Okay, so look at this. So yo-yo. Originally it was a weapon really. Let me just take a hold of string and let go. Okay, so it rolls down. But let's assume that we have a yo-yo that's a disc. Okay, and the strings wrap around the outside and it's tight so at the bottom they would have to stop. Okay, when it got to this string. But let's say before it gets to quite the bottom, let's see what we can find out in particular maybe the tension in this string as it falls a distance H. Okay, so let's draw a picture. Here's my yo-yo and then I'm going to let it roll down a distance H. Okay, and let's just say what's the tension in this string. Okay, clearly we have to use the work energy system here, work energy principle because we're dealing with distances. So what's-in this case we have two choices. We can do the point system, the point particle system, or the real. Remember in the point particle system we treat this yo-yo as just one point and then when we do that we look at all the forces acting on that point as though they were acting on the center of mass. In the real system we use real forces and their real displacements and so a point particle system can have just translational kinetic energy. If it's a point that doesn't even spin, it can't have thermal energy or anything. The real system can have rotational energy, thermal energy, all sorts of energy. In this case the real system can have both translational kinetic energy and rotational kinetic energy. So I'm going to start with the real system first. So in the real system the system is my yo-yo. Yo-yo. So what work is done on the yo-yo? The work is going to be done, a straw free by a diagram. Here's my gravitational force, Mg, and then I have a tension right there. Do you think those two forces are the same? They're not. They can't be. Just think back to the point system. If I have a tension and gravity is the same then the center of mass wouldn't change its motion and it does speed up with the falls. So Mg does work and it moves the distance h and it's in the same direction as moving. So the work done by gravity is Mgh. What's the work done by the tension? Well the tension doesn't move so it doesn't do any work on the real system. So I'm not moving the top part and just dropping from rest. Okay so that's it. Now what kind of energy can I have? I can have translational kinetic and I can have rotational kinetic energy where kinetic translational is 1 half m v center mass squared and rotational is 1 half i omega squared or i as a moment of inertia. Omega is the angular velocity about the center mass. Okay so if I start, let's call this point 1 and this point 2 then I could say work goes Mgh at h. Changing translational kinetics is going to be final minus initial so it's going to be k2 minus initial which is zero plus the final rotational. Let me just go ahead and put in the values for these. 1 half m v2 squared minus zero plus 1 half i omega 2 squared minus zero. That's my change. Okay now I know some things about disks. First I know that for a disk i equals 1 half m r squared where if there's a mass of the disk and there's the radius of the disk. Also I know that if this string is stationary and this is moving and kind of unwinding at the same time without slipping then v to or v at any point is going to be r omega. But that has to be true if you roll something and it doesn't slide then there's relationship between the center mass and the angular speed just like that. Okay so if I put those in I down here I get work is mgh 1 half m v2 squared and then this term is going to be plus 1 half times 1 half m r squared times omega 2 is going to be v2 over r so it's going to be v2 squared over r squared. So these cancel and I get 1 fourth m v squared plus 1 half I get mgh equals 3 fourths m v2 squared and the mass is canceled. So I can solve for the velocity at the bottom and I get v2 equals m v squared root of 4 thirds g h. Now how does that compare to just dropping an object? If I just drop an object it wouldn't have this term and so you'd get if you solve this you'd get squared of 2gh so this is going to be faster or slower than that. 2 is greater than 4 thirds so this is going to be slower which makes sense. The strings are doing something if making them not go so fast. Okay so let me put this up here because it's important. v2 equals square root of 4 thirds g h. You tube gives me a 15 minute window so I have to make this less than 15 minutes and I can do that. Hopefully I have no idea how long this is taking right now. Okay so now go into the point particle system. So now work equals change in kinetic and what does work on this? Well in this case I have two things to work. I have the work done by gravity is going to be mgh plus the work done by tension minus t h because the tension is pulling up so it's going to so the angle between the tension and the displacements 180 degrees so that's going to give me a negative work and this is going to be k2 minus the initial of the c-veroch. Okay well you see where I'm going here. I already know the final kinetic energy. It's the same as what it was before. So I have mgh minus t h equals the final kinetic energy of it was 4 thirds m. Let's see wait. It was 4 third. Let's see. Let me try it now. k2 equals 1 half m v2 squared. So it's going to be 1 half m. That squared is going to be 4 thirds gh. That doesn't seem right but I'll continue anyway. So it's going to be 2 thirds mgh. 2 thirds mgh. So now I want to solve for t. I'm going to add t h to both sides and then I'm going to get mgh equals 2 thirds mgh plus t h. So track 2 thirds mgh from both sides and I get mgh minus 2 thirds mgh is 1 third mgh and that's going to be equal to t h. I'm going to divide it by h t equals 1 third mgh. And that might actually be wrong. I think I might have missed a half a sign somewhere but I'm not going to remake the video because I'm that lazy. But you see the idea. Maybe you can find the error if there is an error and fix it and that will be good for you. But does this make sense? Well this can't be bigger than the gravitational force or it would speed up going up and it doesn't do that. It is small and it has the right units. So I think it's at least close. If this was a test question, even if I got that wrong, I'd get most of the credit correct. But also you see that it doesn't depend on how far it's fallen. So it's a constant tension. Assuming the mass of the string is negligible. Okay cool.