 times the rate at which the object is turning. We did that in physics one. v equals r omega. Look familiar? I hope. If not, assume you're physics one because of it. Now that's true for any other point on the object as well. Perhaps point b. We connect the two, draw the velocity vector perpendicular and then calculate the magnitude as the product of r omega. So this is r vc omega. And we could do that for any possible points on that rigid body. Every point in reality which is simply orbiting around that center and it just depends on how far away they are from the center is how fast they're going. That's review. So the deal is with the method of instantaneous center is if we have some rigid body here and we know how two points are moving. So some point here, let's switch things up a little bit. Say that we happen to know it has a velocity in that direction. We don't necessarily need to know the magnitude for this method but we do need to know its direction. And we know some other point on the object. Point b moving in some other direction that is known. Something like that. If we draw perpendicular lines to those vectors, those two lines will cross somewhere at what is the instantaneous center of rotation for this object. It may be in general motion but for a single instant in time it's as if it's rotating about a single point. It's not unlike what we saw with the wheel rolling along the ground. The bottom, the contact point is not moving at all and it's at that instant every other point was rotating around that point. We draw perpendicular to these where those perpendicular cross is an instantaneous center. An instant later that object will not only have rotated somewhere else it will have moved somewhere else because we're looking at general motion and there may, there will be another instantaneous center somewhere else. But for this instant in time that's where the center is located. We can then use the magnitudes, I don't know how an object is going to rotate like this with the two velocities I drew because it's supposed to be as if they're both rotating around it but the two directions I picked they're going opposite ways so I don't know how a rigid body is going to do that. But that's the technique. I guess the easiest way to change it up is to just put a on the opposite side. That's what happens when you do things on the fly. There it works. Now a rigid body could have velocities in those directions. Couldn't in the other directions not for it to remain a rigid body. Now we can see that at this instant in time the object is rotating around point C such that the two points we know have those velocities. Then for any other point on the object all we need to do is connect it to the center that we've now found. We'll know its direction which will be perpendicular to the insipati and depending on how far away it is we can now find its velocity as well now that we know the angular speed. So the sole limitation really of the method of instantaneous center is can you indeed find that point? Are you able to do the geometry such that you know where that point is so that you can then use it for any other point? If you can't easily establish this point it's very difficult to then use that and it might be in the long run easier to do the relative motion method anyway even with this cross product. But for some problems where the geometry is straightforward enough and you can find point C then you can use point C for any other object any other point on the rigid body. One quick caveat is that if you happen to know the relative motion or the absolute motion actually of two points such as that and well let's see preparing for what I want this to show. Let me change this a little bit. I think that'll work. Let's see. Once we connect the perpendicular to those two points notice that the point of instantaneous center the point of a parent instantaneous rotation need not actually be on the object. It could be off of the object and it appears as if the whole object is rotating about that point but the problem still remains you need to establish that point C and then once you have you can use it to find the velocity of any other point on the object. Alright let's see how it would work for that problem we just did. Roller that runs in a horizontal track trying to find that velocity it's connected by a link arm at 45 degrees that's connected by another link arm which itself was at 60 degrees. You have to be careful when you draw these because there there are times when the inaccuracy of the drawing will lead to what seem to be the right solutions but are not necessarily just has to do with the the illusion of a two-scale perfect drawing and you have to be very careful with where you're establishing these these points. So the we know due to the rotation of this lower arm let's see this was a b we call that b you know due to the rotation of the lower arm that point b is moving in that direction. It must because it's in pure rotation about point a. We also know however the direction of motion of point c which I'm sorry point d which is in that direction because it's constrained to the horizontal track. If we draw perpendicular to those two velocities though we've already got one as it goes right along that arm a b perpendicular to the other velocity d would be straight up and down where those two lines cross is the point of instantaneous center which in this case would be right there. So we can use that as long as we can find this point it's it's not too difficult it's right there on d maybe if I drew a triangle it's a little easier to see so there's point c there's point b and there's point b now we need to locate that point c because we're going to need to know the distance from c to d to figure out the velocity from the angular velocity of that arm which in this case is pretty easy because the angular velocity of bc is the same as the angular velocity of cd so we're imagining this to be a new rigid body rotating in that direction and that will then give us the velocity. So when we find this instantaneous center we treat those three points then as a new rigid body and we can figure out then the angular velocity as needed so let's see how that works. Vb was originally found from r a b omega a b and those were both known if you remember that arm a b is 300 millimeters, 0.3 meters and the angular speed was 4 radians per second velocity is 1.2 meters per second that's the velocity of b. We wouldn't have to do that in the other method as well remember we needed that for the relative velocity of d to b. Now we know that that is also equal to the angular velocity on this new virtual real rigid body of dbc so it has to do with how far it is from this new point c which with the geometry we can figure out might as well give you that it's just a matter of the law of sine and cosine. We know that's 45 degrees. We know this is 105 degrees because that's the 45 plus the 60 of the two arms originally and we know that this one side here is 125 millimeters. With the law of sines and cosines you can determine that point c then is such that this distance is 176.8 this line is 241.5 so that completely establishes where point c is and now we can figure out what the angular velocity then is of this new rigid body. Angular speed of that the new virtual rigid body made up by the instantaneous center and the other two real points is the velocity vb that we know from the set up of the problem divided by the distance that point is from the new instantaneous center per second divided by this new distance 176.8 millimeters. That comes out to be 6.79 radians per second. Notice that the angular speed of this new rigid body with vc even as it lies along the link arm that line vc does not have the same angular speed as did the original link arm itself even though they all lie on each other the points a, b and c all lie in the same line. So you need to find this new virtual angular speed of that new virtual rigid body. Now we can use that to find the velocity of point d. That's r cv omega cd which is the same as omega vc because they're on the same rigid body. The angular speed for any two lines on one rigid body are the same as any other lines on that same rigid body and we've got both of those. That's 241.5 millimeters. The new omega vc that we just found 6.79. 6.8 is close enough and so that comes out to be 1.64 meters per second. Is that what the speed of point d was before? So no cross products required but more geometry required. Depending upon which you prefer if you're happy with cross products then the relative motion method works. If you are intimidated by the geometry then the instantaneous center works as well. Both are somewhat limited especially this one is limited that it only finds the situation for an instant in time. It doesn't necessarily work for the overall motion of machinery which would be probably just as important. It's very seldom we have machinery that runs only one single little angle and does nothing else because if it's moving it's not going to be at that angle an instant later. You could make an integral out of this. Oh well yeah you probably could just don't forget an instant later c isn't there anymore so you have to find where the new c is then all those distances to change all these angles speeds have changed and then if you've got an object with several link arms it even gets more more complex but doesn't mean it's not doable. We're not going to do it just because it is quite involved so we'll stick to just a more straightforward solutions at a particular angle at a particular speed. Alright let you try it with another problem we did on Friday. We had this one where we had a wheel that the center was fixed and then the bottom of the wheel was attached to a link arm that was horizontal at the instant shown and the other link arm which happened to be the same length was also fixed on its end. Remember doing that problem Friday? We did it doing using the method of relative motion relative velocity and so take a look at the same problem in terms of the method of instantaneous center. I think we labeled that point c but now we'll label it point something else. Well you can use whatever letters you want. You're big boys and girls put your own letters. That was the setup we had there. 30 radians per second. Each arm was 0.2 meters. The radius of the wheel was 0.1 meters. Okay if you remember we wanted to find the angular speed of the two link arms. I don't know the link arm BC and the wheel CD but also wanted the speed of point C BC omega CD and the velocity of point C. That's what we're looking for I believe. Alright so give it a look using the method of instantaneous centers. First thing is to determine where the position is for that instantaneous center and then determine if you can actually solve for it geometrically. You can't find that point C or if it's way too difficult to find it it might just be easier to do the cross product in the first place. It might help if you make yourself a simplified drawing. Make it clear which points don't move whenever you want to note that and then figure out where the instantaneous center is. So I want you to do that first. Show me where you found the instantaneous center to be. Once you've done that then you can solve the rest of the problems if we get the same thing. At least one student in here will do a fairly easy thing to do the first couple times through this type of problem this type of solution. It's just very easy to do. So when you find that instantaneous center, oh I'm sorry I didn't know you got one. Show me where you think it is and once you've gotten my okay where it is then you can solve the rest of the problem. I need you to draw this one David. I said there'd be at least one. Joe volunteered to be that person. They're veils. Thank Joey for taking the bullet. Be careful. Follow my very clear instructions for how to find the point of instantaneous center. Each drawing though is a little more complex and so sometimes that knocks students off of yep yep I just showed it here everybody. It's a pretty small job. Phil you getting that? I am still sketching. You guys gotta sketch faster. Where? Nope. Draw on the velocity vectors first. At least two points you know one of them including the point we're looking for which in this case would be a C I guess. Once we find the velocity is C the angular velocity of C d would be very straightforward. And then you don't have to wait for me to tell you whether you got it right because we did this problem Friday using the cross products. I don't think we were on this one were we? No I think on this one I did tell you the BC's horizontal. I think that's the only piece we had. Oh no I'm sorry data was 60 degrees. All you do to check your notes from yesterday or Friday yep okay most if you have it for those of you who don't pay attention draw the the velocity vectors for the points where you know they'll when you know their direction don't necessarily have to know the magnitude. Point B because A is fixed point B will be moving in that direction. There's just nothing else it could do because point A is fixed. Point C must move in this direction because point D is fixed. Once you've got those two velocities draw lines perpendicular to the velocities. So for B that's easy we just continue this arm AB out. We don't know where it's going to intersect so that first line could go to any distance. And then a perpendicular velocity of C puts the point of interest in its instantaneous center right there. And because of the direction of the two others we know that that virtual rigid body has an angular speed in that direction and you can find out what that is maybe call it omega 2. Because we have an omega in here, omega AB. And since you know the velocity of B or you can find it at least since we know how far away it is from the center and that's going to be 30 rations per second. You can then find the velocity of point C and it should be the same thing that we had on Friday. This this geometry is fairly straightforward. Remember this is 60 degrees that's point two meters so you can find out the find the position of point point C itself or point I see fairly easily I hope. So those of you who didn't quite place that center point right you see it now? Same answer Friday? Oh yeah of course it's going to be an answer to us in terms of square roots. Joe you see that center now? Drop perpendicular to the velocities. The two velocities you know it's very common for a student to connect the velocities themselves but connect them perpendicular to the velocities. And now since we know the velocity of B on the arm AB we can then use that velocity on the arm ICB. But that also happens to be r2 I like a bit. And so you can now solve for omega 2. Once you know that you can use that. The term velocity of C. Did it work? Are you stuck? You have to figure out what that is. Now that you know where point IC is you have to figure out what the distance ICB is. For that instant it's as if those two points B and C are both rotating around that instantaneous end. So we can use that to actually find the velocity of point C and then from that any of the other matter of two is you need a virtual instantaneous rigid body of the three points ICB and D. Just for that instant B, C and IC has a new virtual rigid body instantaneous rigid body. It only lasts for an instant in time. A femoral piece. Is it just arbitrary when you say omega CD or would it be different to write it DC? Oh yeah. That doesn't matter. The order on the angular speed does not matter. For the magnitudes on the distances that doesn't matter either. But if we're doing a relative position vector then the orders matter. The point being when you find now omega 2 it's the same for any lines on that rigid body and we're now representing those virtual rigid body made up of that triangle and the two velocities that we know. At least we know their directions. Velocity is B from the other arm it's connected to and we can use that to find the angular velocity. Omega 2 use that to find the linear velocity DC. Did it work? Same answer as before. We know those two pieces that were 0.2 meters and the angular velocity was 30 radians per second. So the speed of point B is six meters per second. That's just knowledge we get from the link arm AB knowing that A is not free to move just free to rotate. And so now we can find omega 2 as the velocity of B divided by ICB. That's now looking at is if it's rotating around point IC. And I believe that comes out to be 0.4 meters just the geometry on this triangle. That's a point 2. This comes out to be 0.4 and the third side 0.35 or the famous 0.2, 0.4, 0.35 triangle. And so now we can get the angular speed of the virtual rigid body. What's that come out to be? 0.1, 0.2, no 15 radians per second. Which is I believe what we had on Friday. Is that right? And now we can find out the velocity of C. DC equals R IC, I guess it's ICC, good to see, instantaneous center to the point C. Maybe just an I would have been more sense. Or E or whatever. And that comes out to be what, 5.2 meters per second. Is that right? Yes. Just what we found Friday. Sorry, meters per second. Just what we found Friday. Are we comfortable up to that point? Phil, all right? Connie, Joe? Joe, you know, what don't you like? I'm still catching up, yes. Okay? Comfortable with that? Connie? All right. That's not all that was asked though. We also wanted to know what the angular velocity of BC is. How do we find that from here? Want to know what the angular velocity of this arm BC is? What do we find that out now? David thinks he knows. Who else? I don't know. Allen knows? What, look back at Friday's problem and there it is right there? But you watched the tape. I will. We will now. All right, David, pay attention. Let's see if Allen's got it. Give him a turn. Allen, how do we find out the angular velocity of the arm BC? What about this original arm here? What about the velocity of C by the distance BC? No. David? I believe it's actually omega 2. Why? Absolutely right. I believe angular velocity is a free vector, right? Yep. So whatever speed C moves in reference to IC, B also moves in reference to C. As long as there are points on the same rigid body, this angular velocity isn't just for these two lines coming down, it's for the entire rigid body. And remember for this instant IC, B, and C is a rigid body. So this is also then omega BC. Nothing to calculate. It's already right there. Not every problem turns out that way. It depends on what's asked. For the angular speed of C d, the wheel, I think we did that on Friday anyway as pure rotation about point d. And that's why you needed the radius of that wheel. So which of the three methods is your favorite? Well, for the types of problems we do, this method is fairly common because we typically do problems where the angles are known at a particular instant. For a general purpose, if you were working on some robot, some robotics application, you're going to need no more than what it does just an angle of 60 degrees. But we're giving you a chance to get started in this before you jump into a fully paid position from what you get fired two months later. Because you're just saying, why isn't everything at 60 degrees? It wasn't class. Alright, everybody comfortable with this one? Well, we now know the velocity of point C. Where is it? It's right here. Here's the velocity of point C. Here's the wheel DC. We now know the velocity of point C. And we know that it's in pure rotation about point D. So we just do what we've done in physics one. This is R, C, D, omega, C, D. R, C, D is just the radius of that wheel. So we can find then omega C, D from there. Oh, I see. When one ends not moving, you can do what I was doing to buy it. Yeah. When one end is fixed, then it's just a physics one problem. You'll find out the velocity of any other point on that line. If we needed some intermediate point, perhaps there's another arm connected there. Remember that the velocity of any other point you can find by similar triangles. It's just a linear function of the distance from the center. Alright, questions on that for a clear board? Just to make sure we have the connection. Remember we did a different problem on Friday. We had a wheel with a center gear as well. Bottom rolled without slipping. And then the top, a rack running on it. Remember that problem? Yeah, it's very difficult to draw concentric centers. What was that problem? Let's see, on that one, we knew the velocity of the center. And we knew the sizes of the gears, but I don't think we knew anything else. Other than, remember, all of these are, unless stated otherwise, and I don't remember that they ever will be, are all no slip situations that we're working on here. So one of the things we had to find was the velocity of the rack up here. And we had to find the velocity of a point there. I think that was D. Oh, no, this wasn't C. This was A, because we used C for the contact point. Right? That was the problem. All we had was the velocity of the set itself and needed to find some other velocities from that. Now we can use this method of instantaneous centers in a bit of a backdoor way. We know, because of the no slip situation, no slip condition, that point C at that instant is not moving. And since everything is connected to it, it's going to be as if everything is rotating about that point. There can't be more than one point not moving on a rigid body in general motion. There can be one point for an instant that isn't moving, but not more than one, because if two points are not moving, then neither is the rigid body. So we can use the fact that this point's not moving, meaning that this, in this instant, everything is rotating about that point. So if we connect any points to that instantaneous center, the velocities will always be perpendicular to that. And their magnitude will be a linear function of their distance from the center. So we could use that to very quickly find VB, the velocity of point B, which is the velocity of the rack, because we know those distances from the center. And we could also use that to find the velocity of D at that time. And if you remember, we found out the velocity of D was at 45 degrees, as it should be if the line connecting it to point C is also 45 degrees. So we could have done that rack and pinion problem that way. But if I give any of this instead of the cross products, you could say, well, I'll do this, I'll give the cross products. And we would pay any attention. That relative motion equation with the cross product is a very important one, because we can keep angles as variables in that easier or easily than we can with this. All right. That makes sense, connecting that to our old problem? Okay. It looks like you're ready for a get out of class question. Over 10 minutes. So I'll give you a problem. Maybe you can get home a little bit sooner before your little brother gets home from middle school and eats all the chocolate eggs. Oh, no. So he's already he's in front of the PlayStation with his belly and chocolate smeared all over his face. I hate that kid. All right. There's a problem. There's a problem. I hate to. All right. An arm, fix it one end that's vertical, connected by a second arm and it connected to a third arm. And by the way, you are not welcome to point out to me how absurd these mechanisms are. And no parent use whatsoever. They are serving their only use and that's to educate you. So deal with it. All right. This is 100 millimeters. This distance here at the moment happens to be 175. So that 100 meter millimeters is this entire arm here. Let's label some points, I guess. Oh, we're gonna be brave. I'm gonna go with just two, two labeled points. You need other labels on your own. And this arm is 75 millimeters. I think I got that right. Not to scale. We'll call you a lawyer. This has an angular speed of two radians per second. Yeah. Well, that's what it looks like. That's better. All right. Let me make sure things are at 100, 175, 72. All right. Fine. Then VA, Omega AB, VA, Omega AB. And, oh, I did label this point D down here. Omega AB. Do we need an angular, I mean an angular measurement, not an angular velocity, but an angular measurement? Yes. Well, that's horizontal. That's vertical. Okay, what about AB? It's not either horizontal or vertical. I know, I mean the angular. You don't mean anything. You know, oh, maybe the drawing's a little bit off, yeah. Point D is actually right there. Is that out, David? Yes, it does. Yeah. I didn't draw it that way at all. Oh, I see. That's the vertical distance from A down to D, which is just located in the horizontal. So that's enough for you. And if you want, you could do it with the cross product. It's doable. But I think there are two-step problems if you do it with an instantaneous center. If you'd like, you can make a simpler diagram of it, fix a couple of points, use that to find an instantaneous center. You need two known velocities, at least their directions, depending on what else is given the problem. You might not have their magnitude. As long as you have their directions, this is possible you don't have their direction for one of them. You might quite have any instantaneous center to a point C if you wish. Start with two known velocities. If you don't get those right, nothing else is right. So revisit those tracks. Two known velocities. Got it? Good one. Where'd you put the center? Instantaneous center. Both of you got it? Let's see. With this arm rotating and fixed at one end, we know that point B must be moving in that direction. The other point is A that happens to be moving about D because D is fixed. So let's see. If that goes down like that, that thing is going to come over. That's going to go that way. And we know point A will do that. Right, Travis? And then you can draw perpendicular to those, which is fairly easy because it's already down existing arms and horizontal vertical directions. So you should be able to place point C right there. That's the instantaneous center making A, B, D, actually C and D are the same point, making that a virtual, instantaneous, rigid body. And so now you can find out the velocity of point A, angular velocity of A, B, and A, B. Got it, David? Yes? No? I believe so. Yep. We've got about two minutes still chance to get out early. Go get in the lunch line first. Get out to recess a little early. Alright, once you've gotten that, then using the velocity of B, which we can get from the arm A, I don't know if you can label it anything, that horizontal arm, 75 millimeter arm, we can get the velocity of B. And then once you have that, you can get the angular speed here and then you can get the velocity A. Yeah? Velocity of A is .043 per second. The angular velocity of A, B, and A, D is the same because they're the same rigid body and that's .86 radians per second. Right? A, B, omega A, B, and omega A, D are the same because they're on the same virtual rigid body, the triangle that we established, finding the instantaneous in it. Alan? How does that give you .043? Yeah, it should be .086. It's just that, you know, .86 times .1 because A, D is .1. Alright, one or the others, oh, check the numbers when you go home. Let's see, I don't know. VB is 75 times 2 is 150 millimeters per second. Right? So then we can use that to find omega BC because that equals RBC omega BC. And if that's 175 and that's 150 then this is .86. So I had that right. Yeah? Just .86. .86. Okay? And then VA is A, C, omega BC, which is the same as omega AD, which we've got. Yeah, so, oh, I've got that number. That's why. So what is VA, top right of the board? Yeah? .086. Okay, this is the one we didn't like. This was okay. That one we didn't like. So what's that come out of three then? .1 times that? Yeah. Oh, .06. I don't know. Probably has to do with the fact that there's chocolate eggs smeared all over the piece. Okay. Oh, now you've got a class late.