 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about dynamics of rotational movements. We have already covered kinematics of rotation, where we have introduced the angular position, angular velocity and angular acceleration. Now in particular, let me just remind you that if you have rotation of this object, and this angle would be our angle of rotation, its first derivative would be a pseudo vector which is always directed along the axis of rotation. Now the plane of rotation obviously is perpendicular to the axis of rotation. And acceleration, angular acceleration also is directed along this particular axis of rotation. Now if you just don't remember it well, I do suggest you to go to the rotational kinematics lecture earlier in this course and just go through this. It's very simple things. And that explains actually why the angular velocity vector is along the axis and so is angular acceleration. Now today, we will consider the dynamical aspects of this movement of the rotation, which means we are introducing concept of mass, inertia, force, etc. Now I'm not going to redefine mass or force for rotational movement. However, there are certain other concepts related to rotation which are playing the same role as the force and inertial mass. But for the rotation versus the concepts of force and mass, as we know them from translational movement. Now in particular, obviously we are talking about three laws of Newton. Now, let me just remind you that the first law is about an object moving along the straight line with constant speed if there are no forces applied to this particular object. Now, as far as the rotation is concerned, there is an equivalent. Now imagine that you are in free space. There are no forces of gravitation or any other forces and you have this spaceship and for whatever reason it just turns around its axis. It will continue to turn if there are no other forces. So basically as far as the rotation is concerned, this first law of Newton has an equivalent that the object rotating around some axis will continue its rotation indefinitely with the same angular velocity if there are no unbalanced forces applied to this. Now as far as the third law of Newton, that the force of action has counteraction, it's not really applicable for rotational movement. So we're not considering this. The most important, actually, is the second law in translational movement. It stands as this. So we have the force. We have acceleration. So let's just assume that we are talking about straight line movement and the force goes along this straight line. So the force is a vector now and it's proportional to acceleration where coefficient of proportionality is the mass of the object. So I would like to come up with something similar in case of rotation. Now what's probably the most important thing is that in case of rotation while force is a source of acceleration it's not only the force which is this particular source. Now in case of translational movement along the straight line while the force, if it's applied to the object and it can be applied only to the object, right? So that's the source of its acceleration. Now in case of rotation we are assuming that there is a little bit more complicated arrangement. So if this is the mass, I probably have some kind of a rod which connects my mass with the axis of rotation. So there is some kind of connection here which allows it to rotate, right? And the force can be applied well either to the mass itself and we will consider perpendicularly to to the radius right now or it can be applied here as well and it will also cause the rotation, right? So the point of application of the mass is important because although this is a point object which has certain mass M, but the point of application of force can can be not only to the object itself as in the translational movement along the straight line, but also to any position on the rod. Or if you wish, rod can be a line, again some kind of a infinite basically in theory of the lengths and we can apply the force here Now if you will apply the force here, it will still force the object to move, right? So my first most important kind of point here is that not only the force can be the source of the movement of acceleration, but in this particular case angular acceleration which we are mostly interested in. But also how I apply this force. Okay, now let's go to experiments. Now we all know that if you have a door okay so this is the door which I am opening or closing whatever. Now if I apply my efforts at the edge opposite to the hinges, so these are hinges that's where it's moving or attaching. If I'm applying if I'm applying my force and the opposite to the hinges side where usually the handle is located, I can open the door relatively easy. Now, what if I apply the door close to the hinges? Well, it will be much more difficult. We all know that. Actually, I do remember certain hotels where the knob is right in the middle of the door and it's always kind of extra effort which I have to make to open the door using the knob in the middle. Not as I would expect it based on my experience with the knobs at the edge. So the application of the force, a point of application is very important. Now it has been experimentally basically determined that the acceleration which my force, let's consider that the force is fixed. So I'm pressing at a certain point with a certain pressure, the pressure of the force. So it has been experimentally determined that the acceleration of the door, of angular acceleration of the door, is proportional to this distance from the axis of rotation. So let's go to this more theoretical picture. Depending on this distance, so this distance is, let's say, R. So depending on this distance, my acceleration of this point will be different and it will be proportional basically to this distance. So if I will, let's say, put my force right at the object itself, I will have one acceleration. But if I will put it at half distance from axis to the object, my acceleration will be half of this. Now if I will put it on, if I will apply my force on this side, for instance, also on the same radius, into this direction. The long tangential line, always perpendicular to the radius, okay? If I'm applying this, it will be the same as, acceleration is the same as this. If I will reduce the radius by two, my acceleration will be half. So basically, my angular acceleration alpha would be proportional to my radius. Now another thing, which is kind of obvious, it should be proportional to the force itself, right? Twice as strong force will produce twice as much acceleration. Now this is actually kind of a simple thing because if you consider it that we are applying force, for instance, right at the point, then during certain infinitesimal period of time, my movement is practically linear, right? So yes, it's along the circle, but we are talking about infinitesimal time interval. And in this interval, my trajectory actually is almost straight line, which means that my linear acceleration, a, should be equal to fm, sir. f divided by m, right? From this second law of Newton. But my linear acceleration is radius, well, let me use capital R, radius times angular acceleration, obviously, right? From pure geometry. From which we go, from which follows that, obviously, alpha should be proportional to to the force, right? In this case. So, we have the proportionality between angular acceleration alpha and the radius of the point where my force is applied and obviously the force itself. So basically I can say that alpha is equal to some kind of coefficient times f times r. Now what's also what has also been observed that if I have a specific object rotating on a specific trajectory of the radius r, and if I will change my force and change my radius to the point where the force is applied, as long as their product is the same, my acceleration would be the same. So this k is a constant, basically. It depends only on the product of the force and the radius of the point where the force is applied. Which brings us to a kind of a similar situation with this. So in the translational movement, my acceleration is proportional to force and inertial mass is the coefficient of proportionality. In this case of rotation, my angular acceleration is proportional to something, not force, but force times r. So this force times r actually plays logically exactly the same role for a rotational as just force plays for a translational movement. And obviously there is a special name. It's called torque. So this product of the force and the radius to the point where the force is applied, this one, is called the torque. Now in some other places in literature and different countries, it's called moment of force, which is basically the same thing. In the United States, the torque is more common term, but in certain European countries, for instance, they just call it moment of force, which basically is the same thing. So we have this type of relationship which kind of reminds us this. So we found the logical equivalent of the force for translational movement. We found the logical equivalent of this to a torque in rotational movement. So we have replaced our linear acceleration with angular acceleration and the force is replaced with torque. Now, what is this coefficient? Well, that's very easy actually to come up with this. Again, let's consider this. It's an infinitesimal time period. What happens? Well, during this period, my force F will will produce, basically will result in acceleration, linear acceleration of A. Now linear acceleration is r times angular acceleration, right? Now, how can I get from this to torque? Well, just multiply both sides by r and I will have F times r equals to m r square alpha. Now, this is the torque and the Greek letter alpha tau is usually used for for the torque and this is a coefficient of proportionality, which I was talking about, right? Between the angular acceleration and the torque and it's called moment of inertia, okay? Now, we have this moment of inertia of this particular object of the mass m which is positioned at the radius capital R. Now, we know that if I will replace one torque with another torque which is equal to this, to this torque, I will have exactly the same result because the result depends only on the torque, right? So if I will take another F, lowercase F, and apply it at radius R in such a way that it's equal to F times r, which means let's say the r is half of this one, but this F is double this one, so the result will be the same. I will have exactly the same result, the same acceleration. So basically, I can say that this formula can be generalized into F times r is equal to i times alpha, where i is the moment of inertia of my mass, which is m r square, where r is the distance, so this is the moment of inertia and this is a torque. So all together, I can put it in one formula which looks like the second law of Newton, but only with different concepts introduced. But basically, it reflects exactly the same thing. It reflects certain effort. Now effort means in this case torque or moment of the force, which is a little bit more complex than the force for a rotational movement. And this represents the moment of inertia. Again, it's not just inertial mass, it's a little bit more than that, at mass times r square, where r is the distance of this mass. But in any case, it's the same logical concept, because this is a resistance to the movement. What is the mass? Inertial mass is the measure of resistance to the force. Right? So my torque replaces for a rotational movement the force for translational and my moment of inertia replaces the plain inertial mass. Right? So this is f, I'll use lower case, okay. f times r and this is mr square. So this is r, the point where we apply the force and this is r, where the object is located. Now, there is one little complication or generalization whatever you want. You see, sometimes, now we were talking before, we were talking about the force being perpendicular to the radius, right? Now, what if it's not? Okay, let's look at this from the top. So this is my circular movement. So my axis of rotation is this one, right? This is my mass. This is my radius r. Now, what if my force is directed not perpendicularly to the radius? Which is, as you probably understand, the most efficient way, right? Now, what's the most inefficient way? Well, the most inefficient way if we direct force right against the rod where the object is hooked on, right? So that would be most inefficient and it will result in zero angular acceleration, right? So if the force is this way or this way, it doesn't really matter. It's unstretchable, weightless rod. The result will be zero. Now, the maximum result will be, if I will be perpendicular, which means tangential to the circle, right? Perpendicular to the radius is tangential to a circle. Now, what if it's at angle? Well, obviously what we have to do is we have to project here and here. Now, this force can be represented as the sum of these two vectors. This vector will give no result whatsoever because it will be compensated with a reaction of the rod. But this will be the source of the movement which we were talking about. Now, and obviously if this is an angle between, let's call it beta. This is an angle between r and force f. Now, I will use lower case r because the force can be applied not only here, but also here or there or anywhere else. So r is basically the distance. Now, what's important is angle between them, right? So if this is angle beta, then obviously this thing is f times sine beta. So we have f times sine beta. That's a projection. If it's 90 degrees, sine of 90 degrees 1, so that would be the whole force. If it's 0 or 180 degrees, the result would be 0 and obviously there is no movement. So, and everything in between obviously like that. Now, what is the torque in this particular case? Well, if I will multiply f times r times sine beta. Now, I hope you remember from the vector algebra that this is vector product or cross product of two vectors. Vector of the force which is here and vector r which is here. Now, the magnitude of this is this. Now, but this is the vector. Now, this is scalar. So let's forget about scalar representation. Let's use the vectors and what will be as a result of the vector product of two vectors. It will be a vector which is perpendicular to them both, right? So what is the vector which is perpendicular to the radius and the force? Well, it will be this vector. Along the axis, exactly the same way as my omega of t and alpha of t of t, my angular velocity and angular acceleration. Well, we are actually interested in angular acceleration. So what we have right now is that the tau as a vector is i times alpha as an angular acceleration because this is along the axis and this is now this is tau along the axis. So now this is a vector equation. Well, I think that for simplicity purposes we will rarely consider the force which is going under at the angle to the radius other than 90 degrees, but maybe, but I just don't think it's very important. What is important, however, to understand that if we will view the force as a vector, which has not only the magnitude, but also a direction and direction can be not necessarily perpendicular to the radius. This is a more general formula for tau for torque. This is more general formula for the torque and this is more general formula, which is kind of an equivalent of the second Newton's law for rotational movement. So that basically completes my discussion about rotational dynamics. So you have to understand that there are certain things which are equivalent for rotation to those which we know from the translational movement. So whatever is the force for translation is a torque for rotational movement, which is a vector directed along the axis of rotation. Now the inertial mass m for straight line movement, translational movement, is equivalent to a moment of inertia, which is mass times the radius where it's located, where it's located on this circular trajectory. And the angular acceleration plays for rotation the same role as linear acceleration for translational movement. Now I do recommend you to find this lecture on Unisor.com in the Physics 14 course. There is a very detailed explanation of everything which I was talking about during this lecture. I do suggest you to read it in details because well it's just another view. One thing is to watch what I'm talking about and other is to read it as a textbook basically. But now the textbook probably after this lecture will be a little bit more understandable and entertaining if you wish. By the way, the site is completely free. There are some other courses on this site. For instance, there is a prerequisite course for this Physics 14. It's called Mass 14, where you can find for instance the vector algebra, you can find calculus over there and I strongly recommend you to be familiar with all these concepts, whether from this particular course or from somewhere else. But these are definitely needed concepts to be mastered before you go to the Physics. And the site by the way is completely free and there are no advertisements. So I do recommend you to go to the site and watch the lecture from this site and read the notes. And there are some problems to solve and there are some exams actually to take if you wish. Okay, that's it. Thank you very much and good luck.