let's do another example. I'm just going to we're not going to solve it completely. Like the last one, we kind of kind of stepped through some highlighted components. And now, instead of rolling on the surface were dropping and I'm just going to illustrate the same thing happens because now we have a gravity force. How do we apply this? So here are coordinates are why the height, I do have a gravity force acting on the ring. And there's a theta. This ring of pebbles has to have constant distances between pebbles. So it's a highly redundant set, a minimal set would just be Y and theta, right? So if you look at the generalized forces of each particle, same thing as before. We have a position, our velocities subject to constant radius, then the y coordinates. All the force is dotted with these partials on y. And the only force acting on each particle is going to be gravity, but it's the M of each particle, M I times G in the minus anti direction dotted with this partial. And the partial of this with respect to y dot is and 2 so you just get minus the summation of all the mass particles, times G factors on which is just a total mass times G, right? So that's the generalized force acting on the Y coordinate is essentially another version of super particle theory and we're rediscovering in this case. Whereas Q theta in this case we have to take the same forces which is just gravity dotted with the partial velocities here and that's going to be re theta eyes which we have here. An end to an E theta I are depends on the E theta. Some might be lined up, some might be opposite, some might be orthogonal. So it depends on the system. I can factor out G times M2 outside. I have m inside r inside and the theta e. So I'm taking this term dotted with this term and the dot product is not necessarily zero, right? Because as I'm rotating those can be anything. The answer is still going to be zero. And the reason is this property, what does this property actually mean? This is the math, what do we call this in words? Yes, center of mass. Remember we had the center of mass property. That was the body integral of little R times DM for every infinite testable particle. If you do a discrete version, it's Mi times little ri and your little ri are going to be R times ERI times the mass. And this summation has to be zero. Therefore, also its derivative would have to be zero, which is where the derivative of ERS gives you the thetas times of theta dots, but that's a constant because of the constraints we have. So or at least, yeah, there's no E theta I individually, they're all the same. I can factor them out. Which gives from this, you can argue this term has to be zero because of center of mass property that we have a ring. Yes sir. >> Why, no? >> So why did we solve accused of y here. >> Yeah, turn before, [INAUDIBLE]. >> I could have but I'm using the classic Lagrange in just with kinetic energy because I have a have a force acting on it. It's an easy one to use, I could just call it F as well. But yeah so but also the nice thing is it's a force acting on every particle made my algebra little bit easier. I could have used the Lagrange in if I wanted to. So I need to the potential function if you did that is here. And so with redundant coordinates we would argue earlier. Basically, we're going where you're saying this QI would have been nothing. If I had the luxury engine you would have the negative partial of the respect to why are kinetic energy does not depend on why itself? It only depends on why not. So if you look at, yeah. So the generalized force has to be written this way. It gives you the same answer which is good. And for the redundant coordinates of a potential, this gives you zero because it's the only forces acting on it is conservative. It's the same arguments, basically. So you could have done that if you wanted to and with the minimal coordinate sets if we did that then we would only have theta. We've already applied that all the theta dots have to be the same. There's one common thing that we can pick out of it that makes everything else unique and off we go so you can get the same results. Okay, any questions here, just wanted to highlight the falling with gravity, you can do this, this infinite particle stuff. And things sum up again to the same force as what we had earlier versus if you pick the minimal set as we had earlier formulation. Let's see, we did do I did have my minimal here and my redundant said you should get the same answer much quicker. So if you can always pick a minimal set especially if it's rigid. Any questions on that one. Hopefully at this point somewhat redundant example, but we get there.