 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about rotation and similarities between the rotation around the axis and translational movement along the straight line. So our lecture is basically the continuation of the previous one. Previous was about the torque and this lecture is about the moment of inertia. Now, this lecture is part of the course physics for teens, for teenagers. It's presented on unizor.com. I do suggest you to watch this lecture from this website. It contains not only the video, but also the very detailed notes for each video and also some other things like exams, for instance, if you are inclined to take them and the site is completely free. It has no advertising, no financial strings attached. So basically use it at will. Plus the same site contains mass for teens, which is a prerequisite for this course. Especially the calculus and the vectors parts of the mass for teens course are very important because I'm using it throughout the physics all the time. So differentiation and integration are definitely a must. Okay, now let's talk about moment of inertia. Now the previous lecture was to introduce the concept of a torque. So torque is kind of an rotational equivalent of the force for translational movement. So if we will multiply force by the radius where this force is applied. So this is our object of certain mass and this is tangential to the trajectory force F and this is R. So basically this product, F times R, plays the same role in rotational mechanics as the plane force in translational mechanics. Which basically seems to be kind of natural, right? Because this is rotational and there is a radius involved. So obviously it should depend. The further you are from the center, probably the easier it is to to move the object. Okay, now the next thing which we also have introduced in the previous lecture was the relationship which very much looks like second law of Newton. Now we were talking about tau, tau which is torque to be rotational equivalent of the force in straight line movement. I is moment of inertia which is again rotational equivalent of the mass. It basically characterizes how difficult it is to move the object along its trajectory. In this case along the circular trajectory. And mass is obviously characterizes how difficult it is to move it forward along the straight line. And alpha is angular acceleration versus A which is linear acceleration. Now the relationship is, as I was saying, tau is F times R. Moment of inertia is mass times R square. And acceleration alpha which is angular acceleration, it's linear divided by R. Or radius times angular acceleration gives you the linear acceleration. So these are relationship between rotational characteristics and translational characteristics of the movement along the straight line. And this is the law which basically ties them together equivalently to the Newton's second law. Now today we will talk about this eye. About moment of inertia. And since it's kind of an equivalent to the mass, let me just stop on something which is very important for the mass. Mass is additive, which means that the mass of two different objects is the same as some of the masses of each of these objects. Now that allows us to calculate the mass of a real solid object as the sum of the mass of all these little tiny pieces from which it consists. And obviously that allows us to do some calculations, etc. Now moment of inertia, which is a mass equivalent in rotational movement, is also additive. And this is extremely important because it allows us to calculate the moment of inertia of complex bodies, complex solids. Because without this property, we could not really calculate the moment of inertia of anything, like cylinder or sphere or whatever. So it's very important and why is it important for us to calculate the moment of inertia? Well, is mass important to know how to move the forward object? Well, then the moment of inertia is equally important to know how it will be rotating. It has certain difficulty in rotating and we have to overcome this difficulty, this inertia, and inertia in the rotational movement is measured by the moment of inertia. So that's why it's very important to show that the mass equivalent in rotational movement, which is moment of inertia, is also additive. Now, how can we do it? All right, I'm just suggesting an approach, which is yeah, you might consider it as not a hundred percent mathematically rigorous, but it's rigorous enough and it explains basically why moment of inertia is additive. So here is what we're going to do. I will take two different objects of certain mass located at certain distance from the axis of rotation and I will show that their combined moment of inertia is exactly equal to some of these moments of inertia of each of them. Here is how we will do it. So consider we have a disk, a weightless disk, which is freely rotating around the axis and there is some kind of a force tangentially applied, which rotates it. Now fixed on this disk, there are two masses, one mass, mass one on the radius r1 and another mass m2 on the radius r2 and let's consider this radius r. So on the radius r, which is let's say the edge of the disk, okay, we apply the force. We are rotating, just pushing this tangentially pushing it forward and it causes the acceleration of an entire disk and it causes the acceleration of these two masses, which are fixed on different radiuses from the center of rotation, right? Now let's consider our angular acceleration is alpha, whatever it is, doesn't really matter. Now in theory, I can say that whatever inside of this disk is, well there are two objects inside, so whatever is inside, it has certain moment of inertia and f times r, which is the torque of this force, should be equal to moment of inertia of the entire system of two objects in this case times angular acceleration a. So I don't know what this is what this i is, I know f, I know r, I know alpha, I don't know what i is but I do know that these two are fixed on the disk, which means they are circulating with the same angular acceleration alpha. Now I know there are masses, right? So the mass of the first one is m1 and they know their moments of inertia, moment of inertia the first one is this one. Now this is moment of inertia, let's call it i1. Now if this is the moment of inertia of this object m1 and it's circulating with angular acceleration alpha, there must be some kind of a torque which basically pushes it along this trajectory, which means that there is some kind of a force f1 applied on the radius r1, I don't know this force but basically this force is the result of this force. So my force f is pushing the disk around and the disk pushes this object and pushes this object. So these objects are basically are pushed by the forces which are directed this way correspondingly, tangential again to their corresponding orbits and they are sourced basically if you wish by the force capital f. So this is the force which causes these forces to be pushing to be pushing the m1 and m2 around their orbits, around their their trajectories. So I can see this same thing for the second one. All right, fine. Now we know from the previous lecture that what's important to produce certain angular acceleration of an object of a certain inertial mass is the torque. If two torques are the same, they will produce the same result. So if for instance I have one object circulating around the axis, I can apply force directly to this object or I can have a twice as much, twice as strong force applied to half of the distance applied here and the result will be the same because again it's the torque which moves the object along the circular trajectory. So the object has certain moment of inertia, eye whatever it is, it has certain acceleration and it's the torque which is important to produce this exact result and two different torques produce different acceleration to two torques which are the same even if the forces are applied in different points. If the torques are the same, my result will be the same, my acceleration will be the same. So I can reduce the radius of the force application. This is a rod and I apply force on half of the distance, not at the end where my object is, but I will apply twice as much strength as my force. The product of twice as much strength but half the distance will produce the same torque and so I will have exactly the same result. So what I will do instead of applying this force F1 at this point, I will apply a different force at this point where the point of application of F, capital F is, but with the same torque. Now what is the force which gives me the same torque here as this force here? Well if I have initial torque this one and I would like to apply different force on the radius R what I have to do is, right, so this force on this radius gives me exactly the same result as this force on this radius, right, because R and R is cancelling out, right, so result will be the same. So I can consider instead of the force F1 applied at this point on the radius R1, I can consider this force applied at this radius. Okay so forget about F1 and think about this force G1 times R, the word G is, G1 is this, this is F1 by the way. Now similarly instead of considering the force F2 applied on the radius R2 which gives me F2 R2 gives me the same acceleration for a different object, the same angular acceleration alpha but for the object I2. So instead of this torque I will apply this torque, so this is G2, the torque is exactly the same, this torque is exactly the same as this, right, because R and R is cancelling out. So what I have accomplished is knowing this and knowing this, these are two torques, two original torques. I will replace these original torques with my new torques of the new forces but they are all applied at the same point. That's my purpose, I would like all the forces to be applied at the same point on the same radius of the disc and that gives me basically some ability to add them together and here is how. So I have one torque which is this one and I have another torque which is this one. Now this torque and this torque do exactly the same as this one, right, exactly the same. So forget about F1 and F2 forces which are applied to different points on the radius R1 and R2. I will only consider forces this and this applied at the same point as my main force F. Now let's just think about it. If my main force F is applied to this point and it gives exactly the same result, the same action, the same movement as this force and this force applied to the same point. So what does it mean? Well it means that the forces are the same because the radius are the same so the forces must be the same because right now, again, all three forces, my original force F and the forces which are basically causing the movement of these two points, they are all applied to the same point which means this force must be equal to the sum of these forces. If torques are the same, right, results are the same and the radius are the same, then the forces must be the same. And basically this is almost everything which I need because right now what happens is what happens is F times R is equal to, if we multiply by R, what I will have? I will have this plus this, I will have I1 alpha, right, plus I2 alpha, which is I1 plus I2 alpha attack. So this is my main equation from which everything follows because what does it mean? Let's consider this system of a disc with two objects. It has certain moment of inertia and they know that it circulates, it rotates with acceleration alpha, which means that this torque, my original torque, is equal to my total moment of inertia of the entire system of two objects times the angular acceleration, right? So what do we see? They are supposed to be equal, right? This is the same, so these are supposed to be the same. Alpha is the same everywhere. So the whole system is always with the angular acceleration alpha rotating. So some of these two moments of inertia give me one moment of inertia of an entire system of two objects. So it means that the moment of inertia is additive, which means that for any complex object I can divide it into small pieces and sum up the moments of inertia of each individual piece to get the moment of inertia of an entire solid. Let's do exactly this in a couple of cases. Just as an example, consider an axis and a rod of mass m of a very thin rod. So we don't have the width of this, just the length of the length l. And it rotates under certain forces around the axis. My question is what is the moment of inertia of this rod? Very simple actually. We divide it into many pieces. Let's say this piece is on the distance r from the beginning, from the axis of rotation. And this piece, small piece, is dr. It's a differential of r. So this is an infinitesimal piece of the rod. Now you know what infinitesimal means. Again, we're talking about the calculus which you have to know. So this is basically the beginning of the integration which I'm going to do right now. So what is the moment of inertia of this thing, this little piece? Infinitesimal is small, but however, what is this moment of inertia? If the total length is l and the length of this thing is dr, it means that mass times dr divided by l is the mass dm differential of mass which we are talking about. So this is the mass. Now what's the moment of inertia of this piece of mass? Well, it rotates around the axis on the radius r, right? So the moment of inertia, di, if you wish, is equal to mass times radius square. So m r square divided by l dr. Now how can I calculate the moment of inertia of an entire body? Well, I have to add together all these infinitesimal things, right? Which means I have to take the integral from 0 to l of this formula and what do I get? Well, m divided by l. Now, integral of r square, the indefinite integral is r cube divided by 3 and I have to put it using the formula of Newton lateness from 0 to l, which means first I have to substitute l and I will get ml cube divided by l minus m0 cube divided by l, which is equal to n3, sorry, n3, n3, which is equal to ml cube divided by 3. So this is my moment of inertia of the sin rod of the mass m of the length l which is rotating around its edge. Well, you might say that you know what we are rarely rotating the rod by its edge. Usually we're rotating the rod around its middle, right? We're kind of rotated around its middle. So let's just calculate this one as well. So we have its axis and we have a sin rod and it rotates around this axis. How can they calculate its moment of inertia? Well, I can integrate it the same way as I did before, but I will do it easier. Now I'll just divide this rod into two pieces. Now each piece has m divided by 2 mass and l divided by 2 lengths and I will use the previous formula which I have already derived because each piece, this piece separately from this piece, each piece is rotating around its edge, right? So my first piece has, so it's mass times length square divided by 3. So it's m times l2 square divided by 3, which is what? It's 4 times 2, it's 8 divided by 3, 24. m l2 divided by 24. So this is one piece and another piece has exactly the same and it's equal to m l2 divided by 12. So this is the moment of inertia of the rod if it rotates around its center. By the way, what was a little bit unusual, I would say, intuitively it was not obvious. You see, it's really much smaller than the previous. Previous was m l2 divided by 3. So if you rotate the rod around its edge, it's m l2 divided by 3. But if you rotate it in the middle, around its middle, it's about what? It's 4 times smaller, which is a lot. I mean, I did expect that this is smaller because we are basically closer to the axis. But frankly, I wasn't really thinking that it would be that small, that smaller. Anyway, so this is another just little problem about how to calculate the moment of inertia and we will have some other problems. I will solve it or I will offer it as an exam or something like this. And basically, that's all I wanted to talk about the moment of inertia. Thanks very much for your time and good luck.