Okay, let's do a brief review. There's a lot of little math steps, that we went through last time. These are definitely steps I feel, you should be able to do from scratch, basic principle. And we're going to cover some nuances that we go through. So. [INAUDIBLE] no actually, it's going to go here, that'll work. So one of the first things we were looking at is, we've got a blob dn. This blob has a center of mass, and we have some inertial frame. This is the center of mass position vector. This little dn element has a little r vector, and then capital R is to find us the inertial position of that element. I think it solved it. That's kind of where we started out from. So let's see. Cody, what is the Super Particle Theorem? >> Not quite sure. [INAUDIBLE] last night. >> When it's fresh? >> [LAUGH] >> But yeah, I watched the video late last night, and it wasn't. >> You slept through the video last night you're saying? >> [LAUGH] Yeah. >> I see, okay. Good, what is the super particle, Andre? Super particle theorem? >> I couldn't tell you. >> Did you watch the video? >> It's just been a long week. >> Okay, Super Particle Theorem? >> The acceleration of the body becomes on the external forces >> Right, so actually close. What did you mean by acceleration of the body? [NOISE] we have to be more specific. All right, this is a continuum. This could be your fuel slashing in space, a jello. You know astronauts eat jello in space, so there's jello wobbling through the space station I'm sure somewhere. How does this stuff? What's the dynamics of these things? So when you say the acceleration of jello, which part of the jello? >> [INAUDIBLE] >> Hmm? >> Every part of the jello. >> No. Can't be right, right? because jello may have left side is perfectly steady, the other side is wobbling, two seconds later it will all shift and change. So you can't have every part of the jello having the same acceleration. Even if it's rigid, does this pen have the same acceleration everywhere? No, right? So right away that statement, you're on the right track. What was it Brett, right? >> It's the center of mass. >> It's the center of mass. So the Super Particle Theorem says, that the total mass of this system times Rc double dot = F. Now David, what goes into this F? >> It's basically an integral over all the particles, the sum of the forces over all the particles in the body. >> Yeah. Robert, can you be more specific? This is true. But it's not, there's more to this. There's some refinements that we can do. Which forces do we have to really consider when we do this integration? >> External forces? >> External. What happens to the internal forces? Russell? When you sum up over the body, what happens to internal forces? >> [INAUDIBLE] >> Sorry. >> They cancel each other. >> They cancel each other, exactly. Right. So this bolt has a force of one newton, because another plate is pushing on it. But vice versa, the plate is having a minus one Newton force somewhere. And these things will all sum up to be zero. So internal force is completely banished. Only the external forces, so this is really dF external that's kind of what you can think of this, integrated over the whole body. This is what we derived last time. And basically it's a very powerful thing. We'll be using it one of the examples, you'll be using it in the current homeworks as well. So without knowing what's going on with fuel slosh and all the other stuff. If you just know the external influences on your space craft, regardless of what flexing is going on inside in deformations, especially if it's rigid as well. Of course, you can predict what the center of mass must be doing. And the center of mass motion for spacecraft, that's basically your translation, your orbit's problem. So we can account for all of that. Now there's ways that attitude couples into this total force. Can you think of an example where, you need to keep track of what the shape is and the orientation of an object, to figure out what the external force is? Daniel? >> Drag. >> Drag, atmospheric drag right? So there are particular types of disturbances, acting on a space craft that depend on the orientation. And if it's this shape in space, or this shape in space as we'll see in a few weeks, gravity gradients make a big difference right? So there are external influences that do matter on the shape. So those become harder obviously to track, because now you have to track orientations, your codes immediately slow way down, because for attitude you have to have much smaller time steps than orbits typically, and so lots and lots of challenges. But this is it. This is the Super Particle Theorem. Now we had the center of mass definition, MRc equal to this. This was one of them, right. Basically, the center of mass of a blob is the mass averaged location, that's one way to look at it. Instead of an integral, putting in a summation sign, it looks just like your classic statics book that you had. Mass averaged location. What was the other definition that we used over and over again, for center of mass of a property? Thibault, do you remember? >> The weighted average of the positions with the r plus small r. That's a good answer. >> Yeah there we go. So this is the average position relative to the center of mass, have to add up to zero. Then basically it's the balance point right? You put that little lever right there in the center of mass, and you'll be able to balance something. I can't I'm not quite good enough this morning all right. But it's something like that. It's not my pen, I don't care. >> [LAUGH] >> No of course I care. So these are center of mass definitions. And then we said, we can take two derivatives, put the dots in here all right. This was one thing we did on the left hand side, if you take two inertial derivatives you just get this. Then we have the second derivative of this integral Rdm right? And from here we ended up with this expression. Bret right, it was you that brought up that question. That was an excellent question. Why can we take these derivatives inside this interval, because the body if it's jello, the boundary actually varies with time. If B is defined to be this boundary of the actual jello. So in the textbook we have, there's actually a statement early in that paragraph on that section saying hey, when you do these body integrals or these integrals over B, the domain has to be carefully selected to encompass all the mass. Basically what it means is, you create B in this case for a continuum is not just the body surface itself. You can make a box bigger than that. This isn't a fluids or continuum problem. There's not a mass flow of air, or some plasma, or something coming with water, coming through here. It's a fixed shape. And it's doing some stuff. So you can define the bounds of these integrations to be bigger, basically make it big enough that no matter what the worse case deflections are, you've got a bounding box that's big enough to cover it. So when you're doing all these integrals afterwords, something times dm, out here the dm is simply going to be zero. If this stuff deforms and encompasses it, all of the sudden dm becomes non zero. That's a way to look around it. This avoids Reynolds transport theorem or lightness stuff, that otherwise you'd have to use on the boundary. But again, this is a fixed shape. We're not adding, losing mass to the system, we're just looking at deformations right now. So there's a discussion ahead with my co-author, John Jenkins and I have to give credit also to Johnny Hurtado, professor at A&M, he writes a lot of analytical books. He pointed me to some good references. This type of derivative, we also call sometimes the material time derivative, or the co-moving time derivative. because you're really following each dm element. And we're not creating new dms. There's not a flow of dms like in a fluid flow, that comes in and out of the surface area. On the boundary, as you change stuff there's no mass flux coming in and out of the boundary either. So lots of different ways to look at it. But the easiest explanation I think for engineers is, just for this kind of problems you've got a space craft flexing, you're getting all these equations of motions, the results I've shown are classic and definitely true. It's just this step there's some subtleties there. So Brett very good. Gold star for asking the right question from analytical- >> That was Hansel actually. >> Hansel? I thought that was you. Well gold star for honesty. So Hansel, he's not here today. >> [LAUGH] >> Well somebody right back there who asked a great question, thank you. But that was an issue I wanted to bring up. This is a subtlety, but this is an easy way to look around that form continuum. Once we turn rigid, all this stuff is completely simple anyway. Then all these results definitely hold. But even for a continuum, the results I showed you are absolutely true. But if you do these derivations in these books, there's lots of little subtleties. And moving a derivative inside this differential operator, is something you always have to do carefully. So I didn't have a good answer last time, but I do have a better answer now. So we did derive this. And this gave us in the end, this was the dFs and this was nothing but a Super Particle Theorem. Good, so we derive that. Now, let's review quickly angular momentum. Brian, of a blob, what do you remember about angular momentum? >> It's. >> When? >> When there is no external torques. >> So, if you're saying if L is zero then H dot = 0. So, it's a vectorial quantity. Right. This becomes important. So when you say it's preserved, we're saying the inertial derivative of this vectorial quantity is zero. So it's a constant as seen by an inertial observer. As we go through equations of motion today, you will see the angle momentum expression, as seen by the body fixed observer, is actually not constant. So the body relative derivative is not zero. But the inertial derivative is zero. Which is a really good integration check, when you write in code. Every time stamp, recompute the derivative. All outside the derivative. Compute h. And then map h, because you're typically computing the body frame, map everything into the end frame. And those three end frame components, should be nice flat lines. Then you get warm fuzzy feelings that you're doing the right thing. Now when is this true? When is h dot equal to L? CK. Is that always the case? >> It's not accelerating? Center of mass is not accelerating. >> No. Center of mass can be accelerating. You're on the right track, but you're talking about the wrong point. When we have moments and torques, we need to specify something. Robert? >> When the about the center of mass? >> That was one option. When we derived this H dot, we found it was H dot equal to L, plus something. And that something had an RP minus RC. So if the point, about which we're taking moments in torques, is the center of mass then that second part went to zero. The other one is related to what CK is talking about. There was something about being inertial. But it's not the center of mass being inertial, but it's really rp double log. So if the point about which you're taking moments is an inertial point, it could be stationary. That's cool. Or it could be moving at a constant speed, also has zero acceleration, then this equation holds. So that's an important thing, today you will see when we get through equations of motion, especially a slender rod, in homework you're doing one that's tipping off a point and has some friction, we'll do one, there's no friction slides in place. But here we'll be using Super Particle Theorem, H dot = L, and when we use H dot = L we have to make sure we do it about the center of mass of the dynamical system, or about an inertially fixed point. And then this will work right? So good. Those were two important subtleties that we needed. But this is about H dot. Let's step back and talk about H itself. Can somebody tell me something about H? For a blob of jello. Right the fundamental definition is, this integral over a domain B. You have position crossed with linear momentum of that one right? Is this the form we typically use then? Evan? >> No we usually use the sigma form, which defines that difference between r and rp. >> Right we can make that one general. There's something else though. We can break this angular momentum, same thing with kinetic energy, with these blobs. There's a way that we can actually separate into two different chunks. Do you remember, Evan? >> I'd have to look to tell you but I think we- >> Just in words. In words what then we can see? >> We can separate it into two integrals of the positions crossed, and then position of acceleration? >> Lucas, help him out. >> I'd have to look at my notes. >> Matt? >> So we're going to define R in terms of little r and R of the center of mass, and then [INAUDIBLE] integrals and one of them cancels to zero? >> Yep some stuff definitely equals to zero, we're left with two parts, you guys are on the right track. But what are those two, angular momentum? Shayla, do you remember? >> [INAUDIBLE] is equal to [INAUDIBLE] across R. So you sub that in. >> When is this one true? >> [INAUDIBLE]? >> Yes? Is this true for a blob? Maurice? I'm full of good questions every morning. >> [LAUGH] >> This should not be a surprise, this many weeks into the semester. Who knew he was going to ask me questions about last lecture? >> [LAUGH] >> I know. [INAUDIBLE] couldn't foresee this at all. >> When is this true, people? >> For a rigid body. >> Rigid body. It's only true for a rigid body, that's in fact [INAUDIBLE] go to rigid bodies. But we're already answering questions about that, right. People sometimes quickly assume to look at this stuff and direct to deriving of the exam. R dot rigid, R dot must be zero. No. Each little particle is going to have some inertial derivative of this vector. It's not growing, it's not shrinking, because it's rigid. But it is non-zero because it's tumbling. So this part is only true if it's rigid. Let me make that not next to it. If rigid. But, as Matt was saying, we do make this thing, you plug it in, you do all the stuff. Okay. Somebody, somebody here surely remembers how we break this up. What are the two chunks? >> That minus. >> Yeah, what is that part? >> There is the cross product. >> No, not dm. Okay. So this is the second part. But again, if you get into this form [INAUDIBLE], what have you assumed? [SOUND] Okay, so I'm hearing a lot of sub-results that are true for rigid. Let's go to a blob. H is equal to R crossed, R dot dm. You can rewrite this. Where essentially Rc, why is that? Okay. Crossed MRc dot plus integral of little r dot DM. Why is this form convenient? What is the first part then? Andrew? >> [INAUDIBLE] the angular momentum of the center of mass about some point. >> Right, so this is about the inertial point or something that we've picked, right. There are moments about that. R is defined relative, it's the Rc not R. Right. So it's important with these problems, and you start from the fundamentals. You're always going to get this, well this is a blob, I have all my position vectors [INAUDIBLE] inertial. We break it up like Matt explained, relative to center, the motion of the center mass plus motion relative to the center of mass. Momentum, you want to write exactly the same way. Why is that convenient? This in essence, gives you the translational side. That's the orbit problem when [INAUDIBLE] flying through space, right? So, you can separate moment to orbital, a momentum is essentially this part. This part contains what type of motion? Griffin. >> Attitude. >> Attitude motion, and what else? This is blob still. >> Deformation. Deformations, right. >> So shape changes, fuel sloshing, flexing of panels, all this kind of stuff would also be accounted in this. So this is the anglular momentum about the center of mass. And this is the angular momentum of the center of mass. That's the important distinction we keep making. So in this class we derive it, we get these terms but now we're going to focus on this part. And if you go look at the kinetic energy, we can write that in very much the same way. We derive that, similar properties that we used, the center of mass properties terms drop out, but you end up getting the kinetic energy of the center of mass plus the kinetic energy about the center of mass, all right? So this is the classic thing we can keep doing, over and over again.