 Hi, I'm Zor. Welcome to a new Zor education. Today we will talk about rotation within the xy plane around the z-axis. So we have a plane and rotation is within the plane, this is xy plane, this is z and it's all only around the initial position and around the z-axis as around the axis. So everything is happening within the xy plane. Okay, that's our purpose of this lecture to analyze this particular motion. Now this lecture is part of the course physics 14 presented on unizor.com. I suggest you to watch the lecture from the website. Not only because it's free and it is and no advertisement. Also it contains detail notes for each lecture and eventually it will contain exams for every section. So you can take it and doesn't require any information from you. So just completely free. Now rotation. We are talking about not just rotation, but about uniform rotation within the xyz around the origin of co-ordinate and around the z-axis. So basically we are talking about a motion within the xyz, within the xy, sorry, plane, which means z-coordinate is always is equal to zero, right? So z is always equal to zero and obviously z of t is equal to zero and z-speed and second derivative, they're all equal to zero and I'm no longer going to mention z at all. So all my co-ordinates are within the xy, my vectors are within xy plane, etc. So now we have to define somehow the uniform rotation. Now before, we we defined what is a uniform motion as the motion along the straight line with a constant speed basically whenever you're covering the same distances in the same interval of time. Now let's define in a reasonable way the uniform rotation. Now, I think it's very reasonable to define the uniform rotation when we are rotating by the same angle for the same periods of time. So let's say within the first five seconds of rotation you you turn by let's say 30 degrees. Then for any five seconds interval you have to rotate by the same 30 degrees. So that's the uniform rotation. Now if now this is x, y, z. So xy plane, I'm no longer talking about the z and I will put everything on this on this board as this is xy plane. So rotation is here. So this is x, this is y and what I'm saying here is that this angle from positive direction of x to the vector to the point where my object is located, this is the function angle as the function of t of time. And it's supposed to be such a function which basically delivers me exactly the same as my speed delivered in case of uniform motion, which means its derivative is supposed to be constant. And I'm using omega as the constant which signifies basically, let me just call it openly. This is angular speed. In the same way as we had the speed as basically the measure of how fast the object moves along the straight line, this is the measure how fast the angle is changing. So the change of the angle the instantaneous basically speed of changing the angle should be constant. That's what means that's what means the uniform rotation. Uniform rotation is the rotation described by such a function whose derivative by time is constant. And the story, that's the definition. From this we obviously can derive as before how the angle actually is expressed as the function of time. Now if its derivative is omega then the function itself phi of t should be omega t plus some kind of a constant which is basically initial angle from which we started. But we can always start the x-axis in such a way that our initial position is on the x-axis with coordinates r 0, right? x going in this radius and y going in this 0. In which case this disappears is equal to 0, initial position is equal to 0, phi of 0 would be equal to 0 which is supposed to be this way. That's kind of easier. And so we have this particular equation for angle as the function of time is such and such where omega is a constant. Well, okay, now we're talking about angle but we need the coordinate expressed equations of motion or vector expressed or something, not the angle. Well, that's very easy because if I have this angle as a phi my obviously x of t is equal to r times cosine of omega t cosine of phi, right? My x-coordinate is cosine r times cosine it's a triangle and my y-coordinate is r sine of omega t where omega is constant r is also given. So that's how I got my equations of motion. Well, that's great however, I would like to spend some time to just analyze the more physical properties of this type of motion. Now, knowing this let's just think about the following. First of all, let's find the velocity, right? We have omega as angular velocity and it's a constant. Now, how about velocity of the body of this object in coordinate form? The regular velocity which we used to. Now, the vector of velocity will change because direction is changing, right? Direction is not a straight line. So velocity, as we understand it, which is vector v which is equal to x of t y of t and the vector is also a function of t. This velocity cannot be constant. Well, let's calculate what it is. Oh, I'm sorry. Okay, now it's equal to okay, derivative from cosine is minus sine and then there is the inner function w. So it will be minus r omega sine of omega t. That's my x-coordinate, x-velocity. And y-velocity from sine is a cosine and also r omega cosine of omega t. So that's my vector of velocity and as you see, it's changing with the time for obvious reason. Now, first of all, what's the speed, linear speed? Speed is a magnitude of the vector of velocity. Magnitude of this is, you remember, if you have a vector, let's call it pq, then the magnitude of this vector is square root from p square plus q square. Now, in this case, square root of this plus, square root of this square plus this square, you understand it would be r omega square and sine square and this is the same thing but the cosine square. Sine square plus cosine is one. So basically I will have omega square plus r omega square times square times cosine square of omega t plus sine square of omega t. So this is one and square root. So I will have that the speed is equal to r omega. Now, what is speed in this particular case? Well, that's a linear speed as we are going along this circle. How much of the length of the circle we are covering in the unit of time? Basically that's what speed is. Okay, out of curiosity, let's do the following. Now, this is my position vector. This is my speed, my velocity, sorry, vector. Out of curiosity, I would like to do this. Scholar product of my position times velocity. What happens? Now, the scalar product of two vectors with components x and y is you multiply x by x and y by y and sum them up, right? Remember this? If you have vector p1, q1, let's do it vertically. p1, q1, this vector, scalar product by p2, q2. It's p1, p2 plus q1, q2. That's the definition of the scalar product, right? And I will explain why I need this as soon as I will do this multiplication. So I have to multiply x by x and I will have minus r square omega sin of w omega t cosine omega t. Plus y by y, r by r, r square omega, again sin of omega t times cosine of omega t. Which is equal to zero, right? Minus and plus, everything else is the same. Now, you have to remember again your vectors. If scalar product of two vectors is equal to zero, then they are perpendicular to each other. We have proven this in the course when we were talking about vectors in mathematics course. Which means what? Which means the position vector, which is from zero to this point, is always perpendicular to speed, to velocity vector. Which means velocity vector should be tangential to the circle. So if we are rotating, my velocity vector is always tangential to a circle of the trajectory. And at any point, it's perpendicular to position at that point. So that's what's very interesting. Again, velocity, when you're uniformly rotating the body around some center in a circle, will always be perpendicular to the radius vector into position of our object at the time, at any time. Doesn't depend on t for any t. So at any time, velocity at point wherever you are located is always perpendicular to the position vector. Now, let's talk about the second derivative. Second derivative is acceleration. Acceleration vector, what is it? That's the derivative of this, right? So I have from sine is a cosine, plus another omega from here, so it's minus r omega square cosine omega t. Now this one from cosine is minus sine, and it would be also omega square and sine. Interesting. Compare this position and acceleration. What's the difference? Only the difference is minus omega square, multiply. So these two vectors are collinear, and the only thing is we are multiplying by some negative factor. Now my vector of position goes from here to the point, and my vector of acceleration at this point is collinear, but with a minus sign and some multipliers. So my acceleration goes towards the center of the rotation. That's a very interesting fact, right? So acceleration goes exactly collinear to my vector of position from center to the point where object is located, but directed towards the center, in opposite direction, obviously. As if the center actually gravitates, and obviously we all understand that for instance if earth is circulating around the sun, it's the gravitation from the sun which keeps it, and it's the gravitation which gives acceleration, right? So that's why we are at the same time accelerating towards the sun, but our velocity is directed tangential to our orbit, right? So these two movements make a circle. Acceleration towards the center and velocity perpendicular to my position tangential to a circle are two different movements, the result of which is the rotation. So you see we have a lot of mathematics, but also there is very important physical interpretation of this. And finally, let's do the acceleration magnitude. We did the speed here. It's the magnitude of this vector of velocity, which was constant because it was sine square plus cosine square under the root. Here we will have exactly the same thing. The magnitude of this vector is again square root of this square plus this square, and cosine square plus sine square would be the same, would be equal to one, and I will have r omega square as the magnitude of acceleration. So acceleration is a vector which always directed from wherever our object is towards the center, and it's always the same, does not depend on time. So these are physical properties which we have derived from mathematical analysis of this movement. Well, that's it about rotation. Now you know all these details. I do suggest you to read the notes for this lecture on Unisor.com, and well, that's it for today. Thank you very much and good luck.