Now I want to go through an example. I'll start this; I'm not sure I'll finish this in one go but it's a wheel, so we're not looking at a particle. We said Lagrange's equation just needs the energy in terms of a finite set of states and a wheel will have some coordinates that track the center of mass of the wheel and it will have some attitude. This wheel is moving in a plane so I'll only have one translational coordinate and I will have one rotational coordinate. I'm ignoring gravity again because that one it's an extra term and the normal force is mg big surprise. We know that's going to happen but I don't have to include it here. If this thing is moving on this surface, the other force we're going to have is this friction force. I'm just curious upfront what you guys think. Does this friction force do work or does it not do work? It depends if it's a no-slip condition or not. Good. What Julian was saying is it depends if we have a no-slip condition or not. Let's assume we have a no-slip condition. Then it doesn't do work. He's saying it doesn't do work. Ricardo, you agree? I agree. Why, can you give me a mathematical reasoning here? Because the velocity of the contact point is always zero. The answer he said was the velocity at the contact point here is zero. Yes, this is an inertially fixed floor. The wheel is rolling on it even if the wheel is accelerating what happens is the point where the wheel is touching the surface that velocity has to be zero because of the no-slip condition. Therefore the point where F_f is acting is this infinitesimal mass element right that touches the surface, and six seconds later it might be a different mass element that has rotated down. But at any instant of time, you do a snapshot and you have to do the summation of all the infinitesimal particles, there's this one that has a force acting on it, and its inertial velocity R dot is going to be zero, so what's the partial of zero with respect to any generalized coordinate rates? That was the partial velocities we compute, the partial of zero will have to be zero. We'll see this in a math where this comes out why this all comes together but that's good thinking. If this point here is not moving, if it did have slippage then all of a sudden this thing has a force acting on it and I'm kind of spinning my tires in place and that will do work on the system and that's different. Because it's in essence if you're spinning your tires you don't have a no-slip condition that means you've lost that holonomic constraint and therefore there is work being done on the system. If you have a holonomic constraint the constraint that enforces that which in this case is the friction coefficient the friction force, it's going to be non-working. But it is a continuous system so I'm going to tediously derive this. Please I ask you, I plead with you don't ever go to your employer and solve a rolling disk like this. They will just fire you on the spot and go why are you wasting the time. I'm doing this as an educational example that we can use our principals that we derived for particles and essentially what you'll find is the summation of all these terms and you make it infinite particles, and with all the constraints this becomes a rigid body, we get back exactly what we would expect. Because it's rigid we really only have a translational degree of freedom and one rotational degree of freedom. But I'm going to break it down into N number of particles. I will have n degrees of freedom, so there's 16 million particles to define this ring and I have a translational coordinate that defines the center of mass of this ring of particles. These particles aren't just flying all over the place, they are constraint because they are a ring as it rolls so if you think of these things rolling all along these particles must maintain constant relative positions. How do we define that? The way I'm going to have it this ring will have a fixed r I'm just going to impose that, so I'm not giving each particle two degrees of freedom in the plane I'm only going to give it one just because the algebra gets a little crazy so we'll do this once with Theta. But if I would let these particles slide closer and further apart on angular sense, there's this Delta Theta. If I have this broken up into six different particles there would be six times, what would that be I need seven no, 1,2,3,4,5,6 so that would be 60 degrees between each particle I believe maybe I'm off by one. That's what your Delta Theta is that I'm using six particles they have to be 60 degrees apart and it has to be a constant particle 1 and 2 must be 60 degrees apart particle 2 and 3 must be 60 degrees apart and then in a differential way they have to add up. That's how I'm imposing these constraints. The first constraint, what is this one? X minus r Theta_1 equal to 0. That's the no-slip condition. That's the no-slip condition. That guarantees of one of the particles and I've just chosen one I could have chosen particle 50, I chose one just because it's the first one. But in an angular sense, the displacements that you have must be the radius times the translation displacement, so our x has to be r times Theta_1 that's your no-slip condition. That's one constraint, the other constraints now I'll have N minus 1 constraints, because if you have six particles you will have five interconnections at some point it starts repeating again, and these inter angles must be a constant so I would say Theta_2 minus Theta_1 must be equal to Delta Theta and you go rip through all of them and you end up with Theta_ N minus Theta _N1 would have to be equal. That's the Nth one and that all connects this stuff. Now I've solved this, so now I have all my constraints that I'm going to apply. To use constrained Lagrangian Dynamics we have to apply in Pfaffian form that's what we needed. That was this augmented cos function I needed a sensitivity, the partials, the derivatives of these holonomic constraints. Here they're holonomic constraints I want to turn it into Pfaffian form that means I need the sensitivities, what's the partial of these constraints with respect to x times x dot? Basically what happens if you take a time derivative of the first one you will have x dot minus r Theta_1 dot. In this matrix here you can see the first column because it'll be this matrix times the set of generalized coordinate rates. The first column is everything that influences x dot, the second column is everything that influences, why did I call those q's that should have been Theta_1 dot because earlier we defined q1 to be x, not Theta_2 so ignore the labeling here I should change that. But here we will just have minus r Theta_1 dot so I get a minus r that's my no-slip condition. Then all the other ones I'm going to have Theta_2 dot minus Theta_1 dot, so minus Theta_1 gives me a minus 1 here. Theta_2 is a third element that gives me a plus 1 and the Delta Theta is a constant that one derivative just goes away and you populate it, so you get this a nice diagonally dominant matrix structure that you would have. That's the A matrix that I needed. Now let's look again at this force. There's no force accelerating this ring I could have added that, I'm not looking at gravity because gravity will just be equal to the normal force just because we're holding on a flat plane sorry I ignore that one here as well, we didn't know upfront that's a non-working force. What about the friction force? This is where we ended so I want to talk about this now the friction force is only acting on the kth particle, so if you have enough little particles you can break it down. However, that wheel is rolling there is only one of the particles at a time touching the surface. The key insight so that's the only the kth one has it and at that point, your angle Theta must be 180 degrees. If Theta is180 degrees erk and e_Theta k are going to be, er is downwards e _Theta k to the left which is the minus n_1 direction. If you look at your classic inertial velocity was x dot plus r Theta dot times these two unit vectors this unit vector is going to be minus the first one, so you can write it out like this. Why does this go to zero, x dot minus r Theta dot? Non-slip condition. Non-slip condition, exactly. That's where it comes in again so because of the non-slip condition that relationship between x was equal to r Theta therefore x dot is equal to r Theta dot has to be. If it's different you're spinning in place while also moving forward and that's not what we have in this system. Therefore the velocity is zero, so now if the velocity of the point where a force is applied is zero what are all the generalized forces is going to have to be Josh? Zero. Zero, because the partial of zero with respect to any of your generalized coordinates or coordinate rates we're doing the cancellation of dot property has to be zero. I just wanted to show that explicitly help drive home that point. That's why often if you have something with no-slip condition. But the other way to look at it is the normal force, let me just close this there we go, so if you look at the normal force is acting in this direction, and then you have something here and maybe it moves but with the center of mass you do have an mg pushing down on it right and here's a normal force. What's the work does normal force is going to do if you think of work It's worked times distance and this is not traveling anything in the vertical directions. There's normal force dotted with admissible variations, the only place distinct can wobble if miserably is left and right, not up and down. A normal force will never make you lift off the surface, it'll just have you stay on the surface and that's it. Here too that will simply vanish and go so the velocity that this force acts in that direction is going to be the velocity in the vertical direction essentially is zero so therefore all those components are going to have to be orthogonal it's going to go away. The same thing happened here. The total velocity in the other case with normal force you might have velocities in the horizontal direction you just cannot have any in the vertical direction so that's why that dot-product those are all going to go to zero. Here we have velocities that are zero regardless of directions, so no matter what this F_f force does it has to go to zero, so good. We found the generalized forces on this for the x-coordinate and for all the Theta coordinates because there's multiple Theta for multiple beads on this ring all have to be zero. That's nice. Let's start to put this together. These are Lagrange's equations equal to Qx and I'm using t here, there's no conservative forces I didn't have to get the Lagrangian t minus v, I'm just using the original formulation we have the kinetic energy that was plenty. I have the kinetic energy, so if you take the partial with respect to x dot take a time derivative you get Mx double dot and the partial of kinetic energy with respect to x, x does not appear just going to vanish so I have an Mx double dot. This term is going to be this and of course, the right-hand side of this is going to be your generalized forces on the x-coordinate plus these Lagrange multipliers for m constraints and one of them was no-slip and all the other constraints were the beads have to have constant inter angular distances, times these A_ij's that's the right-hand side. If you evaluated all of these generalized forces we just argued, the only force acting on it is the friction force potentially but it's on a point that's not moving, so they were all zero We can always ignore these Q's. That leaves us just with this. Now if you look at this A_ij it is on the first column, so we have m constraints, this the number of columns that you have is really the number of generalized coordinates that we have this is the Q_1 through Q_n that we'd have. The x, the Theta_1, Theta_2, Theta_3, so the number of constraints are this way so when you're summing over all the Lagrange multiplier times the sensitivities with respect to the first generalized coordinate you're really summing down this column. That's why this right-hand side simply gives you Lambda_1. This will be one equation that we will be able to use Mx double dot has to be Lambda_1. Now we don't know Lambda_1 yet so we'll have to solve for that. Then we have all the other equations for the angles so you can go through this and this is always going to look the same. Doing this math Theta doesn't appear it's only Theta dot going through the same stuff you end up with mass of that particle 1 times r squared mass times distance squared Theta_1 double dot. The right-hand side Q is a zero. I have the Lagrange multiplier times the 2nd, 3rd, 4th, until that final column. You can see the first one we'll have a minus r and a minus 1 and that comes in this way and then afterward it's always 1 minus 1,1 minus 1 until you get to the last one in which case it's only Lambda_N is equal to this. There's the whole systems of differential-algebraic equations that we've derived these have to be true subject to all these Pfaffian constraints that we can use now too. This one was trivial we found that, and now we're going to start putting this all together. Looking at the last one we had this equation also equal to Lambda_N, so I know right away Lambda_N must be equal to this. Then if you go back one, here if I had Lambda_N, I can relate it to Lambda_N minus 1 essentially. This allows you to step backwards, it's like a Gram-Schmidt reduction, you solved for the final variable and now you have to solve for all the first variables again we're doing the same thing. This will allow us to step backwards and once I have n minus 1 I can relate those two to N minus 2 and rip through the whole thing, so that's what's happening here. This is the first one so then N minus 1 would have to be m_ N plus m _N minus 1 times this r^2 Theta double dots and you go through the series of these and in the end you're left with this equation. This was the second line there that we had. These m's we just keep adding them up so in the end you have the summation of all the m's that must be occurring and the sum of all the m's is nothing but the total mass so that's how we can replace that. Now let's see what else. There is a Theta here there's no Theta_1s, Theta_2s, Theta 3s. How did we get away with that? What do you think Ricardo? Nobody raised their hand, nobody had questions, I'm assuming you guys are all-wise and have answers for me. If you look at this line the last one there is an m_N r squared Theta double dot_ N and then on the next slide, I just say Theta double dot. What do you think Henry? You just looked like you're about to say something. Yeah I don't know mathematically but the problem is so distributed symmetrically so the angle does not [inaudible]. Well, what mathematically says that? Go back. Now if we go back. You're going to see these. These are differential-algebraic equations so we have to solve these differential equations also subject to algebraic conditions. What's going to happen when you twice differentiate these constraint conditions? They're really zero. No. Well, this side is zero yes but the left-hand side you're going to get Theta_2 double dot minus Theta_1 double dot, and what's Delta Theta double dot? It's zero. It's zero, so you basically say Theta_2 double dot has to be equal to Theta_1 double dot. Remember it's the same thing with a planar pendulum we did it and then we had this x differential equation with Lambdas and the constraint equations and the way we solved them as we twice differentiated the constraint equations plugged in that other double dots into it. You have to use differential-algebraic we have to use the constraints as well. This means from here we can argue that Theta_1 double dot has to be the same as Theta_2 double dot. But then as you go down the loop, Theta_2 double dot has to be equal to Theta_3 double dots and it goes through all of them and so, in the end, they all have to be equal that's what we used, that's what gave us the right to just drop this index because in this case it's very trivial again don't do this at a conference I'll be embarrassed if you do that. But that's how it works out so we can just say Theta. I've already used the constraints and I could have done the Pfaffian and differentiated one more time same thing, you've already had one derivative you take a second derivative you go yes they have to all be equal so I'm just calling it Theta. Yes, Henry. Why is Delta Theta Zero? Sorry. Why is Delta Theta constant goes to Zero? Because by definition Delta Theta has to be a constant that's how we set up our constraints, so that was part of the problem statement that we're looking for beads spacings that is constant on the ring. That's why the derivative Delta Theta so by definition of that one. Anyway, that's why we only have a Theta here but you can see we start summing it up going through this very sparse matrix you basically get this recursive behavior so that's nothing but this. Now we have this. We also have this other equation from the first line we had Mx double dot equal to Lambda_1 which we can use as well, and we can relate x double dot Theta double dot how do we do that? Again we use our constraint equation, so I'm going to go back to this form. Or in the Pfaffian form, it's right here. We know that x dot is equal to r Theta_1 dot so therefore x double dot must be equal to r Theta_1 double dot and Theta_1 is Theta in this case they're all the same. That's it. Again to make this last leap, we used the definition so this is r squared Theta double dot, you plug that in, so r Theta double dot you plug that in here and you get an Mr squared Theta double dot which you can bring over to the right-hand side that gives you two times Mr squared Theta double dot equal to zero. That's our final equations of motion. A cylinder rolling on a flat plane, no gravity good grief how much work, how much you put into this to come up with something that's very simple. Why is it zero here? Or another way to ask it, what would make it non-zero? How would I have to modify this to make it non-zero? You could tilt it. If we'll tilt it there basically you need something forcing it in the x-direction, so you can add a force that is pushing on the cylinder and it's going quicker and quicker because if it's accelerating then there has to be a friction force, it turns out here that friction force is zero. After all the work we did to argue that some force started times miscible variation goes to zero because the velocity where it applies it is zero, it turns out that bloody force was zero to begin with. But that's only true because there's nothing accelerating this wheel, if there were a gravity accelerating at some other force you can imagine that the floor has to grip you and so you can push off that floor and move forward, and then you'd have a non-zero term in there so you can do this if you want to in that one. But my goal here is really to show this summation does lead to something we would see and knowing that the inertia of the ring is the total mass, Mr squared, that's the classic inertia of the ring, that's simply going to be 2I Theta double dot. I know everyone in here can solve this equation much quicker with momentum or an area mechanics, than Lagrangian but this is what you get. Any questions here. It just illustrates you just need a finite set as long as your energy can be written in finite coordinates that it's rigid your energy already accounts for all the motion that somebody derived energy of a rigid body. We had an infinitesimal blob and then they'd all the summation and did the vector calculus and came up with this answer, so you're already accounted for an infinity of particles but in the end the energy dependend on a finite set of states translation and rotation. Is it critical when in that process when in that you start substituting these constraints in? Well, it depends how you set it up. If you can get rid of constraints by just doing a back substitution that's always preferred because then you have a minimal coordinate set and that's always going to make your life easier. It's just not always possible or minimal sets like quaternions avoid singularities, Euler angles would do it but now I have singularities there might be reasons to deal with it you could use MRPs but then we also have to deal with switching. But that works fine actually you could use MRPs unconstrained, derive everything, and then just do kinematics at some point you switch to the alternate set, and you don't have to do anything weird with constraints at all, so I still I like MRP's. Let's find the equations of motion as a one degree of freedom system, if we would say this was the center of mass property the kinetic energy of the center of mass plus the kinetic energy about the center of mass and that's simply Omega transpose I Omega over 2 goes the kinetic energy for a 1D problem its inertia over two angular rate squared, you remember that from undergraduate dynamics. How did we go from here well we used to no-slip condition? X dot was r Theta dot and I'm doing everything in terms of Theta so this, in the end, sums up to this or this they're equivalent. The total energy of the system because it roles and rotates is actually I times Theta dot squared not I over 2 times Theta dot squared that accounts for translation and rotation with the no-slip condition. Cool how much shear is that with a minimal parameter set? Again friction force we can argue has to be zero so that missible variations because that position is always going to be a zero but if you do this math in terms of admissible variations you still end up with this term n_1 plus e_Theta and at the point of where the force applies e_Theta is minus n_1 like before and we have zero admissible variations at that point so zero times the rest is zero. Therefore this term and you don't have to include it which is nice. Lagrange's equations this is going to be zero, you take these quick partials there's no Theta dependency couldn't be simpler and off you down here. I get I'm highlighting we don't have to break things up into particles you can just use a minimal coordinate set that describes the energy of the system and you're there in two, three lines versus what I did earlier, but it's good educational practice.