Where we have a slender rod that's falling to the ground. So basically, think of this rod. You sign in this up on a frictionless surface. Think of an ice rink, a very slick ice rink and he's standing it up and for some reason this is perturbed. There's no friction force here holding your lower end in place. So once you slide off you're just going to slide out. So it's basically just going to fall down and slide down like this. That's what's going to happen. And what we're going to try to find here are our equations of motion. And we're going to use these very principles, super particle theorem, HL equal to L. The inertia tenser definitions and so forth to derive all the right quantities. So I'm giving you here from a material statics book you can find easily if you have a slender rod then the inertia about it's center of mass is mass over 12. L squared, L's length of the rod. So slender means we're ignoring the other dimensions. It's just a homogeneous, very, very thin mass distribution, right. So that's what you get there. So we're going, okay, so we have that. Now let's look at the angular momentum. If we want to write Hc, the angle momentum above the center of mass, we needed the inertia tensor times the angular velocity of this stuff. And you can see, I've got a body fixed point, a body fixed frame, E, that I'm defining here. So EL, E theta, where is E3 pointing? Let me just make sure you're looking at the picture, right. Into the board or out of the board? >> Into. >> Into the board, right? eL, e theta, e3, third finger points into the board, that's where it's pointing, okay? So the angular velocity as we're tilting in 1d, it's going to be theta dot about e3, right. So that's what this part right here, those two terms, that's omega of e relative to n for this body. Now I need the inertia tensor about this one, so let's just write that one out. You've seen the matrix notations, I want to make sure that you get all this stuff right. So if we have this in the E frame, it's a principle frame, you will have I1, I2, I3. And, let's just ignore this. E relative to N, theta dot e3. What does theta dot e3 look like as a 3 by 1 matrix? 0, 0, theta dot, right. And yes, you have these other inertias. We said, it's a slender rod so there are some terms about that. But in the end, this matrix times this matrix is just going to give me 0, 0, I3 theta dot, right? In different components, which is equivalent to that I3 theta dot E3. And in this problem I3 is the N over 12 L squared. That's the inertia about that, the third axis that we have, in and out of the board. All right, we didn't need the inertias about the long axis or anything else like that. That's where, so when you've, this is something you want to practice. Once it clicks, you go how was this ever confusing? But you want to be able to go from the inertia stuff to this quick other formulations especially with the plane of rotation. It should always give you equivalent results. So good, so that's how we get this part. We have to angle momentum about C. Now to get equations of motion, one way we often do with rotational side is we use Euler's equation, which means H dot = L and taking moments about center of mass, H dot = L holds. And I need to have L above the center of mass. So let's see what do we have here? What torch is being applied? The only torch that's being applied here is this normal force, somethings holding that tip from sliding beneath the I surface, right? So the I is still pushing up. It's just not pushing laterally. So it's giving it lateral stability, but it is holding that point in place, all right? Why does gravity's acting on this? Why doesn't gravity cause any torques? >> Ax on the center of mass. >> Right, gravity ax on the center of mass and therefore the moment on for the gravity force. That's when you apply it, it would actually vanish. Very good, okay. So, we need the moment arm. What is the point of force relative to the center of mass then that's this vector. So, this length is L, so if going from here to here is minus L over 2 in the eL direction, right? Back to week one, vectors are distances times direction. The distance is eL over 2. The direction is minus eL hat. That's the moment arm crossed with the force, which is the normal force on the surface, right? And you now have an eL crossed in n2. Well you're going to have to map one into the other and that's what you get, a sign term in here in terms of e3. So we can write this out. So good, we're looking for equations of motion for theta, but all of a sudden we have to introduce an extra unknown. We don't know what this force is on the surface. So we know right away H equal to L is not going to be enough. You can't have two unknowns in one equation. People try, they try really hard, it just doesn't work. So we'll need additional equations. Let's first wrap this up, we have H, we have L, now the derivative of this H, inertial derivative has to be equal to this L. Well e3, this is a rotating frame but what can you say about e3? Is that a rotating axis or a fixed heading axis? It's fixed, actually, right. It's the other two that are rotating. So e3 is actually N1, it's the same thing as minus N3. And N3's definitely an inertial frame so that's why taken the intertial derivative of this, I don't have to treat e3 as a rotating axis. It's actually a fixed axis. So this inertia is fixed. I'm not changing the mass distribution of the rod. The only thing that varies is theta. So you very quickly get the inertia times theta double dot. And here I've dropped e3. Because in the end, everything's along the e3 axis. So it gives me my one thing. So H dot equal to L. I brought L over to the left hand side. And that's how that comes in. So this is one equation. I'm looking for always second order differential equations for equations of motion. And so acceleration, that's my theta double dot acceleration equation. But it have an extra unknown. So H dot equal to L was not enough, what else do we have? People. >> Super particle theorem or energy? Energy here is pretty smart because there's no friction. >> Yeah, energy doesn't typically help quite in these equations of motion typically. The principles we use is either H = L, there's different points we could use. We've used for now the point c. What was the other point we could have used? An inertia point that was another approach. And you can try H equal to L about center of mass or about the inertial point and they can give you different information. And that might be a way to get a second equation. But what we're going to do here is actually use Newton's equation. Basically, super particle theorem, right? That said, well whatever the sum of the external forces is, mass times the inertial acceleration to center of mass, has to be equal to that force, right? So if we do that, the center of mass location, I'm just going to call this coordinate y, you can see here. So Y doubled up in two because its own center of mass is only moving in N2 direction. That makes it easy, it's an inertial direction. Take it twice derivative it's just Y doubly NY doubled up in two. That's my big M times rc double dot that we have to review the particle theorem, right. Now that has to be equal to the sum of the forces acting on the system, and this system only has two forces. There is a force on the surface that's pushing back against the pen tip, keeping it just in place. And there's gravity acting on the slot and that's why we're falling. Otherwise, if you have slippery surface on the space station and without any other human slip. So with gravity +N it's in the N2 direction and NG is in the -N2 direction so this gives you your differential equation. But you look at this and go, I can see Brian already going wait a minute. We get an end that's good but now we introduce a y. We just seem to be chasing our tail here. Every time we have an equation we have an extra unknown. So now we have two equations and three unknowns we're not making any progress. And Brian's absolutely right, so what are we missing? We need an extra equation. Are y and theta independent? Spencer, are those two independent coordinates of the dynamical system? Or another question is, this dynamical system has how many degrees of freedom? >> One. >> One, right, there's only one degree of freedom. There's only one angle, right. Now with that angle, we can prescribe what that height has to be. So there must be a geometric relationship between the y coordinate and the theta coordinates, right. Because you have y and theta, this didn't turn it into a two degree freedom problem. It's still a one degree of freedom problem. So what is the relationship? And that's what we do here. It's very simple, this line would be L over 2 cosine, that gets you to heighth y. That's what I have here, all right. So if I have theta, I know y. And now here, I need theta double dot, that means you simply take 2 times derivative of this y expression, chain rule of this cosine, so you get some sines and cosines and other stuff. But that's how your y double dots relate to your thetas. Theta dots and theta double dots, right. So those are not, we had two kinematic variables that was convenient. This y was very convenient when writing f = ma. But in the end, we have to pick one of them and write everything in terms of one of the coordinates. So if you do this, you plug this in here. You can now, actually this depends on y and n, this depends on y and theta, I can now write n in terms of theta. So I get rid of one of the variables, y. And now the last step is this, this has to go back up in here and now we can write our equations of motion for this slender rod falling in place, slipping down like this. Looks a little bit more complicated. It's amazing how life gets complicated quickly, even with simple examples. Now with this differential equation, is this a linear differential equation or a non-linear differential equation? Non-linear. There's sines, cosines of the states, right? There's even theta dot squares, sine squared of stuff, definitely not linear. So who thinks they can quickly solve this differential equation? Darn, I always hope somebody raises their hand because then I want to talk to you about a PhD program. because I can't solve this. Maybe a mathematician can solve this. But it's tricky, right? So how do we solve this? Well, we go back to integration, right? You put this in state space form. x1 is equal to theta. x2 is equal to theta dot. Then you get your sets of equations. You can integrate. So numerically, no problem. But analytically, this can be a challenge. And very often in these problems we want to have at least integration of one time. That means, I'm looking for the relationship between positions and rates. And it happens, you've done this, high school physics problems. Bouncing balls, perfectly elastic bouncing ball, right? The total energy has to be preserved. It's a conservative system. So therefore, we have our potential energy of the ball, and we have the kinetic energy. And as you're up, you have a lot of potential energy. No kinetic energy. Let go, it trades potential energy for kinetic energies, but the sum has to be the same. Potential energy always depends on position. Never rates. Kinetic energy, you see the definition. It's always ex dots, theta dots, it's all the rate of the coordinates. So if you're looking for a relationship, and one of the homeworks asks you for this in part B in particularly, what's the relationship between some states and rates? Your first step shouldn't necessarily be to try to integrate this equation once. That might be quite hard. The trick is you're going to look for is the system conservative, and if yes, now we can look at the energy balance. So people was talking earlier about energy, this is typically where I use energy systems and I can come up with directly in analytic, at this angle what is my rate going to have to be? It's good conservative. Now in this system there's no lateral friction. There's nothing doing work on the system it's just a conservative one. And this point here is not doing work because this point is not moving up and down. Work is typically force times distance moved and this point is not moving in this direction. So this force end also. This system is conservative. And we can use conservation. So this part should hopefully be boring. Very simple, mgh, right? Or this case, y, we have it as in terms of the angle theta. I know what the potential energy is of this rod. I need the total mass of this rod, the moment arm, the height of the surface, I've got that. The kinetic energy, now remember how do I get this? Kinetic energy, like momentum, I can break up as the kinetic energy of the center of mass plus the kinetic energy about the center of mass. For weird complicated tumbling things like this I really recommend that approach. Sometimes there are shortcuts with other stuff that you can quickly do but this is a way that's very rigorous, you won't double count terms all of a sudden. So the kinetic energy, all the center of mass is mass over 2 times the only velocity we'll have is y dot. And we know how to relate y dot to theta dot right, by using that earlier geometry expression and differentiating. And then here is the moment over 2 times theta dot squared. That's the one-half omega transpose i omega but put back into everything's about E3. And it all simplifies down to the simple scalar equation. So we've got that. So you plug in that kinematic relation between y dot and theta theta dot simplify. This is what you end up with. So you do something very similar in the homework to this. So now we have kinetic energy and potential energy and if you sum the two up, I'm seeing initially this rod is standing up, and yes, something has to disturb it but the disturb it's been treating us infinitesimal. So the total energy is basically just the initial potential. And then something turns it ever so slightly and it will start to slip, right? And so the initial one is this. And then we want to look at what is the relationship at a later time so now we know at a later time we plug in this plus this has to always be equal to this. This is the total energy system. So no matter how far you slipped of whatever angle you're at, I can come up with a relationship. And in fact, solve for the theta dot which we do here analytically. So energy's become very powerful expressions that we can use in these systems to actually predict some great relationships to states as well. If you looked at the prior differential equations, I would challenge anybody to just glance at this. And then predict that this is how states and rates have to relate. It's not intuitive, that's why we have these principles, these mathematical processes in there in place, to do that. So anyway that kind of wraps that one up. Any questions on this example? So when you solve this equations, and there's a few in the homework, if you want to look for h dot equal to l, but apply it about points that make sense. Either center of mass, that's often a very good place to start. Or you might find that inertial point as well. And that might, both of those equations might give you all the information you need. The other one that we have is F = m a of the center of mass, that's the super particle theorem. Between the combinations of that you should always get all the states, everything that you need. If you have multiple kinematic states, it's only one degree of freedom look for the geometric relationship. How does theta relate to y or z or whatever you introduced, all right? And if the statement is somehow relates states to rates immediately think energies. Look for the system most likely to system is conservative. And that allows you to exploit conservation of energy in some way and come up with these relationships, okay? Good, so that's good stop, not, well for that topic that ends there. We've got two minutes, that's good. Well let's do a quick thought experiment. So now what we'll move next into is torque free motion. Daniel already mentioned earlier that if its torque free motion certain special things happen. And in fact, spacecraft once deployed from a rocket, the initial stage is always tumbling in one way or a manner. And then you have to recover, maybe we use magnetic torque bars or we use thrusters or reaction wheels, or something. And we want to go positive. Or these all these different challenges. And even otherwise, you don't have a situation typically where you're continuously controlling attitude because you just run out of resources, even with reaction that only use electrical energy. It's a lot of effort. You try to exploit as much you can the natural dynamics, and the natural dynamics is no torque, it's just there. So can we come up with spacecraft that spin? And people do that quite often. They spin stabilize a single rigid body. Or we'll be looking at dual spinners that means we have multiple parts that spin, but without any feedback. And then multiple parts that's basically a stepping stone into reaction wheels, momentum exchange devices, which we'll derive afterwards and stay up there as well. So that's kind of the motivation for why do we want to look at torque free motions? Really a classic problem, even with three axis control these days you still want to understand the different modes. Why are some motions stable? Why are some motion unstable? And one of the equilibria spin conditions. And if I put it in a particular spin, will it just stay there or will it wobble out of that spin and go and get in the way? And one way we can investigate that ends up being angular momentum. The angular momentum vector H of a rigid body, we know if it's torque-free, H dot has to be equal to 0 which immediately gives us a constraint equation. But this first cut at it though is very complicated because if I do everything in the inertia frame, I will get three components. They all have to ce zero. That's great but your momentum in the body frame, if I have a principal coordinate frame, is simply i 1 omega 1, i 2 omega 2, i 3 omega 3, right? We mentioned principal coordinate frame that inertia tension that's just diagonal. So it becomes a very simple product. h is equal to i omega. But then we have to map this into the inertial frame. And here I'm using [INAUDIBLE] in terms of three to one Euler angles. And immediately, you get all these relationships. But you can see there's tons of sines and cosines and omegas and everything varies with time. And true, that's a constraint, but it's very had to wrap your brain around that and get some analytic insight. So next time when we start off we're going to start here and I'm going to show you a different approach which leads to the pole hold methods. Those of you who have taken 3200 have seen a little of this stuff already. But it's a classic approach in how we can discuss stability, but we'll take it all a step further then from that as well.