 So, we looked, you know, Friday at essentially what's 1D motion. So, that should have been just sort of a physics one review. We're going to step up a little bit now. We're going to look at in general what we call plane curvilinear motion. Again, we're only looking at the kinematics of the situation, which is just simply the where things are, what they're doing while they're there, and when they're there. We're not looking yet at how we make sure that it's where it's going to be. That's the kinetics part that will follow later. There's three parts to this, or three different ways we'll look at it. It's still, it's three different ways to look at the exact same problem. Different problems, as you can imagine, are going to be easier to look at in certain ways than they are in others. We'll look at the plane curvilinear motion in rectangular coordinates. This, of course, is what you're most used to. This is just simply the x, y coordinate system where, as we watch this, these objects travel along some plane curvilinear path. By the way, plane just simply means two-dimensional. So, what we're looking at is two-dimensional planar motion. We'll put it into x, y coordinates, a little more comfortable for most students that way, but a lot of problems once you get used to the other possibilities that we look at, you'll find that certain problems are much easier to look at in the other coordinate systems we'll use rather than the rectangular, even though that's the one most of us are both comfortable with. I know that's certainly true with me when I just think about things happening. I think about them happening on some x, y plane, no matter what they're doing. But we will find other problems that do better on others. We'll also look at a normal tangential coordinate system. Again, it's a two-dimensional look at two-dimensional motion, and so we'll have two coordinates to it. If we were doing three-dimensions, we see we had the third coordinate. There's not going to be anything new that needs to be done for that third coordinate. It, of course, does make things an awful lot harder to draw. We'll also look at what we call a polar coordinate system, and that's generally something like r-theta. And so, again, just two, two coordinates to it, but a slightly different way to look at the very same things. All right, so we'll look at general plane curvilinear motion in x, y space. So we'll need some coordinate system. Again, the origin, as always, is arbitrarily chosen to be most useful to you, most useful to that particular problem. It could be a completely different place than some other problems. And we'll have some 2D curvilinear path across that plane, and it's the motion of the object along that path that we're going to concern ourselves with now. It's location at any time, you know, it's going to be a little more easy for me, I think, as I draw this, if I just put it up, same thing, just do it in a white so that the colors then are a little more useful. So if we have an object at time 1 happens to be there, we locate the object with some position vector as described in x, y space. So it'll have some dimension x1, some dimension y1, and it's those two together that will define the position vector that locates that point for us. Not terribly unlike what we did in physics 1, with the standard unit vectors i and j to give direction to those two components there. Presumably at that point it'll have some velocity along the path, so we might call that v1. At any point along the path the velocity vector is tangent to that path. It'll become a little more clear in a little bit just how we prove that, but feeling to your common sense of it, if there was any component of the velocity normal to the tangent, then that component of velocity would actually be taking it off of the path at that instant. And that wouldn't make any sense for it to follow the path. For it to be on the path, it cannot have any sideways component of its own velocity at that instant. We'll be a little more clear just how we can say that when we get to normal tangential components a day later. And then of course the velocity vector itself is the time rate of change of the position vector, illustrating it for a point 1, but it's generally the case anyway for any position along here. This will apply. And then that's the time rate of change of each of the individual components, different possible ways that you might wish to draw this. My recommendation is that you get used to this dot notation. It's very, very useful, especially as we'll see in a second or two when we do a particular type of problem. In case that's not enough, we could also override it, written it as the x velocity vector component plus the i velocity component vector at that point. Specifically, which one of these notational forms you work in, I don't particularly care. I'm not going to put all of them down every single time. That's just going to take too long to be a big way to effort. But I will typically use the dot notation because I find it very, very useful as we'll see in a little bit. Sometime later, that object may have moved here to some point 2, now described by a different vector and now with a different velocity. But remember, always tangential of the path at any instant. As that undergoes that change in both position and velocity, of course, has an average acceleration which is the change in the velocity vector as time goes by because presumably it took some certain amount of time for it to go from point 1 to point 2. That's the difference in these two velocity vectors over some particular time period. And those two velocity vectors, we can get the delta v by subtracting the two vectors. So I'll very accurately bring it over here. That's minus v1 because now I can have v2 plus a minus v1 and I get the change in the vectors. And then that then becomes the delta v vector that we use here. Divide that by the time which is a scalar, of course, and then we get the average acceleration vector. Might look something like this. A different color than the delta v vector because it is a different vector. A different length could be shorter, could be longer. It's going to be drawn to a completely different scale. But they are parallel because the acceleration vector is always going to be parallel to the displacement vector from which it was calculated. That's just the nature of multiplying by a positive constant. The two vectors will always be parallel no matter when they were drawn. That in itself, drawing didn't come out particularly well just by coincidence, will have an x component, a y component to it. It cleans things up a little bit just to illustrate why we have to be particularly careful with curvilinear motion in x, y space. I'll take off all of those preparatory vectors we had to begin with, the position vectors, the two velocity vectors. Just pick one point on here that we might have been looking at, maybe this was the third pointer, the second pointer. It doesn't particularly matter. We could go through this very same analysis we just did to figure out what the acceleration vector is. For the sake of argument, let's say it came out to be something like that. The two components that make that vector up, the x and the y components, their directions are specific to the coordinate system chosen. They have nothing to do with the particular path taken other than they come from it. So in this case, this would be ax3. The 3 just happened to be the point we're talking about, x the acceleration in the x direction, this the acceleration in the y direction for the point of interest we have to be talking about. Even though at times it may look like it, neither one of these vectors is the centripetal force, sorry, the centripetal acceleration vector commonly associated with motion along a curve. You remember from uniform circular motion that at any time we have a point moving on this circular path at constant velocity, there will be a centripetal component of the acceleration directed directly towards the center. If it's not uniform circular motion, then not only is there this component due to the curvature of the path, but there may or may not be an acceleration component along the path as well. That's a different setup than we have here. There are times where this vector, showing us the acceleration at a particular point in time, might look like it's pointing towards the center of curvature, but that's coincidence as much as anything. Same thing for the two components. Every time we go along a curved path, there is some point that represents the center of curvature for that path for general curvilinear motion along any general path, the center of curvature is going to change instant by instant because the curvature of the path itself is changing instant by instant. For example, down here, the center of curvature might be someplace like that. If there's a straight section for a little period which it looks like there might be here, a little bit of period where there's no curvature to the path, then the center of curvature doesn't exist or is at infinity, however you want to look at it. And then later in here, the center of curvature might be down here, and over here, if the curve tightens up, turns over some, the center of curvature might have moved. So the center of curvature is constantly moving itself because the path is a very general path. At any instant in time, there is a centripetal component because that's just the nature of travel on a curved path. But when the component directions themselves are set by an arbitrary coordinate system, then there are times when it might look like any of those are the centripetal component of the acceleration, but it's most likely just coincidence. You have to keep this vector in the X, Y coordinate system when working with an rectangular coordinate system. To maybe further prove the point, at any time we can change the angle of the coordinate system that will change these components depending on what the orientation of the coordinate system is, but the vector itself does not change. Only the component representation of it will change. So be careful, sometimes the coordinate system can give you slightly different appearing results if you're not careful with it. Careful with your drawing and careful with your analysis. So we've got all kinds of other ways and we can show that my personal preference again being this double dot notation. And if this is the average acceleration vector, these are average components as well. All right, quick problem. Jet has some takeoff path, something like that. It's not a coincidence that this jet looks like the car I drive because my car can turn into a jet, it's pretty cool. So we've got some path there and we've described that path in X, Y space as being something like 0.01 X squared. So it's, sorry, it's Y position. It's height above the ground in this case is described as a very shallow parabola. All right, given also that the altitude is increasing steadily at about 10 meters per second. And I want to find its velocity and acceleration at a height of 100 meters. So at some particular point, 100 meters off the ground, we want to find its velocity vector and acceleration vector. Let's see, velocity vector, that's the rate at which the two component directions themselves are changing. Now let's see, how are we gonna handle it? Let's see, Y dot is already given where is that? Altitude is increasing steadily at this. So Y dot is given as 10 meters per second and assuming that it's fairly constant. So the jet is gonna clearly have a changing speed along that path for the vertical component to always be constant. X dot's a little bit different. We're gonna have to come to it in a less direct path. That was just very simple. It was simply given. Y dot is also the time rate of change of this position vector as given. Sorry, that position, the path equation as given. That's also Y dot. Well, we can do this differential. We can do this derivative. The constant comes out. The inside portion, the derivative is, well, we gotta use the chain rule on it. Power comes down, reduce the power by one on the root and then the derivative of what was inside and we get that. So there's our X dot there. So the Y velocity is actually a function of not just position, but X velocity as well. So X dot, we can solve for it, is Y dot which is given over X where X is the position here, but we can figure out what that is as well because we know we're in an altitude of 100 meters. We'll know where the range is when we're at that height of 100 meters. So we can find that value as well. Y is 100 meters. X must at that time then be, what was that, 316 meters, I believe. When you solve for that. Now we'll check down, 316.2. So we can get the X component. We've already got the Y component. We'll have the velocity vector in rectangular coordinates even though the path itself is not a regular path through that. So the pieces go in. X is 316. And that comes to be 15.8, I believe. And so we have the velocity vector as X dot I, X dot's 15.8. Y dot J is 10. And we have the velocity vector. And we could draw it on that point as expected it's, well we already know it's got to be tangential. It's just now we know what those components each are. So that's Y dot, that's X dot at the point of interest in the whole thing is gonna be. So good thing we made you take calculus before we got here. You wouldn't know what the chain rule was. Questions about that part before we go to the acceleration? How are we gonna find the acceleration? How are we gonna find the acceleration? Recommendations? DET of velocity vector. Keep derivorating. When in doubt, derivorating. Okay, we've got that vector. Actually, this is a specific vector. We need to use the changing vector. We need to take the time derivative. Not of that, that's just a constant because it's only at a single point. We need to see how things are changing. So we need to use X dot and Y dot. Let's see, the form we need to use is these two, Y dot. Well, X dot, yeah, X dot we've got and Y dot in terms of that. And we can take the time derivative straight away. Got all the pieces there. This of course being the fiber here. This of course being double dot R if we're looking for it. Make sure you have the pieces there. Okay, we can do that derivative. Time derivative of X is X double dot in the I direction plus the derivative of this term. Oh, sorry, there's supposed to be a J in there. That's the Y component. So make sure you get a J component, a J unit vector in there. The time derivative of that piece, again the constant comes out in front and then we have to use the chain rule on these two pieces. X plus X double dot, that's the first part plus the times the derivative of the second part plus the second part times the derivative of the first part, a J. Now, here's somewhat the beauty of what we are given. What is that equal to? Zero, Y. Remember, this is Y dot, Y dot is constant. The altitude is increasing steadily. Y dot's a constant. The derivative of that is zero. So we can use that then to solve for X double dot. X, remember, is the position at that point. We've already got that, that was the 316. X dot at that point, we've already got, that's the 15.8. The only thing we're left with is X double dot and that is minus 0.791. Put that all together. We already have this, is there a question? Got the acceleration vector. Anthony, you're frowning. Where'd you get the X double dot, X plus X dot square? It's right here. That's from the chain rule on the Y portion. Took the time derivative of the velocity vector. The Y portion, I need to do the chain rule on and that led to this term here, actually this whole term in brackets. And so we have then the acceleration is minus 0.791 I, here's the first second square. Which is completely in the X direction drawn to whatever scale we might draw to. Notice that that is not a centripetal acceleration because one, the center of curvature doesn't lie along there. You'd expect it with that kind of curvature to be way up there somewhere. That'd be the radius of the circle it's on, the curve it's on at that instant. So that's just the acceleration vector in X, Y coordinates. Some other coordinate system will still have that acceleration but the components might be different. Yes, this is X dot, the velocity in the X direction squared, which is this 15.8 squared. So that's 100 meters, no, 316 meters. That's the X position at the point of interest. X double dot we're looking for. X dot is 15.8 squared and those two together are zero. Fair enough. All right, any questions on that one before I clear up? Go a little bit more. Stop the tape, take a break. Recton linear motion, besides just the general type we looked at there, is projectile motion. Always takes place in a plane with the down direction defined of course by gravitational motion. Typically, not typically, realistically, the acceleration in projectile motion is a function of the velocity due to air resistance effects. We're not gonna deal with that. We could do the very same approximation you did with me in physics one if you had me for that. Where we approximated that kind of motion. But we'll look at a more idealistic setup where the acceleration in the Y direction is a constant and of course that constant being the strength of the gravitational field or G. If you had me for physics one, this is the way we did projectile motion problems. It's also part of reason why I prefer this dot notation that we use. So for any given projectile motion type problem, it is a planar rectilinear type motion. It always occurs in a single plane. Only way you could have it occur in a three dimensional space is if there's some kind of sideways component of force, either this projectile is flying through a crosswind or it's a powered flight of some kind. These are completely unpowered, unpowered flights. This is what's known as a ballistic trajectory. The term ballistic means unpowered. If we're using a rocket of some kind, we don't start this part of the problem until the rocket's shut off. Which means the only force in the problem is that of gravity. And hopefully you remember from physics one, what we termed that kind of motion. Where gravity is the only power. The only force in the problem. That's, this is also known as a free fall problem. So we'll take the trajectory problems in this class to be free fall problems. Unless there's some other part to the problem that is very obvious in the problem. A parachute, some kind of jet engine or something attached to it. Or if air resistance is included. All right, we'll take a, you stop the tape for a second if you want. All right, all projectile motion problems no matter what they are. Whether it's from a higher elevation to a lower, to an even elevation between the landing and the launch. Or whether it's from a lower elevation to a higher. None of it matters. They all solve in the very same way. They all involve of course some launch velocity. Often termed a muzzle velocity if we're using guns or cannons or something to do this. Some launch angle or elevation as part of that. And then they all solve in the same way. For my physics one students, what did I always do next? Divide the problem into two parts. Divide it with what? Semi permeable membrane. A semi permeable membrane. And we keep the parts separate from each other. We have a horizontal part and a vertical part. And these two are happening independently. It's not the case if we take into account air resistance, but for a free fall problem they happen independently. And it's all based on the fact that the acceleration in the vertical direction is zero. There is no acceleration in the vertical direction because of the fact that it's both an unpowered flight and it's without air resistance. Vertical acceleration is constant and we take it to be known. Typically these problems are on our planet. So we'll use G. If it's a problem on some other planet then you'll need the gravitational field strength in that way. The minus sign. That minus sign is not a feature of G itself. It's something I arbitrarily add just using the downward direction as negative. I don't have to. I can do exactly the same problem some other way. It's just my pedestrian brain is very, very comfortable with downward being a negative direction, upwards being a positive direction. Every other part of the problem has to agree with that though. As a consequence of the zero velocity in the horizontal direction, sorry, zero acceleration in the horizontal direction, we know that the velocity in the horizontal direction is constant. It must be if there's no acceleration. Not only is it constant, but we know something more about it. It's the horizontal component of the launch velocity. This, I don't care what type of projectile motion problem you're doing. This is always true, what I'm putting up here. And all you have to do is fill in the little parts and solve for the things you don't know. Whatever the unknowns in the problem are. Acceleration, velocity and position. How do we find the position of an object traveling with constant velocity? It's that velocity times the amount of time that was in the air, so it's nothing more than that. That's the entire horizontal side. There's nothing more to do with it. Well, there's some terminology we might need. This is typically called the total distance of travel in the horizontal direction. It's typically called the range. So you may be given a problem that says that a range of a projectile or ballistic missile or something is such and such. You have to interpret that to be the total x travel from where it started to where it finished no matter what the altitude of either one of those points. Because there's no vertical consideration in this. Notice, as we're doing this, the obvious nature of my preference for this dot notation, there are only x's over here. Horizontal motion is the x direction motion if we put a rectangular coordinate system to it. And there's no y's over there. So it helps you keep things straight. You're gonna have a lot of velocities in this problem. It's very easy to confuse them if you don't have a semi-permeable membrane down the middle of your paper. All right, if the acceleration in the y direction is constant, what's the consequence then of the y velocity, the vertical velocity present? How does it change or does it? Of course it does. If there's acceleration in the y direction then the y velocity is going to change. How does it change? Linearly, it's a linear change. If that's a constant, then the derivative before it is a linear and that would be y double dot t or minus gt, not quite. What's missing? You need to add the initial. Yeah, if there's any initial velocity, y dot zero using zero for the original pieces. And that itself is the vertical component of the velocity. We have to put a subscript on that because it does change. It starts out at v zero sine theta but changes from then on in this linear fashion. Constant times a variable plus a constant is a linear description of the motion in the vertical direction. And then the vertical position itself as a function of time, vertical position as a function of time would look like. And that would change exponentially, right? Nope. There's an exponent in it, yes. There are a moment. It's a quadratic change. It's quadratic. One half gt squared plus y dot t plus any initial altitude. But I typically just make this a delta y over there, put the initial height over there. Then you're only concerned with the change in height and just sort of reduces things by a single variable. Any other constant acceleration equations. This is one of the constant acceleration equations. There are three others. They all apply in any way you might need to solve that side of the problem. What about here? I didn't bother writing down and any other constant acceleration equations. Once you put a equals zero into any of those equations, the only one you're left with is this one anyway. Everything else disappears. So both sides are constant acceleration problems. This one by far the simpler because the constant acceleration happens to be zero. This one just becomes a regular 1d constant acceleration equation. Semi permeable membrane because three things can cross either side. The time is on both sides. So is the launch velocity and the launch angle. Those are the only things that can cross the semi permeable membrane. After that, all you have to do is take any problem you're given and find out which of these pieces you have and which you don't. And if you can't find one of the pieces on one side, you usually can by finding something on the other side and bringing the t or the theta or some other part of it over. So let's look at a classic projectile motion problem. And these do very well in x, y coordinates, rectangular coordinates. It's the way we started with them in physics one and it's the way we keep going with them as we finish here. All right. So imagine some kind of basketball shot or something such that the shooter is able to hit the basket 2.5 feet above where he releases it a range of 25 feet. His release velocity, the magnitude is unknown because we got the tapes from ESPN and figured it out from there. Now, of course, once the ball is out of his hand that is a ballistic trajectory because it's unpowered flight other than under the influence of gravity. So given that little bit, we want to try to find two things. We want to find two things. Find v0 and that's not going to be much more than identifying what's known over here and solving for v0 is the unknown. There'll probably be another unknown to find it but that's okay, you'll have enough equations for doing so. And find phi, the angle, curve makes as it goes through the basket. That's important because if it's too shallow the ball wouldn't actually fit through the hoop. It might make it through the hoop after it bounces around off a little bit but it wouldn't cleanly go through and so you need enough angle on it there to make sure the ball would cleanly go through the hoop to increase the chances of points on the board in the adulation of cheerleaders nationwide. It's a pretty low basketball. Now that's just over, that's the height above his hands. Okay, so girls come to the court. You only brought that up because you're going to slant this on here and beat that out of you. Alright, so to find v0 you need to identify what parts of this we've been given and then what parts of that we can use and then solve for v0. It might come straight away from either side of the equation because v0 appears on both. Alright, what things are given besides the two accelerations, what things are given and each one of them will have a place over there either directly or indirectly. Where can we put these numbers? 25 feet. 2.5 feet. 50 degrees. Don't be shy. You know I'm not a camera man. Which is the 25 feet? It's actually this. The total x distance. So delta x is 25 feet. X is 25 feet as well if we take x1 or x0 to be 0. Just start the, we'll put the origin at the launch point which I've already done. What about the 2.5 feet? It's delta y plus or minus. Plus or minus. I heard one vote. Thank you for speaking up. I appreciate it. Is it plus or is it minus? Delta y plus 25 or minus 25 feet. It ends higher than it started. That's all you need to know. It's plus 25 feet. Any projectile motion problem we can handle in any way if we do it in this form. 2.5. Sorry 2.5. You can see it but there was a decimal point there. I just had to emphasize it. All right. What else do you know? We know some good parts of this. We don't know the initial velocity but we do know the launch angle. We can use that in here with this 25 feet as the total distance. So we don't know v zero but we do know the launch angle. So at any time t it's going to be at some position x. At total time it'll be at x equals 25 feet. Over here, what else do we know? Same kind of thing for the launch velocity. Remember we don't know the v zero so we don't know that launch velocity but we can use that in here. We get a position as a function of time and we put that initial velocity in there. Are we getting closer? Notice the basket. We know that x is going to be 25 feet y is going to be 2.5 feet that gives us two equations with two unknowns. v zero and t are the only unknowns. It pulls 2.5 feet. That with the last two equations gives you two equations, two unknowns and it's a system you can then solve. When you do so we won't bother with the here because this isn't a class in linear systems. Actually it's not a linear system. 1.3 seconds. Give or take a little bit of velocity you should be able to solve for 29.9 feet per second. Amazing. Every time you make a basket you do that v zero calculation in your head. What about the last part of it? This angle with which it hits the basket. You take a little bit of time and figure out what that is. We'll let's see. We won't necessarily have time to go into another problem but we'll make that a get out of class question. You get this angle assuming I have it right in the paper. You get this lawn, this landing angle. Travis, do you have it or are you thinking? Okay. Good. I thought I smelled something burning. If you find the angle with which it enters the basket then determine whether or not all would actually make it through cleanly. Possibility of a little swish or you're going to hope for a rebound. Who's got a plan? Zero. Anything we can do? Notice that over here there's nothing in here that says what the final angle is. Only the initial angle. That's what theta is. It's the initial launch angle that we had there. I'm asking for a different angle. What are you right there? David, you got it? No. You know what to look for? See fairly straightforward now that you have those two pieces. That seems relatively busy. That means there's some progress being made. Alan, okay? Got an idea? Give it to him. It's not a test. If you need to ask a neighbor what they're doing feel free because basketball is a competitive sport. Doesn't mean that engineering is. At least not competitive in here. Anthony, got an idea? If we look a little more closely at this piece here we've got the ball coming into some angle and we're interested in what that angle is at that last instant. The 25 feet down range, 2.5 feet up. At that instant it has that velocity. Nothing over here directly gives us anything about that angle that it's making. However, we do have ways to find we call this 0.2 x dot 2 and y dot 2. If we find those two components then we can find this angle because it's that angle. If you find the two final velocities in either direction then you can find the angle. Right, Chris? You did it? You don't like me to come to you. Because you'd like to stay no matter what. You don't have to kick you out. How come you never get what I get? How come you and Travis never get what I get? 0.2 Well, this x dot's a constant, right? And so what did it come out to be? 0.2 Good, we agree on one thing. 19.2 meters per second. Now remember that x velocity is constant the y velocity and I can't see the numbers you use there. But Chris, you put in the wrong g. Did you use 9.8 for g? What's g in the English system that God gave the world? 32.2 feet per second squared. That was on Moses' third tablet the one that he dropped on the way down which is why the metric system is getting in there. You see where it leaves that in the tape. David? Yep, that's what I got too. Better? Yeah. How do you find y dot at this point? Well, we have an equation for y dot. We know the acceleration. We now know the initial velocity. We can find the velocity at that time. Shouldn't be positive or negative. Or does it even matter? Better be negative because it is going down in the y direction at that time. Alright, there's still a little time to get out early. David? Do we go early? Chris, you got it now? Oh, say it out loud. Get out of the class question and just blur it out. You just qualify yourself if you have to stay. But yeah, that's what I got. Stay in. What? Go through the physics lab. What am I doing with that? Because I don't think it's locked from our side. I think it's locked from the other side. Okay? Alright, let's see. Let's put the last little bit together. Then we can all get out the class on time. Alright, we'll need those two pieces that we came up with from the first part of the question. The x velocity component. This is the final velocity. The x component we already know because it doesn't change. Once we are v0, we know what x dot is. What was v0? 299. Times cosine fifth. That's the x component of the velocity. The y component of the velocity, it's always changing and it changes to the tune of this one equation. Minus 32.2. That's the acceleration. Don't fit the minus sign of this down. Times t, which is 1.302 seconds plus whatever initial component we have and that we've now got because we have v0. Remember that sine 50 degrees is not the final angle. It's just the initial angle, the initial component that gives us the launch there. And I believe that comes out to, that comes out to 49. This comes out to be like minus 19, which is very, very close to the other component so this comes out to be about 45 degrees. Give or take a little bit. Hopefully that's the home team, but Timberwolves gets in and that's a wrap.