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12. Rotational Kinematics

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Hey, guys. So in this video, we're gonna talk about problems where we have multiple wheels or objects like wheels such as cylinders or disks or gears. And when we have multiple of these things change together connected to each other by either chain or a belts, much like a bicycle, let's check out how these work. So these problems where we have to it could be more than two, but it's almost always too well like problems. So when I say wheels, I mean things like disc cylinders, etcetera. They're pretty common in rotation cinematic. So let's check them out. Um, there's two basic cases and we have a case where the wheels were rotating around. They fixed axes. In other words, you have something like these two wheels here. Imagine that there's a chain around them, but this where was bolted, Let's say to the wall and this. We was bolted to the wall, so if they start spinning, they're not gonna move sideways. Okay, so in this case, we have W because it's going to spin. But the wheel itself, there's a w um, let's call this W one and then there will therefore be obviously w two here. But this we will have no velocity of the center of mass. This point doesn't move sideways, okay? It does not move sideways. Same thing with this point. BCM, too, doesn't move sideways. Okay, so there's no velocity center of mass on example of this is if these were, um pulleys or gears that are like I said, attached to the wall thing. Example I gave you earlier is a toilet paper that's fixed to the wall. Um, and it rotates around itself like toilet paper usually works. Um, the other one's a static bicycle, so there's two basic ways you can have a static bicycle. One is you lift your bicycle from the floor, and then if you spend the wheels um, guess what? The rotation of the wheels doesn't cause the bike to sideways because it's basically now fixed. It's not free to move the other ways if you flip your bike upside down. So basically, if the wheels are not touching the floor, it's a free access to free wheel, Okay, and then in other cases, we're gonna have situations where two wheels air connected. Um, they're both free to move. That's a bicycle right as the wheel spin they spend together. So we're gonna look into that later. All right, So the big thing to remember the big thing to know here is that whenever chain connects to two wheels or connects two wheels, we have that the tangential velocity at the edge of one wheel equals the tangential velocity at the edge of the second wheel. So let's pick a point here in a point here. And I can say that this V t equals this VT. They're all the same. Okay, so VT one VT two VT two is the same as VT one. In fact, you could pick a point anywhere, And because it's the same chain, um, because it's the same chain, the velocity velocity has to be the same. Now, remember also that when you have a fixed disk So let me write this year Four fixed axis. Remember that VT equals are omega. The tangential velocity at a points away from the center, VT is given by our Omega. In this case, we're talking about one. So it's our one omega one. Okay, So this guy here, for example, would be our to Omega too. Now the VTs they're the same. The VTs, they're the same, but so that's good. But the ours air different R one and r two are different than we just have different sizes. Um, therefore, because these guys were different than the Omegas will be different as well. Okay, the ours air different, so the angular speed will be different. In fact, the greater my are the smaller my omega and vice versa. If you got a tiny wheel and a big wheel, the tiny wheel spin much faster while the big wheel is slowly spinning. Okay, so there's an inverse relationship there. All right, so V one equals V two and V equals R. W. So we can write this. We can write that VT. One equals VT two. So are one W one equals R two W two. This is the big equation for this video. The most important one. Now what I want to do is I want to show how there's four variations or three rather variations of this equation. Okay, and that's because we don't only have W or Omega to talk about to describe how quickly something spins. We also have frequency, period and rpm. All four of these are useful in describing how quickly something spits. So I'm gonna I'm gonna replace W with f t and R p m. And this is going to generate different versions of this equation. So w is to P f. Um, it is also two pi over tea. And remember, frequency is our PM over 60. So I can also say that w is two pi instead of f. I'm gonna say rpm over 60. Okay. Can you see that? Cause you can. All right, so we got these versions, and what I'm gonna do is I'm gonna replace, um w here with this this and this. Okay, so our one instead of w gonna write to pie, have one. And then aren't you two pi f to? What happens is the two pies cancel. So you're left with are one F one our to have to. Okay, if I were to do the same thing with these other guys here, I'm gonna do this one more time. Um, if you want to do this with two pi you would get are one. I'm sorry if you were to do this with t you get are one two pi t one. Aren't you two pi t to the two pipes? Cancel again and we're left with, um, we're left with. Let me put it over Here are one over t one equals R two over teach you. And then the last piece is with our PM. Um, and I will do the same thing. I'm gonna just kind of skip here for the sake of time, but I'm gonna say that you can have r one r p m one equals R two. Our PM to the most important of these four is the first. The other three are just sort of derivative, um, equivalent similar versions of them. But I find that if you know all four of them your best suited to solve these problems very quickly. Okay. Thes problems were very straightforward. This is the basic idea. When you have two connected wheels, thes four equations will work for you. So let's do a quick example here. I got two years of radius 12 and three. So let's draw this. Here are one equals to the little one than a bigger one. R two equals three and they are free. Um, they're free to rotate about a fixed axes. Okay, now, little tricky here. Remember, I talk about fixed axes and free access. Um, eso. Even though I use the word free here, they're free to rotate about fixed axes. So these are fixed. Okay, which is what we've been talking about. Which means the velocity of the center of mass zero. What it also means is that our one omega one equals R two omega two and all the variations. Okay. All right. So it says, you know, when you give this smaller pulley, um, pulley or gears, let's just be right. This these were gears, not pulleys. When you give the smaller years 40 radiance per second doesn't tell me which way. So I'm just gonna spend it this way. I'm gonna say that w here is 40 radiance per second. All right. I want to know what is the angular speed. So this follows this way, and then it spins this way. So if this is W one, what is w two? That the larger one will have very straightforward thes air two connected cylinder. So all I gotta do is write the equation for them Are one omega one equals R two Omega two. We're looking for Omega Two. So Mega, too is our one omega one over our to just solve the equation. This is to over three, and Omega one is 40. So the answer is going to be 26.7 radiance per second. All right, um, that's it for this one. Very straightforward. Let's check out the next example.

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example

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All right. So here we have two police with radio I 20.3 and 0.4. So notice that this one's a little bit smaller. So are one is points three and r two is 20.4 attached the shone. A light cable runs through the edge of both police. Light means that cable has no mass runs through the edge of both police. Um, the equation for connected police. Um, when they're not, when they're fixed in place is that our one omega one are two omega, too. So we're supposed to use our which is the distance to center. Now, when it says that the cable runs through the edge of both pulleys, the word edge here tells us that the distance of the center in this case happens to be big are the radius, which that's what's gonna be most of the time. Okay, so if you're not sure, you can, um, pretty safely guessed that that's what it is, But the problem should tell you. Okay, so that means I'm gonna have big are one omega one big are to Omega, too. Um, it says you pull down on the other end, causing the pulley the police to spin. So if you're gonna pull down this way, this guy is gonna spin with Omega one. And this guy's going to spin with Omega too. And then it says when the cable has a speed of five, what is the angular speed of each? So in this cable has a V equals five. What is Omega one and what is Omega too? Okay. And what I want to remind you is that the velocity here is the same as the velocity here, which is the same as the velocity here to the same velocity at any point here. So we can write that the cable is V T one VT two. So V cable, which is five, is what equals R one omega one and equals R two Omega, too. Okay. And that's what we're going to use to solve this question. So if I want to know what is Omega one, I can look into this part of the equation right here. Okay, so to solve for Omega one, I'm gonna say five equals R one omega one. So Mega one is five. Divided by points three and five, divided by 50.3 is 16.7 radiance per second. And to find a mega chew. Same thing five equals R two omega, too. So Omega two is five, divided by 50.4, which is 12 points, five radiance per second. Okay, so that's it for W W two w one w two. Um, the key point that I wanna highlight here. The key point I wanna highlight here is that not only are these, um, these two velocity is the same at the edge, which allows us to write that are one equals R two, but also that the equal the velocity of the cable that pose them. That's what's special about this problem. It's this blue piece right here that equals the velocity of the cable as well. Okay, so please remember that just in case you see something like it. All right, so that's it. Let me know if you have any questions.

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