 Any leftover residual questions from what we started with on, what, Monday? Remember, we're looking at now much more general motion where we don't have simple translation. We also don't have simple rotation but can have clearly a combination of the two of some. Which is, of course, the way most things move. Certainly, if you look at any part of a car with a lot of stuff, and they're not just the wheels going around, but the distance and all that, all just a ton of stuff going on and all that's a much more general type motion thing. So we looked at, on Monday, we looked at absolute motion and gave a fairly straightforward, I hope, prescription for how to work some of those problems. It's always easier on the board to just say, do this, than it is to get to the problems and do that. But we'll practice hope it gets okay. So if you're all right with that, then, we'll go from our first way to solve these problems which was the absolute motion, to our next step, which is relative motion, which we need to break into two, well, depending on how you count it, maybe three parts. So this first part that we'll look at today is the relative motion velocity part. We'll look at some problems using our relative motion method here to determine the velocity of some of the components of some rotating and or translating gizmos. And then we'll use it in a slightly different way on Monday and then next Friday we'll do relative motion acceleration, which is kind of the same just as you found out with this business is the acceleration, there's an extra step, not an extra step, an extra part to it because the acceleration of anything going in a circle, even if that circle itself is translating, there's a centripetal component as well as a tangential component, the acceleration, not the case with the velocity. All right, so let's do a little better review here. Here's some object we're looking at. Remember, we're looking at rigid bodies known. This could be a gear, a wheel, a linkage, a space station, kind of looks like a fat brother-in-law on the space station. It could be any of those type of things. And any line we inscribe on there, in fact, any triangle, if you remember, stays the same though its orientation might change. And so we might ask ourselves something like, well, if I know where point A is somehow and from some arbitrary origin here, if I know where point B is, or if those are the two things I want to find and don't know, I can use the fact that there's certainly a knowable or at least somewhat distinct relationship between the two that is also a function of where they are individually. In other words, the relative position of these, and we have done this before, so this is hopefully review. We did it when we were looking at particles. We might write it something like this, the relative position of B relative to A. We can do it the other way around too, but I gotta put something, so I'll put that. The position of B relative to A, which means if you're sitting at A, where do you need to point your eyes and how far away is point B? I don't care what they're doing. I don't care if one of them's moving and the other's accelerating or what. I don't care at any particular instant what we're talking about is if you're at point A, where do you turn and look and find point B? How far away is it and in what direction? And in fact, that vector itself is this one. We find that by the difference between the two vectors. The book doesn't quite write it that way. What the book writes is point R, B, and I'll write it down so you can see. If you can do either one, I don't care. I as usual have some kind of a method, my approach that I found a little bit easier for students over the years. That's what the book has. It doesn't say the relative vector between the two. It says if you want to find B, you need to first go to A and then look in the right direction, which of course makes sense. However, in terms of getting the equation correct, I think this forms easier because you just look at it and you've got BA, BA in that same order and you got it. Then you can rotate it around any way you need to get into the shape you need. If you do need B in the problem, then you can take this, you know you got it right, and you can put it in there. If you're good at memorizing stuff, you'll always remember this and then forget it just right. I always screw this kind of stuff up. I forget, well, wait, which one do I put here? No, which one's there and I get all messed up. I can't even say BA, BA, but then I forget where the equals I get all messed up. A very small one. So either one's fine. I don't care which because they're both the same. I just find the first one a bit easier, especially since what we are talking about is the relative position. What we're gonna use now is this, the time derivative of this, which is the velocity. It won't be quite good enough for us to just take the time derivative of it. That, of course, is true. It's just not gonna be as useful for us. So we're gonna go a little bit farther with it to make it more useful because, well, let's think about what this is. The velocity of point B, as it looks to you, sitting and pointing. So whatever these two things happen to be doing, I don't care. Maybe that one's doing this, and that one's doing that. I'm gonna put that up a little bit more as you'll see why in a second as I shoot it up there for a bit. You know that that's what those two individually might be doing. Since they're tied together as a rigid body, then it makes the rigid body go through space in some general way that we're trying to come to understand. So this is the velocity, what B looks like it's doing if you're sitting at A. So imagine you're sitting here at A. You may be moving, you may be not, but you're looking down at B and you wanna see what it's doing as it appears to you. Now as you look down and you're on this body that may be rotating, may be translating, may be doing both, could do all kinds of things. As you're sitting at A and you're looking down at B, B never gets any farther away because you're on a rigid body. So there's no radial or centripetal or normal component of this relative velocity. B cannot be coming any closer to you. It cannot go any farther away from you. As you're sitting at A, all B can do, as you're looking at it, remember you're on this thing, you're not looking at the trees in the background and the clouds going by and all the other things we do in cars when we're moving around, you're sitting there at A, you're looking down at B, it can only do one of two things. As you're looking at it, it can either go straight right or straight left relative to your position. That's all it can do. In fact, this is how you look at the world anyway. Here's you and you think you're the center of the universe and you look at the rest of the universe and all the rest of the universe is doing is orbiting around you. That's all B can do when you sit at A. Can't get any farther away, can't get any closer, can go on side to side. So in the essence, it rotates around you just like you think the universe does. That it orbits around you into the center of the universe. So if you're looking, if you're here at A, the velocity of B relative to you is either perpendicular to that line that way or as maybe is more likely with this example sketch on the board, but it's always perpendicular to the radius line, the connector line between you two. Could be left, could be right, depends upon the problem. But that's all the velocity component of B relative to A could possibly be. Which means it could be then the distance between the two of you times the angular velocity of the body upon which you sit. Well, that needs to be a vector. For our two dimensional problems, that's going to be in the K direction. What I don't know is it clockwise or counterclockwise. As you stood at A, we could have either one of those possibilities in a problem. But a little often a better way for us to get this is if you remember we have a cross product that will give us the same thing. And for a lot of these problems, I think that works out very nicely. These cross products aren't particularly difficult, but when you run through them, you're immediately set up with the solution you need to the problem. And I tend to think they're a little bit easier. In other words, over the years, I found more students tend to get it right if they take the time to just step through these. Rather simple cross products. I have a question about that. Yeah. What is the reason for us being omega cross r and not r cross omega? It doesn't work otherwise. You get the wrong direction in a right-handed coordinate system. So it's not that this is something that was made up and decided, oh, let's do this. It was, I don't know how it was developed. I've never seen the derivation of it. But that is, in our right-handed rule, one-handed coordinate system, that's the only way it's going to work. We'll see it, we'll double-check it in a second. Because in this problem, because we don't really know what this one's doing, it's hard to show in this problem. So let's set up a simple example and we'll check this and you can see that it works only that way. I just can't tell you why it's got to be that way. Because I'm not sure what the derivation was. So imagine we have some simply pinned linkage there and it's connected to another link that is much longer and these are nice, smooth pins there. And the other end of that is just riding on a horizontal surface. It's not pinned, because if it was, then this thing wouldn't move. We'd have a static problem. So dynamic's collapsed. So to give you some of the dimension, well let's see what's labeled as A, B, and C for reference, four meters, four meters. So obviously the arm at this instant, pictured, is at 45 degrees. It may or may not be important. And this is twice that, 0.8 meters. So a lot of these problems, we have to look at them at an instant in time, figure out what they're doing. If you were really doing this, you'd need to know what it's doing at all points of time. For our starters here, we're looking at this instant in time. And this one is turning that way. So this is omega A, B, is 10 gradients per second. At that instant, at the instant that is 45 degrees, the angular velocity of that link is 10 gradients per second. Now, what happens next, I don't know. I don't know which one of those is constant. Doesn't matter, that's not what we're asking from. Also, a quick note that students very often forget. Notice this is not A relative to B, or B relative to A, or anything like that. Everything on a particular rigid body has the same angular velocity. Any line I could draw on a rigid body. Remember, we started the thing with triangles, but it doesn't matter. It can be any part of it, any line on it, any piece of it. As long as this is all one rigid body, whatever angular velocity it has, so does every part of it. What that means to us is, or if you want another way to phrase it, angular velocity is a floating vector, which means it's always in a particular direction, as drawn here out of the board, but it's out of the board anywhere we need to put it. We can put it anywhere and it's exactly the same, exactly the same vector has exactly the same meaning. That's not necessarily true with the velocity. I can't say the velocity of A is up here because point A isn't up there. I could say the velocity of A is like this instead. That gives us the same thing. Velocity is a sliding vector, as omega is a floating vector. It can't go anywhere. Velocity can only be along the same line that's described. Force is also a sliding vector. I can have a force on A in a particular direction, but if I move that force up here, everything's different. But if I have a moment on an object, that moment's true for that object everywhere. You remember that from stats. All right, so here's our problem. A problem due dueer, and we want to find two things. Velocity of point C, the angular velocity of the arm BC. So a simple two-link problem. All right, let's see. Let's start with what we know and see where it goes. Let's see the velocity of C. We know that that's going to be the velocity of B. The fact that point B might be moving, plus the way that C moves relative to B. That's just this relative velocity thing with B's and C's, and then reconfigured a little bit so it's in the form of life, or my life. So that's just our relative velocity equation. Solving for the one of the things we were told to find. Now let's see. We also know that the velocity of C relative to B is like that form. So that's going to be omega BC cross C relative to B C relative to B wherever it is in location. Well that's pretty easy. That's just geometry that we can come up with. That's doable. There's C, there's B. We know what the geometry is here. We can figure that out without too much trouble. We're looking for that, so that's not known. We don't know that. So we're going to have to come up with it. And that'll help us find that. So this one equation has both of the things we're looking for. So let's see, we've got to think about it a little bit more. Oh, there is one other thing that we know about this we can put in. This velocity vector of point C that we're supposed to find because of the problem, we know that point C has to go along that table. Which means we also know that the vector BC is also BC, the magnitude in the I direction. It's got to be. It's restricted to traveling on a horizontal table. All right, so let's see. Let's double check what this position vector is that we need. Because when we have that, we're going to need to do the cross product of it. So let's see. Are C relative to B? There's B, and there's C. So that's C relative to B. If you are at point B, you need to look in that direction. You need to look in that distance to find point C. What is that vector? Well, as a vector, that's its magnitude, but as a vector, and you should keep this in vector form because we're going to need it in cross product, which is a lot better in a vector form. There it is, point A, I. Point A, I. I is point 4J. Yeah? Meant. So our CB was no big deal. Just pull it right off the geometry. Sometimes it's a little more difficult of course, the balance is pretty straightforward. Omega BC, well listen, this cross product itself is relatively easy to write out. And then all we have to do is add it to this. But what is that? What is the velocity of point B? We gotta get that to add it in. Oh, it's the velocity of point A plus the velocity of B. That's our relative velocity equation. We can do that for any rigid body in the problem. And AB is a rigid body. ABC is an rigid body, so we couldn't do it from point C all the way back to point A, but we can do it from point B to point C. That would give us this velocity B. We could put it in there, and we just add them all together. We'd have the velocity of point C. If we know what the velocity A was, if we don't know what this is and we can't find it, we're screwed here. What's the velocity of A? Zero, it's pin. So in this case, the velocity of B, absolute velocity of B is the same thing as the velocity of B relative to A. And that's Omega AB crossed with RB relative to A. Velocity B to A, position B to A. It's gotta be in the same direction or you're gonna introduce a fake minus sign. No sweat, that was given. No sweat, that's geometry. So we can do this, put it in there, add it to the result of this cross product, and get our answer. Let's see, let's do that. Let's do this velocity B first, since it's a cross product and we may need a little bit of a reminder how to do cross product. So I've just taken this now and moved it up there so we can do the cross product. What is the vector Omega AB? I have to have it in vector form so I can cross product it. It's magnitude 10 radians per second and negative K into the board, right hand rule. So minus 10 radians per second K crossed with where B is relative to where A is. And that's 45 degrees. So that's pretty easy. That's pretty easy. In fact, it's those pieces there. It's 0.4 I plus 0.4 J. Let's set up the matrix, do the cross product. These are fairly simple cross products in two dimensions because we have one vector that's one dimension only. So let's practice. I, J, K is the top row of our matrix. The second row is the first vector in the cross product which is minus 10 K, 0, 0, minus 10 radians per second. These vector components all have units and the units that are important in the cross products as they aren't in the other equation. Third row is the second vector. All 0.4, all 0.4, zero, and each of those is meters. So there's our cross product matrix. It always helps if you know what you're looking for in the end it will help you keep the minus sign straight. Let's see, VB must look like that, right? Because B is going in a circle around A. That circle's going in that direction at that instant. So we know that we're going to have a plus I component and a minus J component. So we can get our minus signs right. Cross, cover the first row. We've got VAT minus VAT product which has another minus sign in it which is then meters per second velocity. Cover of minus, now the J of cover of VAT that minus that which has now two minus signs in it plus a third is a minus sign which is also four J and then the K there's all zero. Does that make sense? Four I minus four J, 45 degree makes sense. We know it's moving at 45 degrees so the two components should be the same plus I minus J we're okay. All right, where does that take us now? Now we've got this, we've got that piece. We've got that piece. How many unknowns do we have? VC, is that two unknowns or one? It's a vector, is it two unknowns or one? It's one because we already know the direction we just don't know the magnitude. So this is one unknown. What about this? Is this one unknown or two? It's a vector, but is it one unknown or two? We don't know the magnitude. Do we know the direction of omega BC? Was this vector R? No, no, it's even easier than that to come up with. What would this linkage do if it was actually working? It would be a positive arc. If AB is going that way, BC's got to go that way. It's got to go to, C's got to stay here. B's coming down to here. It's going to do something like that. It's got a little bit, it's going that way. No other way that linkage could work. So we know the direction of omega BC, we don't know the magnitude. So how many unknowns? Two, two total. Magnitude, magnitude unknown. The directions are both known. So we need two equations. Here's one equation. Where's the other equation? It's even easier than that. Remember when we were in very first week or two in statics or physics one, we were summing the forces and we had two unknowns. And the only equation we had was the sum of the force vectors is zero. It was two equations. Vector equations break out into their component pieces just like the vectors do. This is two equations. And I'll show you how we get to it. So let's see, I'm going to need some board space. So keep this in mind and we'll write it up here. We now got VB, which we need right there. We've got that vector. So we need to do this cross product to get VC. So here we go. V, VC is VB, which we've got now plus omega VC, which we don't have crossed with RCB, which we do have. I, J, omega VC, what do I do? Well, you write it in as an unknown. And that's what we're looking for. It's omega VCK, so it's zero, zero, omega VC. I don't know what it is, but this is how I'm going to find it. And RCB, the third line, I know. It's 0.8 minus 0.40. And that's okay. So far, that's just the mechanics of writing out a cross product. I don't know about you. I can't do cross products on the fly. I got to do this and don't carefully, because there's minus signs that disappear. There's multiplication things you can screw up. Just the possibilities for failure are endless. And this, remember, is the velocity of C relative to B, which we know is going to be something like that. If you were sitting at point B and this arm's rotating that way, that's what you're going to see point C do. So we already know what it should look like. All right, so the I direction, that minus that, which is two minus signs. So it's 0.4, omega VC, I. Is that right? Really easy to mess up minus signs here, so be careful. Minus the next one, which is cover the J, do it in the same direction. That minus that, so we had one minus here plus the first minus anyway. So that now becomes a plus, all right. That's 0.8 omega VC. So we expect both components to be positive, which is exactly what we threw and expect. Both components to be positive. There is no k components to those two zeros. So now, I can put it right below here now as we do it. So VB is four I plus the product of this vector, plus the product of this vector, plus this cross product, which we just did. So add the I components. We've got four I from here and 0.4 omega VC from there in the I direction. Is that right? That's the I component of VBC plus the cross product we just did. In agreement? Or just not talking. Plus the J components. Minus four plus 0.8 omega VC. And we have no k component. Is this still two equations, two unknowns? We know VC in the I direction. So this must be VC because that's what we're heading for. But what's that? That's 0. There is no J component in our answer. So that J component must be zero. Well, you can solve then for omega VC, bring it back into here and get VC itself. And yeah, omega VC should be positive. That's how we followed it all the way through. That's why there's no minus signs in here on omega VC. Beautiful, that'll work. Solve these. You know this component J does not exist. So the coefficient must be zero. Solve for omega VC, that's one equation, one unknown. Put it into here, one equation, one unknown, you're done. As long as we didn't mess up any minus signs anywhere. I think we did. Omega VC, the magnitude, we already have the direction. We got that by observation. This is always nice. What is that? 50? And we know it's gonna be radians per second because that's what the whole equation was. So the five comes back into here. Magnitude of VC then is what, six? And we know that's meters per second because that's what all the units were there. All done. Did it come up with it on your own? But if you take it step by step, you know, we started with what we were asked to find and put down the first thing we knew about it. You know that VC, we know how it moves relative to B. And we said, well, okay, what do I know? I can get an RCB, that's not too big a deal. BB, remember, I dropped out the same thing. We figured out what it is relative to A. That's more stuff we can find, we can know. Just don't forget that little step. It's very simple, very obvious when you see it, but students just send it, you know, they're looking for the big complicated stuff for all their answers. And sometimes you won't forget the simplest little things for the answers. Okay, got the pieces, could we try another one? Most students don't like cross products, but these are pretty simple cross products. Just keep your minus signs in tune. Plus, remember almost any time you're doing a cross product, you at least have some idea of the sense of it. And so you should know if your numbers should be positive or negative, you can determine if you're making a mistake most of the time. All right, so let's do another one. I'm getting tired here. I don't want to fall asleep driving the van this afternoon, so you need to do a little work on this one. Here we go. We have a wheel that's not even touching the conveyor belt, it's rolling. So it's kind of like magic. So here's the conveyor belt. The conveyor belt velocity of the conveyor is two feet per second. The radius of the wheel, it's rolling on the conveyor without slipping. I forbid if we had to do problems where there's slipping, yeah, we're just learning. All right, I want to find a, well, I only need to find one thing. At nine o'clock, it's point A. Find the velocity of A. Remember these things are, these things are vectors. If we're doing particle motion, we just say A is moving the same thing as the conveyor belt. We're not doing that, we're not doing rigid body general motion. That means this thing's turning as C is moving. Oh, in fact, we have the speed at which it's turning. Sorry. Omega of the wheel, 15 radians per second. I say it is. I didn't say it's not rolling along the conveyor belt as the conveyor belt's moving. I just said there's the speed of the conveyor belt, here's the speed of the wheel, what's the speed of A? Not the speed of A, the velocity of A. So we need the magnitude and the direction. All right, these relative motion things, we need to link the speed of the thing we're looking for to something we can either know or we can find. We need to link this velocity with the relative velocity equation to some other point that will allow us then to find it. The velocity of A, if we know the velocity of some other point B, we can find it. Is there any other point whose velocity we know? How fast is it going? This point here on the wheel is going the same velocity as the conveyor belt. Very direct, Alex. Call that point B. I mean his proposed point B. Do we know the velocity of point B? He says it's the same as velocity of C. Jake doesn't believe so. Do we doesn't believe so? Doesn't want to play? Is that what that- I don't think so. Bobby? Holland? Well, now let's think. Let's think. Now, hang on. Alex, where did that come from? That idea, it's an interesting idea. Nobody else sees it. Where did it come from for you? It came from the fact that it was in contact with the ground at that point. Well, conveyor belt. Not only in contact, but there's no slipping. Now, Monday, we did this deal of a rolling wheel in contact with the ground such that it wasn't slipping. Do you remember the speed we determined that contact going to be moving at any instant? Zero, what it really was, if you think about it, is that velocity of that point is the same as the thing that's in contact with, which happened to be zero on Monday. It's no longer zero, the ground is now moving. That point is not moving relative to the ground it's in contact with. There's no slipping. That point's got to have the same speed as whatever it's in contact with, which is the conveyor. Alex, you are perfectly correct. VB equals VC equals VCI equals two feet per second. We're gonna take a moment for the class to apologize to Alex, you're not believing in him. Jake won't do it. Look at that Jake. You are not gonna do it. Hey, would one person like to represent a class? All right, Alex, after class we'll just beat him up. We'll make him pay, we'll make him pay. Jake, are you okay with that? It's not an obvious thing. Well, with Alex, I mean, but not the rest of you. You have to remember that the no slip condition guarantees that the bottom of the wheel is moving whatever speed it's in contact with, because if they're in contact and they're not slipping, they've got to have the same speed. No, because the whole wheel must be moving that way at that second, and it's turning too. So point A is gonna have a different velocity. Remember the wheel isn't just moving along, it's moving along and there and turning. So we've got to trust the equations and find that. So let's see, we're looking for the velocity of point A. It's, let's see. It's VB minus VB relative to A. You all right that right? Here's something that might also be a little more subtle that can sometimes help and it should help here. It's this, and don't forget this little rule. It'll help us here. V of one thing relative to another is equal to the opposite of the one thing relative to the other, with those things reversed. If you think about it, that makes some pretty good sense. If you're on something and God knows what it's doing is it tumbles through space, you're sitting at A, looking down at B, B can only do a horizontal motion. I'll draw it that way, but it could be moving in the other direction. Now if you're at B, looking back at A, it's got to be doing the opposite of that. Perfectly clear there's an alliance connected. And if you write out the cross product, you'll prove exactly that. So we'll chuck that in, gets rid of a minus sign, also makes the relative velocities we need a little bit more obvious. Because we're looking for point A, we know what point B's doing, so it's a little more obvious to look at that. What is A doing relative to B? Well, in fact, here's the line connecting them. B is if we are on the conveyor belt ourselves, B wouldn't be moving, and the wheel turns that way. That's got to be A relative to B right there. So B is not moving to A. Sorry? What? Go ahead and ask your questions. Alex, I'll answer it for you. Well, no, remember a split second later, we have a new contact point. This contact point won't exist anymore. It won't be a contact point. But at this instant, point B is moving the same glossy as the conveyor belt, and A relative to that, since the wheel's turning and that point's not moving, it looks like the entire wheel at that instant is rotating around point B. Instant later, things are different. So we even know then what the sense, remember this is a cross product. Let's see, velocity of B, oh, velocity of B, that's the same thing as the velocity of C. So that one's done. We just need the velocity of A relative to B. Sorry? Omega AB cross RDA. Omega AB, it's got to be the same, these points always have to be on A rigid body together. Crossed with R, A relative to B. Have to be in the same order. Omega AB, do we know that vector? What we're trying to do is this piece here that we'll then add to VC. Do we know this vector, first part of the cross product? Minus 15 K, because it's into the board. Minus 15 radians per second. Okay, the matter that this wheel is on the conveyor belt, the conveyor belt's moving? No, Omega, remember, it's a floating vector. Cross, what is the vector A relative to B? Point A's here, point B's here, they must be separated by a 45 degree line. That's got components to the radius. 1.5, get the same thing. A lot of these cross products you can't do on your calculator, because they'll have unknowns in them. This one doesn't have to do with it. Others would, but the first one would do this. That times that, which is two minus signs per second. Well, actually we expect both of them to be positive. So minus this, that times that, which is two more minus signs, that's three minus signs, like this, minus, a minus, a minus, that's four minus signs. They collect. Is that what you got for that cross product? That, okay. Guess what we expected in the picture? If the line from B to A is 45 degrees, any perpendicular to it is also 45 degrees, and that's just what we got there. VB, same as VC, is two feet per second. I added to that, nine feet per second. I. That's VB, which is the same as VC, which is moving in the I direction only, plus the first component, because they're both the I components. Plus the J component, which is only that second part, and that's feet per second. Now, if we're doing this study to figure out, if we put a link arm itself connected to point A, then we need to figure out what that's gonna look like. Now we know what velocity of point A. Okay, Bob? No? Man of me? Yeah, man of me. Not talking to me, not talking to Alex. Your friends are dropping like crazy. It's gonna be you in a rocking chair with a shotgun on your lap someday. Care to get out of my lawn. Ready for your own? Nothing, people like more. At lunchtime on a Friday afternoon, they're not in trouble. Very much the same as those, just the situations all change, so. There we go, here's a wheel. Attached to it is a link arm. So as the wheel goes around, that link arm's gonna go around with it. That end's gonna go around and make this whole linky thing work for us. And that's attached to another link arm that itself is pinned. So as the wheel goes around, this thing's gonna go chubby, chubby, chubby, chubby, like the old choo-choo train used to do. That we love from our childhood. Right, simple picture. Here's some of what else is going on. Each arm is 0.2 meters. And the wheel is half that, radius of 0.1 meters. At this instant, let's see, less likely to say, at this instant, Omega AB is 30 radians per second. 60 degrees. If the wheel's going at a regular speed, like it might be if it's attached to a motor, then those, that Omega would change. Is that angle changes? But that's not what the question asks. We're just asking. At this instant, if you were really making something here that needed to work, you're gonna have to do the full analysis. So do the full analysis when you need to. Let's not do that yet. Fine. The angular speed of BC, that link arm, the other link arm, angular speed of the wheel. We can call that Omega DC, or CD, which are going on. Any idea how we can get to these? And as you've seen, and as you can imagine, that they're linked, so how they work influences each other, but the equation they appear in might be a separate equation. For example, we do have an equation that would have that in it. It might be, oh, let's see. The velocity of C equals the velocity of D plus the velocity of C relative to D. That's true, isn't it? Velocity of C relative to D, that's gonna have in it the speed at which DC is rotating. What's the velocity of D? That's easy. So this will be then Omega CD, which we're looking for, we don't have it, we're looking for it, are C relative to D. Oh, that's no problem. Look how nicely that thing lined up. Okay, so that smells suspiciously doable. The speed of VC is gonna depend upon the speed of C and the speed of B. Speed of C, we would hopefully get out of here, could then use it with the speed of B, velocity of B, to find Omega VC. What's the velocity of B itself, though? B is on an arm that's pinned at one end, and we know how fast it's turning. So we know that B is going to have that velocity at that instant and its magnitude will be R Omega. So when you get the velocity of B, we'll have the velocity of C, those two together will give us the velocity, an angular velocity of the arm DC. So there, this is just like your mom used to do when she says she wanted you to cook dinner. She put out the recipe book, she put out the things. Some stuff's still in the refrigerator, though, because you don't want it out on the counter and where the cat can eat it. The equations I have, Omega DC, would that change at all? This one? Yeah. Would that change at all? Like, move point C up to, like, point. The equation as written would not change. The numbers that'll come out of it would change a lot. But this holds no matter what, especially since D is pinned at the center, it's the center of the wheel, but that general equation holds the specifics of it, changes at some instant and time later. All right, so we work this one out, see how many unknowns are in this equation? What two are they? We don't know the magnitude, do we know the direction? Yeah, at that instant, as things turn, at that instant, that's got to be VC. That's the only thing it could do at that instant. And instant later, it's doing something else entirely different, not entirely different, but it's not doing that. So the magnitude there is unknown. What about Omega CD? Do we know the direction of it? Yeah, you can figure that out from the picture. It may help you if you just draw what things will look like a second later. And then you can tell the orientation of the arm CD and how it's going to change. So yeah, we know the direction of C and the direction of the wheel CD. Yep, it was here, now it's here, it's twisted counterclockwise a little bit. It shifted some, but it twisted counterclockwise and that's what Omega has to do with. So we're supposed to find the velocity of that point C. Why? I mean how? How? By working this, doing this cross product. I don't know if I've ever heard of something for Omega CD. You are? You do this cross product, it's a vector equation, it'll have two equations, two unknowns. It'll have the I components grouped together and it'll have the J components grouped together and you know what each of those should be. You don't know which direction Omega CD should be. And the DC, you know, too. Relative to D, similar to that. Position of C relative to D, which is this position vector, R, C relative to D. Actually we, those are perpendicular. If you remember the cross products when the vectors are perpendicular, it's just the product of the two magnitudes and well that's what we got right there anyway. C, R, C, C, or C, which we don't know. Omega CD cross, let's see, Omega CD cross. So that's here, down in here, down like that. Works perfectly. Omega CD. Wait, glad I got two ears. Do we take one? Alex, you take the other, go. Well Alex is right right here, so you might as well just, you know what I mean. Which ear were you talking in? Wait, your ear. Alex, what? Alex, ladies first. Do we, oh, Lou, he's busy, she drank the popper. No, she's done. It's the same thing you did. What was it? Point one, Omega CD, not point one. Oh yeah, where'd the VZ come from? That came from Magic Math Length. So I, that, minus that, we have two minus arms. I mean, I don't think it was here, it was because of zero minus. Plus we know it should be positive anyway because that's what we got here. You don't VZ, right? No we do, it's got to go that way. We know that, I can't. Because it's attached to the wheel, as this goes that way, that's gonna go this way, the wheel's gonna go that way, that's all it can do. So the second lever doesn't hit the ground? It does, but B is moving. Because A, B is moving. A little bit later, those arms are gonna look something like that, point C will have moved up to here. Point B will move down to here. All right? So it's so confusing, so why is B C pointed horizontally right? Because at that instant, that's what it's doing. Oh. An instant later, it's moved to here, has a little bit up flossy, moved to here, has a little bit more up. Moved to here, it's always going to be moving tangential to the wheel. But, did not ask about those later times. Asked about this time. Huh? Oh, you got some more work to do. Sorry. Yep. Because we have another equation for B C. It's V, V. Because V C is, this point C is connected to point D and it's connected to point B. Also have V C in it. Yeah, V, V, B, we know. Because it's just circular motion about point A at a known angular velocity. It's moving in a circular path around point A. That's all it can do. The instant its velocity is tangential to the circle or perpendicular to the radius, the line connecting the two, we know that distance, which is R, A, B, A. We know the angular speed, omega, A, B. Since they're perpendicular, it's just the product is the magnitude there. And then, we now know this angle, that magnitude, it can give us the whole vector. I was just confused about that cosine variable, any sine variable, what was with that vector? It's this vector here, V, V. There it is right there. So, it's got an I component that's V, V, cosine 30, I. That's the I component, the J component, V, V minus sine 30, J, minus because it's going down. Oh, you just affected the amount. And that's all known. The meters, that's 30 radians a second. We add to it C relative to B. Or C, B, because that link arms were zonal. You don't know omega, C, B, but we do know that this all comes out to be the same as that because that's V, C as well. Should be two unknowns. Omega, C, D, and omega, the two things we were looking for that you write down in that one equation. Oh my goodness, time to go. I wonder if nobody was smiling anymore. Well, my suggestion is, don't do anything else this weekend, just do these. Let's just take a little practice. They take a good, a decent drawing and may not hurt at times redo the drawing because they do get kind of messed up, but don't check what you get for that. We can finish it on.