 In uniform circular motion, we have a particle that follows a circular path at a constant speed. We have two ways of describing that motion. One is with linear variables and one is with rotation variables. For the linear variables, the position is given with the power along the curve, the power along the circle. Which, of course, is its linear speed as the time increases to the next line, the linear speed, being the derivation of that position. We can also give it as a rotational entity along the circle, which should be equal to the rotational speed omega times time. Omega being the derivation of the angle of time. Now, if we consider the time it takes to go around one time as the period capital T, we can say, oh, okay, so this is my linear speed is constant. I have to go around once, so a total path of 2 pi r, one second few, is over the period. Which would give me the linear speed similar for the angular speed. So, okay, go around once. It's 2 pi in radians over T. So, as you can guess, we're going to measure the angular speed in rats per second. We angle in rats for the linear distance with the meters, and then we get the linear speed as the linear speed. Now, if we compare those two times here, is there a direct connection between the linear speed and the angular speed? Yes, there is. See, we have 2 pi r over T. Here, we have 2 pi r over T. That means we can calculate the linear speed as the angular speed times the radius. This has worked out with the units. If you look at it, that's per second times the radius in meters. You see that? And the radius per second. The radius is one of the units that can appear and disappear. So, again, yes, I do that meters per second. Now, the next thing that I want to look at is the linear velocity, which is a vector. So, I will have to describe it in i-hat and j-hat protection. So, first, I do have this e v times... And if I look at my x component of the velocity, when I'm here at zero x component, at 90 degrees, I have my negative moment. So, it's component at 180 at zero again. And at 270, I have my highest possible value. So, if you think about it, you'll have a function to do that. And it will come up that it is sine of the angle that does this, the angle which we say is omega t. However, at 90 degrees, we are at the negative value. It's in fact a minus. And for the value at zero degrees, I'm doing it with my highest speed. Then we have zero speed at 90. Negative speed at 180. And the closer at zero speed, again, in my direction. Once we're at 270, so plus cosine. Because now I can try to figure out what does the linear acceleration look like. The acceleration, the linear acceleration, is the derivation of the velocity. So, if I derive this equation to that, I should get an expression in i hat and j hat for my linear acceleration. So, let's do this. v is the main just constant. If I derive sine, I get cosine. Don't forget, in the derivation, transform it up. Plus, if I derive cosine, what do I get? I get minus sine j hat. Now, let's rewrite this a bit nicer. I could say, okay, this is my v, which is omega r. The omega that I have here and here. So, omega square r minus cosine omega t is sine of t j hat. So, let's have a look at what that actually would mean. My acceleration has a maximum amplitude of omega square times r, which we disappointed. If I look here at my position here, I would have cosine omega t. So, minus one x and sine of zero. So, here my acceleration is on this one. What about here? At 90 degrees, I will have zero for the x component. Minus one for the y. If I'm here, I have minus cosine omega t. So, to plug in 180, I will get positive x component. And again, zero y. And down here, it will be at 270 degrees. Zero x and plus one for the y. So, the acceleration that we have is always pointing to the center of the circle. This is also what we call this centripetal acceleration, of which the magnitude must be omega square times r, which if I plug in this equation here for omega, gives me e squared over r squared. So, that will give me...