 Hi, I'm Zor. Welcome to Unisor Education. I would like to continue talking about rotational motion. This lecture is part of the course called Physics for Teens, presented on Unisor.com. I do suggest you to watch this lecture from the website from Unisor.com, because every lecture, including this one, has detailed notes. And also, you might be interested in taking some exams, which are also on this website. Plus, the website contains the prerequisite course called Mass for Teens, especially useful for this course, for the physics, is the calculus and the vector algebra. So, I do suggest you to be comfortable in these areas of mathematics before you, listen to these lectures. So, today is rotational or angular momentum. That's what we're talking about. Now, you know that I'm paying a lot of attention comparing the translational movement along the straight line and rotational movement, and the reason equivalence between certain concepts in one and another. So, today I'm going to again compare another very important characteristic of the translational movement along the straight line, called the impulse of the force and momentum. I will compare it with corresponding characteristics of rotational motion. Okay, so, let's start from just a little bit of repetition what exactly this momentum and impulse in translational motion. So, if you have a force which moves this object M, let's consider a simple case that at the moment t is equal to zero, this particular object has speed equal to zero, so it's at rest. And then, there is a constant force during certain period of time t acts upon this mass. Now, what we know about what happens? Well, the Newton's second law tells that we will have an acceleration of this type. And if I will multiply both sides by t, I will have f times t equals to m times a times t, but what is a times t? Well, if the initial speed was v, the force is constant, which means a acceleration is constant, a times t would be actually v at the moment t, right? So, it would be mv. This is not exactly a precise formulation of this equality between the impulse of the force and the momentum, because what's really more customer and more precise from physics and mathematical standpoint to talk not about the whole period from zero to t assuming that the force is constant, etc. So, what is much more customer is to have two moments, t and t plus dt, which is t plus infinitesimal increment of time. So, if the speed of this would be v of t, the speed of this would be v of t plus dt, which is equal to v of t plus acceleration times dt. Remember this? This is just simple, where a acceleration is the derivative of the speed by time, right? Now, in this case, it's a little bit more precise to have it this way. Force times infinitesimal time period. So, what happens with the momentum which is changing from m times v to m times v plus v of t plus dt? So, the momentum will increase, right? This is basically madt, which is m times dv, which is d of mv, right? Because m is constant in this particular case. So, this is an impulse of the force f during the infinitesimal time increment dt. And this is an increment in momentum. So, impulse during the infinitesimal times dt causes infinitesimal increment of the momentum. So, that's basically all about translational motion, about the motion along the straight line. Now, I would like to talk about equivalence between this and the corresponding rotational impulse and rotational momentum. Sometimes they say rotational, sometimes they say angular, I will do it interchangeably. Okay, so let's go to rotational movement. And the derivation is actually exactly the same as in case of a straight line movement. Now, we have a lot of correspondence by now between translational and rotational movement. So, we know that the force in translational movement corresponds to torque in rotational movement, where the torque is actually the force times radius from the axis of rotation where this force is applied. It's very important for rotation to know the radius where the force is applied. So, the force plays exactly the same role as torque, force in translational plays exactly the same role as torque in rotation. Now, next, we have the mass. Now, again, mass is important just by itself, if you're talking about the straight line, as the measure of inertia, how successfully, if I can say so, the object of mass, m, would resist the force which is trying to accelerate it. Now, the corresponding characteristic is related not only to mass, but also, again, on the position relative to the axis of rotation. And we have something which is called rotational moment of inertia, which is actually m times r square. All right? Now, they play basically the same role. Now, one more thing. Now, when we have an acceleration, well, you can actually add speed and acceleration along the straight line. In the rotational moment, we have angular speed and angular acceleration, omega and alpha. What do we know about the relationship between these characteristics in translational motion along the straight line? Well, the second Newton's law. Equivalently, we have already come up with this. We have tau equals 2i times alpha. Torque is equal to product of moment of inertia and angular acceleration. So, basically, all the corresponding things, f, m, tau, i, a, or alpha, they're all basically going parallel to each other. All we have to do is properly consider rotational characteristics from translational. Now, let's talk about impulse. Now, here, impulse, as we know, we know this, right? That's the impulse of the force during infinitesimal time period dt and its relationship to increment of the momentum of motion, right? Well, let's do exactly the same here. So, let's multiply it by dt. So, it will be tau times dt equals i times alpha times dt. Now, alpha is acceleration, which means it's a first derivative of omega, which is the angular speed by time, which is d omega of t, right? From which we basically derived, assuming i is constant, then we will have i times omega. I skip this dependence of t, we kind of assume it, right? So, we have exactly the same correspondence between, in this case, rotational impulse of the force and rotational momentum. So, not such a big deal. It's very easy to derive this type of correspondence. It's exactly the same way as in case of translational movement. So, again, everything is parallel. So, whatever the parallelism you have between force and torch, between mass and momentum of inertia, between speed and acceleration and angular speed and angular acceleration, the second load, Newton and the corresponding dependency between torque, moment of inertia and angular acceleration in the rotational movement and correspondingly, the relationship between impulse of the force exhorted during the infinitesimal interval dt and how it's related to the increment of the momentum of the motion. Here, exactly the same thing, the rotational impulse of the torque during the infinitesimal time dt. It's how it's related. It's equal to increment in rotational momentum or angular momentum. That's what we're talking about today. So, this is our rotational or angular momentum versus momentum of motion in a straight line movement. M is replaced by, mass is replaced by moment of inertia and speed is replaced with angular speed. Again, all the correspondence we are here. Now, in particular, as a consequence of this, what follows? Well, if force is equal to zero, or if you wish the balance of all the forces which are applied to the same object are equal to zero, they all balance each other. Then we have, as a consequence, that d of mv is equal to zero, which means mv is equal to constant. If mv is constant, there is no increase. There is no increment of the value. Now, this is called the conservation of momentum in a translational movement. So, if all the forces are balanced each other, then there is a law of conservation of momentum. If you remember, we were solving a few problems like billiard ball was actually rolling and hitting another billiard ball. We had some problems related to this, speed of this, speed of that, different angles, etc. We were using this law of conservation of momentum to determine what will be the result of this one ball hitting another ball, or another two balls, whatever. It's exactly the same here. We have exactly the same law of conservation of rotational or angular momentum. If my tau, my torque is equal to zero, which means all the different torques, or if you wish, forces times radiuses of application of these forces, if they balance each other, then torque is equal to zero, the resulting torque is equal to zero, and the resulting rotational momentum will be constant. From this, it follows that d of i times omega is equal to zero, which means rotational momentum i times omega is constant. So this is the law of conservation of rotational momentum. Now, in practice, how it can be done? Well, very simply, if you consider, for instance, a wheel which has certain mass, and you have two, now it can rotate, but now it's at rest. And you have exactly the same objects here of the same mass. So what happens? We have one force, the gravity of this is trying to rotate the wheel in this direction. So if it's equal to f and this is r, so f times r is the torque. But now this is also the same force f. Now this one rotates counterclockwise, this one rotates clockwise, so their balance, the sum of these two torques is equal to zero. Now, why actually we are summing with different signs in this particular case? I think it's something which probably, it's good if I will remind you about this. You remember that the force is a vector, right? Now, the torque is also the vector. Now, where is it directed? It's not this direction of the force, no. If you remember, you are always considering f times r as a magnitude of the torque. But what is the direction of the torque? This is a direction. It's a cross product, it's a vector product. So they are considering this particular case, they are perpendicular to each other. And that's why the magnitude is equal to magnitude of this times magnitude of that times sign of the angle between them, which is 90 degrees. So it's times one. So that's why the magnitude is correctly equal to f times r in this case. But what's the direction? Well, direction, if you remember, we have two different rules which actually result in the same thing. It's a vector which is perpendicular to both of them since this is the vector product, it's perpendicular to both of them. Now, direction should be from this to this. So in one particular case, you see the r is in different direction. f in the same direction but r is different direction. And that's why the product, vector product of one of them would be in this direction and another would be in that direction. And that's when the torques are added together, that's why they are equal to zero. And that's why in simple case like this, I can just subtract from one another from f times r minus f times r. But why do we have the minus in the second case in this case? Because it's directed to opposite direction as a vector. So let's not forget that the torque is a vector in as much as the force. But here in the rotational case, it's always directed along the axis of rotation. And it's equal to vector product of the force and the radius vector into the point where the force is applied. Alright, that's basically it. I do suggest you to read notes for this lecture. Actually, I suggest it to do after every lecture including this one. And then I will probably offer a few problems. I will solve them myself and then I will maybe offer something for an exam to this rotational dynamics. So I'm trying to basically finish up this particular topic of rotational dynamics. I think theory is basically covered. So I will consider a few problems and the exam that will be it. So thanks very much and good luck.