 All right, so we get up on Friday going to, I think we'll finish today with our last of the 2D planer motion Portrait systems we're looking at. We did the Cartesian system. That's very useful in, well, if nothing else, you're used to it. You're familiar with it. You've been using it since she's probably fifth grade or something, some of you. Also works particularly well for any problem that just happens to line up with that system as projectile motion does because the gravity is all in the y direction. Most of the motion is from buildings that are in the y direction and downrange targets, which are in the x direction. So that works pretty well for those type of things. Then we look at normal tangential coordinate system. Remember what kind of problems that works particularly well for? If the path is no, it works real well for those type of things. However, as we saw, we did a projectile motion problem in the normal tangential coordinate system. So the problems can be done in any of these. It's just sometimes they turn out to be a little bit easier than others. And we'll look today at the polar coordinate system in 2D motion. We'll talk about that one just real briefly about what it means in three dimensions. The Cartesian system in three dimensions is no great change. You just add on the third dimension the z direction and deal with it like that. The normal tangential is kind of also the same. Just add a z component to it in and out of the plane just like you would with the Cartesian. Polar is a little bit different, but we'll just briefly address it without any great explanation. So the polar coordinate system, oh, and remember of course, because these are 2D planar motion coordinate systems, each one of them has two variables to describe the location and then from which changing location we get the velocity and then the acceleration. And of course the unit vectors was i, j for that. I think we used e n and e t for this one. And now for the polar coordinate system, it's only slightly different than the normal. Same kind of idea. So here's some path upon which our object is traveling and it happens to be at some point on that path. The location of that is described as being some distance from some origin. And so we'll call that r. That being some origin, oh, and then that that just tells us the distance it was away from the origin, but that in what direction is described by some angle theta measured from some reference direction. And so it looks very much like you might imagine if you were stuck at the north pole and you were trying to find out where anything else in the world was, you'd say it's so far away and it's in that direction. So the polar coordinate system then would have unit vectors for the direction in the r direction there and then the theta direction, as I drew it with this, would be something like that and again these are all orthogonal coordinate systems that we're talking about. So it's the actual position vector itself then r would be the magnitude r in the r direction. So it sounds kind of simplistic and redundant, but that's all the more it is. And then if we give this thing some kind of velocity, remember that's always tangential to the path. So that velocity put the position vector right back up here again because then the velocity is the time rate of change of that vector, which will take into account both distance from the origin and direction because it's any change in the position vector, more simplistic notation, of course r dot, and then that will be the time rate of change of the vector we just put up there. And that's going to take a little bit of new work on our part. We're going to have to come up with a few new concepts we haven't had to mess with before. All right, so we have two things here that are changing with time. It could be changing with time, both the distance from the origin, but also the unit vector itself is changing with time because if this object's moving along the path, er, one time is pointing, it's always pointing straight out from the origin, but its direction changes as well because the direction from the origin can change. So this actually has two components to it that we get by the chain rule. r, er, dot, the time rate of change of the unit vector itself, plus the other half of the chain rule, r dot, er, so that would be a velocity in the radial direction, purely a factor of whether it's getting closer to or farther away from the origin as it moves along that path. If we had purely circular motion, r would be constant and then that term would be zero automatically. All right, so we have to evaluate, we have to evaluate this time rate of change of the unit vector in the r direction first. So I'm going to blow it up a little bit so we can see it. There's er sort of in that same direction I had there and some little instant time later, actually let me get that out of the way and put it on top, some little instant time later, the unit vector in the r direction is now that way and there's been some change in the r unit vector. Now that in itself we can look at, let's see, that's a vector, so the magnitude of it, the magnitude of that unit vector, well it's just, since this is very small difference, it's really just the arc length that the tip of the unit vector sweeps out, would be plenty close enough for us, so that's a distance out there times some change in the angle that occurred during that same time, so the arc length, the magnitude of this change is going to be r, not r, the distance delta theta is the arc length, the distance is one, that's the unit vector. So this will be one del theta, that's the arc length, and so we've got that, we've got that piece in the direction of however, I already took it down, in the direction of e theta, so delta er is delta theta e theta, the arc length, and it's in a direction perpendicular to the direction of the unit vector itself. Alright, so we need to figure out just what that is, or just what the time of change that is now, which will be the limit, delta t goes to zero of this vector over delta t, but that's the same limit of this new part we've got here now, delta theta delta t in the theta direction. So I've figured out what the change in the radial unit vector represents, I need the time rate of change of that, so I divided it by delta t, drive delta t to zero, and then I made my substitution of what the change in the unit vector in the ironsion is, which is delta theta in the theta direction. Because now we're all done, the limit as delta t goes to zero of delta theta over delta t is theta dot. So the time rate of change of the radial vector is equal to the angular velocity in the theta direction. We can then put that together, r theta dot e theta, that's just making the substitution. I already have r making the substitution for the time rate of change of the radial unit vector, and then the second part I don't have to touch. Let's see if this makes any sense by comparing it to the type of thing when we look at circular motion back in physics one. In circular motion, in physics one we probably wouldn't use theta dot, would we use for angular velocity? Omega. So this would be r omega, and if it's circular then the theta direction is the tangential component, so this would be, we can just write down that as its direction. You remember that v equals r omega from circular motion when we are looking at physics one. And r dot is zero, meaning there's no normal component of the velocity, no radial component of the velocity. In circular motion it's all the same distance from the center of motion at all times, and its velocity in any time is r omega. So that looks very much the same as what we've seen before with circular motion. Okay, so no tricks there really, but we did have to look at things in a different way than we have before, because we have this time changing unit vector. Alright, so just capturing these things again, theta dot e theta, all I'm doing is rewriting it so I can kind of build a summary for ourselves as we go. And this last component is only if it's getting nearer or farther from the origin, and this is essentially as it's passing across the origin, across its angle of view. So maybe I'll write down the change in the unit vector because we're going to need that important. The time range change of the radial vector is theta dot theta. Alright, just using that piece. Okay, so we have the position, we have the velocity which is the time range of the position in the two component directions. Now we need to do the acceleration as well. Acceleration is theta dot, so it's d dt of the velocity vector, which I've got here, r theta dot theta half plus r dot e r. So if you love the chain rule, today's your day. So here we go. First part of this has three parts to it itself. Each of them might have a time variant quality to them. So we're going to do the chain rule on that. r dot on the first one leads the next two alone. The first one alone do the second one. So it's theta double dot. We'll leave the third one alone. Leave the first two alone do the time range change on the last of the three. That looked right for the chain rule on that first part. r theta dot e theta. I didn't miss anything. Looks okay. I have the second part. So it becomes r double dot. Leave the second part alone. Leave the first part alone. We get er dot. I think everything's okay. Look alright everybody? Okay. We can collect a couple light turns. Well one thing we can do is this very last term here, this time rate of change of the radial vector. We've already got that. We know what that is. This piece is that theta dot e theta. What? Is it okay? Yep. One thing we do have that's new is this little piece here, this e e theta dot. So we're going to analyze that in just the same way we did with the other one. So imagine we've got some object, there's r so that establishes for us er theta. And what we want to see is the time rate of change of the theta direction unit vector. So I'll do just what I did last time. Draw a nice big picture so we can see it. There might be an original position. A little bit in time later. It's moving around the path some so now e theta might look something like that with a change between the two. Now e theta right there because of some change in angle. Del theta just like that. I think we're okay. Yep. And so that has magnitude. That has magnitude of one. Del theta just like we did before. It's the arc length. But this happens to be in that direction back towards the origin. So it's as a vector is del theta in the opposite direction of the radio vector minus er half. So that's its magnitude and then its direction is the minus er because that's pointing in towards the origin where the unit or the radio direction is away from the origin. And then let's see we're over there. We're trying to find the time rate of change of that. So we want the time rate of change of this with the limit as delta t goes to zero of minus delta theta delta t er. And that's theta dot again. This is minus theta dot er half. So that goes where is that? That goes right here. Minus theta dot e. Don't worry I'm not going to expect you to develop this just just use it once we've got it. So let's collect like terms which means group them by unit vector. So that's r dot theta dot e theta plus r theta double dot e theta plus the last one r dot theta dot e theta dot. Oh but I already have one of those. So I'll just put a 2 in front of it. So I took care of this e theta e theta e theta but comfortable so far. And then the 2 in the radial direction. Two components in the radial direction. Oh wait this is in the theta direction plus the 2 components in the radial direction. r double dot minus theta dot er in the r direction. That's beautiful. That's a big jump work there already. Let's let's see let's see what it looks like if it looks at all familiar for us. If we're doing circular motion. So r is a constant. This term drops out. We have just r theta double dot. Theta double dot in the theta direction. What did we call theta double dot in circular motion? Called it alpha. Right so this is just r alpha perpendicular to the radial direction which is the path. The path is always perpendicular to the radial direction. So that's r alpha. And if r is a constant we've got r double dot. We have minus theta dot. I lost an r didn't I? Let's see where was it. This is oh and I have the er in there twice. So let's fix this one. Let's fix this is the r direction. Not that one not that one. That is a minus sign so we'll take it second. This one r double dot minus oh minus r minus r theta dot squared. That's in the r direction. That's better. We would have caught it because it didn't work at all right here. And that's it. Alright so if r is constant r double dot is zero. We have minus r theta dot squared minus r theta dot squared. But we didn't use our theta dot. We used omega. So this is minus r omega squared in the radial direction. Opposite the radial direction is in towards the center. This is the centripetal acceleration that we always had in circular motion anyway. Alright so I just have a second while I reset the taper. And the minus sign means it's towards the center as centripetal acceleration is because our radial direction is away from the center in our positive radial direction. Increasing r if that happens. Wow that's a beautiful piece of work. You guys should be proud of yourselves. So let's write it down up here. Then we can clear the board and use all of these pieces in a couple of problems. I'm just rewriting the acceleration part here so I have all three of them together. Okay all three starts with our time rate of change of the theta vector, unit vector as minus theta dot. Now I'm mixing up parentheses. You can note some chalk. Makes it easier. Okay looks a little better. Okay let's do a couple problems. Looks a lot worse than it really is mostly just because it's new. We have never looked at a time varying unit vector before. That's all a new concept for us. But it was pretty straightforward I hope to figure it out. The book presents this very same thing only a slightly different way so either one is fine. Alright so let's do a problem here. Imagine we have some tracking radar on the ground which makes very good use of the polar coordinate system because that origin is fixed and whatever object is being tracked will be very much tracked by a change in the angular position. So here's a plane perhaps we're tracking. It's got some velocity v. It's in level flight. Being tracked by radar, I don't know if you know this or not it's blue. So being tracked by radar such that it's changing components as it's tracking. Well it knows that it's 6400 feet away. That's just simply the same kind of thing that radar does. Send out the signal in time how long it takes to return. R dot and this it would get from Doppler shift is 312 feet per second. Theta at this instant happens to be about 40 degrees. Theta dot is minus 0.0309 gradients per second. Watch your units here because for angles we need those kind of things. R double dot. So the rate at which that velocity signal is changing is 751 feet per second squared. 9.751 and theta double dot. The angular acceleration as that plane is being tracked 0.003807 radians per second squared. Alright all the type of things that you can read right off the screen of a radar tracking system. Alright so we want to find velocity and acceleration of the plane which we get by filling in the velocity and acceleration vectors with all the parts that we know. Quick word is that. So the velocity of the plane is R theta dot. 400 feet theta dot is minus, what's the minus sign here mean? The minus sign on this theta dot number. The angle is coming towards the ground. Yeah this, if that's theta as I gave you then that's decreasing. That's what the minus sign means. Nothing more than that. 0.030, no not 79, second squared. That's in the theta direction plus the R dot component 312 feet per second. Sorry that's not second squared that's second. And so you work out those numbers we already, we know what the vector looks like because I said it was in level flight but just to get the numbers to it I believe the minus 250 in the theta direction plus 312 in the radial direction and that's feet per second. This velocity vector broken into that component and that component. So that's 250 feet per second this is 312. Notice because of the way theta was drawn which was a little bit different than the original layout for this problem it's in actually in the opposite direction of what we had before. This is 60, that's R theta dot which is the first component here R theta dot. Is that an equal 250? 198. Mine was down one line. Oh I see it, no, no. I think I misread it. Yeah I did, I'm sorry. It's not 309, it's 39 at least for the numbers I have but remember just a problem. See everybody's taking notes on pencil anyway so I don't feel guilty. That's 250 then? Yeah. Okay good. Alright you do the acceleration vector in R theta direction and then we'll also sketch that on the little pictures you would have looked like. And again it's a matter of simply taking the acceleration vector and putting the pieces that we know to R dot which is 312. We've already checked the units on everything so we can be a little more cavalier with it because we've got a lot to write R theta double dot theta double dot the angular acceleration that's the angular component and then the the radial component. The radial acceleration 9751 the percentage squared so our chain is working on times theta dot squared which we need to make sure the units work out. They do the right one. R double dot minus oh not quite which would be obvious because the units aren't working out. There's R double dot now I need minus R theta dot squared theta dot squared. Okay that's better and that's in the radial direction. R double dot minus R theta dot squared gives us the two acceleration components in the R and the theta direction. Alright we got them. Got the first one there John for a chance. Yeah that's what I got. That's feet per second squared and that's in the theta direction. Yeah that's what I got. .0166 and that's in the radial direction and this is all feet per second squared. So the acceleration those are both about the same magnitude so a quick sketch should be pretty easy for us. It's got a little bit in the plus theta direction which is increasing theta so maybe something like that. That's A in the theta direction and about the same component in the R direction. Alright this acceleration component in that direction comes because as it's getting further and farther out it doesn't have to track an angle as fast. It's angle decreases by somewhat but it's a matter of taking the parts where we're going and these are very easily read right off the screen for the tracking station. I'll leave those we'll meet them. Have I got it all? Oh are you okay? Yeah. Okay so we can look at a second problem but leave that up there because that's the whole basis for this whole important system. Alright leave this one for you a little bit. Imagine a robot arm that has something right at the tip of it and maybe it's moving apart from one place to another maybe it's making a weld in some way but it follows a path that is something like this. R is 1 plus 0.5 cosine theta so a distance from the origin which we'll put right at its own pivot spot as a basis for the pull accordance system and theta is measured from the vertical in this case. So as the angle changes the distance from the center changes a little bit so it's certainly not a circular path because it's certainly not constant radius. Alright given these little pieces here when theta equals 45 degrees we know that theta dot is 0.6 radius per second and theta double dot is 0.25 radius per second squared. At that point where it's 45 degrees so about like drawn find the velocity and the acceleration vectors in the pull important system we're using. So we've got theta dot we've got theta double dot what we don't and we have R what we don't have is R dot and R double dot. So where will those come from? We've got R if we deliver it once and then twice we've got R dot and theta dot R double dot just the time derivative of each of those. Don't forget to use the the chain rule because theta dot as a part of that also changes the time. First derivative that's constant minus 0.5 sine theta. Does that look right? Does that look okay? And you can take the derivative of that again to get R double dot. And then all of those we know at a particular theta so those can all be evaluated. I'll set the table. Thanks Tom. So R double dot I'm going to add a couple parts to it and look at the same thing. We're going to have minus 0.5 out in front as a constant double dot sine theta plus cosine theta theta dot squared. Is that right? And yeah we won't have a minus sign there so that'll look okay. Yeah not theta dot. And so those are going to be evaluated because we know what theta dot theta double dot R, we know what theta is. So like the problem we had before we now know all of the components of each of those two pieces. So find those real quick and then we can make a quick sketch of what it looks like actually on the on the path. I'm not sure just how the path looks with that piece but we can at least get an idea what to make sure that the component directions we've got indicate the kind of thing we've got before. Theta dot is positive so this angle is growing. We know that it's moving down along the path. Should have a path. No that's okay. The path is getting a little bit closer to R as theta grows. It's the cosine decreasing as theta grows up to 90. Alright so we've got V and R all, A all, identify all the component parts. Make some sense? See once we get V and A established after that first half of the class it's a little bit more straight forward. Filled and okay. Alan alright with this? Yep. Good. See any tears? I don't hear any whining. Some of the numbers as we're going. So at this point, oh by the way I'm sorry this is meters. If you put it in feet that's okay. It's just a unit change. It's not going to affect the values of the numbers at all. 3, 5, 4 meters. R dot though are rated which it's coming to or away from the center, the origin. I have that as a minus so yeah it's actually getting closer to the origin and the acceleration. Those are the same numbers you guys got in those ones? The bill was okay. And that's just putting the known angle in and the theta dot and the double dot are just plugging those in. And that gives us everything we need. The six components we need to write out the entire velocity and acceleration vectors. Then put it with the drawing see if it makes sense. You do know that the velocity must be tangential. That does not mean it doesn't have an R component to the velocity because the center of curvature and the origin are not the same points. No it should be positive because it's moving in the direction of increasing theta because we have a positive theta dot. Radial unit vector perpendicular to that and not necessarily parallel to the velocity itself is the theta unit vector. Remember tangential components, normal tangential components are to the center of curvature. This is to some of the established origin in there. Not necessarily the same. They would be in circular motion the same but not in ours. Alan okay? Some matter doing it. Watching your minus signs? Watching your decimal signs. I think it should be the way it looks. No ER is not negative. ER is always away from the origin. The velocity? R dot yeah. R dot is negative. Yeah we already have that. R dot is negative. So it's distance from the origin is decreasing and so it does kind of fit the picture I've got there of it. And it looks like theta dot should be positive which yeah that's what I have. Question Alan? Okay. Looking my faces back to where I wrote those down. So I can't look at the board. You can't see the board? Do I not write big enough? That's not what I said. I've got all of those formulas written on a couple of pages back here. So if I put them up there I know we're going to need them. Do we have a disagreement on the numbers? Don't we agree with Chris? No? You guys are talking. This is something a professor would become a divorce lawyer. Yeah the velocity that's the right acceleration. Yeah I think so. Yeah negatives as always have been are very important. Negative is negatives and units. Watch them carefully. Chris and Phil are you guys okay? Let's see what we've got for the velocity vector. About 812 in the theta direction. Minus, well we have just that r dot up there. Minus the whole point direction. And that's meters per second. The velocity, then that's pretty much what it looks like with my sketch here. So there's theta component, v theta, radial component, not very big and in the negative direction. So v r. So if we break the sketch of the path coming closer as it comes down, it looked pretty good. So nice big drawing is how it plays well. Not working for the post office. And if that is that we come up also with the acceleration vector. 0838 for the theta direction. 0841. 0841. 08. Okay so maybe a little bit around. We'll call it 0848. Theta direction. That's meters per second squared. And then the r direction. 703. Something like that. Give or take a little bit. Meters per second squared. And that, yeah that's negative. So we have an acceleration component. A little bit in the r direction. Very little bit in the theta direction. Oh, it's just getting messy from sketching over there. Don't you feel grown up? I told you this was just advanced physics one. This should be called substitution 101. Yeah. Well this one, it's kind of messy. It started up. Get it going because we had to do those time changing unit vectors. But once you've got it established, then it's pretty straightforward. You just have to watch your minus signs and your decimals for the most part. Alright, so I think you're crying for a problem here of your own. Your very own. Actually I'll save the problem a little bit. Just to get out of class one. And talk a little bit about three dimensional coordinate systems. Just so you can see them. Just so you got a bit of a picture of them. The, of course the Cartesian coordinate system is just like it was before. Where any point we might locate as a z component, a y component, and an x component to it. And then the velocity and acceleration vectors are just time rates of change of any of those components. We did some of those in physics one, so that's no very great departure from what we had before. The normal tangential coordinate system doesn't lend itself quite as well to three dimensions, but the polar coordinate system most certainly does. And I'll tie it to our familiar Cartesian coordinate system just so you can get an eye on it. The polar coordinate system we just established has to do with a point being at some place away from the origin and at some angle as measured from some reference line. So we had coordinates r, theta. Then to add a third component to it we just add a z component. So we have to have a three dimensional space. Then we can use the same r theta and then add a z component to it. And then we can do this very same thing in three dimensional space. And things don't really change other than the fact that we have on each of these for the position we'll have as well on a z component possibly. For the velocity we'll have z dot and for the acceleration we'll have z double dot. So there's not very much more that needs to be done. The polar system once done this way is called cylindrical coordinates because we take the circle that we had in the base plane, extend it up in the z direction and we essentially get a cylinder that way. So it's not known. In three dimensional space it's not known as the polar coordinate system. It's known as the cylindrical coordinate. And then there's a third one that doesn't lend itself very well at all to two dimensions and that's spherical coordinates. Again referring it in some way to the x, y, z plane which is which. But now the location of some arbitrary point in three dimensional space is described with three vectors, the distance from the origin. So we have r as the same as the original component we had before. But then two other angular components. We drop its shadow down to the x, y, z plane somewhere. We see the shadow makes a line something like that. So that's our first angle theta. The angle the shadow would make in the base plane. So we're trying to locate the point A. And then the third coordinate is this angle down from the vertical direction. I guess that would usually be z then. Some third angle phi. And so the three coordinates are r theta and the angle phi. So it doesn't lend. Well if it's in two dimensions just r theta then we just have the polar coordinate system that we started with. Alright we're not going to do much with that one. It's very, very useful in astronomy where the observatory itself is at the very center or if we're talking on greater scale than the Earth center which we all know this Earth is the center of the universe anyway. Specifically the United States for all my foreign listeners. Okay so one last problem. Get out of class question for you. Looks something like this. Alright then we have an arm that barely sticks out of maybe a table top or something and then the main bulk of the arm is parallel to that. And it can rotate around this part of the arm that's perpendicular to the table. So as the arm rotates it rotates in a plane perpendicular to that direction z is drawn. Now there's a sliding error on the arm whose distance to the base of the arm can change with r and that makes some angle with a third dimension theta such that that angle is thinking with time at t cubed radians. Also its distance from the origin is 100 t squared meters. So find v with one second. So back to polar coordinates because this little collar whatever it may be rotates only in a plane and you can find all the pieces you need. You can find it r, r dot, r theta dot, theta, theta dot, theta double dot all at one second. And then you can put them all together get the velocity and acceleration. Just show me the velocity and you can get out of class early. You've got seven minutes to cash in on that exciting offer. Anthony, you can do it. Check little parts as you go along. That's fair game. So you've got theta and r and you can find theta dot, r dot, theta double dot, r double dot and then put it into the velocity acceleration equation because it's at 13.2 seconds. Of course then you notice the r gets really big real quick since it's where it's pretty big two big t cubed velocity. I don't think we've had a favorite student or a million dollar award on camera yet. Nope. Yeah, that looks right. That one's not right. That number up there is not quite right. We'll double check it. Anybody else? Catch your trouble, Alan? I don't know. We'll find out in a second. Are there still a chance to cash in early? Any statement? Any statement? All right. Excellent. Really? Promise? Promise. All right. Nobody has a lot of statements from the year. Got something, Tom? Nope. They are on just one part of it. Who else? No. You and it. I want to see it the other way around. Alan, did you find it yet? 32. Yep. Different. That one's not. Did you get it, Phil? Minus is right, 1800 right, that number is right. What's your misunderstanding? Huh? I don't know. Come on, I don't even give you one second. It couldn't be easier. You'll be able to do it in your head. Yeah. Got one minute, Alan. One minute to fix it. David, you got something? I don't have acceleration for a month, sure. I found this correct. That's a spirit. No, someone's wrong. Yeah. That one's good. Nope, that's worse. Let's see here. Just so you have them. V is 200 in the R direction. Plus 300 in the theta direction. Is that what you got, Alan? Yes. Yeah, what's wrong with it? Billion of time. I'm having unit time. Same thing here. You did not know me well enough, Alan? I had a funny feeling that's what you were doing with me. You made me do it twice anyway. I had it wrong the second time. Theta minus 700 in the R direction. Meter per second squared. Okay. That's a wrap.