 We'll do one of those as a warm-up and then we're going to look at another way to do some of the very same problems, then you can make your choice of which one you think is the yummiest. So imagine we've got running in a track of some kind, some little roller that happens to be attached to an arm that is at 45 degrees. Now remember, a lot of these problems, just to get through them, to be able to do some stuff with them, we have to say this type of thing where we say at this angle, find a velocity. At any other angle it's going to be a little bit different than these things tend to run through cycles of some kind possibly. So it's a little bit artificial. So if you go into this business you're going to need to know what these things are doing at all times, not just at one single little angle, but we're just getting started here. So this happens to be 45 degrees and this angle here is 60 at the moment shown. And a couple of the lengths, let's see, this is 125 millimeters, this is 300, and that arm happens to be turning at four radians per second. So we want to find the velocity of the roller in the slot, so maybe I'll call that D. You'll find out shortly why I've left C out of it, it's not that I'm prejudiced against the letter C. Gosh, that would throw out 126th of all the great Sesame Street episodes right there. I just can't have that. All right, so we want to find out the velocity of D. All right, what we'll work with on Friday if you remember was the relative velocity equation, the fact that the velocity of any one little piece of something is going to be related to two things, one, the velocity of any other piece of that, and remember this only holds for a rigid body, you can't do this from one body to an X. This is a rigid body equation and it's a vector equation, don't forget those two things. So the velocity of one end is related to the velocity of the other end, plus what the relative velocity is between the two, D relative to B. If you were sitting on NB looking down at D, what would it appear to be doing, remembering that it's, again, a little bit of an artificial construct here in that if you're sitting at B, you'd have to ignore the fact that they do yourself for moving. If you were moving at constant velocity, you would only have visual clues anyway that you were moving, as often happens in the car. You know, you're going down the highway, you don't know you're moving unless you're looking out the window, other than you bump on a little bit. So that is for a rigid body and is a full vector equation. So if we want to find out the velocity of D, then this is as good a place as any first to start. This is what we're working on Friday. The thing to do is start putting together what you know and what you don't know. For example, V, the velocity of D, well, due to the fact that it's in the slot and due to the fact that things are going this way at the time, we know the velocity of D must be down that slot and in that direction. It just can't do anything else. So we can use that. We can say, maybe we want to say V being in the I direction, maybe just keep until our normal orientation. We'll call that the minus I direction. It doesn't have to. It doesn't matter what we're looking for is the magnitude of it anyway, but you have to be consistent with anything else that goes into it. The velocity of B, well, that's the same thing in that B and A make up a rigid body as well. So those are also subject to this relative velocity equation for rigid bodies, except there's a little bit of difference. Point A is not moving now in this problem. So B is going to move relative to A. Well, we can envision just what that's going to be because this rigid body, the arm, link arm, A, B is pinned down at one end. B is only going to go in that direction. It just can't do anything else. Now, you can use the geometry of the problem to come up with just what that angle is, but you can also use the relative velocity equation that we are using on Friday that this velocity, this relative velocity can also be found from the cross product of the angular velocity with the position vector between the two. And if we could do that, that would be put in here for B, B. And we could do the same type of thing for the relative velocity in D to B. The angular velocity vector D, B crossed with the position, relative position vector. And that's the kind of thing we are working on Friday. All right. So if you need to see it visually too, though, this, all of these equations have three components to them such that if you look at that equation in graphical form, it must form a perfect closed triangle. And for some of you, maybe that's a better way to solve some of these problems. For example, we can look at it this way. We know that the velocity of point D relative to point B is always perpendicular to the line connecting the two. So that's got to be the velocity of D relative to B. Because if you were sitting at B, point D couldn't come any closer, couldn't go any farther away by our definition of a rigid body. That means it can only go to your left or your right as you sit at point B, looking down the arm at that. So we can draw that as a rigid or as a vector triangle and maybe for some of you that can help sometimes. So VB looks something like that. I'm not sure just how long it is, but I know it's going to be in about that direction, VB. Add to it VD, which has got to be something about like that, VB plus VDB, those have got to add together to give us the velocity of D, which we know is horizontal. So we know that these two, we know that those two, the original part of these two vectors must have equal and opposite J components for them to end up as a horizontal vector. So either way you can now solve this, maybe use in law of sines and cosines, or you can use it as the components or however you want to do it. Either one, they're fundamentally the same thing. It's just for some of them we need to do this at least to come up with a magnitude. So that's what we were working on Friday, a slightly different view to it. Actually another way you could look at it, just in case you're not confused enough, here's DV. We know that D has to go that way. We know that V has to go this way and we know that they can't get any closer together. That means they must both have the same velocity along its own length, along the arm's own length. I don't know what we call that. Velocity, I don't know, what? What do you want to call it? Quick, I don't know. Something. V, two. I don't know. V prime. We'll call it V prime. Two components must be the same. Otherwise point D and point B would be either getting closer to each other or farther apart from each other. All of these are the same thing, are perfectly equivalent to each other, and may help you see what's going on a little bit that a one may be more of your liking than the other. So if you like vectors and cross products, there you go. You're in hog heaven, right? Vector cross product heaven. This is it. You like the graphics and semi piece of each, there you go. So give it a try. Whatever you like, whatever works, see if you can come up with a velocity VD in the next couple minutes here, working together. What would Jay be talking to anybody? Not his name. Jay, we used to be so close. The order in which these are read, whether it's on R or B, and we're going to have acceleration on Friday, it's if you're at A, where is B? Or if you're at A, how is B moving? So it's always drawn, the vector itself is always drawn from the second letter to the first. So bats are D relative to B. And you do have to have that right if you're doing the cross product. Boy, you don't have to have it right if you're... Well, I guess you don't need it through the diagrams, particularly like doing cross products. But we have found on these 2D problems, they're fairly straightforward ones because the vector in which they result is only going to be in the Z direction. And there's really only one component to it. Can I find this? Not the whole component. Remember, there's two unknowns, because this is a vector equation, it's really two equations. So the two unknowns would be, well, VD, you're going to know from there, VD and omega DB. And you could solve the equation, I think, so that you can avoid even finding that. Do the right ones. Elimination first. These are full vector equations. So don't come to me looking for help if you don't have vector signs over them. Doesn't hurt necessarily to find omega DB, just to make sure the sign is right on it, because you know in what direction it should be rotating. You can comment this in your algebra. Yeah, you do. Because you need... Well, you don't have to if you can find DB some other way. But you need not just the magnitude of DB, you also need the direction that you can get that from this drawing for this one. So no, you do not need to do that if you can do it from the diagram itself. The, what, the 30 degree angle here. Half heads is of, half weights is a full width. Equations. So an i component here, an i component here, and they must be equal to j components, and they must be zero. D has no j component to it. Automatically, what's the vector DB? That was omega, omega DB you have there Bobby? Yeah, yeah, 6.79. Yeah, that's what I had too. So once you've got that, then the DB should be just the next part you pull out of it with the, must be a math error, because you've got the right angular velocity. You've got the right thing here. So as long as that's right, well, let's see, as it, as it moves, this is points going to get a little bit higher. So it's going to have to go something like that. It's going to turn in that way. So omega, omega DB should be positive. Do you think you've got a positive for omega DB? No, these are millimeters. Millimeters squeezed in there now. It's all in millimeters? Yeah. It's just a decimal place though. Sometimes you use them, sometimes you don't use them. Well, sometimes it is DB in the I direction. It's not minus this, it's minus this magnitude of the direction. And then, when you do this side of it, you should have a couple I components you can put together, and they must equal that. And you'll have a couple j components together, which will equal zero, because there is no j component here. And that's the same thing as you see here. When you put these two vectors together, you end up with no j component. Where to get the omega DB from? This is two equations, because it's a vector equation in I and j directions. Two unknowns. VD, the magnitude VD is unknown, and the magnitude omega DB is unknown. So you can solve those two equations for the two unknowns. You might be able to solve this without even bothering my bad unknowns, since it wasn't asked for it. But I don't remember how the algebra shakes out. Yeah, it looks like you're going to have to find omega DB in the j direction, and then use it in the I direction to finish the problem. Okay. So Bobby, did you find your mistake? Is it a little math error? Yeah, I think it is. Does it kill you? I don't think. All right. We're going to come back to that in a second as we look at another method. So we'll start with a clean drawing when we get there. We've seen from other things we've done that anytime we have a rigid body rotating about some particular point, in this case I'm going back to a quick discussion of pure rotational motion, no translation. Then we're going to use what we've got here to go to the more general problems we are working on where there's rotation and translation. So remember back when we had pure rotation. So for argument's sake, let's say it was rotating that way. We know now that any other point on the object, we can figure out its velocity from the fact that it will be moving perpendicular to the line connecting the two, and with the velocity that's proportional to the distance between them and that angular velocity there. In other words, the velocity of A equals omega. That's nothing new, but that's just from physics one. But we've maybe taken it a little bit farther in that we also know that any other point is going to be moving slower if it's nearer to that center or proportionally faster if it's farther by similar triangles. In fact, if we label these points what we're really saying is that the whole body, every part of that body, every line we can describe on it, every two points we could put on it are all moving with the same angular velocity and that that's proportional to their distance away from the center. So if we take that solve for omega that's the same for all of them, we have just where that rA and rB is the distance from the center of rotation. And we can even take it back off the other side, just continue this line straight through and we know then that the velocity of any other point is all found from the same similar triangle. We can do this any place we need. We can even do it some brand new place because they all have the same angular velocity. We've already used this several times when we looked at an object rolling without slipping, we know that the center is moving half the speed as a point on the opposite side at that instant. We've done that in a couple of problems. Is that right? Is that familiar? At that instant. In fact, since that point is sort of ephemeral in that a couple, an instant later that point doesn't even exist. We have a new center, a new contact point. This is really what you might call an instantaneous center. It's only the center of rotation for one instant. An instant later there's a new contact point and it's an instantaneous center. So we're going to use that idea now for our next approach to these general motion problems, the idea of an instantaneous center. The object that's in pure rotation, it's obvious not only is it an instantaneous center, it's a constant center. But for an object in general motion like a rolling wheel, this idea of instantaneous center can be quite advantageous for us if we can find it. Turns out it's not too big a deal. Any rigid body at any instant in time is rotating about an instant center somewhere. We just have to find it. So imagine we've got one point going in that velocity, some other point moving with that road. Just want to get a decent picture so things will work out right to start with. We'll say we've got some other point. It looks like something like that. We want to figure out something about the motion of that object at that time. We know that at any time any point is moving perpendicular to the center point. It's obvious if it's pinned at that point but it's also true if it's in general motion. So if we draw the line perpendicular to these velocities, we know the instant center must lie somewhere along that line because V, A, A is moving perpendicular to that line at that instant. The same is true for B. Those two cross is the instantaneous center and we know that those two lines connecting them always have the same omega as each other. If we can geometrically find where that point C is, thus get the distances AC and CB, then we can figure out the velocities. If we know the velocity, we can find out the angular speed of the body or whatever it is that we need to find for any point. You do a little sketch. It looks similar to this. It's bound to be that the instant center is. Some sketch like that, we just figure out where the instant center is. We'll see if we come up with about the same thing. Notice it does not depend at all on the magnitude of those two velocities. A location of the center doesn't. Well, it kind of does, I guess, because the farther they are from the center, the faster they're going to be moving. But in terms of just sketching it out now, in terms of where that center is, it doesn't matter. You're in technical free-in sketching, too. You don't need that kind of felt. Great, Jake. You did an awesome fool when you didn't hold the hand at all times. You got one of those, didn't you? Oh, yeah, definitely. All right. Very simple again. Perpendicular to the velocity, nothing else matters. And it's very easy, as you'll see in some of the pictures we're going to work on, that it's very easy, that it's distracted by other things in the diagram. Perpendicular to the velocity only. You should have gotten a center maybe somewhere down there. Notice, too, that it does not necessarily need to be on the body. It's just an instantaneous center. Nothing changes if I simply change the shape of the body to do that. It's still the instant center. So it need not exist on the body itself. Once we find that, you can see the direction of rotation. And then using the geometry there, hopefully, we can find the rotation. A question about the first line. How do you know what's going to rotate on a clockwise? That's the...on the first one. First one I wouldn't. I guess it couldn't move like that. The velocity arrows I picked, because these two arrows are moving towards each other, and they have a component such that the distance AB would shrink. So this one was just a bad sketch. It needed to move B out more this way. I guess, just a bad sketch. There. Now, we're rotating about there. That looks a little better. But, do we... So the original drawing would work if the two velocity vectors were really different sizes here. If, like, B-A was really big, B-B was really small. So you don't get any... No. Because the location of the center has nothing to do with the size of the velocities. It only has to do with their direction. And as I had that one originally, A moving this way, and B moving something like that, when you find the instantaneous center, there's no way a rigid body can be rotating about a single point at any instance such that they're doing this. Because this one has got to be going that way, and this one's got to be going that way. And that's not possible for a rigid body. They must have the same angular velocity. These lines must have the same angular velocity as each other, because they're on a rigid body. So that was, unfortunately, and I apologize, just a bad sketch. All right. Let's go back to the original problem we're doing today, see if we can get this to help us then. That was D. Now you know I didn't have a point C, because that's going to be the instant center. We know D has got to move like that, because it's down the slide. You know, B can only move like that, because arm A is pinned at the other end, where these two perpendiculars must cross. You've got to be careful with the geometry that it doesn't make you think it's someplace where it isn't. I don't have a protractor for this. I'm just sketching the board, so it can't come out in pretty different spots. So be careful with it. You know that's 45 degrees. That side's vertical, because that's perpendicular to a horizontal velocity. It may go down to A. It may not. It does look like it in my cartoon, but it might not look like it in your cartoon. And we know this angle, or this one if you'd rather, is 30 degrees. And then somehow all these points come together. Remember we also had these, we had the distances, how long were those arms? 125 to 300. This was 125, 300. This is 125. Now I have that triangle there. I was doing omega inch per second it was. So both of those are also turning with that same speed. Because they're at that instant rotating about the same center. So if we could find out this distance, we know it's angular velocity, we can then find the velocity of D straight from that. Lobb signs, lob cosines. On this thing, sometimes the geometry is a little bit easier. Sometimes you need something like that. So you have two angles. Usually the lob signs I guess works better for that. You figure out what this angle is. Good triangle. There's two right triangles as well. Now whatever geometry works for you, as long as you get this distance cd. Because the velocity of D is equal to the angular velocity of, not a, cd times the distance between those two rcd. Omega cd we've already got. It's got to have the same angular velocity as bcd. So we only need to find that distance there. And you can do it by lob signs or lob, lax and lob cosines would be easier, probably. So what dc over the sine of the angle at corner b, what, 105 degrees? Is that just like that? Yeah, well this arm was at 60, so this angle must be 30. Because together they're 90. Remember, this is vertical. This is vertical and this is known to be 60. So that must be 30. It doesn't go directly to A, as a matter of fact. But it runs through that member of A. But, see A, this was, again, these are sketches on the fly. So A is really over here somewhere. D is not necessarily right above A. It just kind of looks like it in my drawing. But it's not. Vb must be perpendicular to that arm because it's pinned down at one end. To make this more to scale, that's 125, that could be. A is really much farther back down here. That's more what it looks like if it drops to scale. Because this is well over twice as long as that. So using the law of signs, which remember is the magnitude of one side over the angle opposite it. That ratio holds for any of the three sides in the diagram. See the importance of technical freehand sketching? You got to be really careful that your sketches don't tell you what answers there that isn't. Just because your sketch had to come out however you want it to be. It's a lot different. It will originally be in the 4.64 meters per second and the negative 5. Yeah? That'll get 4.3 right. Well, let's check DC. So solving this, db is 125. We have sine b, which is 105 over sine 30. That's the side, the length of DC. What do you get? This angular velocity is not equal to this one. That's why this is in trouble. Because they're not on the same instant center. So those two together then represent with this point c, a separate rigid body. So actually this is omega db. Because it's the same angular velocity as that. So db we know, rbc we can find from the other sine 45. We already know the velocity of v. There's millimeters away when we have four readings per second. DC, what's this come out to be? It's 76. Sometimes when I break that up into two right triangles, I just add the extra moments. I get different answers. By breaking it here to get that distance? Yeah. Okay. I don't know how you're going to do that on this triangle. Because you don't know this distance or that distance. Oh yeah, because that's not today. I corrected that already. No, I know. That's the velocity here of this, which is this whole arm moving at four readings per second. That distance is 300 millimeters. So it's the product of those two. That's vb. Pictures can throw you off mightily. So let's be careful. Where did what come from? Let's try another one with a little more terrible drawing. Sorry about that. Make sure we did that right this time. 100 millimeters long. We will say c for the instant center, wherever that is. And that arm, bd, is rotating it two seconds per second in that direction. Find the velocity of a. Buy the method of instant center. Check it. That's okay. B and d have to be level with the midpoint. Remember, you're not paying the attention necessarily to any of the individual arms, other than to get the velocities. And use the velocities only to get the instant center. The mean itself. Look at the velocities when you find the instant center. What's point b going to do? Straight down because the other end is pinned. That's all it can do with that instance. So there's vb. In fact, we can figure out how big it is because we know the distance. We know the angular velocity. What's a going to do? It's got to go horizontal to the right because that's the way b is going to pull it and it's pinned at the other end too. What's your next step then? Perpendicular to what? The velocities. So here's one velocity, the perpendicular to that. So we know the instant center lies somewhere along there. Perpendicular to the other one and where they cross is the instant center. The velocity of b, how do you know that it's not perpendicular with a member of ab? Because the r and bd is pinned at the other end. But it's also pinned at ab? This is, these are separate rigid bodies. This rigid body is moving at both ends. This thing is in general motion. This is in pure rotation. So b cannot do anything other than follow in a circle around point d. Now don't make the mistake of thinking well if this thing kept running sooner or later it's not going to work. Because like I said, we're looking at these at a particular instant in time. Not as if they're going through an entire circular motion. Which they couldn't. Well maybe they could. Maybe when bd got all the way over here there's still enough length left in everything to reach general. But that's not the question. The question is at this instant at this instant find the velocity of a. Do you need any? Now with these velocities you know that this rigid body has some angular velocity like that. At that instant where ab and c make up an instantaneous rigid body. And that must be equal to points now make an instantaneous rigid body. This is the same sense of rotation that the piece ab itself would have. For bA you can solve it for omega if you need it. It's perpendicular to the velocity no matter what the arm you're doing. Very tempting to just make the velocity. Thanks. Nope. Let's see. Do you know what vb is? Or can you find it? We know what vb is because it's that angular velocity by that distance. So solving for vA. What's vb? 2 times 75 0.15 per second. RBC. BC is well if this is see this has got to be 50. How do you know that thing? Because I said that this was level with the midpoint. So now we have a triangle 50 there. How else do we know about it? Missing something? On the page. Numbers in? I think something's missing. Because all I have written down is that 50. I have written down. My solution, that's 175. Did I give you the distance between the two points? I think what I had was the length of that. I must have recopied my diagram and not drawn and written that down. So how long would that be then? Does that come out to be a nice round number? If you square that and square that and add them to square root that. Does that come out to be a nice round number? Nice round, doesn't it? So what's this angle? Is that nice round? We always have nice perfect round numbers in engineering. All angles are 45. All dimensions are multiples of 5. Yeah, go with one set of 5. I think it was just probably on the other diagram when I didn't really copy it. Did you contract how many mistakes of that today? I think I'm confident it was at the midpoint. We've got to see if we can do this on tape. For 3 meters per second then. The simple elegant method it presented. It's right. See now I've got to do this again next year. I've got to teach the hands again next year so I can fix this tape. Hold up, hold up. What's the omega equals the A of RAC? We have RAC. Omega is not even 2 meters per second. No, that's the angular speed of this arm. It's not the angular speed of what we really have is a rigid body that's like that at that instant where one side is the other arm whose angular velocity we don't know and made up of the connectors to the instant center. So that's a different angular velocity than the one that was given which is what, the E. I'll let you choose by whichever method you want. Here's the your get out of class question. Have a wheel rolling such that it's center 1.2 meters per second. Let me make sure that's right. Nothing else has been right today. Yeah, 1.2 meters per second. Here attached to it upon which is riding gear and rack system. Outer wheel is 150. Radius. Find three things. No meters. Meters. The size of a football field. A football field and a half. I want the next club to make that for extra credit. Find B of the rack and the angular of the wheel velocity of point D level with the center. And you can do by whatever method you want. Instantaneous center relative velocity or, I don't know, any other thing. The velocity of the wheel or in other words the center of the wheel. If that was on an axle attached to a cart or something, that's the velocity. Is there an instant center here? There always is. Every rigid body has an instant center at any instant. Is there one here? Talk about that being an instant center. Anyway, plus you know that the center has got to go level with the ground. This upper rack has also got to go level with the ground. And so if we draw perpendicular to the two of those they line right up somewhere on that line has got to be the velocity and we've already done that type of problem here. And then B, I'm sorry, D must be perpendicular to that one. And notice J sits that distance is not the same as that distance. That velocity can't be the same as that velocity. Point C at A in the rack B. You can find out what the angular speed is and then you know that ratio is the same for any other object at that instant. Velocity for the entire rigid body is the same. Yes, because they're all in the same body. All of those points are rigid bodies on the wheel so all of them must have the same angular velocity. The angular velocity isn't the same when you go from that to some other rigid body. For example, the rack itself is a rigid body and it has a completely different angular velocity. In fact, it's zero. Any instant on the well, any object on the rigid body all have the same angular velocity to any point, not just C, but to any point. Because the whole wheel has that same angular velocity. Now if they're slipping that'd be all together different. Then point C would have its own velocity that would be at the rigid somewhere else. Yep. Velocity at D. Yep, that's right. So you get it going early. How's that? Hold on, get moving. I'm just sticking away. To make sure you label that. That's two. Two meters per second. Omega is eight. Eight gradients per second. Yep. Always got it.