[MUSIC] Hello. Now, let's learn another coordinate that will describe the rotational motion, which is polar coordinate. Polar coordinate is a very, very popular to describe the rotational motion. So suppose that you have a point A has a velocity of v vector here, and then note that in t coordinate your et vector is this is defined along with the v vector. But in polar coordinate you have an origin, and then from the origin you have a distance from point O to the r that's the direction for the r vector. And here we are describing the counterclockwise data, as a direction for the theta. So in this case, we define the as a vector is outward of the r direction and then e theta is perpendicular to this. This is how we define the point a in polar coordinates. So, your r in polar coordinate is going to be r, the r, and the vector. To calculate the velocity, you can take the derivative of the position r vector, and that means you have r.er plus r dot. So, similar to the t-n coordinate is this dot vector, the unit vector times derivative is not going to be 0 anymore. So lets take a look at the case where a has been moved to the a prime over infinitesimals time. So we now turns it out to be v prime, and then the r vector from the origin to the a prime now has been changes from the solid line to dashed line. And in this case, your vector is actually extension for the r displacement. So this is going to be now a new vector prime, and this is a new e theta vector. So as you can see here, even though the magnet doesn't change, the direction has been changed. So your vector over time, this is not going to be 0 anymore. So let's take a look, vector originally and prime, with the amount of d theta. And then since this is a unit vector, the magnitude of the is only d theta is equal to d theta. And then direction is normal, the counterclockwise normal to the vector, which is e theta vector. So take time derivative of is going to be theta dot e theta vector. So, plug that in into here. So what you can get for the velocity is, r dot vector in r theta dot e theta. Now, acceleration take the derivative of the velocity as well. So, take the derivative of velocity which has two components, have many, many components here. So let's take a time derivative one by one. So, r double dot and r dot unit vector dot, and we know this is theta dot e theta. And r dot theta dot theta, r theta double dot e theta and r theta dot in e theta unit vector time derivative. So as you can see here, if theta vector also change its direction. So, time derivative is not going to be 0 anymore. So e theta originally and e theta prime, with the amount of delta theta. So de theta is magnitude of d theta with the unit vector. And the direction is, this is another counterclockwise to the e theta vector. So it's going to be minus here. So here, we didn't have a minus sign for the dot, but for e theta dot, we have a minus theta dot term. So if you plug that into this time derivative form, you will have total one, two, three, four, five terms, and then if you rearrange it as a in terms of term and e theta term, you have a total of four components of the acceleration. r double dot minus r theta dot square, which is centripetal acceleration. And 2r dot theta dot, which is Coriolis acceleration, and r theta double dot e theta. So, compared to the t-n coordinate, the polar coordinate, acceleration is a little bit more complicated. And you are supposed to practice many times, so that you're most to memorize those four terms. So I'll give you a chance. So this is a blank sheets, a partially filled in blank sheet. So, with the point a polar coordinate defined by the r why don't you just derived how you can obtain the v as this one, and the acceleration is that one? Why don't you just take a practice now, before you forget it. Let's work on an example for describing the motion for the point, using the polar coordinate. There's a gripper is rotating counterclockwise direction with constant angular velocity omega. So it's rotating this way, with the omega. And why are you extending the arm with the rate of l dot and l double dot. So, and then this arm is getting longer and longer. Well, I'm not sure if this is a positive sign or negative sign. But anyway, the length has been buried. And you're supposed to find the velocity and acceleration of the pivot point P here. So let's use the polar coordinate and defining the vector in the theta vector. And let's recall how you can formulate the velocity and acceleration in polar coordinate, whenever your petition is defined by the r. Can you remember? Yes, it's r dot plus r theta dot e theta, and acceleration as a four term, which is, yes r theta dot r. Double dot minus r theta dot square, to all the theta dot plus r theta dot square for the e teta term. Now, this r dot term and r are given here, so r dot term is l dot, r is going to be distance from the origin to the point p. So that l naught plus l. And same for the acceleration as well. So, we have most of the everything is given, and constant angular omega. So theta double dot turns out to be 0, so you can just plug that in what's been given in the problem, to obtain the velocity and the acceleration. So pretty easy. So, that means for the polar coordinate, rotational motion, you'd better either be able to derive the v and a right away. Or, just practice many times so that you almost memorize the formula. Let me ask you a question. This problem we solved using the polar coordinate, but can you express the acceleration in terms of t-n coordinate? Yes, why not? So to do so, first, you should set where which one is the et-en unit vector? It's not going to be same as and e theta here,because t-n coordinate is defined the from the velocity. If the people point p has a finite distance, e theta is going to be equal to the e theta vector. However, since this arm length has been changing, maybe velocity will be a little bit tilted off the e theta vector. So in this case, this one is et vector and this one, whatever normal to that en vector. And then based on that, you can find a formulate the acceleration in terms of vt and en direction. So here is this v squared over ro, is the ro same as l naught plus l. The distance from the p with respect to the pivot point O here. Now, because since en vector is not coherent to the direction of the arm. Those ro is totally different from, has nothing to do with the l, not an l. Then, when we should use the polar coordinate and we we should use the t-n coordinate. Well, for some cases, if you could apply for the polar coordinate, usually the case where you can exert a specified where the origin is. Compared to that, the t-n coordinate, since the radius of coverture the varies over the trajectory. So sometimes, when you describe the motion using the t-n coordinate, are the cases you cannot specify the origin. Let's solve another example. This is a rocket vertically fired, and there's the observation station. And then to describe the motion of this rocket in terms of polar coordinates r and theta here. And at this instant, I measured r theta radial acceleration double dot and angular ratio of theta dot. So what's going to be the magnitude of the rocket, the velocity vn a. So let's start with defining the r theta coordinate here, and the relationship for the r the v is what r dot plus r theta dot d theta. And the magnitudes will be the r dot square plus r theta dot square, and this is what we've been observed here. But this wasn't, so this is unknown actually. How can we find out the magnitude without knowing this value? There's no more information from the problem. In this case, let's look at the situation and maybe the motion constraint will give you the answer. So, this r dot is our direction to our components of the v. And then this one is the theta component of the v, like this. And the total velocity which is a vector sum of v theta and vr, is in vertical because it's a vertically fired. So the v theta and vr is not going to be independent anymore. And it has some constraint by the kinematics. So vr which is r dot is here, this angle which is th theta. v theta tangent theta, so if you're using that relationship, you can find out the magnitude of the v, same for the acceleration acceleration has a four term. Magnitude will be the square term for the rr component in the square for the theta components. And since those parameters are observed, like a you have r double dot or, and theta dot term, but these theta component of the acceleration has not been measured, like a we don't know what's going to be r theta r theta double dot. So, by looking at the rocket motion, those acceleration of a theta and ar or constraint, because the total acceleration should be directed in a vertical way. So even though we don't know the a theta, those information could be obtained from the knowledge of ar, which is observed. Using the cotangent theta in a relationship. So by plugged it in, we were able to find out the magnitude of the acceleration as well. Okay. So in this chapter, we learn how to describe the motion in the rotational motion in the curved path using t-n coordinate in the polar coordinate. The TN coordinate, we actually define the coordinate a cell in terms of the parallel to the velocity vector, et. And whatever normal to that is en, whereas the polar coordinate we have an origin, and then origin distance from here to the point is going to be the r, and then extension will be vector. And no perpendicular to that is e theta vector. So there are two different ways to find the unit vector, in t-n and polar coordinate. So why don't you just filled in the blanks, how we can derive the velocity and acceleration. And during those process, you must be able to handle the, how you can handle a unit vector time derivative. And check your answer from the previous slide, if you could have got the velocity and acceleration correct. Okay, so I strongly recommend you should practice this slide by your own. So in this chapter we learned how we can describe the rotational motion using the polar coordinate. The polar coordinate acceleration has a four term. So, more complicated than the t-n coordinate. And then I hope you practice many times, so that you almost memorize those four terms. So this is the wrap up, and next time we are going to briefly go over what how we can describe the motion in a relative coordinate. Thank you for listening.