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Conservation of Energy with Rotation

Patrick Ford
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Hey, guys. So in this video, we're going to start solving rotation problems with conservation of energy. Let's check it out. So you may remember that when you have a motion problem between two points, meaning the object starts here, ends up over here somewhere where either the speed, the the height H or the spring compression ex changes any combination of those three guys changes we can use most of the time a the conservation of energy equation to solve these problems. So we're going to do that now to rotation questions. The only difference is that in rotation, your kinetic energy can be not only linear but also rotational. So that's the new thing that you could be spinning and you could actually be. It could also be both right. It could be that our total kinetic energy is linear plus rotational. So we're going to use the conservation of energy equation, which is K initial plus your initial plus work. Non conservative equals K final Plus you find a I want to remind you that work non conservative is the work done by you by some external force, plus the work done by friction. If you have some. Now, when you do this, remember, you write the energy equation and then you start expanding the equation. What I mean by expanding is you replace K with what it is, and K used to be simply half MV squared. But now it could be that you have both of them, right? It could be, Let's say that. Or let's say instead of have from the square, the object is just spinning. So you're gonna write I Omega Square? Okay, the key thing to remember and you'll do this for the rest of them. The key thing to remember the most important thing in these questions to remember is that you will rewrite the end omega in terms of each other. What do I mean by that? What I mean is that when you expand the entire equation, you might end up with one V and one W or to V s and one w whatever. If you have a Vienna W, that's two variables. You're going to change one into the other so that you end up with just one variable. For example, most of the time V nwr linked by this or sometimes they are linked by this right? Sometimes they're not linked at all. But most of the time they're connected by either one of these two equations, which means I'm What I'm gonna do is rewrite W as V over r and whatever I see a w I'm gonna replace it with the view over are So instead of having v n w. I have V and V, and that means that instead of having two variables, I have just one, and it's easier to solve the problems. That's the key thing to remember is rewrite one into the other. Let's do an example. All right, so here we have a solid disc solid disc. Let's stop There means that the moment of inertia we're gonna uses the same of a solid desk, which is the same as a solid cylinder. And it's gonna be half M r squared. And it says it's free to rotate about our fixed perpendicular access to its center. Lots of words. Let's analyze with staying here free to rotate, it just means that you could rotate right like you can actually spin. Some things can be spun around others, Kintz Um, but even though it says that it's free to rotate. It's around a fixed axis. Okay, remember, it's the difference between a roll of toilet paper that is fixed on the wall, and it's free to rotate around the fixed axes versus a free roll of toilet paper that can roll around the floor. Okay, here were fixed in place. So we're gonna say that it spins like this, so it has no view, right? Like the actual disc has no velocity V because it's not moving sideways. The center of mass doesn't change position. Um, but it does spin. Okay? Actually, we don't know which way it spins. Let's leave it alone for now. But I'm just gonna write that Omega is not zero because it's going to spin now. What else? It says that the access is through its center. So it's spinning around the center like this. Okay. And it's saying that it's perpendicular perpendicular means that it makes a 90 degree angle. Okay, Perpendicular means it makes a 90 degree angle with the objects. So I got a little disc here. Um, this sort of looks like this, and I want to show you real quick what perpendicular means. So imagine the this is the face of the disc perpendicular means 90 degrees to the face of the disc. Which means I'm gonna stick my fingers in here, and it looks like this Cool. So perpendicular. Looks like this. This is the axis of rotation, which means the disk spends like this. Okay, Hope that makes sense. So you're gonna see this all the time. Perpendicular is just going to mean 90 degrees with the face. Just means that the disk spins like this. Which is how you would imagine the disk spins. A desk isn't gonna do this right Or some weird stuff. So just spins around its center like that. Cool. So this expends like that, This is sort of a top view off the disc. Um Okay, so the disk has mass five and he goes five radius six, and it is initially at rest. So omega initial is zero because it starts arrest, and then you have a light along the light cable that's wrapped several times around the cylinder. The disk. Okay, light means that the cable has no mass, so I'm gonna right here. Mass of the cable is zero. And you wrap it up a bunch of times. You got a lot of problems like this, and basically what we're doing is we're saying we're setting it up to say there's all this rope around this thing So when I pull on it, it's going to unwind. So it says, here you pull on the cable with a constant 10 Newtons. So let's draw cable right here. And then it says Force of 10 in such a way that the cable unwinds horizontally at the top of the disc, the cable unwind horizontal at the top of the disk is exactly what I just drew here, right? So the cable is unwinding horizontally. It doesn't say if it's the left to the right. I just threw it to the right. Um, now the word unwind here is repeated. Sorry about that. And then without slipping, this is key because it's unwinding without slipping. I can say that the velocity of the rope equals little Are omega okay? And this thing, this rope will have a velocity V this thing will have on omega of the disc. So V rope is our omega of the disc, where R is the point where the rope touches the disk so it's where its distance between center. The axis of it, I should say axis of rotation distance between the axis of rotation and the point where the rope touches the disk or where the rope pulls on the disk. Okay, let me show you what quick example here just to be very clear here. Let's say this disc has a radius of 10. So the distance all the way to the end here is 10. But the But let's say I'm pulling right here at a distance. Five. So what I using this equation the rope equals are Omega. I would use the five. Okay, just to be clear, when you write these equations here, we're gonna do this a bunch of times. The are is not the radius. That's why it's a little are not a big are. It's how far from the center the rope pulls. In this particular case particular case, the rope is pulling at the edge, so your little are happens to be the radius, and by the way, that's how it almost always happens. But you could have it. You could have a different situation like this, so you should be ready just in case. Um are is typically the radius, but it doesn't have to be. The radius in this case rate is is 6 m. Okay, so without slipping is what tells us that we can use this equation right here. Okay. Without slipping is what tells us that we can use this equation right there. Alright, ignore any frictional forces. If you don't see that, you can just assume that you're supposed to ignore friction unless it tells you what the friction is. And then we're gonna use conservation of energy to find the angular speed of the pulley. So I want to know what is Omega final? What is Omega final after you've pulled the rope for 8 m? So you're gonna pull the rope with a force of 10 with for a distance delta X of 8 m pictures a little tight here, but basically would look like this until 8 m of rope unwinds from the disc. Alright, So let's use conservation of energy and we're looking for W final. So kinetic initial potential initial plus work, non conservative equals kinetic final plus potential final. Okay, the is their initial kinetic energy here. There is no kinetic energy the beginning because the disc isn't spinning, there's no linear energy and there is no rotational energy. The disk doesn't move sideways in the beginning, doesn't spend. Um, there's no potential energy. And that's because the remember potential energy is relative to a change. It depends on your change in height and the height doesn't change. Delta H zero. The disc keeps its same height, so you can just cancel out these two guys, right? You do have a potential energy because you are both the floor, but the two potential energies are the same. Okay, now, what about work? Non conservative work, non conservative? Is the work done by U plus the work done by friction. There is no friction here. Told us to ignore frictional effects. But you are pulling on this thing right? And the work done by you is the work done by a force F which is F d co sign of theta. Okay, your force is 10. You do this for a distance. D of 8 m and cosign of fate. A remember is the angle between your displacement and the force. Now here you push this way and the rope moves this way So the angle here is zero. So I'm going to do the co sign of zero in the CO sign of zeros one. So I have 10 times. Eight times one 80. Okay, so this is 80 Jules. And at the end, we have kinetic final, um, kinetic final could be linear. And it could be rotational is their linear energy at the end. There isn't because the disc spends around itself, but there is rotational kinetic energy at the end. So let's expand that 80 equals Half I Omega squared will make a final. This is exactly what we're looking for right here. Omega final. Okay, so let's expand the I so it's gonna be half gonna put the I in here. I is half m r squared because we No, it's a solid disk. So I'm gonna put half the mass is of five, and the radius is a six. Let me just put a six there. Cool. What we can do is we can move everything to the other side and solve for omega. So here we have 80 equals, um, things whole thing here. It gives us a 45 Omega final squared, so we'll make a final is 80 divided by 45 and then you take the square double sides. You get this. And if you solve this, get out of the way. You get 1.33 radiance per second. Okay? And that's the final answer. Just took a little while. But it's because I wanted to introduce some of the terminology for these kinds of questions. Some of the language you're gonna see now. In the beginning, I mentioned how if you have a V, um, if let me write this here, if you have a V and A W, you're going to rewrite W in terms of V, right? Well, in this question, when I expanded everything, I only had a w. So I didn't have to change one into the other. And I was looking for W. So I just solved for it. Okay, so you do that if you have the two variables so that you can simplify, that's it for this one. Let's keep going. Let me know if you have any questions.