here this would be plus 3. 60. Okay, So if you multiply all this crap on the left, you get a big negative number, a smaller, positive number. And if you combine those, the left side combines to be negative. 135,000. Um, and the right side is 2025. So our PM final times 25. So I'm gonna move the 2025 to the other side and divide the two, and we get to the final answer, which is negative. 67. So the final rpm is negative. 67. This means that it's also clockwise. Let's talk about this real quick. What this means is, so this is the second answer here for part B. Very similar set up to part eight. Just the only thing that changed was this, um here that instead of zero, was 3 60. Okay, so it let's talk about this real quick. It was spinning at 1 20. If you added a disk that isn't doesn't spend it all, it just makes it heavier. So it's gonna go from 1 20 to 107 But if instead you get a disk that's spinning in the other direction, right? So one disk spinning, Uh uh, clockwise. And then you add a disk that is counter clockwise. Um, the final of the two will be lower, and that's because the bigger disk has to slow down a lot to cancel out the opposite rotation of the other objects. All right, so this hopefully this makes sense. But let me know if you have any questions and let's keep going.

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