In this problem. A rope is wrapped around a cylinder with the radius R and a mass Big M. The rope has a mass of little M, and the rope and the cylinder are rotating about the cylinder central axis at an, um, at an angular frequency of omega, Not we are asked to find the final angular frequency after one rotation of the cylinder where the rope is hanging vertically downward and traveling downward at some speed V. We need to figure out what the final angular frequency, which I will call Omega Final is. And we also need to figure out what V is in this case. So, um, this is a conservation of energy problem. Um, as we go from our initial set up to our final set up, there is a loss of gravitational energy associate ID with the changing position of the center of mass of the rope. And, um, because of that, there is a gain in the total kinetic energy off the system. So what we're gonna do is set up in equation that will equate the energy of our initial set up to the energy of our final set up, and then we can use that to solve for Omega Final. Alright, So first, our initial set up the energy is entirely given by, um, the rotational energy of our rope and of our cylinder. Um, there's some gravitational energy associated with the position here, but because we only care about difference of gravitational energy, I'm going to include a negative potential energy in this term here, and you'll see that later I'll show you. But first, the total rotational energy of our initial set up is given by one half times I omega squared for the rope and plus one half I omega square for the cylinder. All right, we can ignore the thickness of the rope. We're told that we can ignore the thickness of the rope and so we can approximate that the mass of the rope is entirely concentrated along the surface of the cylinder. And so the and I'm sorry. The moment of inertia of our rope is then given by a little m r squared. All right. And the initial angular frequency is Omega. Not then for our cylinder. The moment of inertia is given by one half. Begin R squared and then omega not squared All right, So this is the total rotational energy of our initial system. All right, so now let's figure out the energy of our final system. In this case, we have some rotational kinetic energy associated with the cylinder. So I omega squared plus one half m v squared the kinetic energy of the rope with mass m minus. And this one I do minus. Because in this situation are gravitational energy is less than in our previous than in our initial, um, in our initial set up. So the gravitational energy here is lost. We have a the mass of the rope. The center of mass of the rope descends by a distance of pie Are we know that because the whole distance of the rope is two times pi r. It's the circumference of the cylinder. And so the center of the mass, which is halfway along the rope, descends at a height, descends a height of pi r. And that means the associate ID change in potential energy is mg pyre. All right, so now we replace All right, so this I just replaced I with the known moment of inertia of the cylinder for V v has a given relationship between between itself and are in Omega Final. We know that this relationship exists for Omega Final and our because that's the relationship for tan, gentle, tan, gentle velocity of the of the surface of the cylinder, which is going to be the same as the velocity of the rope. In this case, because the rope moves around the cylinder without slipping. So we can say that one half MV squared is one half m r squared omega f squared and then we subtract mg pi r Yeah, yeah. Now we set these equations equal to each other and we can solve. So first, my first step is to rewrite both of these equations. Oh, and by the way, you'll see that now we have one equation and our Onley unknown value in this equation is Omega final. So we can use this equation to to solve for mega Finally just need to do a lot of complicated algebra two isolated on one side and express it in terms of, um, other quantities. All right, so this whole complicated equation we just need to solve this in terms of Omega. Not so in my first step here. I'm going to do two things. I'm going to multiply each term Times four. And I'm also going to divide each term by R squared. All right, so I'm multiplying each term by four divided by r squared. And this term becomes to em Omega squared. Plus this term and Omega squared equals and I'll make a final square toe Plus to em make a final squared minus four m g pie over are all right. So to repeat myself in this first step, I divided each term by r squared multiplied each term by four. In this next step, I'm going Thio I'm going to combine like terms and on the left side and on the right side. Alright. And our next step is to divide every term of this equation by to m plus m so that becomes omega. Not squared equals omega Final squared minus four mg pie over our times. Too little implicit. Big in perfect. All right, so in this last in this last bit, I'm going to do two steps. First I'm going to add this term to both sides of the equation, and then I'm going to take the square root of both sides toe get omega final by itself. And what we'll see is that Omega final equals the square root of omega not squared. Plus for mg pie over are to and plus yeah, all right. And this is our answer. Wait. I have toe put a box around that. So before we move forward, um, one good thing to check at this point is toe make sure that the units are going to be appropriate. We see, on the inside we have, uh, a frequency squared here and then on the top, it's mass times acceleration on the bottom. We have mass and a distance there. So this whole term here will also come out to give us, uh, units of frequency squared and we take the square root of that. We get units of frequency, and so our units do match up, and that's good. Um, the last step is to figure out what V is. And I already said that V is given by this relationship here, so we can simply take what we just found out. Multiply times are so the equals, our times, omega final, and we just show that omega final is square root of Omega, not squared. Plus or mg pie over are to m plus and all. Right. And there you have it. That's the whole problem. Good job, everyone.