 Alright, we're, we had a reintroduction on, last week, I think both days we took a reintroduction looking at rotational motion. Now we're going to go to general motion, which is a step beyond the stuff we did in physics one. We're going to look at not just translational motion, that we started the class with, and started physics one with. We're not going to look just at the rigid body rotational motion. That's where we ended up in physics one. Actually we did the kinetics of rotational motion. But now we're going to look at the kinematics of general motion, which is the two of those together. The fact that things can rotate and translate at the same time, of course is a little more realistic than just one or the other in terms of getting a big picture of the things that are going on. We're going to do this in two flavors. The first is as an absolute motion analysis, that's what we're going to do here for the first part of today. In that, we define the constraints, which means come to understand just where and where cannot this object move. For example, we're looking at a common car tire rolling down the road, which is clearly an example of translation and rotation at the same time. One of the constraints is that the tire has to follow the roadway. It can't go into the roadway, and as long as we're not looking at any kind of major bounce of any kind, it won't leave the roadway. So the constraint is that it would follow the roadway. Then we'll look right after that at relative motion analysis. This is where we use the fact that any points on these objects are an equidistant apart, so that if we know the motion of one point on the object, we can use that to find the motion of some other part point on the object because of the relative motion between the two. We might get to that today, depends on how things go today. We have lots of detailed drawings to do, so as we start getting into these, the first couple sketches aren't going to be a big deal. After that, we're going to have some pretty good detailed drawings that make them kind of big when they get started. For the absolute motion analysis that we'll start with, we want to do a couple things with it. Generally, what we do is establish the known path of a point. This is one of the critical pieces of putting the absolute motion together. For example, if we take the very simple idea of the wheel rolling on a road, it's fairly clear, I hope, that the center must follow a straight line. That's due to the constraint of the wheel maintaining its size. Remember, this is rigid body studies, so the wheel is not going to change size. This is due to the constraint that the center of the wheel must follow a path that's parallel to the roadway itself. Once we've established something about the position, we'll find S as a function of either angular position of the wheel or a function of time, and then from that we're going to establish then the velocity and acceleration of the wheel and of any of the points on the wheel that we might need to find, depending on whatever the problem is. That's our general picture, a little bit more specific of how we're going to do this. Then we'll start in on the real picture of just what this is going to involve. Imagine we have a wheel here. We'll label that initial point of the center as G. It's in contact with the ground a distance r, o, a. We'll label that initial contact point A, and then some little bit of time later, and I'm kind of exaggerating because I've got to fit everything in here. The wheel's rolled to that spot, so G has moved here to some place. We can call G prime. A will have moved to some other place. I don't know where I could, depending on what the distance is, maybe A prime up there. That will mean that our radius marker went through an angle of theta. Because of the arc length we can identify how far it rolled because it rolled a distance that was represented by this much of the track and that's a distance r, theta, when theta is in radians. That's the kind of thing we're going to do. Only we'll be a little bit more specific with it. That's just a general idea. So our first step will be to identify the position as a function of theta and that's going to involve this idea of arc length, S equals r theta. Now with just saying that, there's a very, very important idea that's here, but it follows through the other parts that we're going to do because then we'll take the derivative of that to find the velocity, which will also be a function of theta, but because of the time varying nature of theta, that will also be a function of time or theta dot, which is omega. So we'll relate S with theta, S dot with omega, and then of course we'll find acceleration, S double dot, and that will be a function of alpha and all the other little parts that will come into a theta. So we'll have these three concurrent ideas as well. Those three simple little things there are absolutely crucial to everything we're going to do, but they're also so simple that I could probably not even mention them and things would still work okay. What these three things together are called is the no slip condition. Those do not apply if there's any slippage at all between the wheel and the road. So we've got to maintain those. For the most part it's nothing more than us saying, well it's nothing more than us using these relations at any time. That immediately establishes the no slip condition. But they also are inherently defined if we happen to have a gear running along a toothed rack, a gear rack opinion type system, because gears by definition don't slip. We use that the other day. So we had that belt running around the pulleys. We had to assume that those pulleys, those belts were not slipping on the pulleys. If they were slipping, then there's velocities between the two and we can't establish all the pieces to that with at least our first steps through that. You have to take a more complex analysis with it. Alright, so we're going to start this picture over again and make it quite a bit larger. Alright, so there's our initial position. We're going to establish a couple things. One of the things we need to do here is when we identify the constraints. In this case it's the constraint that the center has to follow a horizontal line parallel to the roadway itself. That's an easier condition than using the roadway itself as the constraint because no one point is always on the roadway as the wheel rolls. There's a different point on the roadway at all times. But that center of that circle, the center of the tire does indeed have to follow a path parallel to the roadway. So that's a good constraint for us to use and we'll lay our coordinate system in line with that because then it makes things very easy. So our coordinate system will be that in the x direction and then that in the y direction and it's got to remain on the roadway there. So our initial position is such that, let's see, we'll name a couple points here. Just for reference, there's point A. Point B is the current contact point. That will be our angular reference line. We want to see how that line changes as the wheel moves a little bit. We'll have a secondary reference line that we'll put at some, a little bit of an angle here to it just because we're not going to go through even parts of a revolution. We want to look at some general condition where we roll to any other possible position or angle. Not forgetting that there's no slip of the wheel as we go along here. So that's our original setup. The other thing we just need to label for the sake of being obvious is the wheel has radius r and is rolling with angular speed omega. So that's our set up there. We've got that at least. Some little bit of time later the wheel has moved to some point there. That's not bad. Those look to pretty much the same side circles. You guys, you must be in subdued state of awe. I can tell subdued because you're not standing clapping right now. Usually when you use your head as a template. What? What are you saying? Use my head as a template. You're not understanding what you're saying. We'll watch the tape. Alright, so it rolls to some other place where point A is now moved down to somewhere here. I'll call that A prime and of course B since it's a rigid body is still perfectly opposite that. So A went about, what's that, about three eighths of a turn and so D, that'll put D at the top. Not for any particular reason just so we always have some point we're looking at at the bottom of the top and we can see how things have moved. So it went through that much of an angle. So A was straight up. There's the angle of rotated theta. Remember it's rotating with some angular speed omega and that causes it to move some velocity V which is part of what we're trying to establish. What is the velocity based upon the angular velocity and those kind of things. So we've got all those parts there. So it moved, don't forget this is our origin, the original center. So it moved that far. It's moved now to point X where that some distance X along there. But more than that point A has moved from up there is now located somewhere like that. So that's the position vector of point A. And A is just some general point so we'll be able to establish that as a general condition. So let's see. The position of point A is made up of two parts. One part is this component all the way out to here. So that'll be rAX. From the origin it's the horizontal distance that point A has traveled. It's just X plus this extra little bit here farther than point A travels. The center traveled the distance X but point A traveled this little bit of distance farther. And that little bit of distance is r sine theta. No r cosine theta. No r sine theta. Let's see. It's a little hard because it's a little bit over an angle but if it was only that far. Yeah that's the angle. The side opposite the angle. So plus r sine theta. See remember what we're trying to do is establish the position of any point as a function of theta and then we can deliver it that as we need. And that's in the I direction. So that's the I component of this position vector rA. And the X component is that little piece there. How much it's dropped below the original origin. And all these apply even if it was over that. Plus minus r that's just simply cosine theta j. So that's the position of point A at any rotational distance theta. We can fix things up a little bit. We still have an X in there and that's a little bit problematic. It would be nice if everything was a function of theta. But X is the very same distance as this arc length here. Those are one and the same. That arc length is the same as the distance X because it's the amount of distance that was rolled out here as the piece rolled. And B went from down here to up there along that arc length. So that's r theta. So we have sine theta. Put the minus there. We can pull the r all the way out I guess. Doesn't matter. It's a constant. So it's not going to be too big a deal as we go through the pieces there. Okay. So that's the position of any point A. Well not any point A. I guess the point is started there. Well I guess it couldn't be generalized at any point. What we want to find out next is what's the velocity of that point. The wheel as a whole is moving sideways with some velocity. But because of the rotational nature there's another component to the velocity of the point A itself. And that's going to be the time derivative of that position vector because that locates point A. So that's d r d t. But we don't have that position vector as a function of time. We have it as a function of theta. So we're going to have to do d r a d theta d theta d t. That we can do. This last little piece. That's no trouble because that's the angular velocity of the wheel d theta d t or theta dot or omega. Alright so let's take the derivative of this r a. See if we can get all the pieces together. Alright so we have r times d theta or that which would be this part with derivative with respect to theta which would be 1. Gotta get the minus signs alright as well. Yeah that'd be just cosine theta. That's the d r d theta of the first piece. But times d theta d t which is simply omega. So that's d r a x d theta times d theta d t. And then that's in the i direction. Little pieces right? Yeah. And then the second one it's a little bit easier because it's not a compound in any way. It's an r's a constant. So this becomes the derivative of this with respect to theta is minus sign becomes plus because of the minus sign already there r sine theta d theta d t which is omega. So put that in and then that's in the i direction. And then just to clean things up a little bit we've got an r omega on each term times 1 plus cosine theta i plus sine theta. That's just pulling the r omega out. The r omega sort of a familiar term which I was nice to have it out there. So there's v as a function of theta because everything else is well if it varies with time. We can take care of it in a second. Now let's see. Here's another very important point in all of this. It goes with the no slip condition. It pulls 180 which means the wheel has simply done half a rotation. A is now from the top all the way down to the bottom. That's the one rotation of 180. What do we get as the velocity for point A? Theta 180. What's cosine of 180 degrees? Not one. Negative one. So this is zero because we have one plus a negative one. So there's no i component to the velocity and the sine of 180 degrees is zero. So the velocity for point A is zero. At any instance the contact point, because that's what A is after a rotation of 180 degrees, the contact point has an instantaneous velocity of zero. Any time there's no slipping of the wheel. The contact point has an instantaneous velocity of zero. That's why I don't guess I want an A on there. I guess I could put Cp for contact point. That's going to be very useful to us a little bit later and actually several times here but be very, very useful in a couple days here of weeks or two. Also though it's one of those things that's very easy to forget. It just seems a little counterintuitive. You feel like every point on this wheel should be moving because in an instant later it's somewhere else. But if you also look at it in terms of the roadway itself if we have the no slip condition this point is in contact with the road. The road's not moving and this point's not moving either at that instant. So as the wheel moves the contact point moves but this is not the velocity of the wheel and the point that just happens to be in contact with the road has zero velocity. An instant later it doesn't. It starts changing velocity. I didn't say it's not accelerating. I just said it has no velocity. And as you can see it's sine and nature as we go with that. Alright so let's let's clean up the board a little bit. Let me put our position vector there r times theta plus sine. They all do it for rewriting this so I clear some board minus cosine theta j. So they're pulled r out. Now we'll rewrite the velocity up there because then I'll have a little bit of rule of mind that we can do the acceleration. So the velocity is r omega times 1 plus cosine theta i plus sine theta j. Alright that's just a little bit of housekeeping for me up here at the board because now we need to deliver that again to get the acceleration of point A which is the time derivative of the velocity but we don't have it as a function of time. We have it as a function of theta. So we're going to have to do the same thing we just did. Do dv d theta d theta dt and remember d theta dt is omega. The angular velocity. Okay so once let's not belabor the taking the derivative of this too much we can get down to r times alpha plus quantity 1 plus cosine theta. Remember we've got to do a couple of chain rules in here and they're like minus omega squared sine theta and that is the i component. When you have a little time and those of you who love taking derivative of trig functions and compound functions can sit down and do this. And the j component plus r times alpha sine theta plus omega squared cosine theta j. Let me make sure I get all the minus times right. That's the only one. Alright that doesn't really clean up very much so I'll just leave that as it is. But we can now figure out what the acceleration is at 180 degrees. The 1 plus cosine theta drops out again as does the sine theta so the entire i component disappears at theta equals 180. Looking at the j component the sine component drops out. The cosine at 180 is just one so we have r omega squared j omega squared j which should look somewhat familiar. This is using our no slip condition also b squared over r which doesn't matter if it's point a or point c it's whatever the contact point is at that instant it's got a centripetal acceleration component. So there I did it for point a here it just happens to be for point c so I'll call that a c prime and that's the only component of the acceleration at that instant. Which kind of makes sense if this wheel rolls just a skosh farther than that wheel that contact point there'll be a new contact point down here but that contact point will have risen up a little bit as that edge of the wheel starts to rise up as the wheel turns so you can kind of see a piece to that. This is the one we're going to need a little more often but this other piece is also important in its own right. So now we have the whole general motion for a wheel rolling along a level surface and we did what we planned we get r as a function of theta and then take the derivative of that to get the velocity and the acceleration. Alright any questions of delight for that? Get all the pieces? Okay so we'll look at a couple a little bit more specific problems. That's a pretty general problem just wheel rolling around a level surface but it did give us that notion that the contact point is not moving at any instant in time. Okay so let's do another problem. Imagine a circular can it's essentially a disc however it's mounted off center it's center of rotation will put right there that way you get an eccentric path by each of the pieces and there's a follower arm that rides against that at one point level with that so there's a follower arm there that's you know runs in some kind of rollers just to keep it from flopping around and it's spring loaded so it remains in contact with that with that wheel. So as this wheel goes around this arm's going to move in and out following the edge of it as that wheel as that can turns. Alright we'll give it a particular angular velocity and acceleration and remember it's about that point O so this wheel is not rotating about its own center it's rotating about a point attached to the edge and then that arm is meant to follow that for whatever reason so let's label a couple points for reference oh what we want to do is find the velocity and the acceleration of the rod itself so in this case our general motion is we know that wheel is or that rod is constrained and move horizontally we're just not sure how fast it's going to move at any one second. Okay so that's yeah that's the picture. The end of the rod follows a straight path we just have to figure out what it is it's this point is of some interest as we're going so we'll label that point B point A is the center and so R is the distance A0 and okay so we need to establish how the position of this bar as a function of the angular position of the wheel so that's the angular position of the wheel and the position of the end of the bar at any instant is that distance x and how that changes will give us the velocity and the acceleration alright so that's that's our picture so x we need to relate to theta and then take the derivative of that to get the velocity and the acceleration so let's x is twice this distance which is our cosine theta and then the velocity of the rod will be the derivative of that and we don't need to do this a vector because it can only move horizontally due to the rollers but we have x as a function of theta not t so this will be x d theta d theta dt but that's omega so we're going to relate it's the horizontal velocity of the bar to the angular velocity of the wheel itself by doing this derivative now so without belaboring it either we get minus 2r sine theta d theta dt this first part that's just dx d theta so this is minus 2r omega sine theta so that's the velocity of the bar related to the angular velocity and the size of the wheel itself so we can figure out something about the velocity of the bar and the acceleration is dv dt but we don't have v as a function of t we have this function of theta we have to do that with the same kind of thing we did before and that's a little bit more involved because I have both omega and theta in the equation there but we can finish up with oh we can pull out r as well it's constant that's the size of the wheel not the distance from the center of rotation to the contact point of the bar alpha sine theta plus or minus square cosine pull the minus 2r out and so that's all we're looking for on this piece we established the position of the point of interest which is that contact point but that defines where the entire bar is as a function of theta then take the derivative of that once and then twice to find a velocity and the acceleration so that's just sort of calculus one stuff it's setting it up that always takes a little bit more of a step a little bit more of an intuitive step through it but we're not going to deal with this one as much we'll be getting to relative motion analysis a little bit quicker enjoy that when you took the derivative of this you ended up with one part that's d omega dt and d omega dt is out there so if you step through the derivative of this you've got these two parts that are both time variant we're not assuming this is a constant velocity rotation we're allowing the fact that there could be acceleration okay but that's all we're looking for on that point well occasionally in the problems they'll say at a particular angle what is the velocity and the acceleration so you just put the angle in what would change in this if that it wasn't a wheel it was a regular shape like a lot of cams aren't perfect then r wouldn't be a constant r would change the theta as well I guess if we could make it more simply though it's always going to be simpler it's a circle but with any cam analysis you need to know the equation of the cam itself and so that would become a part of that that would then be also our function of theta if you don't think these problems are hard enough Alan you feel free to make them harder you usually do too okay I do I don't know at some point these are gonna get harder no we're we're gonna do another one or two of these and then we'll get on to the relative motion analysis which generally is more useful and students find a little easier I do this this kind of well they're taking the derivative trig functions is always confusing I think you've got minus signs and and trig changes and all kinds of stuff going on but just setting up getting that first part you get the first step and you're fine it's just sometimes that's the hardest part of these problems it's been my experience for the students and was my experience going through these classes so we'll do another one garage door opener type problem so we have a door hinged at one point and supported by an actuator of some kind a piston if you wish it goes up to the midpoint so that's a maybe a hydraulic piston of some kind that opens and closes the door and it's attached at the midpoint of the door which is a meter on either side and the cylinder extends at a rate of a meter per second so that's actually been the speed of the middle of the door not the velocity because as the door opens these angles change that's the speed of the middle of the door and then we can figure out then what the rest of the door is doing we want to find actually the angular velocity and acceleration of the door at 30 degrees where theta is the amount of the angle at which the door is open so the one point that's traveling along a prescribed path is the center of the door and it's going to go along a circular path like that a radius one meter as the as the pneumatic cylinder extends and so we can set up a triangle based on that that be the origin of that point a and this point B so we have the speed of point B that's the point five meters per second so remember we've got to establish that where that point is as a function of theta and then take the derivative of that to get omega and alpha and as they follow so let's do this as well we'll call the the length of the actuator that distance AB which is variable by the point five meters per second we'll just call that s for abbreviation sake because now we have a triangle whose sides and angles we can relate with the law of cosines so that's a oh a is the is the side of yeah I already have that so the distance a oh or oh a where well that's that's just two meters this where the door is when it's closed yes I should show that as well so this these distances here are also one meter so we know that a oh and then the law of cosines a oh squared plus b oh squared minus 2 a 2 a oh b oh cosine theta does that look like the law of cosines for that triangle make sure I don't mind yeah okay and this is two meters squared because that's a fixed distance a oh and the oh is also a fixed distance that's one meter so those two together is two meters squared because those are fixed distances so we just have this theta we now have s as a function of theta which is the first step we needed in all these the position of the door which is this distance s a d the position of the door the position of that prescribed point that establishes where the door is and is following a circular path and then we can now take the derivative of that to find the velocity and the acceleration of the door and do it at particular point there so we can we can do these easier said than done of course with those values put in and taking the square root because we want s not s squared it simplifies to 5 minus 4 cosine theta that's putting in the values 1 meter 2 meters as as the simplifying okay so we need to take the derivative of these so the velocity of that point is ds dt which is ds d theta d theta d t there's our omega so we take the derivative of that I know you guys want to be doing all these trig derivatives yourself I'm kind of stealing a great fun of this particular class let's see it's square root so the power comes down get the same quantity the power reduced by one it was to the one half power we take the derivative that reduces the power by one and then times the derivative of the quantity inside which is the 5 minus 4 cosine theta 5 is a constant so we get the derivative of the cosine which is minus the sine that becomes plus times d theta dt which is omega that look great I think that's all the parts we've got we can simplify this even more or see this is omega what is it as obvious is that little piece there is 1 over s because that was s there and that's to the minus one half coefficient so we can simplify that even a little bit more it comes to sine theta remember we're going to take the derivative of this so the simpler we get it the simpler we can make it and so that all reduces I believe to 2 sine theta yeah because the one half one half times the four leaves us the sine theta omega so that's pretty simple and we can evaluate that at 30 degrees if we wanted to leave it in general terms we'd leave it like that oh we already know what this is because this is the velocity of that the speed of that point b so at 30 degrees actually at any time let's put it this way that equals the point five meters per second that that bit of moving remember that being this s is all set up on where that point b is and how fast it's moving so now we can find what omega equals at 30 degrees by dissolving that and it's point six two gradients per second just putting in 30 degrees and solving for omega nothing magical about the 30 degrees just a chance for us to put some numbers in and see what the values are but based on the speed of that point b we now have the angular speed of the entire door so we've got s we've got B we now need the acceleration so we'll take the derivative again we don't have v as a function of t we have it as function of theta but we we've been doing that whole thing along as we go along here factor omega is going to come back into this one again as it has before alright so we're now taking the derivative of this piece here two sine theta over s times that's just this quantity times river that quantity which is omega dot which is d omega d t or alpha so there's our acceleration component in there two omega s cosine theta dot no that's not right theta times theta dot d sorry d theta d t that's just this piece on there again so this is what we're doing is of course the the chain rule business the thing is we have theta omega and s are all time variants so that's why this one's getting kind of strung out a little bit we've got these three parts going into it omega sine theta minus s square that's not minus s quantity squared that's s squared minus so that minus stays with her and then d s d t which is v which we already have so that's the whole smear of that piece the whole derivative of that it's multi part because remember we've got three parts in here that are all time varying s theta and omega are all changing with time so we can then put all these pieces together this is s double dot if you'd rather oh we can we can we can simplify this a little bit that may help so this is 2 alpha sine theta s because alpha is d omega d t the second part here that d theta d t is omega so we have an omega squared this becomes 2 omega squared over s cosine theta and then the last piece the little piece on the end that gets really easy that's minus v square because notice we have 2 omega sine theta s 2 oh no not v square d over s square it will be that's v squared over s that takes care of the extra s there that's a little simpler and we've got all the pieces then so then we can evaluate it at 30 degrees then the acceleration is minus 415 radians per second oh wait that's got to be alpha no that's that's a oh dear s dot in radians per second square so some of you can check it your eyes haven't lost over yet Anthony your brother miss which part yeah that's that's right here yeah that's what that's what this is that's v okay so I don't know if this is a I think this I think this is supposed to be meters per second not radians per second but you can double check it when you pull the pieces together when you sit down and go through this two hard parts one is just establishing the original position equation I've raised it was the one over s of that part then the hard part you gotta derivative it with respect to time and get all the pieces right you can't can't lose any minus signs or squares and that just gets to be complicated so we're not going to make a big deal about this one this is this is a much more of an academic exercise than is the relative motion which students generally tend to gather better and enjoy more because of that so was that question I was just looking at the derivative the same that if it's the derivative of the velocity with respect to time is the same exact thing this velocity with respect to theta times theta with respect to time and data with respect to time is omega then you should be able to replace this step here no or the acceleration right here if you erase d theta of dt and put omega in the tallys then the answer you get is a lot simpler it's only like the second term of that whole thing just this down down one more right there but these these terms drop out well if you have to do it wrong if you just if that's the velocity of their two sine theta over s omega then you take the derivative of that and then multiply it when you're done multiply times omega and you get two omega squared cosine theta over s well it can't be right because you've lost these two terms because don't forget even though you have omega here it is time bearing it as well and you have that on that d theta dt on each of the terms no I think well oh we're looking for alpha so yeah but that is alpha so I had the units right symbol yeah that makes more sense oh yeah you don't do this in and you don't well when you're doing on your calculator you can do sine of 30 degrees okay you know you set your calculator appropriately if you're putting in degrees or if you're putting in radians it's just most people I don't know about you guys I can picture what degrees are angles and degrees are much better than I can in radians so that was just you know not return of chosen value for that all right so hopefully that scared you off absolute motion analysis well enough couple homework problems do your best with them but we'll do most of what we do with the relative motion analysis so we'll establish that next you'll find that a lot easier I think which means so do I so this stuff kind of makes my points I like the real stuff a little bit better which the relative motion analysis seems to be a bit more and there are some problems it can be done either way so if you're not specifically given a way to do it then then don't worry about it so first a little bit of a reminder on relative motion we've already talked about it before but if we have two points a and B and some arbitrary origin then we can locate the two vectors or those two points with their position vectors r a and r b and the relative position vector this is just a reminder that we've done this before the relative position vector b relative to a if you're sitting at point a where do you have to look and how far away do you have to look to find point b in this with this vector describes the direction you look and the distance you'd have to look to find point b is defined as the difference of the two position vectors remember my my offering that it's with those subscripts in the same order b a b a and then you can always get the vector in the right position with that and then the relative velocity vector is the derivative of that which just becomes the difference in the velocities of the two and the subscripts keep the same order as they before so if we know something about the motion of one point on a rigid body we can keep we can get an idea of the motion of another point don't forget on a rigid body the distance between any two points is a constant so we'll use that doesn't matter which but it's the fact that the distance between the two is a constant now what that means if we've got some rigid body and it's got these two points a and b on it and we know something about how point a is moving let's just for example say point a moves along that path then we can figure out more about how point b is moving using this relative motion here's the idea don't forget we're looking at generalized motion so there's translation as well as rotation so not only is this object moving along here because point a is moving along that path but it's turning as well so some little bit of time later just because of the way egg where a is so some little bit of time later point a having followed along that path but because of the rotation point b is now down here and we can figure out the position of point b and particularly the velocity and the acceleration of point b if we know something about the motion of point a and the relative motion of the two between them so if we're looking for the velocity of b that means solve for this that's going to be the velocity of a plus the velocity of b relative to a that's just solving this velocity equation for bb I'm trying to figure out something about the velocity of point b the whole object is moving along and it's rotating so there's some complex more complex notion of velocity of b at this one point than there there is to point a point a's just falling along that line that constrained line of some kind so this is the piece that we need to figure out how that business works and here's the simple idea that'll give it to us that's the velocity of b relative to a if you're sitting at point a what does it look like b is doing that's what this question is that's what this term is b relative to a you're sitting at a you're looking out at b what does it look like b is doing if you're just looking out at b you're not looking at anything else you're looking out at b you see b is off in that direction at first that way it be a little bit later you're looking that way and be and then sometime later maybe point a is moved to here point b is over here you notice as you sit at point a that point b never gets any farther away from you is that true yeah it's a rigid body that's our definition of a rigid body but none of the distances between any arbitrary points will change so you're sitting at point a you look out at point b and notice it's not getting any farther away but it is moving always to the side it looks as a matter of fact like it's orbiting you doesn't it no matter what that object does if you're sitting at point a it looks like point b is orbiting around you you happen to be moving as well but when you're sitting at point a you don't sense that moment or you don't pay attention to it you're just looking at point b it looks like b is just orbiting around you of course you are young enough you think the world over or if it's around you anyway it's not the whole of universe but here's a here's actual justification for that are we taking a to not be turning them a is just a point a is on the body it's right there the body later turns now a is right there now a is right there the thing that we're saying is that we we know something about that velocity of the velocity of point a that tells us what the whole body is moving when we are doing particle analysis we only look at part a and say the whole object is doing that but now we know the object is turning around it I don't know which particular direction looks like maybe it's that direction but you can't say point a is or isn't turning because point a is just a point it has no dimension points to have no dimension so point a is going along a known path the rigid body is also rotating around it in some way we're trying to figure out what the velocity of point b is due to the motion of a and the fact that the object itself is turning that'll get risk no the the only time you're worried about the reference frame on the body is when you're trying to figure out what this is because this is relative motion bees relative motion to a so in this case your reference frame is point a but no we don't consider that reference frame to be turned the the reason this is good is we already know that that's as far as point a is concerned this is pure rotation and so if we know the velocity of point a we can add to it the fact that relatively speaking the object is rotating about point a and we've already looked at that we looked at that last Friday that's omega cross omega a b which is the angular velocity of the body times the relative position vector and that we've already looked at we looked at that back in the earlier parts so we want to know how b b is moving we look at how a is moving because that determines where the whole object is going but we also look how the object is turning around point a as point a moves itself and so we can figure out now the velocity of being with the relative motion analysis here's another point that students often forget notice I I didn't bother to put a slash between this when we have any rigid body with any number of points on it that we can connect so if I had an a a b and a c and this object has some angular speed that angular speed is the same for any of these lines they all have the same angular speed students sometimes forget this this angular velocity is for the entire body and any reference line we can scribe on it so all of these pieces are also rotating with angular velocity omega where the center of rotation is we don't know it doesn't matter if that body has that angular velocity any part of that body also has that angular velocity so this is a much easier number to come up with than students a lot of times think they think this often is something different they have to determine if you're told the angular velocity of one part of the body you know the angular velocity of the entire part of the body okay so we'll draw it to a close there and we'll put this to work on Friday given the velocity of a and how the object is turning we'll be able to find the velocity of any other point and then of course the acceleration as well boy that's not interesting enough to get you back on Friday what