Hey, guys. So here's another classic rotation question that we're going to use conservation of energy to solve. And it's a yo yo question. Right? So we are. We have a simple 100 g, Yo yo, that we're going to release from rest now. Simple just means that you're gonna be able to make some assumptions to simplify the yo yo yo. Yos are actually more complicated than how we're going to solve them. But here we're going to simplify it. That's what simple means. It just means, you know, go nuts with simplifications. Alright, so mass equals 0.1 kg. It starts from rest Venus and go zero. It falls and rolls. Yo, Yo's do that right? So they're falling and rolling on the way down. Um, unwinding the light string around. It's cylindrical chef. So as it falls, it unwinds. A light string. Yo yo has a string around it. Light string means the mass of the string is negligible around it's cylindrical shaft. And that's because the yo yo has, um, I also has a thing in the middle. The cable. The string is here, but then the oil usually has sort of an outer casing like that, right? Three idea is that what matters is this inner radius, not the outer radius. The radius just covers the outside. So effectively, we're gonna just worry about this and say that I owe you. Looks like this. Okay, let's actually put the little string here. Now, if the string is here and you release a yo yo, it's going to fall and it's going to roll like this. So if the string is on this side, it's going to go like this. This is the velocity of the center of mass, and it's also going to spend with the Omega. Okay, So the radius of the inner, which is what matters, is two centimeters. So 0.2 m. It's as if the union can be modeled after after a solid disc. In other words, treat this thing here as a solid disc, right, which is what we're gonna dio. Um, in other words, I will be half m r squared because that's the eye for a solid disk. I want to know what is its linear speed after drops 50 centimeters. So the drops 50 centimeters. Um so I can think of this as the heights Initial, right. This is 0.5 m. I can think of This is the height initial was 0.5 m and the heights final since it falls, um, being zero. Okay, Even if the floor is farther below, you can just move the floor up and say, I'm gonna call this 00.0. Therefore, this point is 0.0.5. Okay, that's how you're supposed to do it. All right? I want to know the final speed. It's released from rest. So the initial velocity zero, I want to know what is the final speed. That's part game for part B. I wanna know what's the final angular speed. So what is the linear speed and what is the final angular speed? So what is Omega Final is what Part B is asking. We're gonna use conservation of energy to solve this because I have a changing heights change in velocity. But before I do that, I want to remind you that whenever a rope pose on a cylinder on a disk, um, or in a police, something like that V rope equals are omega disk, where R is the distance between the center and where the rope pulls in this case, the rope is pulling all the way at the edge of the of this inner cylindrical shaft room. There's another one here, but we're ignoring that, Um, the the rope the string is pulling at the edge of this cylindrical inner political shots or is gonna do this The radius, the distance is the radius because it's at the edge. So you can say that the equals R Omega. Okay, so that's gonna be an equation that we're going to be able to use. We're going to have to use. All right, So let's go up here and we're gonna write Kinetic Initial plus potential initial plus work, Non conservative equals Kinetic Final, plus potential final. Okay. Is there a Connecticut issue? No. There's no kinetic in issue, Andi. That's because it's at rest. Initially, there is a potential energy because you have some heights. Um, there's no work non conservative. You're not doing anything. Even though you're holding the string, you're not actually giving energy to the system, right? So your work, you don't do any work, there's no friction. There is Connecticut, the end, and there is no potential at the end because we're moving the ground up, so to speak. And that's the lowest point. And at the lowest point of emotion, your potential energy is zero. So, basically, you have all your gravitational energy goes into potential audio gravitational potential goes into kinetic. So let's expand this MGH initial equals. Now here, kinetic final, you have to figure out, Is it linear or is it rotational? And in this particular problem, it's both linear and rotational. That same object has linear and rotational energy. So what we're gonna do is we're gonna do half MV squared Final plus half I Omega Square final. Now, the next step is to expand I in Omega, right or rewrite Omega. I should say so. This is gonna be m g h. I equals half MV Final squared plus half, and you're going to see how this cancer is really nice. And the answer is gonna come out really simple. And I is gonna be half m r squared. And if you hold off on the temptation to just plug in numbers, this is actually gonna become pretty neat. Pretty simple once you see it. So if remember when you have V N w when you have V and w What you wanna do is rewrite W in terms of V so that instead of having a d n a w have a V and V, that's fewer variables. That's better. So from here, I can write the W Omega must be V over R. So I'm gonna plug V over R over here. And when you do this, notice that the ours will cancel. Right? The artist almost always canceled. Um, pretty much always canceled. But you gotta be careful about that. Don't get trigger happy and start canceling stuff. Notice also that the masses cancel. Right, Because I have a single objects. So all of these ems air referring to the cylinder, uh, to the cylindrical shaft here to the yoyo itself. Um, and that mass shows up in all three terms so I can cancel the masses. So the EMS canceled in the arse, canceled. Alright, so let's let's clean this up a little bit. I'm gonna multiply both sides by four because I want to get rid of this quarter right here. Okay, so I'm gonna get four g h initial. This half becomes a to V final, and then this quarter becomes one So this whole thing is gone, and the only thing that's left out of this whole thing here is the V. So it's V final squared. This is obviously three V final squared. And if we're solving for V, he's gotta move the three over and take the square root. Okay, so the final is the square root of 4/3 G h initial, and this is pretty cool. Borderline cool. Um, in that notice that it doesn't depend. I'm gonna write this here because it's really important. Actually, it doesn't depend on M or R. So the final velocity of a yo yo has nothing to do with its radius has nothing to do with its mass. It on Lee depends on obviously the gravity. It depends on the initial height. The high you are, the faster the more you drop, the faster you will be. And this 4/3 here comes from your moments of inertia from your shape. So if you have a different shape, this will have a different fraction here. Okay, So it could have been I don't I'm gonna make something up. It could have been 5/2. If you have a different shape. Okay, but so that's one point. A second point that I want to make is that in these rotation problems, the form of the answer is very similar to their linear counterparts. What the hell did I just say? Imagine if you drop a block and after the block falls the height of H, we can calculate using motion or energy. But you may remember this that the velocity at the bottom is the square root of two GH. You might remember this, right? This is for linear motion. Okay, this is like a separate thing. Um, now, look at how similar these two guys look. This is very similar to this. The only difference is that this has a two and this is a 4/3. So that's what I mean by the form, the shape the equations look similar, um, to what you would have expected in linear motion. The difference is that the coefficient is difference. Another thing is that the coefficient is less so in rotation, same form the equation. The answer will look similar. Two linear. It has a different coefficient, and the coefficient is not just different, but it's a lower coefficient. Quite efficient is just a number in front of something else. Okay, 4/3 is 1.33333 which is lower than two. What this means is that if you drop a block and a yo yo, the block will fall faster than the yo yo. And that's because the yo yo has two types of energies. The block on Lee has linear the yo yo is converting its potential energy into both linear kinetic energy, which makes it fall fast and rotational kinetic energy, which makes it spin. So because it's spending energy into spinning, it actually falls slower. In theory, if this yo yo were to spin at this like an infinity speed, it would actually never fall because we'll be busy spending all of its energy in rotation. So hopefully that makes sense. The reason why I want to point this out is because you can use this as a way to verify if you're correct. For example, this fraction could never be greater than two. And if you ever do a question like this, um, where you end up with a number that's greater than the coefficient and linear version. If you happen to the middle layer version, then you know that something's wrong. If I gave you, for example, a question where instead of the yo yo being a solid cylinder but instead we had a hollow cylinder, the solution will be exactly the same. The only difference is that this I over here, this number in front of the I would be different. And that difference would sort of work its way through the problem. That's this half right here. This half is this right here, right? And you end up multiplying, blah, blah, blah. So you end up with a fraction that's different. Four or three would be different, but whatever that number is, it would be less than two. Okay, so just trying to give you sort of an intuition, intuition in terms of how linear and rotational problems that look very similar also have very similar answers. Just slightly different Cool. That's it for this one. I mean, if you have any questions, let's get going