 The short of course, it's not right. Do you realize it's a free phone? Uh, no. Just, let's go ahead and just put that in. Let's go ahead and get that into the final draft. A little less work for me, a little less work for you. We like both of those. Why did anybody have it all done? Now they're meant, well, if you had it almost all done, it almost goes straight into the report. So there's not necessarily any great shakes of work that's been lost. All right. We were looking last Wednesday at some way to look at systems of particles. We've spent, we've spent all the first weeks here. Oh, by the way, don't forget, five weeks left. And it goes by very quickly. So if you're missing work, or postponing things a little bit, you could, could be sorry. So get the stuff done, not just this class, any class. It goes by very, very quickly. Before you know it, we're going to be on the beach in our Speedos. All right. So we've been looking at systems of particles. We started that yesterday. What we found out was, was kind of nice. Well, we were looking at very, very simple collisions. And we're going to look at more and more involved in elaborate collisions as we go. We were looking at simple collision. I can't remember who was driving behind me. An awesome red sports car got hit. A couple things from that. One of the most important things we found was that even though each of the objects, which we've always been taking to be particles now anyway, each of the objects very, very clearly undergoes an acceleration. As anything is going to happen in the collision, there's going to be a big change in speed. The car going behind that was going faster is going to slow down quite a bit due to a force that car felt from the car in front. However, we found out that we could analyze almost the entire collision without ever even worrying what that force was. That's what we came out with on Wednesday. We were able to take a look at the whole collision. And by that, I mean, find out what the after-collision velocity was of the two vehicles. We assumed for at least the timing that they collided, stuck together, and then therefore we're doing the same speed. And we found out when we looked at it that way, looked at it as a system, we could ignore what these forces were, just look at the velocities before and the velocities after. Actually, we have to look at it a little bit more than just velocities, wasn't it? As we took these to be a system, that meant these forces were internal to that system. Thus, they're always paired together and they're action-reaction-pair and they canceled each other so we didn't even have to worry about things. Right? That's what we did on Wednesday. Long time ago, bless you, long time ago. Everybody was here except for Samantha. Amanda was here but Samantha wasn't. Since those forces were internal and during the collision, those were the major forces far and away, more involvement of those forces than any other, then we could make the assumption that during the collision, there were no external forces. What did we do with it from there? Do you remember? I didn't derive that because of MA. I never derived it. God gave that to us. That's on the 11th Tablet. Moses dropped it on the way down. Newton had to discover it. Actually Galileo. Now what? Since we make a system that makes all these forces internal, unless they are equal in opposite pairs, they cancel each other, we didn't have any other forces going on. Remember, we're only looking at the instant of the collision. We don't talk about any of the skidding afterwards. That's a whole different deal. Since there are no external forces, what came next? Check your notes. It's there. My notes aren't on your phone, are they? You thought they were for a second? You got confused? I think my notes are on my phone. I have the note-taking app, the physics note-taking app, from the app store. Is it in there, Joey? Is it in there? Well, that is what came next. From that, we have this business that involved momentum. What was our symbol for momentum? P. It is a vector because it is made up of what? How do we define momentum? That's called velocity. Now, since mass is always a positive number, that means that momentum and velocity as vectors, what can we say about them? Those two vectors, they're different vectors. They're very different, not very different vectors, but they are different vectors. But since mass is always positive, and for most of everything we do, it's just a constant, what can we say then about these two vectors? Something about the quality of those two vectors. They're always the same direction, not only the same angle, but actually pointing in the same direction. They just differ by a simple constant. So if we know the momentum, we'll know something about the velocity. If we know the velocity, we'll know something about the momentum either way. But then on Wednesday, I put those two things together for the analysis of a system of particles with no external forces on the system. What then became our working equation? Our working... No? No, on Wednesday. We looked at kinetic energy, but I just asked you to calculate it so we could see what happened to it. It had to do with momentum. For a system with no external forces, which is why it's nice to call this as a system because it made the forces internal, and we didn't have to worry about them then, which is great because we didn't know what they were, and they're very hard to come up with, so we had some pretty good instrumentation to do that. If we have a system with no external forces, then what happens to the momentum of that system? It's the same before as it is after. Then a couple of different ways we can say it, but system momentum doesn't change, or we can say that system momentum before equals the system momentum after. Either one of those is the same, but that was pretty useful, especially in things like car crashes and the like, which we'll spend a lot more time looking at in the next chapter. When things... Two cars crash, it's pretty easy. There'll at least a little something about the crash, and from that we can get out a lot of the rest of the information, which is exactly what they do in accident analysis. The police on site do just that type of thing, and some engineers are making a good living being court experts where they'll take all those pictures and measurements that the police took, and they'll reconstruct the accident using the basic physics of what goes on. So there's an alternate career for you, possibly. Do you have a security guard? Yeah, I don't think somebody halfway through physics one wants to go in and claim they're a court expert and then face the defense on cross contaminants. So, all right. So that's what we were working on Wednesday. So we're going to extend it a little bit here. We'll use our continued example of two cars. Though it need not be. In fact, we can have a system made up not of separate items, like two separate cars, but it could be a system made up of several things joined together. Either way, it's going to be the same to us. What would be nice is if we could extend this idea of these two things grouped together as a system, especially a system where there are no external forces, because then we know that the momentum is concerned no matter what happens to those two cars, whether they crash or not, whether they are changing speed or not. None of that matters as long as there's no external forces that we need to worry about. We can figure out a whole bunch more of what's going on. So what would help is if we could come up with some way to represent more of this system than just the fact that we've got some parts in it and we kind of add those things together. If we can come up with some other way to represent the system, well, the things we need to consider. We need to know where this system is Well, how do we talk about a system being some place when it's made up of two or maybe more cars and they're all moving around with each other, all kinds of things could be going on? Wouldn't it be nice if we could talk about the system being some place and not individually break out each of the little parts? Imagine the space station made up of millions of components all put together. Yet there's one place, one spot that represents the position of that space station so that they know where it is and they know what orbit to put it in. They don't calculate when we got this little arm sticking over here so we'll calculate where it is and then we got another arm sticking over there and then we got a fat brother-in-law who just walked across the space using the other side and the whole thing tilted like that and they could see it happening from Earth. It'd be nice if we could just say this one spot represents where the space station is and then all the other parts do whatever it is they do but we still know where it is and what it's doing. Once we know its position, of course, then we'll also know something about its velocity and acceleration. So it'd be real nice if we could come up with this position that represents the position of the system then, well, we could take the time derivative of that. We know the velocity of the system. Not the velocity maybe of every single little piece but the velocity of the system. Like they know the speed at which the space station is moving but they're not constantly calculating where each little arm is, whether or not one of the astronauts did that thing when they spit M&Ms at each other or globs of juice and all the other tricks they do that use up taxpayer money. If we could do that, that'd be nice. Nobody's volunteering? We can do it. It's really rather straightforward. What we'll do is to establish the position, we'll do what we've always done with position before. What was the first thing we did when we were worried about the position of something? Not, what do you mean, find the origin? Yeah, we establish an origin and it doesn't matter where it is just as long as we all know about it, it's very useful then. So we'll establish an origin. So let's say right there, that'll be my origin. It could be anywhere, it doesn't matter because everything we do from here on is going to end up in the same result that doesn't depend on where that origin is. So here's what we're going to do. Let's see, we've got these two objects each of some mass, just to simply make them m1 and m2. And with respect to our origin, each one of those objects is at some certain place. I'll just simply call x1 and the other one's at some point x2. What I want to do is replace those individual x's with some x that represents the position as a whole because then if I have no external forces I don't care about the individual parts, I just need to look and see where the system is and I can tell a lot more what's going on. So there's got to be some spot, I don't know where, that we can put the entire system. So maybe I'll call that x system and instead of imagining the individual pieces are where they are, which could be quite confusing if there's a lot of them, I'm going to imagine the entire mass at this one spot if I can just figure out where that is. That way I could take out the two individuals, replace them with one system spot and I cut my work in half. Even more if there were more parts. So let's see, let's look at it like this. This is a mass out at some distance. So that's kind of like a mass sticking out in an arm somewhere. This is a mass on a bigger arm sticking out somewhere farther. And I want somewhere where I can imagine the total mass of the system is and where would that be that has about the same kind of balance to it? Mathematically, what I'm trying to do is something like this. If I can take each mass and where it is, I don't know why it starts with two, let me put one in there. If I can take each mass and where it is, not V. See I'm worried so much about what that girl's name is that I came right at the board. I'm so concerned. So I've got each mass and where it is represented like that. And I want to replace that with all of the mass at one spot. It represents the whole system. It's as if we went back so far, stepped back so far away that we no longer see the individual spots. We just see the two kind of come together as a dot, sort of like when we do look at the space station. We see it as just a little dot. We don't see it as a bunch of solar collectors sticking here and there and a bunch of armed laboratories built on the Japanese laboratory, the Italian laboratory, the American laboratory, all of that. We just see it as one little spot. That's what we want to do. Step back so far, and we want to know where that one thing is so we can figure out what that one thing's doing. Well, we've got the answer right there. The thing I'm trying to figure out is where do I place this total mass that represents the whole system so I can just divide through by that. Total mass, that cancels. Well, what is the total mass? It's just m1 plus m2, isn't it? Now I know where to place that center of mass that represents the entire system. In fact, you've heard that word before, the center of mass, where you use one spot to represent a much larger, much more complex object. We now can put it at some spot such that the simple calculation works out. Notice, too, that once we know where that is, we don't care about the origin anymore. Yeah, we measured it from the origin, but then once we've done that, that system spot represents those two masses and it doesn't care where the origin is. From then on, it's exactly the same relation between the two things that we would have had there. So let me just rewrite this here on the board. So we've got this. In fact, let's label it cm for center of mass. You're pretty familiar with that term before. Maybe you don't know how to calculate it, but it's not a foreign term to you. So for a two mass system, it's as simple as that to figure out where that is. So we'll do several examples here in a second. And if there's more than two, big deal. You just add them on. Just keep going. Just add them all on. No big deal. All right, let's see. You now know where the system is. We have a single spot that represents where the system is. Now we want to know how fast that system might be moving. What did we do weeks and weeks, not months ago, when we first had position and we wanted to know how fast that object was moving once we knew its position? What? Xm prime. Xm prime. Oh, take the derivative? Oh, yeah. We don't use prime in engineering physics. Because then it looks like an exponent. It looks like cm to the 1. That's the stupid mass feature. We want to know now the velocity of it. So we take the time derivative of it. And I'm not editing that out. Well, let's see. The bottom part, that's just a constant. So that's not going to be a derivative at all. The top part, the constant, what's the time derivative of x1? What's dx d1? Sorry. dx dt. dx1 dt. What? It's the velocity of object one. What is ddt of m2x2? Mind you, we've all taken calculus. That's about as easy as they get. What's ddt of m2x2? m2v2. Plus m2v2. Wait a second. I'm going to rewrite this a little bit. Just a little bit of algebra here. I'm going to move that top part over here. I just moved it over there all by itself. Then I've got this bottom part here. m1 plus m2 times now vcm. Does that look familiar? It looks exactly like the momentum equation we had for collision on Wednesday. However, I didn't say anything about a collision or anything. But we can look at it like this. This is the total momentum individuals. If you want total momentum before, this is the total momentum after. After we combined everything as a system. That momentum wouldn't change because there were no external forces on it. Why should that momentum change? All we did was make a simple representation of the two objects as a single one object. Irrespective of what the two were doing inside the system. I don't care if they're colliding. I don't care if they're passing. I don't care if they're driving up at McDonald's. Well, I do care about that because it's not healthy. But it's not my business. That's even further encouragement. I would hope that what we've done here by determining the position of the system since it now looks exactly like it did from last Wednesday that what we did is correct and useful. Anybody want to take a wild guess of what the acceleration of the system will be? Of the equation we just had. And that even has to do with whether the two are making each other accelerate when they hit. Because a force will cause each one of them to have some MA. But if it's a force between the two of them, it's internal to the system, and we don't care about it. So that's pretty useful to us. We don't need to know what every little piece is doing and then account for every little piece individually. We just throw them all together. And heck, we're laughing at it. We're just having just no trouble. And in fact, we could even put external forces on them because when I had that system there, anything I just did for determining the center of mass of the system doesn't really matter if there's a force on it or not. It would still apply. So we're going to do that more in the next chapter, I think. Well, let's see. Let's go ahead. We'll put on external forces now to have the system. Oh, sorry. I was using Cm now instead of cis. So let's continue with that designation. I only have to know the acceleration of one spot. This M here is these two M's put together and brought over. So this is just going to be M1A1 plus M2. That's nothing more than the application of that last little piece to it. Well, that's the sum of the forces on Object 1 plus this Object 2. I'm going to take one step and then we're going to come back to that to finish things. I'm just going to take one step just to see in general what the sum of the forces on something is. And this could even be force on one from two. It may not be because when we add them all together anyway those two pairs would cancel each other. But just to be completed, it's every force on one, every force on two. But let's see, what do we got here? MMA, which is DVT. I'm just trying to do some of those sum of the forces. See if I can come up with something else here on this type of thing. Well, you know that if there's a constant in a derivative that comes out, I'm going to do the reverse. I'm going to take this constant back into the derivative because that gives us that pair again, MV, which we also know as momentum. So this is DP, DVT, or if you like, you can write it as p dot. Either one's fine if you want to stop at the DP, DVT, no trouble. That's sum of the forces acting on an object. The sum of those forces is equal to the time rate of change of the momentum. The rate at which the momentum changes. SDT, nowhere. So let's take what we just came up with here and put it here. This is the sum of all the forces on the system. So that's going to be the time rate of change of the momentum of the system, the center of mass, if you will. So let's do that since that is the designation we've been using now. Same thing we had, right? And that must be equal to the time rate of change of the momentum of Object 1, DP1, DVT, plus the time rate of change of the momentum of Object 2, DPVT2, and so on. If there's more and more objects, we just add them all up. These things are happening at the same time, then we don't even need the time designation and we're just back to the change in momentum of the system is equal to the change in momentum of all the individual pieces, even if the forces don't sum to zero. We can just add up all the momentums even for those situations where the forces don't sum to zero. We looked at the case when they do come equal to zero, the momentum doesn't change. Even if the forces don't sum to zero, we can still look at our individual particles as a full system. So that takes a lot of the burden off. We don't have to babysit each little piece. We just have to know where the system is and what's happening to the system then the individual pieces will follow that as they will. So we need a little practice then, finding out where this center of mass is so that we can then figure out what's happening to the system, what that represented. So let's look first at a system you're all familiar with, the Earth-Sun system. Oh, sorry. The data I have is for the Moon. So Earth-Moon system, I think for homework, you have to do the Earth and the Sun. So we'll do the Earth and the Moon that way you have no excuse for not getting at least one of the problems right in the homework. So here's the deal. There's the Earth. We call it the Blue Marble. South America, Florida, where Santa lives. There we go. There's the Earth. It probably looks like a photograph from space, doesn't it? It's about the picture of the Moon rising or the Sun rising over the Moon. They took from the Moon when they were up there. That's what that looks like. Somewhere nearby. Too bad I didn't bring my green chalk because everybody knows the Moon's made out of green cheese. There's the Moon. Some distance away. We're going to find out where the center of that system is. The center of mass of that system because as the Earth orbits around the Sun, it's the center of mass that orbits that follows the orbital path, not the center of the Earth that does. As the Earth goes along its orbital path around the Sun, it's the center of mass that follows on that orbital path. So the Earth actually wobbles a little bit as it goes around the Earth, and then the Moon does even more than that as it circles around that Earth. As the two of them together follow the orbit of the Earth around the Sun. So I want to find out where that center of mass is. That's what's orbiting around the Earth, that center of mass. We can do it just like we did with the other picture with the two cars there. We can pick an arbitrary origin. Any suggestions? As good as any. No reason it can't be actually fixed to something we already know. What's nice about that is we know certainly how far away the Moon is from the Earth, and we're going to need that distance. So the location, the location of the center of the Earth measured from our arbitrarily chosen origin will be... How are we going to find that? Remember? Just had it up a couple of minutes ago. It's gone now, but what was it? The mass of the Earth times the location of the Earth with respect to our arbitrary system plus the mass of the Moon with its location with respect to the arbitrary system divided by the total mass of the system. Notice, once we come up with this location of the center of the mass and it puts it wherever it is, we don't need the origin anymore because that point is fixed. It's just that now we know what it is. We know where it is. We can measure it from there, or from there, or from any other place we want. We know where that origin is, so it's not artificial that we're putting in this arbitrary origin. It's going to allow us to find a spot and then once we know where the spot is, we don't need the origin. We know where the spot is from that on. What is the mass of the Earth? Nobody knows offhand? Mike, is that on your awesome cell phone app? No, you didn't get that app? Darn it. So you put your physics notes on there but you can't get anything off the mouse. Len, what are you doing? Looking in the book. Good job, that's where it is. It's in there somewhere. Yeah, it scared you, didn't it? It's a sweat. So, Len, what's the mass of the Earth? 5.98 times 10 to the 24 kilograms. What's Xe? The location of the Earth with respect to our arbitrary origin. It's zero. See how convenient it was to put the origin there? It's zero. So we didn't even need the mass. Well, we do need the mass of the Earth because we need it down here. That happens to be zero. So choose your origin carefully and you can sometimes make things easier. So that's that mass times zero. What's the mass of the moon? 10 to the 26th? 22nd. 22nd. Can I say I don't think it's heavier than the Earth? And what's the Xm of the Earth? What should we use? Sorry? No, I mean, what are we going to get and Len to look up for us? Is that the radius of the moon? This Xm, this represents the location of the moon with respect to our arbitrarily chosen origin. What we call the orbital radius. It's not a perfect circle. It's a little bit ellipsoidal, but we'll use the average and that's what should be in the book there. They have the radius of the orbit of the moon, Len. Yeah. Divided by these two masses, add it together. 5, 9, 8 times 10 to the 24th. Plus, that's zero. So we put the origin right at the... Oh, by the way, even saying that, was using this whole idea of locating a big object with a single little point. So we're doing... You're used to doing that anyway. Nobody even flinched. Len was flinching at other stuff. Worked that out, of course. You got to figure it's going to be somewhere like what I drew. It's going to be a lot closer to the much heavier Earth, the much more massive Earth, much closer to the much more massive Earth in our little system as M gets bigger, X gets smaller, so they all balance. In fact, that's pretty much what we're doing, and you're just balancing the masses so that they balance at the same spot. What do we got? Bill, you got something? Yeah. Not going to share it? Do we have to buy it from you, take up a collection? Maybe. We'll wait. We'll see if we can get it for somebody for free. Maybe we can get it from somebody with a little more generous heart to come ask them if they're in favor of the term. Seven. Seven what? Is the answer? Seven what? Got it? What is it? 4.7 times 10 to the sixth meters. So, pretty close, pretty close to the Earth. In terms of the other distances we're talking about, 10 to the eighth, all the way out to the moon, this is about a hundredth of that. So, it's one-one-hundredth of the distance from the Earth to the moon. Len, can we trouble you for one other thing? What's the radius of the Earth itself? 0.4 times 10 to the sixth meters. Notice that then. The center of mass of the Earth is actually where? Inside the Earth. No reason it can't be. That's the spot that orbits the sun in a smooth orbit. Other than that, the moon and the Earth both wobble around that spot as they do their orbital motion in orbit around the sun. It's that spot that's doing the orbit. But that's nice. Now we know where that is. We don't have to calculate the Earth and the moon separately. If we want to know the momentum of the Earth and the moon system or the acceleration of the Earth and the moon system, we can calculate all that based on that one little point. Your turn. Here's a molecule made up of a couple atoms. It'll make it very simple for us. It's not a chemistry class. We'll say all these atoms are the same size. It'll make things real easy for us. Our first time through this. Five atom molecules made up of five identical atoms. C5. This distance. 0.1 nanometers. What's a nanometer? 10 to the negative ninth. So this is 0.1 times 10 to the negative ninth meters. It doesn't matter. We'll keep it all in nanometers and not mess with the 10 to the ninth tables. This distance. 141 nanometers. This also 141. This angle here. 130 degrees. Once you figure out where the center of mass is. If we built this and we could do it out of spaghetti and marshmallows. Where's the spot where this whole system would balance? That's where the center of mass is. Just eyeballing it. You've got to figure somewhere in there, wouldn't you think? So that's good. Because when you come up with an answer, if it's wildly off of that spot, you've got to figure you did something wrong. Alright, what's our first step? Let's do this together and then you can do the rest on your own. What's our first step? We'll use that atom right there as the origin. Where the center of mass is. If you don't want to calculate it or find the one person in class you're in speaking terms with. All five atoms are the same atom. We're extremely close. We didn't draw that it's tilted on this axis. But with respect to the moon, I don't know. Oh yeah, it is tilted on this axis. That's the earth? Perpendicular? Or in the plane? That's the sun. It's fiery and yellow. And that's the orbit of the earth. And there's the moon orbit in that same plane. Or does it orbit? The orbit's around this point. It does? I don't remember. I just know that sometimes it's there in the sky and sometimes it's not. I just don't have to remember the specifics of that. Alright, let's see. We're looking for the position of this center of mass. Once we know where it is, then it's fixed to the molecule and we don't have to worry about where the origin is anymore. Because then we know where it is. Just add up all the little pieces. Just do exactly that with it. I always ask somebody sitting beside you before you ask me, I charge a lot more. It's like a hundred bucks an hour for my control. Don't forget, this is a two-dimensional molecule. So we also need to know where YCM is as well as XCM. It's just got a lot harder, isn't it? Now you're sweating. And the stakes just went up now. Just like that, with the stroke of shock. It's doubled the amount of work you have to do. YCM and let him do XCM. Right, Tyler? Which one would you volunteer to do? YCM. YCM as well. YCM. Since it's symmetric about the X-axis, then YCM is zero. It's right on the X-axis. We know that. In fact, I ought to drew that. So always look for some kind of symmetry when you're locating the center of mass. Because by definition, the center of mass is going to be right on the axis of symmetry. In this case, the X-axis is the symmetry axis. All right, so let's, we'll number these one, two, three, four, and five. Now we have to do that for five of them. M1X1. M2X2. X3. What did I say about the mass? Mass are all the same. So in fact, the mass, oh, now it will have mass distributed out of each of the pieces. It will also cancel the X's and divide by five. Actually, it's kind of like the average distance then. Well, even that makes sense. We just got to make sure we're using the right X's for each of this. What's X1? We have to account for the fact it's to the less of the origin. That's the only way we do that. So negative 0.1. X2 is zero. They're all the nanometers, so I'll put that down at the end. X3 is 0.141. What's X4? The origin. So it's 0.141. That little bit of distance there, whatever that is. So that's the 141 times the cosine of 65 if you want. Does anybody have that number? It's got to be a little bigger than that number because those are as far as three and farther. What? 0.2005. Is that confirmed? 005, I'll just call it 201. Is that correct? That's four. What about five? Same distance. So I'll just multiply that by two. That's over five. And we figure it better lies somewhere in here, probably a little bit closer to three than it is to two. What's that come out to be? Four or something like that? Somewhere in there. Which is just to write just about where we'd expect it to be. Anybody not get that? 0.88. Okay. Oh, that's what Joey was trying to say. All right, Joe. There you go. Now, no matter what else that molecule does, we know where the center of mass is going. If we put some force on this molecule, all we have to do is worry about where the center of mass is going. We don't have to worry about each of the individual pieces separately. It could be that in some situation, you actually need to, but we're still working pretty in general. Anybody not get that for a reason? It'll give you the world's hardest problem. I'm going to come back in the morning because you'll still be here. There's a Q. L on the side. Center of the Q, because it's symmetric on... The center of mass is automatically at the... on an axis of symmetry and it's got axis of symmetry in all directions. So, this one has a little bit of a trouble though. Not a trouble. It's got a whole drill through it. Something like that. Now it's no longer symmetric in all three directions. Just to help with the sketch, I'll call that the x-axis. That's the z-axis. That's the y-axis. So, looking at it in the x-z direction, which is looking right down this hole there, you see this. There's the hole. To locate the center of axis, sorry, the center of mass, we're going to need xcm, ycm, and zcm. Three-dimensional solid. To locate a single point, we're going to need three coordinates. Anybody want to volunteer for any one of those first? It's centered in the z direction. This is the y one. What's the y dimension? If I put the origin down here at the corner, just give me a quarter of a distance. It's got two planes of symmetry, which is going to put the center of mass for whichever two, right at the center. For example, if we look at this, in the z-y plane, that's y, that's z, there's the hole. If we look at that from the y-z plane, so where's the center of mass, at least in terms of both y and z, it's right in the center. But you have each, just have like x, y, z, so you have to have each one. Oh, how do you write it down? I don't care. That's just no taking. We don't have xcm right now, so... because we need to find out what the center of mass is. This is a hole, so there's more mass of this stuff to the right than there is to the left, so the center of mass for a quarter. You've got to figure there's going to be something like that. So how do we figure out how big xcm is? The other two, just by symmetry, they're right where they were, they're right in the center. How are we going to find it? Well, we do it just like we did it before. x1m1 plus x2m2, where what does the 1 and the 2 stand for? Not even halves. You can make it even simpler than that. Because if we made it a half here and a half here, we've got to figure out what the mass of this part is minus the cylinder drill down. That's starting to get a little bit messy. Minus the... That's probably the easiest way to do it. The mass of the cube minus the mass of the cylinder. Because that's pretty easy. Anything that's missing, we'll just give a negative mass because it's taken out of the system, so we'll take its influence out of the system. What's the mass of a volume, though? The cube's not a whole. I said, what's the mass of a volume? What's the mass of any volume? Let's see. Rho equals mass over volume. Know what that means? That's density. So, however... whatever volume they have, times the density equals the mass. So, let's see. Let's let 1 equal the cube and 2 equal the whole. Where's the center of mass, x1 of the cube, without the whole missing? Because we're going to take off the whole second. x1 is L over 2. Its mass is its density times its volume. What's the density of a cube? I mean, the volume of a cube. L cubed. So, there's x1 and 1 for the cube as a whole before we drilled out the whole. It's x2. Sorry, what is x2? The center of mass of the whole. You're right there. It's one radius from our edge L over 2. Sorry, no, L over 4. What's the mass of the whole? Density times the volume of the cylinder. It's a whole, so we're going to subtract. We're taking the whole out. We're going to take out its influence. Density times the volume of the cylinder. Density of... sorry, the volume of any cylinder, some h, of the base times the height. So, it's pi r squared h. It's the radius of the whole. So, there's pi r squared. What's h? The height of that whole. It's 2n2. Minus because it's a whole. It's something we took out. And then the mass... Well, the mass is these pieces we have here. Again, we just put them back there without the x part. What is this material Huh? Is it aluminum? It doesn't matter. It's canceled out. Any object of one single material that looks like this is going to have the exact same center of mass no matter what it's made out of. About that, and you may work it all the way out. We know it's going to be a little bit over halfway. So, it should be a little bit bigger than L over 2 0.56L. A little bit over halfway. Now, if that was some part of the space station we needed to move it, we know how to move it now because we know where the center of mass is. If that was some object that was going to float, we know how it would float now because we know where the center of mass is. All we had to do was use the negative for any part of it that's missing. Alright, see you tomorrow. Be ready to work on your reports.