 Hi, I'm Zor. Welcome to Indezor Education. I would like to continue discussing rotational movement and I will try to build some parallels between translational movement, the movement basically along the straight line forward with certain speed and we will usually consider the constant acceleration and the corresponding movements around certain axes, the rotational movement. So translational and rotational, moving along the straight line and moving around the axis. Okay, so what's common and what's different? Well, first of all, let's think about how we identify the position when we are moving along the straight line. Well, usually we have certain point, which is the beginning of our motion. This is something which we can have a zero point and in most of the cases we are measuring the distance from time equal to zero to certain time and the distance is function of the time. So from s of zero to s of t, this is the distance covered during the time interval from zero to time t. Now, what's the equivalent of this position relative to the beginning of the motion or the distance relatively to the point where the motion started? What's the equivalent in rotational movement? Well, obviously if we are talking about rotation, we probably should do exactly the same thing. At t is equal to zero, we measure the point where exactly the motion started and then as our rotation goes on, at the time t we will have certain angle. So the angle of rotation is in some way an equivalent of the distance during the translational movement. So this angle of rotation phi of t is basically the equivalent of the position. So in this case, position is distance, let's say meters or something else. In this case, it's an angle in regions for instance. So right now we are talking that translational and rotational motions, something which is position or a distance from certain beginning point, is parallel to angle of rotation in case of rotational movement. Okay, next. Next concept which kinematics usually considers when we are talking about translational movement is the speed. Now which is the speed? It's also a function of time and by definition this is the first derivative of the distance from the beginning by time. Well, obviously when we introduced it we have something like infinitesimal time period from t to t plus differential of t and our distance increased from s of t to s of t plus differential of data, differential d of s of t and obviously the average which is basically this during this period of time, whenever my infinitesimal time period dt is shrinking to zero, that would give us the first derivative. Okay, now let's do the same in case of the rotation. So what I will do is I will have the first derivative of my rotational angle of time and I call it rotational or angular as they say angular speed. Okay, so we have angular rotation and its first derivative is angular speed. Now, obviously this function is given, we can very easily calculate its first derivative. Now, here is a very important aspect of rotation. You see in translational movement the speed is always associated with the vector character of the movement. That's why we're talking not about speed but in most cases we have to really talk about velocity. Velocity is a vector which accumulates in itself two components of the speed, the magnitude and direction. Now, the vector character of angular speed is not so obvious. However, if we will look deeper into this particular problem of rotation, we will find that it's not just the first derivative of the angular position which we are interested in. In theory, if we want our angular speed to represent the rotation in as much as velocity represents the translational movement as a vector, we have to specify a little bit more than just the rate of change of this angle as the time goes on. We have to specify the axis around which rotation actually occurs. Well, in the same way as the vector of velocity specifies the basically the line along which the movement is occurring. Now, in this particular case, in case of rotation, what's very important is where exactly is the axis of rotation. And also another very important characteristic is direction because it can be this way or this way around this same axis. Now, for this reason, we represent this first derivative not just as a scalar without basically any kind of direction. We will represent it as a vector and here is how we will do it. So let me wipe out this which is actually the view from the top of our rotation and I will specify the rotation in three-dimensional way. This is this is the axis of rotation. This is a plane of rotation. This is a center of rotation. Now, this is our angle phi of t. And now what I will do is I will put this vector from this point along the axis as a vector. Now, it has the magnitude equal to, well, the first derivative obviously. And also it's very important. I can position this vector on this particular picture up or down. So what I'm going to do is if my rotation is this way, then I'll position it upward. If rotation was in an opposite direction, I will position my omega of t, my angular velocity downwards. So what's the rule? Well, here is the rule. If you look from the top of this vector or the end of this vector to the plane, if rotation seems to be occurring counterclockwise, which is positive direction, okay, that means that we are looking correctly. Now, if rotation is in this way, I have to put it down. And if you look from the bottom on to rotation in this direction, we will also have it opposite to the clock. So it's always from the view from the view from the top. The rotation occurs always in the positive direction. So that's basically the rule. Right? So from the top of the vector of angular velocity, looking on to the plane of rotation, rotation seems to be counterclockwise in the positive direction. So now we can say that this particular vector completely determines our rotation because it has the direction which corresponds to the axis. It also has direction along the axis which specifies the direction of the rotation and the magnitude specifies the angular speed. Now, there is another rule which basically can be applied to find out what is the direction of my omega up or down on this picture. It's called the rule of the right hand. So if you will position your right hand, put it on the plane where the rotation occurs in such a way that your fingers are going around the axis and they basically direct where exactly the rotation, which direction rotation goes. Then the thumb will go to direction of the omega. Now, if just on this picture, if direction is not from here, but in a different direction, this direction, then this would be an invalid position of the right hand, right? Because my fingers would point this direction, but it really goes another way. So I have to from underneath, I have to position my hand like this, my right hand. Then my fingers and the corresponding movement would coincide and my omega will point downwards in this case. So that's basically called the rule of the right hand. It's equivalent to whatever I was just saying that from the top of the omega rotation should always be viewed as positive direction, counterclockwise. Okay, that's all about angular velocity. So this is the vector. Now, what is the relationship between angular velocity? Or I can say angular speed, if I just forget for a second about the vector characteristic of omega. So what's the relationship between angular speed and the speed along a trajectory? Well, along the trajectory we can always calculate speed basically using this particular rule, right? So if my angle is changing as phi of t, then during the infinitesimal period dt from t to t plus dt, during this infinitesimal time interval, my angle will change from phi of t to phi of t plus dt, which is basically d, well, I should put from this, it should be a difference from phi of this, my ending minus beginning. And this is df of t. This is my change of the angle. So this is my d phi of t. So if this is an angle, now what's the length of this arc? Well, obviously the length of the arc is radius times the angle. So the length of this arc, which is the differential of s of t, where s is basically the distance from the beginning along the circle, it would be equal to radius times my angle. So this is basically the relationship between distance covered from the beginning and the angle by which we have moved. Now, obviously if I will divide it by dt, I will have linear speed along the trajectory, which is v at point t, and it's equal to radius times omega of t, because this is the first derivative of the angle, which is this. So this is the relationship between linear speed along the trajectory, the circular trajectory, and the angular speed of the rotation. Okay. And the last one is what we usually consider in mechanics in kinematics is acceleration. So what is acceleration if we are talking about translational movement? Well, that's the first derivative of the speed by time, or second derivative of the position or distance by time. So we have a of t, this is equal to v derivative of t, or if you wish s second derivative of t. Well, we can do exactly the same with acceleration of the rotational movement, and in this case we're talking about angular acceleration. So angular acceleration usually is alpha, alpha of t. It's the first derivative of my angular speed, or if you wish second derivative of angle of rotation. This is alpha. Maybe it's not clear this is alpha, this is alpha. From which we can derive exactly the same kind of relationship between linear acceleration and angular acceleration. Now, in most of the problems related to translational movements, we're talking about, if we're talking about acceleration, we usually talk about acceleration which is a constant. Like for instance, we are moving with the same acceleration, constant acceleration under for instance actions of some permanent constant force or something like that. Now, we also can consider a movement around a circle, the rotation with constant angular acceleration, which means my angular speed is increasing, increasing, increasing. When does it occur? For instance, you turn on the engine from the state of rest, it starts picking up the rotations up until certain rotations per minute, let's say the maximum. That's how your engine actually works, your electric motor or something like this, or your engine in the car. It happens during certain interval of time and the speed of rotation changes from zero basically because it was rest to certain maximum for which this particular engine is supposed to be. So, this period, during this period, the speed is constantly increasing, which means my omega is constantly increasing. And we probably, with a certain degree of certainty, say that, okay, it's increasing with a constant rate, which means acceleration, in this case, angular acceleration is constant. Now, in this particular case, what we can do, we can do exactly the same which we had for translational movement. Let me just remind you. If you have rotation, if you have movement along a straight line with a constant acceleration, then you can say that, first of all, you can say that the speed at moment t is equal to speed at moment zero plus acceleration constant times t. Remember this formula? Well, where does it come from? Well, first of all, it comes from the definition. This is a definition and this is a constant, right? So, integrating, if the first derivative is equal to constant, then the function itself is equal to this constant times t plus certain constant. What is this constant? Well, if t is equal to zero, this constant is equal to v of zero. So, my speed at time t is equal to speed at time zero plus constant acceleration multiplied by time. So, that's how this is obviously derived. Now, whenever you're talking about position s of t, now this is, again, beginning position plus speed at the moment zero plus at square divided by two. Remember this formula? Again, how can I derive this formula? Well, very simply, since my acceleration is second derivative of the position, of the distance, then we integrate it twice. Now, integrating it twice, I will have that my s of t is some kind of a polynomial, one constant t plus another constant, right? So, what are these constants? Well, obviously, if t is equal to zero, now this is zero, this is zero. So, this one would be equal to s of zero, right? So, this is my constant. Okay, fine, we got that. And then, the next one is, how can I find out what's the value of c one? Well, we can just take any other moment of time. Let's say time is equal to one, let's say, and you will get this another constant, this one would be equal to this one, right? Now, as far as the coefficient here, we need the second derivative to be a. So, I have to put it over two, because the second derivative would be two, the first derivative would be a t square, first derivative would be a t, right? And the second derivative would be a, right? So, that's why we have this. And the first derivative is supposed to be a speed, right? So, if we will take the first derivative of this, this will go out, this will be a t, right? So, the first derivative of this would be v of t is equal to a t plus c one, right? And this we have already covered, if you put t is equal to zero, you will get c one is equal to v of zero. So, this formula is very simple, both actually, this and this, and they are derived by integrating the concept of acceleration, which is the second derivative of the distance or the first derivative of the speed. Now, we will do exactly the same in angular case, in case of rotation. So, what do we know? We know that by definition, my first derivative is my angular speed, and my second derivative or first derivative of the speed is my acceleration, which is constant. So, alpha is constant, right? We'll do exactly the same as before, we just changed the letters. It was a instead of alpha, it was v instead of omega, and it was s instead of phi, because everything else, all the logic which I did by integrating and checking what's the state at t is equal to zero gives the constant. So, what we will have is this formula. Alpha t squared divided by two plus, not v now, it's omega, omega of zero times t plus initial angle. So, this is how my angle of rotation changes with the time, if my acceleration, angular acceleration is constant. And obviously, from this, we have this, and that's how my angular speed is changing. So, again, absolutely the same type of derivation of this as before. If you take the second derivative of this, you will have alpha, right, because this will be zero after the first differentiation, and this will be zero after the second differentiation. And you will have only alpha, which means that the second derivative is alpha, which is constant. Now, if you take only the first derivative, then you will have basically this formula, because this will disappear, this will be a constant, this one, and this, after the differentiation from this, you will get this. So, everything fits. So, what my point was to parallel a translational movement and rotation around the axis. So, we have introduced the angular speed and angular acceleration. And by the way, when we will go to the dynamics, we will learn regular dynamics and forces and masses and Newton's laws. And then we will try to parallel the theory to rotational movement, and we will introduce rotational dynamics as well, based on rotational acceleration, angular acceleration, and speed. And whatever the equivalent of the force will be in case of rotational movement, the equivalent of the force and the equivalent of the mass, etc. So, the dynamics of the translational movement is really, can be somehow put into a correspondence to dynamics of rotational movement. But so far, we were just learning pure kinematics. So, it's a pure movement without any kind of forces or masses or inertia, etc. Okay. Now, I suggest you to maybe read the notes for this particular lecture on physics for teens. This is the course available at Unisor.com. 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