 If you remember from Monday, we're looking at two-dimensional or planar motion right now. And there are three different coordinate systems we use to describe these things. Certain ones are better for certain problems. The Cartesian coordinate system that we reviewed a little bit on Monday works great because you're so used to it. It's a very, very familiar coordinate system. Today we're going to look at normal and tangential components, coordinates. However, we'll find as we look at this that using these coordinates is not exclusive of using Cartesian coordinates at the same time. We're going to do some problems in normal tangential coordinate system, but we'll also look at the same problem in certain ways with the Cartesian coordinate system just as a way for us to look at something new and still have some familiarity with it. And then we'll look at polar coordinate systems on Friday. Again, all of this is two-dimensional motion. For the Cartesian coordinate system, as we've seen before from physics 1 and statics and the like, it's not a terribly big deal to add the third dimension other than it makes things very, very difficult to draw, both on the board and on paper. Normal tangential coordinate system is a little bit more difficult to use in three dimensions. It's generally left as a 2D coordinate system. And then polar coordinate system is very easy to also use for three dimensions. Actually, then it becomes much more like what's called the cylindrical coordinate system. So the game goes something like this. This is very good. This normal tangential coordinate system is very good when the path is already known and even better if the velocity is known along that path. So imagine we have some path upon which our particle, whatever it may be, can travel. At any instant, its velocity is tangential to that path as it must be because if there is any component of the velocity perpendicular to that direction, that would be a component of the velocity actually off the path. So there's always a tangential velocity, no normal component to it. At any point in a curve like this, there's some curvature that can represent, some circle that can represent the curvature of the path at that point and that may change as we go along the path, both the center of curvature and the curvature itself might change. But at any instant, there is a circle we could draw that represents the curvature of the path, perhaps very locally, but certainly is a good approximation of what the path is doing in that little spot there. Obviously at other places, there are other curves. So that center of curvature is moving almost as much as the object itself is, whether it's a vehicle, a plane, or a particle. That then defines for us the two directions that we have in the normal tangential component system. Normal is always directed towards the center of curvature, the local center of curvature. And we use the unit vector sine E, just what our authors use, and then put a little hat over to represent as a unit vector. And then we have a tangential unit vector normal to that. And those are the two coordinate directions for a normal tangential coordinate system. As defined by both the path and the direction of travel of the object itself. So it's a moving coordinate system. It moves with the vehicle and depends upon what the curvature of the path is at any instant and the direction of travel is at any instant. If our little particle is back here, then the normal tangential coordinate system would be like that and so on for wherever place it happens to be. So this is one that's useful if the path is known because otherwise there's no idea where these coordinate systems are because you have no idea where the path's going next and things can get terribly messy terribly quickly. So that's our basic setup there. I guess I can add to it that the normal component of the velocity is always zero by definition of travel along a set path. So don't forget that little piece to it. Let's then start looking at what we do have. So the velocity itself we'd write as, the velocity vector itself, we'd write as whatever the magnitude happens to be and then always in the tangential direction. Now a little bit bigger picture of our object at a point here, that velocity and a radius of curvature which for some reason we give the symbol rho rather than r, just to spice things up a little bit. As the object moves along the path, that radius of curvature is going to be moving itself with some angular velocity, whatever the angle of measure is, it'll be moving with some angular speed theta dot. So the velocity we could write as rho theta dot in the tangential direction. That's no different than what we saw for rotational motion in physics one where we had v equals r omega for motion along a circle, a rotational motion. That's just v equals r omega from physics one. So that's not like a big break from something we've done before. Now the acceleration vector of that depending upon what its particular motion is at any instant can be made up of two parts. It could be moving with a time changing velocity along the path which would be then in the tangential direction. That's no more than what you see when you go around a corner in your car if this pedometer needle is moving in some way then you have this component of the acceleration. You can observe this directly in your car on any corner you might be taking. There is however a normal component to that which is nothing more than the centripetal acceleration that we also saw in physics one for any curved path motion. And that is, if you remember, v squared over r, so we're not using r, we're using rho, so it's v squared over rho and it's always in the normal direction which by definition is already towards the center of curvature so this is indeed a plus sign itself. So those are the two components of the velocity vector, I'm sorry the acceleration vector itself just to separate those two. This being a linear, well sort of linear translational acceleration along the path and then this being centripetal acceleration no different than anything else we saw before. In fact what we did in physics one with uniform circular motion is v dot was by definition zero, remember the uniform and uniform circular motion meant at constant speed so v dot by definition was zero and I never even brought it up, we didn't look at that component because in uniform circular motion it doesn't even exist so we never brought it up in physics one. For any other form of that we can also use this v equals r omega v equals rho theta dot we can also change little parts in there if you so wish for other forms of that. v squared, so with a rho on the bottom would be rho theta dot squared as one form or another form is v theta dot itself where remember v is the translate of velocity itself so what other parts do we have there oh just this one this could be rho theta double dot which is the same we as we used in physics one of this being a equals r alpha for some reason in dynamics here we don't tend to use omega and alpha we use theta dot and theta double dot instead but they do mean the same thing alright there's the setup let's apply it to real life and find out why teenagers crash their cars so often so imagine a car your car not my car because I don't drive recklessly is following a circular path with a radius of curvature of about 2,500 feet which is fairly typical for a curve in an automotive curve very very large radius we'll give it a speed of 60 miles per hour want to look at three cases constant speed around that curve increasing speed around the curve to the tune of 2.75 feet per second squared and decreasing speed around the curve of the same amount find the acceleration alright do me a favor real quick because we're going to need that velocity but we got to make sure the units work right so we need it mile feet per second so that it goes with all the other units we've got there how many feet in a mile you're not from Colorado Denver of course is the mile I see 600 seconds so we get all the feet per second alright for part one constant velocity around the curve path at that instant is 88 feet per second what then would be the acceleration given what I gave you a second ago that it's the dot in the tangential direction plus v squared over row in the normal direction push it back to paper for setup one what is V dot it's zero time rate of change of the velocity the velocity is constant the tangential component of the acceleration is zero only component we have is in the normal direction which is towards the center of curvature and of magnitude v squared over row where v squared is the 88 feet per second and a row is the 2500 feet so you could have just done that physics one there's no big shakes there comes out to be I think 310 3.10 feet per second squared in the normal direction which for an answer we can do nothing more than make a little drawing because it depends on where it is and how fast it's moving at any time not how fast but what the path is defines the normal direction in the instant but to the second possibility where we now have a component of V dot what now is the acceleration same velocity at that instant but there's now a translational acceleration of the velocity change time and a change of velocity itself acceleration still has the same basic shape it's just some of the numbers change down which changes the overall acceleration so if you come up with those components if any we can take a look and see what it means who wants to do the easy one and do the normal direction it's already done because this hasn't changed any oldest changes the V dot has changed and it's now 2.75 and for this part we're saying that it's speeding up along the path this pedometer needle is going up to the right so that is positive that isn't the tangential direction and that's 310 because that hasn't changed any and if we draw something to scale they're not too different in length so it's approximately 45 degrees and so the total acceleration is more something like that the trouble of course being if the grip with the road if the friction with the road is not sufficient to supply that kind of acceleration in that direction then the driver is going to skid out doesn't doesn't take a big stretch of imagination to realize if you're going around the corner and accelerating while you do it you're more likely to lose traction and spin out what's the magnitude of that comes out to be something like 4.14 feet per second squared at at about 40 40 something little over 45 degrees if we use that as our angle so you were at 310 feet per second just going to constant speed just picking up speed a little bit and the acceleration is is climbing and it's now not normal to the to the path remember the friction has got to supply that the force to supply that acceleration all right if you would please you do part three and sketch it like i did with that on my left turn signal just sit there maybe you've seen me all right what i'd like you to do is this but also the sketch for what it looks like for this case case three where there's deceleration along the path almost there should be fairly quick shouldn't it should be fairly quick still going though that should have been about a three minute problem right Phil they see it how has the normal component of the acceleration changed we're still looking for very same things we were looking before just to know for change a little bit how has the normal component changed not at all the velocity hasn't changed the radius of curvature hasn't changed the normal component hasn't changed so it's what it's the very same thing that was before and has been for all three of the parts how about the tangential component same one we had before it's just not in the positive direction but in the negative direction because it was a given as a deep acceleration an acceleration opposite the direction of travel so that's the tangential component and the magnitude the same as before it's also a deeper sector in fact the angle would have been the same only sure so either of those cases if you feel like you're starting to slip out of a curve accelerating and putting on the brakes either one might make things worse if you're already slipping out you should probably stop doing the physics calculations and settle things with your maker so good luck i'll read about it in the paper on monday no don't do that that's the worst thing let's say for a professor find the student's names in the paper it's never good so stay out of the paper all right any questions about that or life in general since i'd like to expound on both okay another problem then all right nothing changes other than the the geometry of these problems but it's still the case for any one of these that the velocity vector is whatever velocity that has and it's only in the tangential direction and that the acceleration vector is made up of the two components if there is tangential acceleration and the only way we wouldn't have a normal acceleration is if we either are on a straight path or have no velocity either one of those are fairly boring problems for normal tangential components all right so here's another problem kind of a ups type problem looking from above we might see a package delivery system some kind of conveyor belt that could look something like this where packages move along the conveyor belt get to a point have to do a bit of a corner and we want to make some analysis about that again the normal tangential components are fairly nice for situations where the path is known so we'll call that point a and this point b some of the specifics what's say that distance is three meters and the radius of the turn is two meters starts from rest given a couple things has a acceleration there that is constant throughout of two t meters sorry point two t meters per second so as time goes by the acceleration is increasing so not a constant acceleration situation and we want to find the acceleration at b if friction isn't sufficient at that point you're going to lose the package all right i'll do the easy part because no matter where it is that part's already given for this first section what's the second part of the acceleration for this first three meter section y zero it has velocity it's going to be moving all the way along there the radius of curvature on a straight section is infinite so we need to come up with the velocity there so that we can then put together the acceleration at that point and it will have a normal and tangential component to it and it doesn't really matter if the curve keeps going around at that point it has curvature so we need to figure out those two parts one acceleration and b we need to come up with the velocity and b curvature is given um that's the only part we need i think oh well we'll actually need the acceleration of b in the tangential direction because that does change so we'll need that this is given but as a function of t so we have to remember that's changing as well we'll need both of those by the end acceleration and tangential directions at b that'll just be a function of of the time whatever time it takes to get to point b okay how can we do that how can we find either one of those pieces starts from rest we know it's acceleration as a function of time so we need to figure out its speed at the end of the track and its acceleration at that same time because it is time bearing in there so do what John do a river rate we have v dot as a function of t point 2t and we know that that's dv dt so if we integrate zero the velocity of v dv zero to whatever time i guess i could call it tv we don't know that yet so we can do that integration will that help us find tv that will help us with the velocity at b but only if we know this time at tv so will this help us remember this this given acceleration is all the way along the path so this would certainly help us find the velocity at b in fact this side just integrates to that since v1 is zero we can't finish this integral because we don't know the time at which it reaches the bottom so what's next when you're thinking we're going to stop and restart my tape chris you're busy you got some idea to share with us integrate what integrate again integrate well this will give us the velocity as a function of time i guess if we want to keep it a little more generic we take the actual limit off and just put it on when we're ready we can integrate that again because that will give us then the distance from zero to s as far as it's going to go we can solve for that because we do know how far it's got to go it's got to go three meters plus a quarter of the turn of a two meter circle so this distance sv is known or easily findable and then that can lead us to finding that just that time tb at which it makes the comes out of the loop so a little bit of integrating to do is all you don't mind you love that this one integrates to 0.1 t squared evaluated between zero and any time t and then that you put in there and integrate again that's your test that distance i'll give you why you're working at 6.14 meters that's the entire distance from the starting point all the way to the exit point at v is 6.14 meters so the first one's integrated for you we know the velocity is a function of time that you can integrate again for the distance if you get that time feel free to check with me on it so you don't go farther because my own numbers are never wrong about something yep that's scary if we agree no that's not quite right what do you have david uh no double check double check because there's no sense taking the wrong time farther yep good i'm feeling so confident the number i actually got was right i don't know if that happens right this integrates to what 0.333 t cubed evaluated between zero and tb i don't need to do that i just put b on it and then that gives us the total distance was this 6.14 meters so you can solve for the time it takes 0.03 yeah so you have left space therefore so you can solve for the time it takes to make the run from there to there with that given acceleration starting from zero that'll give you the time then you can use that time to figure out what the translational velocity is the tangential velocity you can also find the velocity itself by putting the time back into here and then dividing by the radius of truth we can get all all the parts of it then and then sketch what it would look like at this point to blow it up a little bit we've got the end of the curve coming down points we'll have some velocity bb we can figure out once we have the time using the middle equation over there we'll have some acceleration tangential direction some acceleration in the normal direction somewhere you don't look happy okay yeah that's right you got the derivative all right so now since that equals 6.14 you can solve for the time yeah 5.7 that's how much time it takes to get down in here now that will give you the tangential acceleration at that point you can also use that to find the velocity at that point which you use to find the normal acceleration tangential is 0.2 at 5.7 seconds 10 to 11 1.4 1.14 yeah so that's this component is that the right direction it is it's an acceleration along the path rather than a deceleration then is v squared over rho at that point v square v we can get sorry so you can figure out the velocity in there 3.24 meters per second if I get the different numbers in these they're right so far 3.24 meters per second square over the radius of curvature two meters and it comes out to be 5.24 yeah some things right on any of these things anyway I'm going to build on a factor of safety so about five times more acceleration to the center of the circle then there is to the along the path so pretty severe normal component compared to the tangential component all right but again the sketch in terms of the coordinates at the point of interest which if we wanted to since it is so regular could be expressed in x y coordinates as well but need not be it's a very regular problem okay back there David all right with that one yeah I forgot to make it make this conference okay okay all right a couple more quick problems will be all set so imagine a jet takeoff path of something like that might we use my jet 400 feet per second and acceleration that they got maybe from radar or something of 70 feet per second squared of about 60 degrees so find two things rate of increase of its speed and find the radius of curvature all right with those first words rate of increase of speed what are we looking for does that agree that is the speed that is the v we're looking for the change in that we're looking for v dot there and of course here we're looking for a row now this is this is one that you can overthink if you're not careful because you're given some it's pretty useful information in that at that point where the plane is the acceleration is already given and then you can break that into tangential and normal components the tangential component is v dot the normal component is a function of the radius of curvature itself the trigonometry to find the two pieces get lunch to the cold okay gonna start looks good for the trigonometry all the tangent ones very easy the normal ones just as easy that is the radius of curvature then working Tom then you can check with Travis this is okay maybe you get it to check you what you don't care about things i'm never sure about this that should be the change well i'll give you credit don't have units on the screen but there you go of course this is 35 feet per second squared the other part comes oh i need to write it down what's 70 times sine 60 you have that this is 60.6 and then you've got the velocity that was given the 400 feet per second and you can solve for row the radius of curvature and you get about 2640 and if you're talking about jet pilots you need both components of the acceleration when you're trying to figure out the g's that a pilot is going to experience giving both components of the acceleration because both of those are acting on the body you may even have to take no account of gravity as well though uh well those those are about the same magnitude of gravity same order of magnitude all right one last problem all right i'll take my plane use it to make relief drops of vital supplies this is one of the just one of the heroic things i do every summer take my jet plane now drop things on people so given that the height is 1500 feet from where the object is released and the speed at release is 150 feet per second want to find two things so the drop point is a the i guess the well the release point at a drop point at b find the acceleration at b and the radius of curvature at b and as usual for these projectile motion ones and neglect their resistance once you concentrate on this one a little bit because they're a little bit more subtle piece to it that might not be this obvious some of the problems leading into this presumably it'll hit the at the ground with some velocity won't be vertical hit the ground with some velocity tangential to the path of that moment well that's the kind of thing you're going to need to find the radius of curvature a bigger view of v v all right what are your plans let's see the acceleration i'm asking for the acceleration vector so that's a unchanged from before so if you can find this normal component and the velocity at that time that will give you the radius of curvature and if you can find the acceleration in the tangential direction that'll give you the component of it there and then you can find both pieces let me ask you this maybe this will help shake it loose a little bit at the instant of release what is the acceleration vector in normal and tangential components at the instant of release no it's not zero it's no longer connected to the plane the pilots just hold the mechanism it's just at the start at the very very instant uh initial point of its projectile motion what's its acceleration right there just grab it in fact that is the normal component at that instant tangential component of the acceleration is zero there's there's no reason for it to accelerate we're neglecting air resistance remember it's released from the plane the plane was traveling at constant speed anyway what is the what's the general acceleration vector the instant before it hits the ground can you tell me anything about the acceleration in general forget forget the normal and tangential components for a second what's the acceleration in general just before it hits the ground still g i'll g grew a lot from one picture to the next so the normal and tangential components that's the tangential component the normal component is something like that so the tangential component you can find from the general acceleration vector if you can find that angle maybe not because that's what we're using for the radius of curvature if you can find that angle you can then find either one of these components from the gravitational vector how do you find that angle how can you find this angle beta let you concentrate on that right three piece keeping over here but how you could find this angle because in the two the normal tangential components just come from the gravitational acceleration vector and I think back to physics one a little bit find this angle you can find these two components right off the given vector we're running feet per second and that's 32.2 not 9.8 be afraid to make a nice big drawing be shy here unload or don't know what to do either or go get out of class question we're getting kind of close we've got time oh we've got something what's that oh I think it was down down to the rest of your all workplace back uh yep that looks good you have the acceleration there too don't say it out loud well you must have it if you have that number somebody can take Chris for a little bit of his expertise or you move that kind of table this wasn't working for you Chris yeah I guess it looked like it was a very thick line in places you know the trouble with this thing is wherever I slip the marker down the actual mark it puts on the screen is about three seven inch away so it's very difficult to put a mark where I wanted to do 201 that's that thing what do you got Tom you're okay with what to do if you had beta right because then you could just it's just trigonometry find these two components then the two components are the parts we're looking for that would give you v dot this would give you v squared over row it's kind of right there in front of you if you can find the velocity vector at impact you will have that angle as well because the velocity is always in a tangential direction so at impact just just before impact hasn't actually hit the ground yet because then it's a different acceleration problem that velocity is made up of two components a horizontal component if you can find those two you'll have the velocity vector and you'll then have this angle beta you can do it off the other angle if you wish there's no big difference how do you find the horizontal velocity of the box just at impact if you don't remember it's just you might want to sue your physics one teacher for for malfeasance straightforward isn't it yes yeah so keep it sit on it for a bit okay i'm keeping a seat yeah somebody gets desperate and asks offers you cash did that help a little bit um this is just good old ordinary projectile motion from physics one what's the velocity horizontal horizontal velocity of a projectile neglecting air resistance you can tell by the picture it's got some horizontal components because uh neglecting air resistance this would never actually be following vertically there'd always be some horizontal component is it not this that's the velocity with which the box was released and in a ideal projectile motion that velocity never changes remember that so your other task then is find the vertical component then you can find the angle then you can find the acceleration and the radius of curvature but i'll do it for you guys how do you figure out the vertical velocity acceleration yeah it's a constant acceleration problem in the vertical direction you know the distance you know the initial horizontal velocity you know the acceleration you want to find the final final velocity v y squared equal v y initial squared which is zero plus 2a delta y there's everything i can't give you any more than just hand it to you so now it's a true get out of class early question you give me the uh what was i looking for the acceleration vector and the radius of curvature and your weekend begins christ you already had it did you have to get Travis not that the radius of curvature something wrong you have something david i used another equation but turned out the same consequence yeah you can you can solve for if you're doing it in a constant um acceleration equation you can use one go come about it in a roundabout way what am i looking at is that in here that your velocity vector no here's the velocity vector i can't read that okay not later sorry got me out you want to check any intermediate numbers as you go for example uh you can check v y with me we have to take the square root take the square root of my head what's that there that's the radius yeah what went wrong yeah remember we practice strict segregation of those uh horizontal vertical components right 25 yeah a little over 25 so yeah and then your round out so so from this you can get the data is about 25.8 then you can use that to find the two uh components that'll give you the uh acceleration vector you already have the angle you can then find uh radius of curvature a good bit two week two minute longer vacation you know Travis you'll do something stupid in that two minutes a wasting you earned it nope something wrong besides the overabundance of significant figures for something now no well these were my those all of those yeah radius of curvature right so probably everything else is too uh not a unit some it's hard to tell but i think those are the digits i had yeah should get a bb of 345 feet per second you need that to find the radius of curvature after you've found the magnitude of the normal acceleration we can slip it away we don't one minute to get out early because i have to see you in class or something so it doesn't really matter you are going to see me just something very very excruciating not excruciating just something very routine routine not excruciating there's a bit of a difference yes not error excruciating lost in the y direction 3 11 this yeah i'm in is 18 you're saying no no i was wondering what your value is for the tangential i have 30 feet per second squared that's not right david i think i'm reading and yeah because that'll be uh b times cosine beta where beta is 25.8 it's almost right and then the normal component i had 14 feet per second square and so there's all the pieces then that gives us everything we get a radius of curvature at this point remember it's it's a local radius of curvature just something that would match the curve for a little bit there we get something rather large like that 85 10 what's that a mile and a half all right questions before we wrap up then okay notice that that one worked in both corner systems partition and normal tangential normal tangential works very well if you happen to know the path almost seems to work very well when you have when there is a system to account because uh yeah it would because if you leave this in x y you get acceleration in both directions where a normal tangential you'd only get acceleration well you still have acceleration in both directions but you don't have open wind resistance acceleration in one direction because the drag is always opposite of the velocity