Hey, guys. So in this video, we're going to start talking about Tory acceleration problems, also known as rotational dynamics problems where we have two types of emotions. What that means is that on top of having a rotation, we're also gonna have objects moving inland linear direction, so it's gonna have a combination of both of them. Let's check it out. So remember that when we have problems where a torque causes an angular acceleration where torque causes an angular acceleration Alfa we use the rotational version of Newton's second law. So instead of Africa's, they may. We're gonna write torque. Some of all torques equals I Alfa Okay, so that we have torque Alfa questions. That's how we solve it. But in some problems, we're gonna have more than just Alfa. We're going to have rotational and linear motion. So, for example, here in this picture, um, if you you have a block that's connected to a pulley. If you release the block, it's going to accelerate this way. But because it's connected to the pulley, it's also going to cause the pulley to accelerate this way. So the block falls in linear motion and the pulley accelerates around that central access in rotational motion, we have both. And in these questions, we're not going to use just, um, Tourky, Cozy Alfa. But instead, we're gonna use both Torque equals that Alfa and efficacy may. Okay, so we're going to write some of our forces equals m A for each acceleration we have, and we're going to write some of all torques equals Alfa for each Alfa that we have. So for each A, we write articles I may. And for each Alfa we write, torque equals Alfa. Now, what do I mean by each day and each Alfa, we have to count how many emotions exists in the problem here. I have one object with one acceleration, one type of motion, linear motion. And then here I have another object with another type of motion. So there's two motions in total. Okay, But if you had Mawr types of motions and I'll get to that in the example below, you would use more than just two equations. All right, we'll get to that. Um, when you do this, when you combine f equals a made with torque equals Alfa, you end up with an A and an Alfa. That's true variables. One of the techniques we're gonna use to solve these questions is instead of having A and Alfa, we're going to replace Alfa with a And by doing this instead of having A and Alfa, I'm going to have a and A imagine in this second equation here. Imagine if that Alfa somehow became an A then you would have a there in a here. And that's good, because instead of having two variables, you now have one variable. This is a key part in solving this question is going from, um, Alfa Teoh A. Okay, And the way we do this is by remembering that a and Alfa are connected. They're connected by this equation a equals R Alfa, where r is the distance between, um it's the distance between the force and the axis of rotation. So it's our our vector from the torque equation, if you remember. So where are is distance? I'm gonna call this distance to access. Okay, distance to the access from the force. Okay, But this is actually not the equation we're going to use, because what we're looking for is we're trying to replace Alfa. So what? We're gonna do is we're gonna say Alfa equals a over our Wherever we see an Alfa, we're gonna replace it with a over our and that's going to simplify. Okay, so we're actually going to using these questions a combination of three equations because it may torque equals Alfa. And we're gonna use this one here to link the two. The two first equations. Alright, The last point I wanna make here is that the signs for a an Alfa as well as the signs for V in omega must be consistent. Must be consistent. What do I mean by that? So I'm going to give you one example that allows me to talk about these four variables, and it's this one. You have a disk that's rolling. Um, actually, let me draw it. You have a disk that rolls up a hill. Okay, so imagine that if you are going this way, right, and then you're going this way. Okay, you're drive over here. You are spinning like this and then you go up the hill spending like this. So that's your V. And this is your omega. Okay, but if you're going up a hill, gravity is pulling you down. So your acceleration is downhill. It's going to be like this. And hey, and this means your Alfa is actually like this. Okay, Because if your velocity if you're going up like this, means that you're Maiga's like this, then I'm acceleration. That's down means that you're alphas like this, Alright? And all of these signs have to be consistent. So in most of these problems, because these are dynamics problems, they're problems about acceleration. We're going to say that the direction of positive will follow acceleration. Okay, so let me add that here, um direction off. I meant to make a little star direction of positive will follow acceleration. So this would be positive, right? Which means V is negative. And if these negative omega has to be negative as well, because via Omega must have signs that are consistent with each other and Alfa must be positive. OK, so an Alfa positive together in this particular situation, V is opposite way. So it's negative and Omega goes with V. So omega goes with V and the sign of a goes with Alfa. Okay, It's very important. Those signs of consistent All right, let's do an example here before we solve an actual full length problem. What I wanna do is show you some of the things that I've talked about here in a variety of different examples. Let's check it out. So here, for each one of these, we want to solve for acceleration. First thing I wanna know is which equations would you start with? How many equations? What kinds of equations. Alright, so and then we're going to sketch a diagram. So remember what I told you. You write. Remember what I told you? That you write this point right here. We write f equals M A for each A and we right, 20 goes Alfa for each for each alfa. So what that means is that you look at this on the system here. Let's say that this mass is bigger than this so that the system tilts this way. Right? This means you have an acceleration here. Um and this thing is gonna accelerate up. And this thing is going to accelerate this way. These air linear accelerations, the pulling they're also gonna accelerate in this direction. Notice that all the accelerations are consistent with each other. Okay, I'm gonna call this positive, Which means this is positive. Which means this is positive. Positive, Positive. So the direction of positive is this Okay, you end up some weird stuff like, this guy is positive down and this guy's positive up. That's okay. This is the same direction because it's a system and they're moving together. So because I have three days and two alphas, I can write five equations. Alright, So I have three equations that I can write three Africans. They may one for each one of these blocks and to torque equals Alfa, um, one for each of the disks. Okay, One last point I wanna make here is that all of these aides are the same. A one equals a two equals a three and all of the officers the same. So it's gonna call this a and all the officer, the same Alfa one equals Alfa too. So I could just refer to Alfa. And lastly, not only are all the is the same all the office the same they're also connected A equals r alfa. So all the accelerations are connected somehow. Okay, Now what I wanna do quickly is for part B is sketched a diagram showing forces acting on these things. So here I have. Um let's start with this guy over here. Let's call this one, 23 one to Okay. So the guy at the bottom here you have m one is being pulled down by M one g. And then there's a tension here. T one. Okay, if you are this pulley right here, I'm gonna just draw it over here. Um, you have attention. The same tension here, pulling him down t one. And then you have this tension here pulling the other way. T two. Okay, Those are the only two forces that matter if you are the block over here, Um, let's call this guy. Mm, too. Then you have t to pulling you to the left and t three, pulling you to the right. If you're the disk, you're being pulled to the left by t three. And then you're being pulled down by, um, tea four. And then if you're the block, you're being pulled up by t four and then being pulled down by mm. Three G. This is entry. Obviously, there is normal and m two g here. But these guys don't really matter because they just cancel each other. They don't really affect motion. Okay, so here's all the stuff now, these are all the different forces. If you want Thio, we're gonna quickly put some signs here. So remember the direction of acceleration is this Okay? So when I write, f equals I made for this guy, this is positive, and this is negative. Okay, this is positive. And this is negative because because it follows this direction, this is positive, and this is negative. Okay, Now, that being said, when you look at the torques look at these guys here. These discs, the disks who have torques acting on them. So there is a torque pulling this disc this way. T one right here. If you got a disk, T one is pulling the disk like this down. Right s. So this is the torque due to t one. And this is the torque duty to we have to be consistent with signs spinning this way is negative in this problem, because everything is going that way. So this is positive and negative, and we're almost done. Got to squeeze this stuff in here. Um, I got the torque of t four barely see that there. And then I have the torque off T three. The t 41 is positive. NT three is negative. Okay, Hopefully you don't actually get a question like this, but you might get a much simpler version. I'm just trying to show you all the you know, for a particular case like this, what matters? Use the direction of the signs and all that stuff. OK, eso next one is much simpler. It's a simple yo yo. So the idea is that you have a cylinder here that you let loose and it spins so it accelerates down this way. Now imagine if you're holding the If the string is here and the yo yo is here and then it starts falling. It's gonna roll like this. Okay, so it's going to roll like this now. That's because the strings on that side, if you had a yo yo here, then you let it go. It's going to accelerate like this, and it's going to accelerate like this. In this case, they're two different situations here. In this case, this is positive and this is positive. In this case, this is positive. This is positive notice that in both cases, a down is positive. But here the acceleration is positive. This way. Here, the accelerations. Positive. This way. All right, so it depends on the situation you have. But the key thing is that these two guys match each other. Okay, you got a match. So the acceleration have to both be positive together. All right. So how many equations do we have here? Well, how many types of motion I have? One object with two motions. One object with two motions. I have the same object has an A in an Alfa. So that means that I'm gonna write f equals in May once for that object. And I'm gonna write some of all torques equal Scialfa once. Okay, Those are the number of equations. Now, let's look at all the forces. You only have one object. You have a force of tension that's pulling you up. Um, acceleration is down. Alfa is this way. Okay, so this is positive. This is positive. Noticed. Attention is going up, so tension is negative and tension causes a torque. Tension causes a torque in this direction. Um, let me draw this a little bit differently. Um, how do I put this trying to move this away from here? Let's do this. Let's do tension over here. It's negative, and it causes a torque of tension this way. Okay, Now does the talk of tension spinning the same direction is the Alfa. It does Alfa spin like this, right? And the torque of tension spends that way as well. So this is positive. Okay, so getting the sign here is very important if you scrub to sign to get the wrong answer, Okay, Something interesting that happens here is your attention is negative, but the torque of tension is positive. That's okay, right? So it's supposed to be like that. Don't freak out. Just because tension is negative doesn't mean that the torque of tension is positive. They're going to be different. So that's that's that's set up for this one. And let's do this last one. Here I have a cylinder rolling downhill. Um, it's a similar situation as the simple yo yo and that I have one object with two motions. Okay, so I have one object that has both in a in an Alfa. That's because it's rolling and moving at the same time, right? therefore, Aiken. Right? Um, some of our forces, because it made once and I can write some of the torques equals I Alfa once. So before you can start solving these questions, you have to know which equations how many equations you have to draw. You have to, right? Okay, now, here we have this object here, and I'll tell you, we'll talk about this more later. But I'll tell you that the forces you have are m g X. Obviously you have an M g y as well, but MG just canceled with normal. These two guys cancels, they don't they don't really do anything. And there's a friction that goes this way. And the friction is responsible for the Onley torque you have. So there is a torque off friction. Now, in this case, you are going this way. The ends. Actually, you're like this. You have a V this way, and you have, um, on acceleration this way. Okay. You're following this way. You're getting faster this way. So your omega is this way, and your Alfa is this way as well. Okay, so down and spinning this way is positive. So this guy is positive. This guy's negative. Um, this torque here, this torque here is in the same direction as Alfa. Right? All of these guys were positive. So this torque is positive as well. Notice that. Just like what we had here. Tension was negative, but the torque of tension was positive here. Frictions Negative, but the torque due to friction is positive. Okay, well, check out these two examples later, but for now, I just wanna introduce this cool. So again, just a big introduction. We're gonna doom or of these? Hopefully, this makes sense. Let me know if you have any questions and let's keep going.