 We're looking at general motion, and we took the terrible tasting medicine first. The general motion is this business that things can translate and rotate at some time. Car tire being the obvious example, but we're going to see that lots of other things do just this. And they're not wheels. They're not even circular. They're just able to move about some point. So we're now combining the two of these motions into what we call general motion. A more, I guess more realistic, a more flexible type of motion because a lot more things can be done with it. We are going to look at it in two ways. We begin with absolute general motion analysis, I guess. And that's what we did on Monday, and it's harder to see. It's harder to set up. Once you get it set up, then it's just a matter of taking derivatives, especially if you're good at taking derivatives of trig functions. But it's getting that first part of it set up. Remember, we had to set up the position as a function of theta of angle. And then we take the derivative of that because the derivative of the position is the velocity. So we have the velocity as a function of angle. And that was combining the translation and the rotation part. Because the velocity was the translation part and the angle was the rotation part that went with it. Today, oh, and then we started relative motion analysis. We started that on Monday. We're going to continue with that. Students generally find that a lot easier to see, a lot easier to manage. I know I certainly do. So we'll begin with that. It's based on the fact that if we have two points on a rigid body, and we'll make it about as simple as a rigid body is possible, just maybe a link on some kind. It's got two points on it, A and B. We can figure out something about the velocity of B. If the arm has some kind of angular rotation omega, then B will have a velocity relative to point A that would do exactly that. That's the kind of thing we looked at in Physics 1. If one point is rotating about another, then it at any instant time has a velocity that is tangential to the circular path that point is making. That's just as that point rotates about A. It's just following some circular path centered on A. And at any instant it's got a velocity tangential to that path. What we're adding to this now, of course, is the possibility that A itself might have some velocity. So now the complete velocity of that point B is not just that relative velocity as it goes around A, but the fact that A is also moving. So we would add that to it. And that's what we then set up on Monday, that the velocity of B is the velocity of A plus the fact that B is moving around A. And that's exactly what I've got there for that drawing. So the velocity of B slash A, that's pure rotation, and that's all the farther we went in Physics 1. I'm going to add into it the fact that the object as a whole might be moving in some direction as represented by VA. And we need that plus the relative motion of B around A to get the complete idea of what this velocity of point B is, whatever point B is on whatever object. And we've done that before. We set that up, that velocity of B relative to A. We set that up when we took the day or two that we went just through the pure rotation. And that's the cross product of omega cross R, where R is the position vector that locates point B relative to point A. All of these subscripts, the order in which the subscripts come has meaning. This is always B relative to A as we read it, and so that's true for the position vector. It's the vector that locates B if you happen to be sitting at A yourself. So if you're sitting at A, to look at B, you need this position vector. That's RB relative to A. And so we're going to be doing several cross products today as we step through this business. Not terrifically complicated ones, but we need to do them nonetheless. Alright, so let's step into our first example problem. Imagine we've got a two-link structure with one link there and anchored at one end. And then twice as far into the scale when drawing. So there's another link on there and they're freely pinned to each other. And then that arm itself, maybe it has a little wheel on it that has to ride in a horizontal track. So its motion is constrained to a horizontal. So this is a very simplified picture of pretty much what goes on with the pistons in your car. The piston rides on a camshaft and there's a piston rod that connects to the piston. The piston can only go up and down in one direction in the cylinder itself. So this is a fairly useful, simple type picture. So I'll give it a couple dimensions. This is four-tenths of a meter. So all these dimensions will be in meters. This is eight-tenths and this is also four-tenths. So that first arm happens to be at 45 degrees at the moment we're looking at it. And that arm has a rotational velocity omega. In fact, we're going to assume that to be 10 or 18 per second. So let me also label some points just for reference. That's going to be real important as we do these things. So I'll just make it simple A, B, and C. So we'll call this the angular speed of that link A, B. And that happens to be 10 radians per second. That would be maybe the angular speed of the crankshaft in your engine if this wasn't a piston cylinder thing thing. All right, so we want to find a couple things. Find the velocity of point C. How fast is that piston moving in this? It's not a constant. If that's one arm AB is rotating at that constant speed. The speed of BC will not be constant. It'll actually be sinusoidal. But we're just looking for it at one instant when the measurements are all as given there. And want to find the angular speed of the arm BC itself. So we have an input speed, if you will, on AB. And we want to find the resulting speed of the other arm. All right, we're looking for BC. So let's start there and see what we've got. BC, the speed of it is going to depend upon two things. One is the speed of V because they're connected as a rigid body. The relative speed of C to B, the fact that that arm BC itself has some angular speed. In fact, we're supposed to find that out. And the way we're going to find it out is because we know that the relative velocity is omega CB. So we're looking for that cross RC relative to B, which is no big deal. That's just the vector that runs down that link arm BC itself. So we'll be able to come up with that. We don't know the velocity of B. We don't know the angular speed of CB. The position vector of C relative to B, that we can get when we need it. It's just off the picture. But these other two parts are unknown. So let's work on those. The velocity of B, let's see. By the way, let me emphasize any time we apply this equation that we want to find the velocity of one part that's connected to another part, it's very important you realize that these two points have to be on the same rigid body. Because we cannot do this if RCB itself varies in length. And if it did, then that violates our definition of rigid body. So always apply this relative motion equation to a single rigid body only. We couldn't apply this point C relative to point A because they're not connected as a rigid body. Only C to B. Then we're going to work back B to A because with A we know more things. So velocity of A, just setting up the very same type of equation to find the velocity of B that we can use in here. Plus omega AB cross R. So if we can do that, then we can use the velocity of B that we find here in the first equation. Use that to find the velocity of C and the angular velocity of C B. Notice that A has no velocity. It's pinned, which is just our way of saying that B is in pure rotation about the point A. So that little bit we can do, let's go ahead and do it. Let's see, omega AB. If we use our coordinate system in the usual direction, X that way, Y that way, then positive Z is out of the board, so we'll take it as that. So this is a minus 10 radians per second K for omega, and that's what comes first. It has no I component, no J component, and minus 10 radians per second K. Minus K direction into the board. As confirmed with the right hand rule, put your thumb right into the board when you put your curl, your fingers in the curl of omega AB. The position vector, R, B relative to A, that's this vector here, B relative to A. Not too difficult. Well, in fact, it's these pieces that we've already got right here, the 0.4 and the 0.4. They're both in the positive direction, and they have no K component. So there's our matrix. All it needs is the units, and the omega row has units of radians per second, and we have units of meters on the R row. So there's the setup of the cross-marker. Jill? How do we know it's only negative time for K direction with the omega? Well, I told you the magnitude was 10. The minus is that as drawn, it puts the angular velocity vector into the board when positive K is out of the board. Because this is an x, y, z system. If we ever have x and y that way that in our usual convention puts z positive out of the board, this is z negative into the board, which is where the direction of this angular velocity. Okay with that? You need to get your right hand out. You can get it up there, get it to work. Put your pen down and do some pearls here. All right, so we can do this cross-product. Should we step through it, or are you comfortable doing cross-products without the review of them? What? Keep going? Or do the cross-product now? You're good. They elect you as spokesman? Chris is okay with you being spokesman. We don't have a choice. He speaks out. Okay, but everybody's behind him. So if you want to shake your head, Anthony? Just put the answer down. Okay. All right, you can double-check it. Cross-products all go the same way. You just have to get used to them, get a little bit of a rid. So it becomes 4i minus 4j meters per second. And we can look at that. We can confirm that. Remember, let's see this, due to the layout of this, we know that happens to be 45 degrees. The velocity of B relative to A is perpendicular to the line connecting the two of them. That's actually what the cross-product is saying for us. So this is the velocity of B relative to A right there. If that arm is at 45 degrees, this is at 45 degrees as well, because they're perpendicular. So we have a plus component in the I direction, a minus component in the J direction, and they're the same magnitude. That's exactly what we've got. So be careful with these cross-products. If you mess up a single minus sign, you've got the wrong vector, and we need this because this is the velocity of B that we need up here. So we can look at the picture and check to make sure we got it all right. And the result matches what we expect that it should have looked like. All right, so that's the velocity of point B that we were trying to find right there. That, then, we need, because we're going to put it in there. All right, but we need more importance, please. So now we have this. We don't have that. We're going to have to carry it through as an unknown. So we're looking for the velocity of C, but it has omega as an unknown in it. So we're going to have to get it. So velocity of B, which we now have, 4i minus 4j meters per second, that's the velocity of B that we just got. And then now we need to do that cross-product there plus omega Cb, which, remember, we're looking for. With angular velocity, there's no relative angular velocity between B and C. That R of B and C moves at some angular speed anyway. So mixing up the indices on omega doesn't harm anything because that vector's the same vector for the entire object. So we'll be a little looser with that. And R of Cb, that we do have, that we can draw out. So let's see what's the easiest step to take next. I think it's easiest for us if we do this separately, just so we do a piece at a time. So omega Cb cross R Cb looks like this. Omega Cb, we don't know its magnitude. We're looking for that. We're to find that. However, we do know, because this is a two-dimensional problem, that it's going to be in the k direction only. If this is a three-dimensional problem, we couldn't say that, but it's a two-dimensional problem. So we know that all of our gravitational vectors are either plus or minus k direction. We just don't happen to know which yet. So we'll put that unknown down in the k column there. And then R Cb, that's this vector down the length of that link arm, down the line connecting the two. And that's 0.8 in the i direction, minus 0.4 in the j. And we've got the whole thing. 0.8 in the i, minus 0.4 in the j, none in the k. And then all we need are our units there on the cross product. Let's see. If A B is going this way, then B C is going to be going this way. We can at least get a sense of the direction it's going to turn. Because they're both heading towards being flat, so it's going to have to turn that way. So that will help us as we work through the cross product to check things. We know that the result should be a plus k vector. All right, so if we do that vector, it's a pretty simple one. Just make sure to get the plus signs, right? Everything's plus. We get 0.4 omega C B i plus 0.8 omega C B j. We don't know the magnitude, but we do know that's the direction. Both of those are positive. All we're going to put in is the magnitude omega C B. So we can find that now. Let's see. Well, not now. We're getting close to us. Let's see. I'm going to erase this B, the velocity of B. Since we've used that, we don't need an integer. That's what that piece is there. All right, so let's put, now that we have this relative motion, the relative velocity vector, remember this is the velocity of C relative to B that we just did here. We can now add it to the velocity of B. The velocity of B, well, we'll just do it. We don't have to rewrite it. We have two parts that are in the i direction. 4 plus 0.4 omega. That's VB in the i direction plus VCB in the i direction. Just adding the two i direction components. Plus minus 4 plus VCB in the j direction 0.8 omega. So that's the velocity of the point C. The only trouble is, it's based on the unknown angular velocity of C B, which we don't have. So what do we do? Equation.com. There's a smart phone to help us out. Shoot, we need an equation.com app. There's something we need to realize. We know more about the velocity of C than we've taken use, taken a use of, make a use of because it can only go down this, it can only go down that track. So it has an i component to it only, which means this i component must be that velocity VC. So we can say VC equals 4 plus 0.4 omega CB. That's balancing the i components of the velocity we calculated and the velocity we know to be the case in this picture. This point C can only move along that track. So there's one equation. We've got two unknowns. We don't know VC and we don't know omega CB. That's our one equation. Where's the other equation? David? Velocity in the j direction, is equal to 0. This component must be 0 because point C cannot have any j component of velocity due to the constriction of that track. So we can also say 0 equals minus 4 plus 0.8 omega CB. So this was balancing the i component. This was balancing the j component. That's two equations, two unknowns. You can now solve for the magnitude of VC. We already know the direction. That came right out of the picture. And now we know the magnitude or can find the magnitude of the angular speed of the arm VC itself. We already know the direction but that I'll carry through anyway. So without laboring the point making you do the algebra, 6 meters per second on that and 5 radians per second. So we now know the full vector translational velocity of C and angular velocity of the arm VC. It's this one very useful tool of the relative velocity equation that does all of this stuff for us. We just have to figure out how to apply it. Just remember you can only apply this between two points on the same rigid body. You can't apply it to any two points in the problem. It must be two points on the same rigid body. So that's why we have to do it twice. C to B and then B to A because then using the fact that A was pinned we could finish the problem walk all the way back through it. Questions on that? Before I clear the board, we do another one? All right. Now we get to practice our circle drawing skills unless somebody wants to buy Samantha's circle template from her. What do you got a chance? We're going to look at a two-gear system. That's terrible. That's a little better. So there's one gear. There's one gear that rides without slipping. Now it doesn't matter if these are geared or not. We do not look at wheels that slip as they roll over tracks. So gearing would just guarantee that but either way we don't have a, we have a no slip condition. And then mounted on that is another gear, a smaller one on which rides what we call a rack. So since you're taking notes with chalk you can erase that little bit on your picture. And so we've got that picture there. So we've got a wheel that rolls along its own track and then another track that's connected to the inner and smaller wheel. So let's put some pieces to this. The holes, remember these two gears, concentric gears are mounted rigidly to each other. So they turn with the same speed omega. That's unknown and in fact one of the things we need to find. However as they do so, they give a velocity to the center to the axle. We'll call that point A. Point A has a velocity of 1.2 meters per second. And obviously the axle's going to roll, the whole thing's going to roll, such as the axle's always moving parallel to the lower surface. So of course that'll help a little bit. We want to find a couple things. What's the velocity of the upper rack? We also want to find velocity of the rack, upper rack. We want to find the angular speed of the gearing itself, the whole gizmo. And we want to find the velocity of a point D. Where point D is right here, point D. Find the velocity of point D. It's not going to be 1.2 meters per second. Because it's in translation with point A, but it's also rotating about point A. So we have to put the two together. We need to find those three things. For reference we're going to need a couple other points. So we'll call that point C that stands for contact point. And we'll call this point B between the two, between the upper rack and the smaller gear. So the last thing I think you need is the radius, the radii of the two gears. So the radius of the bigger one is 150 millimeters. And the radius of the smaller one is 100 millimeters. And the whole problem changes if these aren't actually circular. Because A is gone. But we're doing circular ones. We have to be generous with each other's drawings. Alright, so let's see. What do we need? We need to find the velocity of the rack. That's going to be same as the velocity of point B at that instant. Since there's no slip between the two, they're intimate and connected, we can find the velocity of point B, because that will be the velocity of the rack. Obviously it's got to be in the I direction only, so we're going to use that bit again. The rack is not going to be going up in any way just over. So let's see. We're looking for the velocity of point B. Let's relate it to the velocity of point A, because that's the other point we know things are going on. So it's going to have some component to the fact that A itself is moving, which represents the movement of the entire system, and the velocity relative to A. So we'll put in our relative motion part of that equation. Okay, VA is no trouble. We've got that. Let's see. The position vector B relative to A, we've got that, it's right off the picture. And this omega AB, well that's the angular speed of the entire thing. The angular speed of a rigid body is the same as the angular speed of any two points on that rigid body. So that's just omega. And we're looking for that. We know the direction. It's got to be into the board. So let's put this together. VA, 1.2 meters per second, I. That's VA. We already know that. We're given the speed. We know the direction must be in the I direction. Plus the cross product. Actually this is one we could probably pretty easily just do in our heads. Let's see. It's turning that way. Omega's into the board. So omega cross R. We know that that's got to be in the I direction. Let's put the whole thing together just for practice. So omega is only in the K direction. So that's 0, 0 omega. And we don't need subscript as the only piece that's moving in that direction. We expect that to come out with a minus sign as well because we know it's into the board. But we'll just make it simple and just have omega there and let it solve for itself. And then the position vector of point B relative to point A. Point B is 100 millimeters above point A. So it's 100 millimeters in the J direction. That right there is the position vector of B relative to A. So that has no component in the I direction, of course none in the K direction, and is plus 100 millimeters or a meter in the J direction. And our units, radians per second, that's now in meters. I just did that because the velocity is in meters so the radius in meters as well. And so that's pretty simple. We have the VA we started with plus now this relative velocity of B relative to A which is minus 0.15 meters omega in the J direction. And that's meters, that's radians per second, this will be meters per second. David, did that see a hand up? Omega's supposed to be positive or negative. Since we don't always know, just put it in as a variable and then the right sign will come out. So we know that it's in the minus K direction and that's not J and that's a K. We know it's in the minus K direction and we just got that. We got exactly what we expected. So that's the velocity of point B except we don't know what omega is. So we still need to find that. However... Where did you get that sign? It was minus 1.5. This should be just positive. What is? Oh, I must have jumped down a line on my notes here. Yeah, that's supposed to be 0.1. Sorry. Well, hang on, yeah. Let's look. Because this is the velocity of B relative to A. Yeah, so if we can draw that arm. Here's A. The velocity of B relative to that has got to be like that. So yeah, it should have been I. So I'm sorry about that. There's D relative to A. So that's I. Good. We would have had a lot of trouble solving that if we didn't have that. So we can put those two things together and then we've got all the pieces. Let's say yep. So that's the velocity of B which must at that instant be moving in the I direction and all it can do is if that represents the rack, if you remember, and it can only move in the I direction. So we know then that 1.2 minus 0.10 omega I putting those two together has got to be the velocity of the rack. Okay. Trouble is we can't solve all that. There's still an unknown and that's only one equation. So we need to do a little bit more with it. So we need to figure out something more about maybe point B would do. So we could find, let's see, velocity of B relative to equals C. The velocity of C plus omega cross R C B. Let's see if that would help then. Because then if we have that, we can set it equal to this and we can solve for omega. So the velocity of C is 0. Why? It's in contact with the floor and the floor is not moving plus we prove that on Monday and we'll see that the contact point is instantaneously 0. A bit later C's moved up to here and the wheels moved on farther but that's a moment later. We can't look at several moments at once. We have to look at one moment at a time. So this will then be I, J, K, 0, 0, 0, omega, just as it was before. And R C B, hang on, that's backwards. B C, we need B C. Because we're looking for the velocity of point B, we need the point B where it is relative to the point C. So that's 150 millimeters plus another 100, is that right? In the J direction. And that's radians per second meters of C units. Is that right? For the velocity, the position of B relative to C. So that vector's, well it's actually no different than this one, it's minus 0.25 omega in the I direction and that's the velocity of point B as established relative to point C and those two must be equal because it's the same point, velocity of point B. So bringing these together, 1.2 minus 0.10 omega equals minus 0.25 omega and so we can solve that now. So omega equals, and we're expecting it to have a minus sign on it because we had just the symbol there. So I believe comes out to minus 8 radians per second if we have all of our cross products and signs and everything, right? Does that work? Okay, so there's omega. There's other ways we could have done it just depends on which of these three points you want to work with and what order, there's different ways to do it and you should get the same answer. Instead of basing this on point, what we do, we did B relative to A first but then we could have done VA relative to C and then put the two together because that also had omega in it. So a couple of possibilities there, number one works that you have to see. There's no one solution through these because we have several points moving relative to each other. Okay, so that's the velocity of the row. We can now have to actually finish for that I guess VB is, now that we have omega we can put it in meter one of those and it equals two meters per second I and the minus sign on the omega even works out to give us the right direction of the velocity of point B. Okay, so we've got the velocity of the rack, we've got omega, now we need the velocity of point D. So with your permission, I'll erase this and we'll do that now. The velocity of point D. We know the velocity of all the other points we can pick any one of them we want but we'll use A since we are given the velocity of point A. We could do it relative to C, we have been doing all kinds of things with it. Any one will work. So velocity of D relative to A, velocity of A plus omega across R, we know the velocity of A so all we need is that cross product then. A is 1.2 meters per second I, that was given and the cross product is, let's see, omega we now know is 0, 0 minus 8. Remember the minus represents the minus K direction that we knew it should have. So it was a good chance to check our minus signs. And then the velocity, the position of D relative to A, that's that vector going from, no, D, D relative to A, that's D relative to C, too busy anticipating what we're doing next. That's the vector in it which is good, that's a lot easier. That's D relative to A which is simply 150 millimeters minus I and so that would give us meters per second. So we have 1.2 I, that's the velocity of A and this cross product comes out to be minus 8 K, I know that is the cross product so this comes out to be just simply 1.2 J. There'll be no I component because all the zeros, there'll be no K component because the zeros will only have a J component and it'll be the 0.15 times the 8. So it looks like it's moving at 45 degrees. So let's see, the line connecting them is 45 degrees and it's moving at 45 degrees which makes sense. If point C is not moving at that instant it's as if all the other points are in rotation about point C. That's a little harder to see that's a little more subtle but that's what we just proved there with that last little piece. So be careful with the minus signs be careful you get all the directions right take your time doing this right big, if you're right small and you make goofs but then I want to decompose this motion in a second. So is everybody all right with that? Okay. Let's see where that comes from. We have an object here that gear system that's in both rotation and translation. So in rotation we'll do translation first it's a little bit easier we know that point A is moving that way so if we look at just the pure translation part if point A is moving to the right at that speed so is every other point. So point B is moving to the right with that speed point D is moving to the right with that speed and even point C is moving to the right with that speed if we look at just the pure translation part And this is how we would have looked at it in the first part of the class, where everything was a particle. Where one point went, all the points went. Combining that with a rotational part. So let's look at the pure rotational nature of that motion. The pure rotation, well, it's rotating about its own center. So VA is not moving at that instant. VV has some velocity like that at that instant. We'll call that VV relative to A. Because that's where we've got our rotational part. And we know how big that is. That's r omega. We know what it's meant to do is r omega. The point C is also rotating about point A. So that looks like that. Point D is also rotating about point A. In fact, we'd have the same magnitude as that one. VD relative to A. So there's the pure rotation picture. And the actual motion that we get is the combination of those two. Let's see. VA in translation plus VA in rotation is just VA. Just what it was before. Point B is part VA plus part VB relative to A. Velocity of V relative to A. Let's see. What was the magnitude on that? I think we have it. But it's just from the pure rotation part, it's just r omega. So this is point one times the eight. That's point eight. The magnitude on that is point eight meters per second. Is that right? Just r omega in pure rotation. So this is, remember, 1.2 meters per second plus point 8 meters per second. This is moving 2 meters per second, which is exactly what we figured out. Remember, all of these are 1.2. And this is r omega because this piece is rotating and we're looking at just pure rotation of that piece and then putting the two together to get the general motion. All right, let's see what point C is doing. Point C is moving to the right with VA in the pure translation part. It's moving to the left because of the rotation. So those two are going to cancel each other somewhat. So it's going to be r omega. R is the 150 millimeters, that's A down to C. Omega, we already figured out was 8. So point one five meters times 8 is 1.2, right? So we have point C moving to the right, 1.2. We have it moving to the left, 1.2. In other words, it's not moving at all. Which is something we've done, proved on Monday anyway. Plus, it makes simple sense. If there's no slip here, and that's the contact point with something that's not moving, then it can't be moving either. And in fact, because of that, we can now figure out any point and its velocity along this vertical diameter. Even figure out what the top piece is now. It's just a bunch of similar triangles. Because at that instant, they're all rotating about point C. And point D is moving to the right at 1.2 and moving up 1.2. And remember, that's exactly what we found. It was moving at 45 degrees, and we found exactly that. There's the velocity of point D. And so all of those points, everything works out. The two motions put together give us this very same picture of general motion. In fact, we could have solved the problem that way. But I'm not sure. This was obvious. The other one's pretty mechanical. You just have to do the cross products. And just make sure you get all the signs right and you're generally okay. But it really does come out to be a decomposition of translation and rotational motion put together. Is there a question? Is this Morris triangle? No. I don't know who had dips on this. Yes, so don't try that in a bar. They'll throw you out so quick out the back door. Okay, so a couple other problems. Let's see what we can get. Hopefully, you're seeing, too, that they benefit from nice, large drawings. Let's see. Hard one easy one, hard one easy one, hard one easy one. We'll do this one. So, here's a cam of some kind which is fixed at the center and free to rotate. To it is attached a link arm. So, as the wheel rotates, that point will move around. And there's another link arm attached to it. It's the same length. And the end of that point is fixed. So, as this little wheel goes around, these two pieces are going to kind of go in and out like that a little bit. For whatever purpose, maybe it's a, maybe that's a motor to make a paint can shaker or something. Who knows why they do these objects. So, let's put a couple more numbers on there. This will be A, B, and C. We can label that point D if we need it. We're interested in the operating point when this is 60 degrees and B, C is horizontal. If you were really analyzing this, you'd want the full motion but we'll just do a couple points so we can get some answers we can check. The length of these both arms are 0.2 meters. And the radius of this is 0.1 meters. So, we have an angular speed on that arm, A, B. Oh, that's given, 30 radians per second. And we want to find the angular speed of the other arm, B, C. We can figure out the direction as A, B goes this way, B, C is going to go this way. So, we can check the direction as we work through the solution, although plus and minus signs should all work out. And the angular speed of the CAMCV as well. Both of those, I want to know them just as the instant when, this angle is 60 degrees, that angular speed is that, and B, C itself is horizontal as well. All right, troops, what are your recommendations? How can we find the angular speed of either of those? Well, if nothing else, let's start from where we know and work into the part and see how things solve. So, we know the velocity of A and B is connected to that. So, let's start there, moves with respect to A using our relative motion equation that works for any rigid body, any two points on a rigid body. Well, let's see, A's not moving. So, we could find the velocity of B. If we know the velocity of B, we can then hopefully use that to find the angular velocity of B, C, and then use that to find the angular velocity of C, D. That's where we're going with this. Let's see, omega AB is minus 30 K, ratings per second. The position of B relative to A is, let's see, it's at 60 degrees at 0.2 meters. So, it's whatever those numbers come out to be, the 0.2 cosine 60, 0.2 sine 60, I'll just give you the numbers. So, this one's 0.1. That's 0.2 times cosine 60. 0.2 times sine 60 is 173. Radiance per second and meters are the units on those two lines, respectively. And we know that it should give us a result that would look something like this. Let's see, velocity of B has got to move perpendicular to the arm that connects them because the other hand's not moving. So, that's got to be the velocity of B, like that. We better get that kind of solution when we do the cross problem. This is a way to check all the minus signs. And we do, we get 5.2i and 3.0 j meters per second. And that's just what we think B should be doing, is something like that. A little bit more in the i direction because that was a 60-degree angle down here, not a 45. So, just what we expected should be doing. But now we know the velocity of B, we can relate that, so that we can get the angular velocity of B.C. next. So, let's move down the piece. Velocity of C is the velocity of B, which we now know, plus the relative velocity between the two. Omega B.C. crossed R.C. relative to B. Again, just our relative motion tool applied to A, rigid body. In this case, the rigid body B.C. Velocity of B, we now know. We just figured it out. And so, now we can do this part. And in there will be our unknown angular velocity and we can solve for it, hopefully. B we just got. This will be the matrix, Omega B.C. is purely in the K direction. We expect that to come out with a plus sign because we already know the direction is moving, we just don't know the magnitude. And the position of C relative to B is 0.2i. And that's it. And we can do that cross product. Add to it the velocity of B, which we already have. And we have the whole picture, which should look like this. So, the velocity of C, when we combine all these parts, is just the 5.2i here. That was the only i component. The J component comes out to be 0.2 Omega B.C. Which remember, we don't know, minus 3, which was the part of B in the J direction. And that's J and that whole value is meters per second. Is what plus? Here? Here? In the J direction, so it's that one. Remember, there's a minus r in there, that, cross that, minus that, minus minus that, so there's 3 minus signs. So, yeah, that should be a minus sign. Plus, it matches what we expected. This is a stronger indication. It should be a minus than that is, so if I saw, if I thought that should be a plus, but I saw this, I'd go back to that and find out why I had a plus there. I shouldn't have, something wouldn't be messed up. There's 3 minus signs on that J component with these numbers. Okay. How do we solve for Omega B.C.? Zero at this. We know point C at this instant can only move in the I direction. In fact, there's a velocity of C right there because this whole part must equal zero. We can use that to solve for Omega B.C. There's no J component of the velocity of C because of the instant shown in the picture. That's all point C can be doing at that instant. An instant later, it starts to go around the corner. Things are different, but at this instant picture, with that arm horizontal and the point C right at the bottom, we've got the last little bit. So, we can solve for Omega B.C., which is 15 radians per second. That's its instantaneous angular velocity. And some other instant is going to be going at some other speed as all these angles change. And we need to find now the angular speed of the wheel D.C. How do we do that? Last thing, and then we're done. Yeah, actually the easier way to do it is realize this is in pure rotation. There's the line connecting the two. There's the speed. Then we know that V.C. equals R. D.C. Omega D.C. because it's already perpendicular to the center rotation. So, it's just pure D. The wheel itself is in pure rotation. So, we know the magnitude of V.C. is the 5.2 that was left. We know the diameter of the wheel, it's 0.1. And so, you can figure out that the angular speed of D.C. is 52 radians per second. Now, depending on whether that's constant or angular speed of A.B. is constant tells us what's going to happen next, but we're not going there. This is our big tool here for this relative motion. And you've seen, you may need to apply it several times on one object, however, do not forget. This is on A rigid body only. It does not apply connecting two rigid bodies. So, I should say one rigid body only. Those two points must be on the same rigid body to apply that equation. Do otherwise, add your own peril. Questions? We know the magnitude of V.C. because the J component is 0. We know the I component is 5.2. So, we know that this, the velocity of point C, it's in pure rotation about point D. So, we use the rotational motion, the rotational kinematics equations that we got in Physics 1. V equals R omega. That actually is the result if you did the cross product of omega across R. You'd get exactly that anyway. Any time two vectors are perpendicular, the cross product is just the product of the magnitudes. And these two vectors are perpendicular. Right there. All right, that's it.