Intro to Angular Collisions - Online Tutor, Practice Problems & Exam Prep

Intro to Angular Collisions (Two discs)

Hey, guys. So in this video, we're going to start talking about angular collision. An angular collision happens when you have either 2 discs that are rotating and if you, like, push them against each other so that they now spin together. So that's an angular collision. Or if you have an object 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 bar is fixed here so the bar will rotate. Okay. That would be an angular collision problem as well. Let's check it out. So angular collision happens. Angular collisions will happen when one of the 2 objects, at least one, so I should say one or more of the 2 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 3 different setups of collisions. You can have a collision where 2 objects have linear motion. Like for example, 2 boxes going towards each other. In this case, this is a linear collision. And we're going to use conservation of linear momentum, p to solve this problem. If you have 2 objects that are rotating, it should be obvious that this is an angular collision or rotational collision. So the example I gave, 2 discs that are spinning and you push them together so they spin together. And we're going to use this as an angular collision. We're going to use conservation of angular momentum to solve this. So the non-obvious case is if you have v and ω, which is the example I gave with the bar. Right? So this object moves with the v, hits the bar here, causes the bar to spin. The object has v. The bar will get an ω as a result. So we use this, to solve this. This is actually an angular collision. Even though there's one of each, it's not a linear collision. It's considered an angular collision, and we're going to solve it using the conservation of angular momentum l. You can think of it as l basically supersedes as long as you have one ω, l will take over for p. Okay. So this I already mentioned briefly. Similar to linear collisions, we're going to use the conservation of momentum equation. But we're going to use, obviously, the angular version. Okay? That's what this is. We're going to use conservation of l instead of conservation of p. So instead of p initial equals p final, I'm going to write l initial equals l final. Remember the conservation of momentum equation for linear momentum, if you expanded the whole equation, you would have m1v1, m2v2, m1v1, m2v2. It's the same thing here, but mv will be replaced by I ω. So it's going to be I1ω1 initial plus I2ω2 initial equals I1ω1 final, I2ω2 final. Cool. 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 up 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. It hits the bar right there at a distance little r from the axis of rotation. We're going to use the equation that l=mvr. Let me put this over here as well. l=mvr. Okay? I guess it might make more sense to put this over here on the other side so these guys are hanging out together. Cool. l=mvr is what we're going to use. Now what that means is that for that object, instead of using I ω instead of using I ω, you're going to use this. So going to write here instead. Okay. We're going to do this. Don't worry. We'll do an example. Now in this equation, r, as I mentioned here, r is the distance between where the linear object collides right here, red dot, and the axis of the rotating object, the blue dot. Right? So it's just the r vector between those two points. And the last point I want to 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 disc, that is technically an angular collision problem. Though 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, you know, different implications of linear collision mixed with angular collision is that these questions are 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. Alright. So here I have 2 discs. The blue disc, if you read the whole thing here, the blue disc is spinning. Notice I have a disc of radius 6 and a disc of radius 3. So this is the 6, obviously. So I'm going to call this, r1=6 meters and it has a mass of a 100. So mass 1 equals a 100. And then this is I'm going to call this r2=3 meters and he has a mass mr2m2of50. Okay. It says here that the 100 kilograms, so the outer one, the bigger one, spins clockwise. Clockwise looks like this, at a 120 RPM. So I'm going to say that the RPM is 120, 120. I'm going to call this negative 120 because it's clockwise. Clockwise is negative. Okay? And it spins around a perpendicular axis to its center. What that means is if you have a disc, perpendicular axis just means that your imaginary axis line, runs 90 degrees to the face of the disc. So it just means that the disc is spinning like this, standard rotation for a disc. Okay? A second solid disc, which is the darker one there, is carefully placed on top of the first disc, and it causes the disc to spin together. So imagine one disc is already rotating, right? The blue disc is already rotating and then you add the, the gray disc on top of it and now the blue disc is essentially angular speed, the speed, the angular speed, the ω or the RPM of the blue disc? I hope you're thinking if you add some stuff on top, it's going to spin slower and that's what's going to happen. Okay? We're going to have a lower RPM. So this question is asking us to find the new rates in RPM that the discs will have, in 2 different situations. So first, we're going to add a disc. We're going to add a smaller disc here at rest so we just lower it slowly. And in another situation, we're going to have it where the disc on top was actually rotating in the opposite direction. So now we're going to have a disc spinning in one way and the other disc spinning the other way and we're going to land the one disk into the other. Okay. So let's do that. On the first one, we're saying that the initial ω of disk 2 is 0, but disc 1 has an initial RPM, this is RPM of 1, of negative 120. There's 2 ways you can go about this. You have ω and you have RPM. I'm giving you RPM and I'm asking for RPM. The question here is what is the, this is initial. What is RPM final of the whole system? They will rotate together. Right? So what is RPM final of the whole system? I'm giving you an RPM, I'm asking for an RPM, but remember the momentum equation, the angular momentum equation has ω and not RPM. So you have two choices. You can convert from RPM, you can go from RPM to ω, do your calculations and then come back to RPM. Or you can just rewrite the, the angular momentum equation, the l equation, in terms of RPM instead of ω. I'm going to 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 going to start with l I equals l f. Okay. In the beginning, all you have is you have the blue disc spinning by itself, and the gray disc doesn't spin at all. So all I have is I1ω1 initial. At the end, they're going to spin together. They both have rotation. So I'm going to have I2. If you want, you could have written it this way. Right? I2ω2 initial and just say that this is 0 because that disc is at rest. Okay? And then this is going to be I1ω1 final plus I2ω2 final. I hope you realize that this is going to be the same. Okay. ω1 final equals ω2 final. So we're just going to call it ω final because they are, they're going to spin together as a result. K. You may even remember that these situations where 2 objects collide and after the collision, they move with the same speed is called, a completely inelastic collision. So this is technically a completely inelastic angular collision. Cool? Fascinating. Alright. So we're going to be able to do this here, ω final *(I1+I2). And this here is just I1ω1 initial. And then what I want to do we don't have ωs. We have, we have RPMs. So I want to rewrite ωs in terms of RPM. So I1, instead of ω, I am going to have 2pi RPM1initial divided by 60 equals 2pi RPMfinalf divided by 60, and that is times I1