 Hi, I'm Zor. Welcome to Unisor Education. I would like to talk about rotation, in particular uniform rotation, with constant angular speed. I think this particular theme is not paid a lot of attention during the regular courses of classical physics, so I would like to spend some time on concepts of this rotation, and especially rotation, which is relative to another rotation, which is kind of complicated, but nevertheless it needs to be addressed. Now this particular lecture, and many others actually, are part of the Physics 14 course presented on Unisor.com website. I suggest you to watch the lecture from this website, because it contains lots of notes. Every lecture has notes, basically like a textbook. The site is free, no advertisement, so feel free. Okay, so again, our ultimate purpose of this particular lecture is to learn something about rotation around some axis, whereas this axis itself is rotating around another axis, something like this. Well, these axes are considered to be parallel for simplicity, and obviously I basically consider the plane, which is perpendicular to these axes of rotation, and everything will be actually happening inside these planes, which I will use my white board, obviously, to represent. So z coordinate is always equal to zero. Now before going into complexity of this relative rotation, let's just again, maybe one more time, maybe it's a repetition, but still it's important, address a couple of concepts related to rotation. In particular, angular speed, velocity vector of rotation, and linear speed. Now, first of all, angular speed. If you are having an object which is rotating, let's say there is a radius r here, it rotates uniformly. Now what does it mean? It means that the lengths of the arc covered during the time t is always proportional to time t with the same coefficient of proportionality, and that coefficient is called angular velocity. Now, in other representation, if this angle is phi of t, which describes location of the point on the circle, on the trajectory, then I can say that this is position, basically, it defines the position of the object. Now, obviously, this function can be differentiated by time, and that means that I'm basically trying to find instantaneous average speed from moment from t to t plus delta t. During this time, my angle moved from phi of t to phi of t plus delta t, and if I will subtract one from another, that's the angle difference, and that's basically how much my angle has increased during this delta t moment, and if I divide it by the time, which is delta t, actually, I will have average speed from t to t plus delta t. And if I will use the infinitesimal delta t, I will basically get my first derivative of the phi function. Now, I assume in all the problems, which we will consider, I assume that this is equal to constant, in which case phi of t is equal to omega t, if at the moment t, I was at angle zero, which is here. So, this is our normal assumption, and this is basically what angular velocity is. We will measure it in regions per unit of time. So, in this particular case, for instance, if my full circle is accomplished in the time t, then my angular velocity is this. So, this is the time which it takes for objects to go one full circle, and one full circle is 2 pi regions. So, this is my... Now, if, on the other hand, object makes n rotations per unit of time, that means the angle would be 2 pi times n and per unit of time, so divided by t. So, if I know that I'm making, let's say, 25 rotations per second, then t is equal to 1 second, and 25 rotations would be 25 times 2 pi would be 50 pi, and that would be my speed in regions per second. Okay, so angular speed we got. Now, going from this picture, if this angle is phi of t, then from geometry, you know that the length of the arc is equal to r times phi of t, where phi is angle in regions, always regions, remember, not degrees or anything. Okay, so that's plane geometry, obviously, of the circle, that we have to know. Now, if this is true, then I can always find the average speed during which I have covered this particular distance, l, right? So, that's the distance divided by time. Now, again, as in case of angular speed, what I should actually do, I should take, what's my l of t plus delta t? This is the distance I covered by the time t plus delta t, subtract l of t and divide by delta t, which is t plus delta t minus t. So, my distance increment, I divide by time increment, and that's my average linear speed as I'm going along this curve, how much real distance I have covered along this curve, if I measure it during the time delta t. And if I will divide one by another, that's an average speed. And again, if delta t goes to zero, I will have my instantaneous speed along this trajectory. Now, obviously, since this is r omega t, and I have, this is my derivative of this function, so my linear speed would be equal to r times angular speed. So, this is my linear speed as I'm moving along the curve. Now, what is the real velocity vector in this particular case? Well, velocity vector is a vector which is directed towards my direction of the movement, and it has certain magnitude, and how do I obtain my velocity vector? Well, first of all, we are talking about the system of coordinates. So, let's say this is x, this is y, this is position. So, what's my position vector of time t? Well, if this is r, and this is phi, or phi is equal to omega t, then I can say that my x coordinate is equal to r times cosine of angle phi, which is omega t, and my y coordinate is equal to r sine omega t. So, that's the position vector. Now, we talked about many times, if I have a motion, this is basically a definition of the motion. This is how my motion occurs. I have a function of time for both coordinates. Now, if I want to know the velocity vector, well, that's the derivative of the position, right, which is minus r sine, and inside would be w sine of omega t. That's my derivative from r cosine omega t. Now, in inner function, outer function, so the derivative of the outer function is minus sine. This is just a coefficient r, and in the function, I have to have the derivative of omega t, and that would be omega. And here, similarly, I will have r, I will have omega and cosine of omega t. So, that's my velocity vector. Okay, first of all, what is the norm, what is the magnitude of the velocity vector? Well, that's x square plus y square, right? The velocity is a vector, and if I will have x and y coordinates of the vector, the square root of x square plus y square is the magnitude, right? We have a vector, this is x, this is y, the magnitude, the Pythagorean theorem. Okay, so r square, this square plus this square. Okay, obviously, it would be square root of r square omega square sine square of omega t plus cosine square of omega t. Now, this is one sine square plus cosine square, regardless of the angle, and the square root of this would be r omega. Now, look at this. So, the magnitude of the velocity vector is equal to linear speed of the vector as it moves along the curve. Well, that's not a surprise, obviously. Now, another property, again, very obvious is, now, if you will do the scalar product of these two, what it will be, minus r square, minus r square omega sine cosine and plus r square sine cosine omega. So, that will be zero. Now, what does it mean? It means that vector of velocity is perpendicular to vector of position. Position vector is here to the point. So, velocity is tangential. So, that's basically all we wanted to know about the rotation. Angular velocity, linear, sorry, angular speed is better. Linear speed, because these are scalars. They're not vectors. And finally, the velocity vector, which can be expressed in these coordinates, and we know that this vector is tangential to the circle because it's perpendicular to the radius. Now, let's talk about relative rotation I promised in the very beginning. So, here is the picture. So, this is a fixed center, fixed to the earth, let's say. Now, this is the road, which is rotating. It has radius r and constant angular speed omega. Now, let's call this point A. So, I will use omega A here. Now, I have a point B. Now, this is a point around which my disk can actually rotate. Now, it has a radius r, lowercase r, and the angular speed is also constant, but it's different. So, my task is basically find out the position and velocity of point B. Now, what I will do is just use exactly the same approach as I did with rotation of one particular object. But now, I will have to do basically the combined motion, which I can define one of them separately from another. And then, if you remember, if you have one, let's say, system of coordinate and another system of coordinate, which is moving relative to the first one, I can always find the position of the point in the original one if I will add this to this. Right? So, all I need is how my origin of the new system, of the moving system moves relative to this one as long as the axes are parallel. So, that's what I will use. Okay. So, as the time goes by, what is my position vector for A? It depends on time. Well, obviously, my system of coordinate is related to this O, right? And let's assume that in the very beginning, my A is on the x-axis. So, initial angle is 0, in which case it's equal to r cosine omega A t r sine omega A t. That's my coordinates. Okay. Now, let's talk about point B. Again, we will assume that point B in the very beginning as t is equal to 0 is here, horizontal. So, if I will have position of the point B relative to point A, point B relative to point A. So, I assume there is this system of coordinates. This is lowercase r because that's the radius of the lower cosine omega B t. Now, I'm using its own rotation and r sine omega B t. So, that's position of point B relative to system of coordinate related to center of the disk with axes parallel to these ones. So, all I have to do is, if I wanted to find coordinates of point B in original system, unconditional, it's basically coordinates of A plus coordinates of A, B. Because vector from here to here is equal to vector from here to here plus vector from here to here, right? Which basically gives me the expression of coordinates of the point B in original fixed system of coordinates. So, this basically represents my relative rotation. It's rotation over rotation. This rotation is relative to this rotation. Now, this combined motion gives me, well, obvious formulas. It's equal to what? r cosine omega A t plus lowercase r cosine omega B t. That's my x coordinate and the y coordinate correspondingly with the sine. Now, I can find the velocity of this by just differentiating by the time t. And I will have the velocity. And let me try to explicitly do this. So, velocity would be velocity of point B relative to original system of coordinate. Obviously, it's, again, it's equal to the sum of velocities. But anyway, let me just differentiate this. So, this would be r minus, because it's cosine, A sine omega A t plus actually minus r lowercase r omega B sine omega B t. That's my x coordinate of the velocity. And my y coordinate would be, so I will have sines there, right? So, it would be r omega A cosine of omega A t plus lowercase r omega B cosine of omega B t. What is this at moment zero? So, whenever my disk is here, at moment t, my disk is here. That's what my A point and my B is here. So, what happens in this particular case? Well, my velocity vector of zero is equal to zero and zero sine of zero is zero. Cosine of zero is one, so I will have r omega A plus r omega lowercase r omega B. That's the magnitude. Well, that's not just a magnitude. This is basically the vector itself. I have x and y coordinates. But what's interesting is what's the direction of this vector? The x coordinate is zero, right? So, the direction is obviously this. What's also interesting is that this is a sum of two things. This is the vector of this movement relative to zero. And this is this movement of B relative to A. So, whenever we start the motion at t is equal to zero. I never just started. My connection between O and A move almost vertically up, almost straight line, right? And connection from A to B also move straight up. So, basically it's like we have an old uniform motion. Let's say you have a platform and moving, let's say, straight. And then the person on that platform moving in exactly the same direction. You basically add the speeds, right? We were covering this before in relative motion material. So, this is exactly the same thing, but applied to rotation. Because whenever they are aligned this way, I can always say that these are directed along the same direction. And that's why it's just plain addition of their linear speeds. Because the linear speed is vertical. Direction is vertical and linear speeds are basically adding together because they are in the same direction. Well, that's all I wanted to talk about as far as the relative rotation is. This is maybe a little bit more involved than whenever you're considering the uniform motion along the straight line with constant velocity. But in many aspects, the uniform rotation, which means omega is constant as we are considering and probably all the problems which we will deal with, we will have rotation as a uniform motion, most likely. It's equivalent in some way, but instead of the regular velocity you have to use the angular velocity. And then the analogy would be quite obvious. All right. That's it for today. I suggest you to read the notes for this lecture. It's like a textbook. After you have listened to this lecture, that would be very easy reading for you. That's always helpful. And that's all on Unisor.com. The direction to this lecture, by the way, is from Unisor.com. You have physics 14, then mechanics, then part of the mechanics, which is called kinematics. And then you have frames of references. And that's one of the lectures in this particular section. All right. That's it. Thanks very much. Good luck.