 Okay, so look at this. It's a yo-yo originally. It was a weapon really Let me just take hold the string and let go Okay, and 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 not it's tight So at the bottom it would have to it would have to stop okay when it got to the string But let's say before it gets so quiet to the bottom Let's see what we can find out in particular maybe the tension in the string as it falls a distance H Okay, so Let's draw a picture. Here's my yo-yo, and then I'm gonna let it roll down a distance Okay, and let's just say we know what's the tension in the string. Okay? 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 we do that we look at all the forces acting on that point as Though they were acting on the center of math 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. It doesn't even spin it can 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. Let's draw a free body diagram. Here's my gravitational force Mg and then I have 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 it wouldn't the center mass wouldn't change its motion And it does speed up as it falls So Mg does work right and it moves the distance h and it's in the same direction. It's 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 Okay, so and I'm just I'm not moving in the top part. It's 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 one half in the center mass squared and rotational is One half I omega squared where I is a moment of inertia omega is the angular velocity about the center mass Okay, so if I start let's call this point one and this point two then I could say work goes Mgh H Changing translational kinetics. It's gonna be final minus initial. So it's gonna be k2 minus initial, which is zero Plus the final rotational Let me just go ahead and put in the values for these one half Mv2 squared minus zero plus one half I Omega two squared minus zero that's my change Okay, now I know some things about discs first. I know that for a disc I equals one half MR squared Where in this the mass of the disc and R is the radius of the disc 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 it's going to be R Omega it has to be true if you roll something and it doesn't slide then there's relationship between the center mass and anger speed Just like that. Okay, so if I put those in I down here I get work is Mgh One half v2 squared and then this term is going to be plus one half times one half MR squared times Omega two is gonna be v2 over R. So it's gonna be v2 squared over R squared So these cancel and I get one fourth m v squared plus one half I get mgh equals three fourths in v2 squared And the masses cancel So I can sell for the velocity at the bottom and I get v2 equals Square root of four 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 square root of 2gh So this can be faster or slower than that two is greater than four thirds So this is gonna be slower which makes sense the strings doing something. It's making them not go so fast Okay, so let me put this up here because it's important v2 equals the square root of four thirds gh YouTube gives me a 15 minute window So I have to make this less than 15 minutes and I can't do that Hopefully I have no idea how long this is taking right now. Okay, so now go on to the to 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 doing work. I have The work done by gravity is going to be mgh plus the work done by tension minus th Because the tension is pulling up So it's going so the angle between the tension the displacements 180 degrees so that's going to give me a negative work And this is going to be K2 minus the initial zero Okay, well you see where I'm going here. I already know the final kinetic energy I it's the same as what it was before so I have mgh minus th Equals the final kinetic energy of it was four-thirds m Let's see wait. It was four-thirds. Let's see Okay, let me write it down k2 equals one half m v2 squared. So it's gonna be one half m That's where it's gonna be four-thirds gh That's doesn't seem right, but I'll continue anyway. So it's going to be Two-thirds mgh two-thirds mgh So now I want to solve for t. I'm going to add th to both sides And then I'm going to get mgh Equals two-thirds mgh plus th subtract two-thirds mgh from both sides and I get mgh minus two-thirds mgh is one-third mgh and that's going to be equal to th divided by the sides by h t equals one-third mgh And that might actually be wrong. I think I might have missed a half 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 would 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 would get most of the credit card 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