Tangential and Normal Components of Acceleration as Derivatives of Position Vector + Example by @mes on 3Speak
▶️ Watch on 3Speak - YouTube - Telegram - Summary - Notes - Sections playlist - Vector Functions playlist - MES Links --- In this video, I rewrite our earlier equations for the tangential and norma...
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Now look at acceleration vector, the same one we're dealing with, but we're going to now write it in terms of derivatives of the position vector. So instead of dealing with all of this speed and so on and so forth, yeah, we will write it strictly as derivatives of the position vector and its magnitudes and so on. So let's look at what the acceleration vector formula says. So that's this one right here. So this is the one in a box here. right, actually this one here. So this one right here. So v prime, this is one in terms of the components, like that. So I mean, you have v, yeah, the v prime, and then kv squared, n. So we have the acceleration vector is equal to, now the speed, the derivative of the speed, times it by the unit tangent vector, then plus the curvature Kappa times it by the V speed squared, times it by the unit normal vector. All right, now the first thing to notice from this acceleration equation is that the binormal vector B is absent. So there's no the unit binormal vector. So right now we just have the unit tangent and unit normal, but you don't have the unit binormal set up there. And again, recap on that as I went over earlier. So here is, if you have a space curve like that, and there's a unit tangent, it's tangent to the curve, there's the normal, it's tangent, I mean, it's perpendicular to the curve, so it's perpendicular, and it's actually going towards where it's curving in, and then the unit by normal is just a cross product of these two. So you take the cross product of unit tangent, a cross product with unit normal, and then you get, and it's also going to be a unit length of one, and that's also going to be, yeah, that's the B1. So there's none of that. So, Yeah, yeah, so if it's none of that, so then this acceleration vector is always going to be in this plane. Yeah, so it's always going to be there. Not necessarily just in this area, but you can have negative one, but it's always going to be in this osculating plane, and I'll illustrate that with a better thumbnail. All right, let's go back over here. All right, so right here. Yeah, so, yeah, first thing to notice is that the binormal vector is absent, so no matter how an object moves through space, its acceleration always lies in the plane of T and N, that is the osculating plane, which I covered in my earlier video, and recall that T gives the direction of motion, and end points in the direction the curve is turning, as I just explained. So, yes, basically what you have is, there's the osculating plane is right here. So this is oscillating plane. So if you have a curve, and this is from an example, there's a tangent vector, and then the normal is where it's curving inwards, so it's perpendicular to the curve, but pointing towards where it's curving, so it's curving inwards like that, and then this blue line is the oscillating plane all over that. And the other one is the normal plane. That's the normal in there. That's with a binormal and normal there. So basically, acceleration is always going to be in this area. It's always going to be in this blue thingy. All right, and again, recap here, so wherever the curve is going, so the unit tangent is always pointing in the direction of motion, so particle with increasing T goes always across there, and then there is the inwards. This one is direction of the turning, so the unit normal is going inwards. That's pretty epic stuff. So going further, so next we notice that the tangential or tangential component of acceleration is v prime or the derivative of the speed. And again, that's the tangential one here, this one right here. So v, v prime. The other one is curvature times v squared. So curvature times speed squared. All right, so, yeah, notice the tangential component of acceleration is v prime, or in other words, the rate of change of speed. And the normal component of acceleration kv squared is a curvature times a square of the speed. This makes sense if we think of a passenger in a car. Yes, a passenger in a car, a sharp turn in a road means a large value of the curvature, so the component of acceleration perpendicular to the motion is large, and the passenger is thrown against the door. Yeah, because remember, it's acceleration, rate change of the speed. So, for example, I'll just draw this all out, actually. So if you have a road like this, and then, yeah, so then this is the, you have a car, you're driving. Yeah, I'll draw a better car. All right. All right, here, I just pause it and fix up my car. So it looks like that, and you're going this way. So you're going to be curving, or there's a, I'll just pick a road like this, this is a road. Yeah, so you're going to be curving across. So the acceleration, I mean, going forward, basically tangent to this, this is the V. I'll do the velocity. Yeah, I'll do it like, this is a tangent vector there. Actually, well, this one's all zero, I'll just put, this is going to be, let's say you have, this is the rate change. So you have basically the acceleration in this direction, and the tangent one's going to be actually the speed, derivative of the speed. And then now you have, you're going to have a much sharper turn inside here. You can have, I'll put it like this. This is going to be much sharper. This is going to be curvature v. Square. So as you're getting closer in here, yeah, you're going to be thrown all right. And that's because when you go in a straight line, the curvature is zero, so in other words, acceleration is zero. So then right when you start turning, the acceleration's going to be huge. I just wanted to draw that out. Yeah, and then, again, when you do a sharp turn, hence passenger gets thrown against Cardor. Anyways, so high speed around the turn has the same effect. In fact, if you double your speed, so if you double A, and is increased by a factor of four. Yeah, for example, if you were to go much, much faster, so if you have v equals to 2, and then if you have, yeah, let's say v equals v equals 1, you'll have v squared is equal to 2, but then if you have v equals to 2, the speed, v squared equals to 4. And this makes sense because the faster you go, so if you go super, so if you go super slowly, you can curve this with barely any acceleration, you're not going to get thrown out, but if you're going super, super fast, your car's going to flip over. All right, anyways, so although we have expressions for the tangential and normal components of acceleration, it's desirable to have expressions that depend only on R, R prime, and R double primes. In other words, the derivative of the position vector. Yeah, so to do this end, we take the dot product of the velocity v equals VT. Yeah, and this, this formula is again, this is the velocity is equal to the speed times it by the unit tangent vector, that's the one we had earlier. It's scrolling all the way back up, that's this first equation. So we have this first equation right here, so we take this dot product of the velocity with acceleration. So with acceleration as given by the acceleration vector equation, or not just added components equation, just because, yeah, Kako's book reference to, like, the equation number, but I just like referencing it by describing it. So basically, we want to take that dark product of the velocity. with this right here. This is the vector acceleration vector components formula for this here. so that, yeah, you can just search it in the notes, and you see exactly what I mean. So, if we take a dot product with that, and here's quickly pause video, add some more notes here, and simplify using the geometric equation of the dot product from my earlier video, as well as the properties of the dot product, namely property 3, and I'll show you done. So you can recall, recap on the dot product, you just take the two vectors, dot product, is just the multiplication of the components, and then add them all up to get a scalar number. And the important thing is that you have this setup here where it just equals to the magnitude of each dot product multiply by, I mean, of each vector, multiplied by itself, and then multiply cosine theta. And when we're dealing with tangent or normal, what we get is we're going to get a theta, we're going to get a theta that is equal to zero or pi for two. Yeah, pi for two. And where pi for two, again, there's 90 degrees, so things cancel up. All right, so that's why we're going to be using the dot product and going to cancel it a whole bunch of stuff. And also, again, recall the dot product properties, namely, we're going to use this one here. So you can basically expand it out. Dot product of this with B plus C, the vectors, you can expand it out like that. For that. All right, anyway, so what we have is, V dot with A. In our case, again, a V member is just the speed. times it by the unit tangent vector. Dot product with A, A is going to be the speed derivative, times it by the unit tangent vector, and then plus our k curvature, v squared, and then we're going to have a normal vector, the end like that. All right, so that's what we have. All right, so now that we have this, and now we can just use property 3, just expand this all in there. So we're going to get a, you're going to get the setup, it's going to equal to v, v prime, to multiply this inside, and then we're going to have a, it's going to be basically, because those are just scalers, you could take those out, again, as by this one here, you could take the scalar out or N, it doesn't, you can move anywhere you want. So we're going to have now, times it by, I'm going to have, I just put it like this, a T vector, our product with itself. All right, and then plus, next setup is, well, we have the V inside there. We're going to get a Kappa v cubed. Then we're going to have a dot, this one's going to be a unit tantrum vector, dot with unit normal vector. So again, just using this property there, and then this scaler's, you just get them all out of that. So that's what we have. All right, so yeah, that we have that, so what we have is, well, look at this setup right here. So these are a tangent and a normal, so those are basically perpendicular to each other. So what we're going to get is a setup like this. We're going to get a tn dot with n. This equals to just using the geometric equation, this one right here, the magnitude of each one. magnitude of T, and magnitude of their unit vectors, they're just one. Yeah, like that. And then cosine, well, it's pi over two. And that's because they are right angles to each other. So these are basically right angles. Yeah, these are just right angles like that. This is going to be the normal. And this is going to be the unit normal there. No, I mean unit tangent, and I just move it over there. So again, that's why we have it there, like that. So that's a perpendicular. And what that means is, well, this is equals to zero. Yeah, this is going to be equal to zero. Yeah. And lastly, and I'll draw the, I'll recap on that graph for cosine soon. But then this one here, this is going to be t. Yeah, this one's going to be, well, that's going to be the magnitude of this squared, and it's equal to 1. So this all equals to 1, because it's a unit normal, I mean, it's a unit vector. So they're all, because you're dividing it by the magnitude. So you have the vector and divided by the magnitude of it, you're going to get the length 1. All right, yeah, just multiplied by itself, and then cosine, well, it's 0. Yeah, it's cosine 0, and that is, and this one's going to equal to 1 now. Yeah, this equals to 1. And again, it's cosine 0 because you have, well, this vector times by itself, they're both parallel. So you have T by itself, yeah. So the angle between it is 0. So the angle between these two vectors, there's the angle, there's data. Yeah, angle between the two vectors. So if it's perpendicular, it becomes pi or 2, and if it's zero, if it's parallel, just equals 0. So in other words, equals to 1. So this whole thing equals to 1. Box that out. and yeah, this whole thing equals zero. So in other words, this, I'll put it above here actually further away. This equals to zero. And again, just recap, yeah, recap on the cosine, I think I've done it before, but I'll do it again. Anyway, so recap cosine, you have theta, this is cosine, then I'll just recap. And then it looks like this. Yeah, so there it is. It starts off at 1, and it goes to negative 1 here. So low is negative 1, and it starts up at the origin, 0, and this is going to be pi over 2. Yeah, and yeah, that's pretty much what we have. P5 or 2 is 0, and it goes, it goes like that. I know draw this bit better. It goes like this. Yeah, something like that. All right. All right, yeah, so now that we have all that, well, we could just solve this now. Well, this goes to zero. So all we're left with is just this, and the other one is equal to 1. So we're just left with v times by v prime. So what I'll do is I'll move this whole thing down, so I'll put an arrow. Let's do it like this. Yeah, so this equals 2, I'll move it like that. Yeah, so that equals 2, we're going to have I'm going to put an arrow, like this is going to be V acceleration, I mean a velocity, dot with acceleration. So the dot product of the velocity with acceleration, this equals to just v times v prime. So then we're going to move this over to the other side, because what we want is, well, thus, we want the acceleration tangent component, which is equals to v prime. This just equals to, well, . All right, so we have this, and now we know what these all are, because number, we want these in terms of derivatives of basically, yeah, that only depend on the position vector, derivative of the position vector, and the second derivative of the position factor. All right, so this just equals to, well, the velocity, I mean, the velocity is just, well, we know the velocity, that's just the derivative of the position vector. So that's just going to be R. prime of t, so we're going to do the dot product width, and then acceleration is just the derivative of the velocity, or in other words, the second derivative of the position vector, then divided by the speed, which is just the magnitude of the magnitude of the first derivative of the position vector. Like that. This is going to be like this. All right. And now, yeah, now we have all that, what you can do is box everything. I'm just going to box this in, actually. Just going to box this in. All right, so that's what we have. All right, so now that we have, this is the tangent component. Now let's take a look at the normal component. And so basically, using the formula for curvature given by the earlier video thumbnail above, we have, and I'll get to that in that bit, because, remember, this is going to be our normal component of the acceleration. This equals to curvature. times the by the speed squared. So, well, we know speed is just going to be this one here, then we're going to square it. But let's get to the curvature, recall the curvature formula from the thumbnail in my earlier video all the way back up above, think here. Arc length, yeah, this is the arc length, and curvature, yeah, this is curvature right here. So there's a link to that, and then there is the formula. So curvature is equal to, is just defined as a derivative, the magnitude of the derivative of the unit tangent vector with respect to the arc length. and then I show that it equals to this setup here, it equals to the magnitude of the derivative of the unit tangent vector with respect of time, divided by the derivative of the position vector with respect to time, and then I show that is equal to this giant thing here. So equals to the cross product, the magnitude of the cross product, again, these things outside mean the magnitude of the length. magnitude of the cross product of the first derivative of the position vector, cross product with the second derivative of the position vector, divided by the first derivative of the position vector, magnitude of it, and then all cubed. Alright, so that's a good mouthful. I'm going to put this all the way down. So what we end up having is, so this equals 2. So notice this one is dot product, this one's going to be now. is a dot product. Now it's going to be a cross product, so we're going to have a, this is going to be the same thing, same thing as this one here actually, but cross product, so it's going to be like that of T, cross with this, the second derivative of the position vector, RFT, divide this by, and now there's a cube part, so this is going to be, divided by the first derivative of position vector and its, yes, it's magnitude, then cubed, and now we're going to multiply by v squared, and v squared is equal to just this part here, and then cubed, yeah, so I mean squared. So we're going to have the multiplied by the first derivative of the position vector, its magnitude, squared, and in other words, the same thing as this one, so this just cancels with this, we left with the one. So I'll just erase these, just so it looks nicer. and then this does just cancel out, we're just left with equals to, yeah, equals to actually exactly the same as this, but the magnitudes, and it's a cross product instead of a dot product, so that's fascinating stuff. Yeah, but they also give scalers, because dot product gives a scalar. Cross partner gives a vector, so we need to get the magnitude of that. So this is going to be just our derivative vector, cross with the second derivative of it. All right, and divide this by the magnitude of this one here. And yeah, that's it. So that is just going to be our AN. Normal component like that, so box this whole thing in. All right, and yeah, we've got this now. We can go to the last example of this video, and this can be on example 7, tangential and normal components of acceleration. So if we have a particle moves with position function r of t equals to t squared, t squared, and then t cubed. Alright, now we're asked to find the tangential and normal components of acceleration, all right? So let's put this all down there. So let's write these in their iJK formats. So here's the solutions, we're going to get r of t is equal to t squared plus, yeah, this is going to t squared i. vector plus t squared j vector plus t cubed k vector versus we have and now the derivative of this because we basically need to know all these we just need to know yeah basically we need to know the first derivative second derivative and then the magnitude of this one right here and then afterwards we're going to take the cross product and then dot product of those and so so that's what we need yes let's take the second the first derivative so our prime of t this equals 2 this equals to put to 2 t yeah so derivative of t squared 2 t i and then plus again 2 t j and this one's gonna be now plus 3 t squared okay like that last yeah the next setup is put an arrow like that next one's gonna be I'll ignore the arrow, let's just put it like that, just double, now is the second derivative, it's going to be derivative, this one now, it's going to be just 2, i, derivative, and this one's going to be, well, 2, j, and the next one's going to be, well, this is put the 2 down, so 6T, k, like that. All right, and lastly, what we need is this part here, this is the magnitude of the first derivative, so that's going to be, like this, like this, of T, this equals 2. Well, you just take a square root of the components. So 2T squared, so we're going to go 4T squared, plus now, again, another 4T squared. Then plus, this is going to be, well, 9, 3 squared is 9, and then this is T4. Now these just add up, this is equals to, well, 8 t squared. So this is equals 2, square root 8. t squared plus 9 t to the 4 all right so now we have this so therefore the earlier formula for the tangential component of acceleration gives and that's this one right here so this right here dot product of the first…