 Huh? When things get a little bit busier as we're going on, I forget to pose the solution I used to it, so just send me an email. For a minor, I'll put them up. What'd you say to the new kid? What'd I say to him? I don't know. Did I say it? I said, welcome, didn't I? Kind of. Who the hell are you? What? Who the hell are you? Oh. Check the tape. See if he's crying. As you hopefully remember, we're looking at things just on the kinematics of what's going on, the same business as before. We looked at rectilinear motion, which is 1D, now we're looking at curvilinear, on Monday. On Monday, we looked at regular good old Cartesian coordinates for certain problems that are completely sufficient and extremely useful. For example, we did a little liver review from the physics one that we'd done with projectile motion. I even put up on Angel a summary of projectile motion to just sort of condense it for you if that helped a little bit. That tends to work very well in Cartesian coordinates because we do have the completely separate motions of the faculty of acceleration in one direction and nothing else in any of the other directions. So it's nice if we have a coordinate system lined up with that acceleration. It makes everything very much easier. We could do the problem with a coordinate system that wasn't lined up with G, but it just would make things a tremendous thing. Today we'll look at what are known as tangential and normal coordinate system. It's not two different coordinate systems, it's one coordinate system that has the two different directions, normal and tangential. We looked at this before, you're familiar with this, with circular motion that we looked at in physics one. Whether it's uniform or not circular motion, it doesn't really matter. The idea is still the same. At some point we have an object that's moving with some velocity around the circle such that that velocity is always perpendicular to the radius. We always looked at those two perpendicular things. If you remember this also came with centripetal acceleration and those two, that velocity and that centripetal acceleration were always perpendicular. And that's the basis for the tangential and normal coordinate system that we'll use. Tangential being always in the direction, I guess it doesn't have to be in the direction of motion. It usually is, it's not often we have things that are going to change the direction they're moving around the circle. But the normal then, the normal direction is always directed towards the center. In some books or some other studies that would be called the radial direction. Though often times that means away from the center, whereas the normal will take you towards the center. So that's our basic setup for what we're going to be doing here. So if we look at some path upon which our object is moving, at any instant we can define with it a normal, sorry, a tangential direction. And we'll use as we have before a unit vector to give us that direction. And our book uses the notation of e hat as a generic unit vector and then t will imply this in the tangential direction. For Cartesian coordinates, of course we use i which would have been the same thing as ex and so on in the other coordinate directions. So for the Cartesian coordinate direction, the Cartesian system, I mean we make it a little simpler just use i. But for more general coordinate systems like this, we'll use the tangential or the unit vector symbol e hat. And always directed towards the center then is the normal that with this coordinate system those vectors will change their direction themselves as we move along the path. In the Cartesian coordinate system, the i and the j vectors never change direction, never move no matter what we did to the origin, what happened in the problem. But as we move along this path, the tangential direction along this path and the normal direction along the path are always changing themselves. So we have to, we have to account for that. So we're going to look most specifically. Physician is just taking along the path somewhere. We can do it in normal tangential directions if we want. Of more interest are the velocity and the acceleration. The velocity is kind of interesting in that there is no and cannot be any normal component. If there was a normal component to the velocity, that means the object, whatever it might be, would be actually moving off the path in some measure and it can't. It's got to stay on the path. The tangential direction itself is always along the path. Therefore, there is no normal component to the velocity. Let's just believe it as v sub n. It's identically zero. It must be because of the nature of the tangential direction always being along the path itself, just as the velocity itself would be if we were actually describing motion along that path. These are v's. U's look like u's. V's look like v's. That's how I describe it. That's how I remember that v's look like v's and u's look like u's. They look like u's. It doesn't. It's an on camera, I bet. It doesn't. Oh no, that is such a quality of venous to it. It's amazing. Yeah, that's a big point at the bottom. I did see it. Well no, it's where the rest of the stroke goes. It describes the vanedives in a v and u. Alright, so what we will have, though, is a possibility that there could be tangential acceleration. I'll draw it going backwards, though. It could be forwards. This is nothing more than the kind of thing you'd see if you're going along a highway and you notice your speedometer needle moving. That's tangential acceleration. If it's drawn like it is there with the backwards acceleration, that'd be a case where you're driving along the highway and the speedometer needle's dropping. So you have a tangential rearward acceleration. As an example, I either draw it either backward or forward, one or the other, and the possibility that there will be a normal component to the acceleration. And the two together give us some total acceleration for these examples. What we always did in uniform circular motion is the tangential component was always zero. For uniform circular motion, the tangential component was always zero. That's what we did in physics one. We're going to look at more general cases now where that doesn't have to be such. Let's just for emphasis put on some other points to show that these unit vectors themselves are changing directions with motion along the path. That's something we didn't have before, but it turns out to be very useful, especially along very regular paths like these ones would be. Alright, so let's look at a little more depth at what we have going on here. So here's some path. We have an object at some point moving with some velocity u. Just kidding. Velocity v. Well, in any sense it was velocity u. That just makes no sense at all. And we have the coordinate direction at that point as something like that. It may or may not have a tangential acceleration. That tangential acceleration could be either forward in the direction it's moving, could be backward. It doesn't matter. It depends upon the particular problem. So I'll use what I had there and just do the same type of thing. We may or may not have a tangential acceleration. Since we're on a portion of the path specifically as drawn that has some curvature to it, there will most definitely be a normal component to the acceleration. That one there is going to take a little bit more study than with the tangential acceleration. The tangential acceleration itself is nothing more than some change in the tangential velocity. Again, for emphasis. Since there is only a tangential component to the velocity, we don't even put a subscript of a t or an n on there. Let's emphasize that there's just no other possibility but us moving along the path tangentially than the instant. The normal component of the acceleration is the very same type of centripetal component that we had in circular motion where at that instant the curve has some radius that we'll call use R because that's what we usually use for a constant radius curve which is a circle. This path has a curvature such that not only will the radius be changing at different places, but the location of the center from where that radius is drawn might be changing as the curvature of the path itself changes. So these are things that need to be looked at on almost a piece by piece basis depending on where you are on the curve, where the radius of that curve is and what the radius is. This is very much a concern in civil engineering as they lay out roadways, especially freeway curves where speeds can be kind of high and accelerations can be a concern and especially entering and exiting a curve. You've got to ease someone into the curve rather than just abruptly have the curve happen. You can imagine if you're going along a road and then you instantly hit a corner as you would exiting a freeway, you go from a situation of possibly no acceleration to a situation of instant acceleration and that's not always easily well controlled by a driver or by the car. So you have to instead ease the car with a gradually decreasing radius curve so that you don't have an abrupt change from a non-accelerating state to an accelerating state. So this kind of thing is a big concern in civil engineering. But the typical centripetal acceleration that we had in uniform circular motion in Physics I still applies. This is exactly like the v squared over r centripetal acceleration we had in Physics I. Only we don't have r, we have rho for the instantaneous radius of the curve itself and that's v squared over rho. If you remember from Physics I though we had other forms of this when we were looking at rotational motion. If this is changing or if this is some angle theta then v itself is r theta dot if you remember from Physics I. We didn't use theta dot there, we used more commonly r omega. That's the notation we used in uniform circular motion. So this becomes then rho omega squared or rho theta dot squared. Either one's acceptable and recognizable. What is theta? Theta is whatever angle this radius is making from some reference point. We're not interested in the angle itself as much as we are in how the angle is changing. So theta is just some reference angle. We would, I guess we could use that if we're looking for location along the curve but that's not nearly as much of an interest as is speed along the curve. Alright, so then tangential acceleration might look something like this then in generic terms. We may or may not have some tangential acceleration. Remember this is very easy for you to see as a movement of the speedometer needle if you're going around a particular curve. It could be backward from the direction of motion if we're decelerating forward if we're accelerating. And because we're on a non-straight path there will be a normal component as well. We can say equal to zero on straight sections. Well that's then just rectilinear motion where we didn't have any normal component. We're back to 1D motion so of course there's no normal component. If we're not on a straight, if we're on a curve of some time this is then always towards the center of a curve. Even if the center of the curve itself happens to be moving as you go along the path because the curve itself is changing. At any instant there is a center to which it is directed. This one. This is also V dot. Remember we have no normal component to the velocity so if we have the time rate to change that it's got to be a tangential component. And we could also look at it as our theta dot. That's the very same thing we had in circular motion in physics one. Only we didn't use theta double dot per dot but we used alpha. So that was when we said A equals R alpha from physics one. Right Alex? Sound familiar? A little bit? Alright so we've got those two components and a couple different ways to write all that jazz. Any questions on that? Clean them up and go on to something else with it? Comfy? Look kind of friendly from physics one? Like we're visiting an old friend? You bet our new friend's not weird. Did you piss him off? Guess he figured, oh boy videos. I'm not coming to class. Alright so let's look at a couple problems. One is that you pertain to the very type of thing you do. Especially you knuckle-headed boys, right Doobie? Which is redundant. Say knuckle-headed boys. So let's look at three possibilities of something a car can do on a circular portion of road. So let's say at the point of interest whatever that might be of radius 2,500 feet. That's almost half a mile. These high speed curves, especially coming off of freeways, have to be pretty generous. Well, partly we're going to see in a second if we can put some numbers to it. So we'll say a speed of about 60 miles per hour, about 88 feet per second. And we'll look at three situations. Find A for three cases. One is that velocity is constant. You pull off the freeway at 60 miles an hour. You keep that speed around the corner a little bit. And then come out of the corner or wherever else. Also look at a possibility and acceleration of 2.75 feet per second squared and deceleration of 2.75 feet per second squared as well. We'll sketch in the same relative position. Maybe we can label this A and C. We'll sketch in the same relative position we have there what this acceleration looks like. So for case A, what's the acceleration as we expect to see it in normal and tangential coordinate system we're using? In certain curved situations we have two components. Is one of those zero? It says, yeah, what? Anything, any part of this zero for case A where V is a constant. Is that a nod? Tangential acceleration. Any change in the tangential velocity for a situation where this is constant? Then that's zero. Because the tangential coordinate system changes with it. In the tangential coordinate system, the velocity vector isn't changing, I guess. But the coordinate system itself is. So is that normal component zero as well? Now that's, we have now for that instant, at least the instant, we have a centripetal acceleration each part of the row. That's nothing more than the type of stuff that you're doing in physics one. Got it, Jake? Check with anybody? Do you use to acquire any social skills whatsoever as I force you into the two groups? And I wish them all a lot more. There's two of them. It's trying to get excited by both groups. That's the unit vector in the normal direction, which is always directed towards the center of the curve. So that gives the normal component of the acceleration, whose magnitude is pretty easy to figure out and gives it its direction. Because the tangential component is always in the direction of motion. Its magnitude is one with no units. Its direction is always towards the center. So what did you get there? See for a second, the acceleration vector is in 3.1 feet per second squared in the normal direction. The normal direction defined entirely by the curve itself has no relation to horizontal, vertical, north, south. Though if we are talking about a specific curve, I guess it could be referred to in north-south directions as well. All right, same instant. Only now we have a tangential component to the acceleration. What is the acceleration? Also, figure out with it an angle with respect to the tangential direction. What's this normal component? Is it zero? I just had to sketch it this way, but maybe it should be zero. In fact, it is the same as before. It's unchanged. This normal component is unchanged. We've already got that part of it. What about the tangential part? That's given. That's right there. You've got very little to do on this one. However, what's the magnitude of the acceleration? It's angle with respect to the tangential direction, 1-4, at 48 degrees. The angle is as shown with respect to the tangential forward direction. I don't even care who's there. Do you need something? Is there an outlaw on it? So I'm not required to act upon it. I'm capping you can abuse. I don't know what to do. All right, so notice, of course, that the magnitude of the acceleration went up a little bit. So we'll squeeze in here. Then Kc. It's too long. So what's different here from what we had in this situation? You're what? All that's changed is the fact that we have a deceleration with respect to the tangential direction. Now what's the acceleration? Just like the others, we wrote down the full vector form. What did you write down this time? Speak up. Don't be shy. If you say something stupid, I'll edit it out. What? Well, no, give it to me first in the normal and tangential components. The tangential direction is always in the direction of motion, which in this case is forward, but it happens to have a rearward acceleration. So if Frank's correct with the minus. So he's off duty for a second. Then what? We still have the same velocity. We still have the same radius. We still have the same normal component to the acceleration. His other famous quote, who's the god loves us? Of course, don't know. But now what? What's the magnitude? What's the magnitude of the acceleration? Still the same. We have the same general component sizes. shown here with reference to what's marked, it would still be 48 degrees. What else can we put? If this is with the drawing, it's perfectly clear what the situation is. All right. Notice what this means to you as an automobile driver. We know that acceleration from physics one, we know that acceleration is only caused by unbalanced forces. If there's an acceleration towards the center, it must be supplied by some force. Well, in physics one with uniform failure motion, we call it the centripetal force. What is supplying this force here, though? From whence does a centripetal force come to supply this centripetal acceleration? Friction with the car tires. The car wants to go in a straight line, which means it's going at increasingly greater radiuses. Since you've got the car tires turned a little bit, that supplies a frictional force directed towards the center, and that supplies the acceleration. Notice what happens when you're speeding up. Your acceleration increases. Even though there's a tangential component, this is still supplied only by friction with the car tires and the road, more friction is required because more acceleration is required, which means, and it shouldn't be no great surprise to you, if you go into a corner with your speed increasing, you're even more likely to spin out because you lose the adequate friction to supply the acceleration you need. However, you're just as much in trouble if you go into a corner and you're on the brakes. If you took a defensive driving course or a driver-ed, you were told to try to get your braking done before you go into the corner so that you're not on the brakes in the corners. You lose this tangential component, the friction, the acceleration drops back and could be that you're in a safer situation. So, don't hit the gas. Don't hit the brakes in corners if you can help it. Most of you guys go into corners too hot. Don't realize it's up too late. You're in the gas in the first half of the corner, on the brakes in the second half, quickly getting religion in terms of the situations like those. Jake, you got a hand up? Yeah. What about like, when you go into a turn, maybe like zero tangential acceleration, you get to like maybe like, I don't know, it's like the apex of it, and then you can like kind of like accelerate and it kind of like powers you through it better? Part of that has to do with what I was talking about before, that corners are not constant radius. They're a little bit bigger radius here, a little bit tighter at the apex, and then they tend to open up a little bit more there. So what you're actually doing is now the radius is starting to increase the normal component. If the radius increases, the normal component drops, and you can hit it with a little bit more velocity there. As the roll goes up, so could be not to the same proportion, but by some measure. And in fact, that's why race car drivers, when they want to do a corner like this, they follow a path that does something like that. It's a much bigger radius following that corner than it would have been following this corner if the radius goes up so can the velocity so they can do a lot more speed through the corner. Plus it pinches off the guy who was just a little bit behind him as he tries to get through on the inside. If you're on a bicycle, if you're on the elbow, come through the corner, laugh and you win thousands and thousands of prizes at the finish line. All right, any questions with that one? Now you can drive more safely on the way home. If you're not texting at the moment, you can think about it these days. What are you going to do on the way home, Bob? What are you going to do on the way home? I can send this tape to the county sheriff, you know. And you can't say, oh, I didn't know that would be used against me in a court of law. If you know that camera is running, there's no excuses here. All right, here's a more engineering type problem with a little few more twist and turns in it. Looking from above, here's some kind of industrial conveyor. You know, you've seen them if you've ever had to go into a UPS facility or look through the back doors. There's all these ramps and things running everywhere and they've got those rollers on them so that the boxes can nicely roll down the... Look at that. It's a present for you. It's all wrapped and everything is then exciting. It's for you. You thought Christmas was over. All right, so here's the deal. Starts from rest at point A from where we're going to measure things. So we'll call distance along the conveyor S. Starts from rest at that point A. About three meters, three meters later, it starts into a circular portion of this conveyor and that happens to be a radius two meters. We'll call down there at the bottom of the corner. We'll call that point B. Tangential acceleration is 0.2 T. Here's per second squared where T is in seconds. So where are the units on 0.2? That will work. We're to find the acceleration when it's at point B. One quarter of the way around the corner there. That's your engineering task. As I head on up to the Sagamore for a minute to retrieve the whole weekend. That acceleration, that component, is in effect the entire way around. Is the acceleration at point B right here? All with the total? Yeah. Because if the friction between the box and the conveyor surface, those roller things, isn't sufficient, the box won't stay on the conveyor. It doesn't matter. Starts from rest, has that acceleration, follows that path. Whether it's the little wheels spinning and gets it going or some guy in a brown shorts and brown socks gives it a shove, who knows. Think about this one a little bit more. A little bit more involved than some of the others. Of course, it's going to be made up where you have to figure out something about what those are. Once you need to find, work back a little bit. Sorry, acceleration. Is that this? Position. The acceleration we have is a function of time. So we can't put this acceleration in there directly unless we know the time it took to reach point B. If the acceleration is a function of time, what we need is the acceleration is a function of position here. So how are you going to do that? Plug it in there and we'd have that acceleration. The time it took to get to B, you would call that just TB, the time it takes to get to B. We had that, we'd plug it in there, we'd know what the tangential's acceleration was at B. How could we figure out the time it took to get to B? We know where B is. We know the distance to B. It's the three meters straight away plus one quarter of a turn. So if we had the position as a function of time and we knew the position, we could figure out the time it was there. That's a fair thing to say. Where could we get the position as a function of time that would allow us to find the time to any spot like B? Where could we get the position as a function of time? We can, well sort of, get the position as a function of time from the velocity as a function of time, which we can get from the acceleration as a function of time. So there's a solution to map for us. All of those things, certainly velocity is always in the tangential direction, but the distance along the conveyor, that's also a tangential thing, so we can do that step through business. We've got the acceleration as a function of time. Do what to it to get the velocity as a function of time? Integrate it, integrate it, and then solve that for the time to that position. How about the normal component? How are we going to get that? Velocity, once it reaches that curve, but we can take that velocity and square it by the point B. If we knew the velocity at point B, we could then find the normal acceleration at point B, just be square it over R. Where can we get that? That we'll have here because we'll know the time, we can back it up to the velocity, put it in there, and then we set it up. There you go, find those things. There's a solution map. Do if you don't like it. You don't like me, you don't like Fridays, you don't like school in general, you like Jake, you know? Jake's starting a buggy after all these two years together? Terrasel. Terrasel, this is one terrasel, and that's one terrasel. You haven't heard of terrasels? No. You're an academic cell. Is that better? Yeah. A little academic group. Get your own t-shirts and have a secret handshake. Alex is trying to get into both groups or need a group plan on a wage. Alex, you okay with that? Why aren't we laid out help? For the first one, tangential acceleration. Yeah? So mom, we sell for t, and we're going to play it back in 0.2 inches. This, you've actually been given. So if you take the given tangential acceleration, you can find the tangential velocity as a function. From that, you can find the position as a function of time. Since we know the position at point B, we can find the time it takes to get to point B. Yeah. What about the constant? What constant do you think is important? Point two is a constant. Oh, when you do the integral? These are not indefinite integrals. I know, but I don't know if they're bound. Because it's a function of time. Well, that's what we want. Here's how you do it. Set this integral up for me. What is it you're going to integrate? Remember, we talked about the case last week where you have acceleration as a function of time. What do you do with it? Remember? The integral is 0.2t dc. Let's see. We're talking about the tangential component. Acceleration is dv dt. Since we have the acceleration as a function of time, then we'll integrate. It is a function of time, dt. But then your question is, what are the bounds? Well, we know it starts from rest. So we'll call v0 at t0. So there's the two lower bounds. That will integrate up to any velocity v, which changes as we go along because it's accelerating. And that's always a tangential velocity. And that will happen at whatever time t you integrate that way. That will give you then velocity as a function of time. Then you can integrate that in the same way for the position as a function of time. Do you have some sense, Alex? A little bit? A little thumbs up. Frank, is that kicking in yet? Things are clearing. What did you have to get you going? Whatever it was, three hours ago. Four hours ago at 8. You took your hour in that stuff. That's how it went. It's a way down in the academic cell. It's close. UPS, they do enough boxes. They might need to know where they are within every 10th of the second. Do you have 5.7 pips round off or see if they made a goof somewhere? Could be. You might have just had some digits reversed in your calculation or something. Shoot. Shoot. Is that what you heard about? Is that what he said? I don't speak English. You've never heard of it in the same time. Maybe I haven't. What's in the type of it? Try not to. Have you got that? Nobody check with? No, I've done it. I've done it today. What was your time? Did you get a 7.8? It sounds like you had a cold but I can't imagine any germs who live in that stuff. I mean for the rest of the day, I've got meetings all afternoon. Nothing more fun than meetings on a Friday afternoon. And the only people ever on my committees are people that evidently have no desire to go home whatsoever and say, alright, let's see how fast we can knock off this work. They jabber and argue and these are fuss. These battles are over there. At least stuff. There's not much to fight over at the college. I'll tell you. So, is that the same integral I have written up here? You're doing this, this first step? There is. This comes directly from that. There is no V in here. I'm talking about this. That's this step, yeah. Yeah. This might be too much. Okay. Once you get that, again, T now that you have a point, what did you do? I thought, do that integral. There's still no V, V, V in there like you're trying to integrate. You don't have velocities in function of velocity that integrate. So, you're integrating from the initial point we'll call zero to the initial point we'll call zero. The initial point we'll call zero to some point S, B, but you actually know that distance. You can figure it out. That's a known value what's the six meters or something a little over six meters I think, remember? And you're integrating from zero to some velocity. Since this is known, then you can figure out T from that. Oh, sorry, sorry, sorry. That's not B, that's T. Integrate to the time at B and then from that you solve the same thing these guys got and you did too. And then that information flow stops right here. That's why it's good to have Alex in class. He's in the bridge between these two war infections. The whirling dervishes against the terrorists. The academic terrorists. They're written down as vector but then that's not a they can't be equal. That's T, that's 5.9 seconds. Okay, 7.7. Yeah, that's just the way it is. So, now put a direction on it. Yeah, I had that as the magnitude. You got to come up with a direction for it, too. You may need to design these rollers to specifically supply that acceleration. Okay, are you talking about this distance right here or this arc distance? Oh, alright. It seems like that would be our part is we're going to multiply by 2. We're going to multiply by 2. Measure from where? You tell me. I asked for the acceleration so you have to have direction on it somehow. Now you can either do it in this form because the tangential and the normal directions are automatically defined here. The tangential direction and the normal direction are automatically defined by the travel. So, you take whatever tangential acceleration you have. Do you have some tangential acceleration? How much? How big is the tangential acceleration at point B? Alright, same thing I got. So, that's 1.14 meters per second squared. And the normal component is 5.24, right? That's the same thing I got. So, you know the acceleration to be something like that. And then you can mark either one of those on a little drawing as the angle you find. It sounds DJ like you found this angle because it's kind of tiny. But you can find either one. Just let me know what it is. Like your design team know what that acceleration is. Because that may depend on the material or the orientation of the rollers along here or some simple guide plate of some kind of an else. Pat, was that a lot easier way to do the problem? Just sit there and whine until Frank and David do it. That worked. I was following everything we were doing. And that was pretty efficient. Oh, for sure. So, if you occasionally buy one of those drinks to get the work done maybe two or three of those in the test. Alright, any questions? We'll look at another coordinate system then on Monday. Oh, 3.7.