Okay, so let's continue that normal tangential Fourier system. Just refreshing your mind guys that we got the velocity as rho beta dot et, right? And by derivative we got acceleration as rho.
beta double dot et double dot et plus rho beta dot square en and we calculated the absolute velocity or magnitude of velocity as rho finally ended up as beta dot 4 plus beta double dot square 5 And how we had the ET dot was we got a triangle that I showed to you like this. Remember what triangle that we showed to you was beta dot EN. And I asked you to go on as homework, use a triangle and show EN dot. The direction of EN is changing.
So it's pretty easy because EN was like this. Remember from previous class? After what? EN was like this, right?
So this is en, en prime. So delta en, delta en was like this. So it's in direction of, I'm going to skip what I have done previously, but the en dot, calculate as I did previously, is minus beta dot et.
Why minus? Because ET is in this direction so it is in the opposite direction. And we remember that based on the right hand rule we had the sequence of ET, EN, and K. Right?
If you are moving on the curve of your car and here is EN, ET, normal tangential to trajectory and K is out of the page. Based on the right hand rule we can follow everything. But the question is not for this coordinate system.
We're going to see later how to generally calculate time derivative of any. Unit vector, any unit vector. So its formula is this, E any unit vector.
That is the angular velocity vector as I taught you based on the right hand rule, cross product of that unit vector itself. That's the golden formula. Okay, say if you want to find IJK time derivative, you can use this ETENK time derivative, you can use this ERE theta K that I'm going to teach you after normal time initial coordinate system after solving problems.
E theta, E phi, and E theta, E phi, and E r. For spherical coordinate system, time derivative, we can use this one. This is super important for finding the time derivative of the unit vector series. Clear? For refreshing your mind, can I go to the next page?
Yeah, for refreshing your mind. If you have two vectors, A cross product of the P, Can we calculate that is corresponding IJK for example for the Cartesian coordinate system AX, AY, AK, BX, BY, BK. And this can be written as I if you forget about that. I'll say ay, ak, by, bk minus j determinant of ax, ak, bx, bk plus k, ax, ay, bx, by. Simply, it's going to be I AYBK minus BYAK minus JAXBK minus AKBX plus K is AXBY.
minus a y b x. This is generally what you have in high school. So when I say omega cross product of e, you have to develop based on the coordinate system that you do that.
For example, in the previous class I said in normal tangential coordinate system, omega was beta dot. Remember you have a not. And your thumb shows. Direction of that. K.
So let us validate what we derived. I want to calculate ET dot as we previously developed. It's going to be ET, EN, K. Remember the sequence?
Right hand rule. EN, ET and K is out of the page. Right hand rule.
Right now, omega is beta dot K, 0, 0, beta dot. And E T, because it's going to be E T dot, is omega in E T. E T obviously is 1, 0, 0. You got it? Huh?
So if you calculate that, you're going to get theta dot em as I did by triangle. You got it? So if you calculate...
E dot n is going to be again omega in en normal. It's going to be et en k. Again 0 0 beta dot. En is 0 1 0. If you calculate that no doubt you're going to get minus beta dot et. Guys, in normal tangential coordinate system, what's without calculation?
What is k dot? Change of unit vector coming out of the page. What is k dot? If you want to do that, if you don't want to do the calculation.
I told you derivative is change of value and magnitude. Value is one. It's direction.
value and direction the direction of k is changing by time no it's out of the page always so it is zero here we go you got it because dot derivative means change as i said Change or future remember I told you and the integration means past Simply you're gonna use it in any five-step control class You got it. Yes Good. This is so hard to problem-pump it.
Haaa... Problem 2... See guys, these are exactly based on two cars. Determine the maximum speed of each car if the normal acceleration is limited to 0.8 g. The roadway is unbanked and leveled.
Maximum or last. Because you know that an is how much you turn in front of you. a n is how much tell me tell me what is a n i told you a n is e squared over rho so it's given for the say for car a is 0.88 g goes to va squared divided by the row of that is 16 meter bingo and for car b is the same vb square over row is 21 bingo okay you got it simple like that so you see the applications when you are on the curve Ram, wherever you go.
Look at this. The driver of truck has an acceleration of 0.4 g as the track passes over the 0.8. So acceleration of 0.4 g, a pass from 0.8. At the hump in the road at a constant speed.
Constant speed. Okay? A pass from 0.4 g.
So V of truck is constant. okay and add accelerations the radius of curvature of the road at the top of the hub is 98. his radius of the curvature row is 98 meter and the center of mass g of the driver considered as a particle particle is two meter above the road so here is here here is two meter above the red that is the speed we have to talk so i'm gonna give you guys Five minutes if you can give me how to solve this problem. How to solve this problem, yes. Exactly, because it says, exponentials...
point for G as to a plus over top A and the velocity is constant velocity of that is constant remember velocity is Rho Beta dot it stays constant when this is constant what is Beta double dot because Rho already is constant what is Beta double dot tell me tell me how much is that is zero exactly so the acceleration which is in the form of rho beta double dot e t plus rho beta dot square e n that's right this is zero zero acceleration is only this or it is v score over row it gives you the acceleration just put it here and calculate see what i'm saying yeah it's simple like that um the same as you racing car two cars of the ambition at unbanked and level turn They cross the line CC simultaneously. Here we go. CC. Everything is based on what they write down. Same.
Same. See this is exactly what happens to you when you come from say this is College Avenue. You come on the ramp and go merging to I-5 toward which is right.
Here we go. The car is traveling at the speed of 60 miles per hour at the point A. Here is velocity at point A is 60 miles per hour. Until it gets to point B.
After which its constant rate of decadence of speed is this. So it says acceleration after point B. is acceleration is three feet per second as it runs interchange ramp is a decreasing acceleration inside the ramp okay determine the amount of total car acceleration a just before it gets to b it's very easy before it gets to b okay after just passes b and point c It's very easy.
Just put it in there. Do that. Okay. Let me solve this problem for you. I'll also solve this problem for you.
Consider the polar axis of the Earth to be fixed in space. Okay. This is ms. And compute the max of the velocity and acceleration of a point P.
It can be u. Here. On the Earth's surface at latitude 14 degrees north.
Here is 14 degrees north. The mean diameter of the earth is this, 12,742 kilometers, and its angular velocity is this of the earth. So let's go for it. So problem 2109, normal tangential coordinate system.
So as you can see, here we go. So I'll have it just sagittal view, okay? I'm looking at this is NS, north, south, and this is U, guys, point P.
And this angle is 40 degrees, as it says, right? So to calculate what? To calculate what we want to calculate? Let's see.
Calculate velocity and acceleration of point P, okay? So we select normal time G at the coordinate system because this is rotating. RFD is rotating around NS.
So this is going to be, this is going to be, I'm going to find the projection here on the ground here. It's going to rotate around here. That's right.
So I'm going to find this and here is RFD. Right. And I'm going to say this is rho here.
So cosine 40, cosine 40 is rho over radius of the planet. So from here, rho is r e cosine 40. It's going to give you some value. Okay, just check that out.
Then we know that the velocity. When you find from here, it's going to be rho beta dot square E t. Right?
We got rho from here. And beta dot is 0.029 in 10 to the power of minus 4. So we can rho in E t direction. You got it?
Why do that? Because... ET is inside the page, EN is towards the center of our planet, and K is outside the page.
See what I'm saying? Let me just upload it. See, if you put the projection here, it's going to be ET, which is rotating like that.
And this is EN, and the K is this. Your thumb shows what? Direction of rotation. Right hand, your thumb.
Off right hand. Right. So your V is going to be in the direction of... So acceleration of your particle point.
The acceleration is based on the formula rho beta double dot dt plus rho beta dot square. It says the planet rotates with the constant angular velocity. So this is zero. So we are going to put again rho or. you can write like this, rescore over, just calculate, you have row, from here, from en, here we go, we have it, okay, we got it, so, Pendulum.
You're going to see this in my classes, three or four classes forever, specifically vibration class and control class. Write the vector expression of the acceleration A of the mass center G of the simple pendulum. Super important example in both E n and E t and X and Y.
For the instant when theta is 60 degree, theta dot is 2 radian per second and theta double dot is this. Means we have angular and loss of the angular acceleration at the same time. Okay?
I'm going to do that for X and Y. I'm going to assign a respect. Problem 2, 1, 2, 1. This is the pendulum. This is ET. This is EN.
And the K is also off the page based on the right hand rubric, right? And the problem says here is X and Y. Y is like that because based on the right hand rule, you rotate four fingers from X to Y.
K is outside of the page. So both K's is outside of the page. Right hand rule. So based on the coordinate system. So this is G, point G.
We set R of G with respect to point O. It's going to be, here is 4 feet, right? I'm going to say A, and here is 60 degrees, right?
So, based on this triangle, so it's going to be A sine 16, okay? Because this is plus and plus, it's going to be J, problem solved, A. Cosine 6th deliberately in I. Problem says X and Y is this. Or you have to write generally.
You have to say A sine theta J plus A cosine theta in I. Now you know it. Tandrit of velocity of point G in I.
IJ is going to be A. I'm factoring out theta dot minus theta dot sine theta in I plus theta dot cosine theta in J. You have A. A is 4 feet.
Theta dot is 2 radians per second multiplied by. Sine of 60 degree or pi over 3 in I plus again 2 cosine of pi over 3 in J. Bingo.
This is velocity in I and J. And the acceleration of that, acceleration of point G is going to be A minus. theta double dot sine theta plus minus minus theta dot square cosine theta in i plus theta double dot cosine theta. Y is theta dot square sine theta and J is total of S. You got it?
Now you have the numbers. 4 feet minus the total is how much? Problem says is 4.205. 4.205.
Again, it's not enough when you put on the calculator. 4.205 over 3 minus the total square is 4. cosine of pi over 3 in i plus 4.025 cosine pi over 3 minus 4 sine pi over 3 in j in j here we go And this is the acceleration in INJ in X and Y as well. Interesting guys. Why?
Because you're going to find acceleration in ET direction and EN direction. Just I'm going to transfer this INJ inside here. Look at this guys.
I'm going to say this is ET, this is EN. You already calculated the expression, right? This is EN. But now I'm transferring AX that we just found it here and AY here.
See what I'm saying? Here is 60 degree and here is also 60 degree. You can find a n and based on the vector a t from here.
Just find the projection of this. Here we go. Here is 60. And I told you here is 60. So AT is going to be, you have to decompose in the both sides. and you have to decompose also this other side ay and we have two components see what i'm saying so at is going to be minus because these are in this direction that's right I'm going to take the absolute value.
It's going to be ax cosine 60 degree plus ay sine 60 degree sine and cosine. And an is at. an is ay.
cosine 60 degree and see an minus but this is between the other side minus ax sine 60 degree and this i want to sign here for am so we can calculate at and am We already calculated these, right? We already calculated these. This is your Ax, and if you multiply by 4, again here, and say your Ay.
Here we go. We don't need to recalculate everything. Or you can directly put on the formulation. We can say a row.
Beta double dot. I just taught you. This is short way. Et plus rho beta dot square En.
Both must have to yield the same result. Double check. OK.
You have rho. You have beta double dot. OK.
Now let's go, let me see if I can find another problem. Interesting. A football player released a ball with the initial direction shown in the figure.
We know the angle. Determine the radius of curvature of trajectory just after release. Okay, just after release.
And at the apex, at the maximum, the radius of curvature. Just put in the formula and find it. Okay?
So, let's go to polar coordinate system. Normal tangential coordinate system, when we had radius of rotation was constant. C constant. What if the radius of the rotation is not constant? So here it becomes the polar coordinate system.
I'm going to solve many problems on this. Many many problems. And cylindrical coordinates. So suppose this is your car moving and the radius respective here is changing. Right?
Radius is changing. slower bigger and you call so being said we define exactly like that but this time er is outside going outward of the center of rotation okay e theta is look at the end But the ER, this is R, like a lowercase r, because the opposite direction of EN. So R is not constant anymore. Here is R, radius of rotation is not constant anymore. First, we have to set up the sequence of unit vectors.
So ER is the case out of the page. So going to be ER. E theta and K.
Right hand rule. Protecting from R to theta, K is outside the page and follow the sequence. Okay? Remember, for normal tangential it goes like this. Goes like this.
E and E T. Here E R is outside and E T becomes E theta. good And you're rotating for the angle of theta. That's the sequence.
Okay. So after setting up, we can go ahead and work on it. The derivative of the velocity and acceleration term in the polar coordinate system.
So this is the origin of the central rotation. Here the angle theta and I said here is er here is e theta tangential to the trajectory so when you have that you know that we have vr why we have vr you previously had only one v was in direction of et because R is changing so R is changing. So you have velocity in direction of the R at the same time. So you have V theta.
You have both. You got it? So again I'm working right hand rule. E R V theta and K.
As I said radius of. rotation is changing with time as a function of time based on the right hand rule I said omega angular velocity or theta dot is right hand rule is outside of the page's direction is outside of the page in k direction angular velocity okay I'm writing K is out of H. Out of H.
Out of H. So, if you see your car on the curve A, I'm going to say position vector of point A respect to point O, based on the polar coordinate system, is RER. Here we go. Having derivative.
So velocity of point A with respect to point O. Because R is changing with time. R dot ER. See? There are two terms.
Plus R ER dot. In the normal tangential. Because R or rule was constant. We had one term of velocity.
So now we have to calculate this. What I taught you. General rule.
E dot r is omega in ER. We established our right hand rule sequence. Here we go. ER, E theta and K. Omega 0 0 theta dot.
ER is 1 0 0. We can write very simply as er in 0 theta dot 0 0 minus e theta 0 theta dot 1 0 plus k 0 0 1 0. This is 0, this is 0, and the result is theta dot. E theta. Bingo. So in polar coordinate system, because radius of that is changing, the velocity becomes, velocity of point A becomes r dot er plus r theta dot, we say r omega, right? E theta.
Bingo. We found it. We have two terms.
In normal tangential only we had rho beta dot. Because rho was constant we had no time derivative of that. In the polar coordinate system we do have.
Okay. Now that we have it we can go for acceleration of that. I'm going to solve the next class many problems on this problem.
So we a. R dot E R plus R theta dot. E theta. Conductor.
Acceleration of point A is going to be r double dot er plus r dot er dot. We already calculated. Plus. Now these terms.
r dot theta dot e theta plus. We have three terms. r theta double dot e theta plus r theta dot e theta dot.
Now we have to calculate this. How to do that? Simple.
Is omega again in e theta. Our right hand rule sequence. e r. E theta k omega is 0 0 theta dot is 0 1 0. Just write it down here as E r 0 theta dot 1 0 minus E theta 0 theta dot 0. zero plus k zero zero zero one this is zero this is zero the answer is minus theta dot er bingo just put it here so the acceleration coordinate system look at here guys you have one er here and you have one er here gonna be R double dot minus R theta dot theta dot square. ER plus E theta.
R theta double dot plus E dot R is R theta dot E theta. It's going to be 2R dot theta dot E theta. I don't expect that you know the process. I want that you only know the final answer. This is the expression in the polar coordinate system.
Compare these guys with normal tangential. Was Rho beta double dot E t. Your E t was E theta here, that's right.
Plus Rho beta double dot, beta dot square in E n. Your E n is minus E r. So you have an additional term here and an additional term here. You got it, everybody? So for summarizing, I'm writing everything here.
R of A is RER. V of A... is r dot er plus r theta dot e theta y.
Obviously, the magnitude of the velocity is going to be r dot squared plus r theta dot squared. And the acceleration is going to be r double dot. minus r theta dot square ar plus r theta double dot plus two r dot theta dot a theta.
This is ar, this is a theta. So, the magnitude of the acceleration is ar square plus a theta square. The next class I'm going to go over again the pendulum problem to show you how polar coordinate system effective using the Cartesian coordinate system and what and polar coordinate system.
How is effective the polar coordinate system. Okay, I'm going to enjoy it. And then I'm going to go to class problems. There are many class problems I have to solve for you.
Like problem 142, problem 136, and 156. i'm going to show this next class after finishing the pages of prog okay guys any questions ladies and gentlemen next next class i'm gonna go over this because it's gonna be too much for you i'm gonna set up er d theta so let me do it I assist the pendulum polar coordinate system here is l and theta I'm saying l is changing with time length of the pendulum like as a piece of plastic and a piece of kind of kind of elastic prove that its length is changing with times right I want to find this acceleration of point P with respect to point O. Simply, its velocity is going to be, look at here, r dot er plus r theta dot e theta. Your r is l. It's going to be l dot er plus l theta dot e theta.
It's so simple if you select the polar coordinate system. Or the acceleration of that is going to be L double dot minus L theta dot square er plus, if L is changing with time, plus L theta double dot plus 2L dot theta dot e theta. This is the acceleration of this particle pendulum.
Particle tangent. It's pretty straightforward. Well, if you want to expand this in the in the X and Y, good luck.
I'm going to do it next class. We're going to be ready with it. X and Y. And I'm going to compare these with that and see how effective is the polar coordinate system.
And indeed, I'm going to use this in the vibration class. In the vibration class, I'm going to use this when I'm driving the equation of motion. And eventually, at the end of this course, I'm going to drive the equation of the motion using this formula.
Okay? Clear everybody? Yes?
Alright, let me see if I can find something. No, almost done.