 Hi, I'm Zor. Welcome to Unisor Education. Let's solve a few problems related to rotational dynamics. 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 because it contains also relatively full notes for each lecture. There are even exams for certain topics. Also on this website on Unisor.com you can find a Maths for Teens course, which is mathematics for advanced level of mathematics for high school students, which are considered to be a prerequisite for the physics for teens. Okay, now, yeah, and the site is free, by the way, for all. Now, so let's just talk about these problems. Problem number one, let's assume that we have a wheel on this axis. It can freely rotate. It has certain mass, M, but what's interesting is the mass is concentrated only in the rim. So let's consider that the rim is very, very thin, but it contains all the mass of the wheel. Now, if you can imagine, for instance, a bicycle wheel, but it's not exactly like all the mass is concentrated in the rim of this, but it's more or less corresponds to reality. These things which connect to the wheel are really very, very light. So we assume that the whole mass is only in the rim. Now, we have a thread, and we put this thread around this wheel, and hanging on this thread is an object of mass M. Now, let's assume that the radius of the wheel is R, and now, after we rotated this thread, we just let this mass go down under the force of gravity, and our purpose is to find out what will be the acceleration of this mass. Now, obviously, there is no wheel, there is no inertia of the wheel, and this mass just goes down by the force of gravity, then, obviously, it's acceleration with the G, which is a known free fall acceleration. But this is not a free fall, because obviously, this mass has certain inertia, certain momentum of inertia, and it slows down. So the question is how much it slows down. Okay, now, let's just examine the details of this. Now, what exactly is the force which is acting on this mass and on this wheel? Well, obviously, since the mass of the wheel is not zero, there is some kind of a tension here. This is the tension of the thread. Now, this tension of the thread slows down this mass, which means that the total force which acts on this mass is its weight, and I will consider this T to be positive for this one and for this one. But I will use the negative sign to indicate that actually the tension is, the real tension vector is directed upwards while the gravity directed downwards, so that's why I put minus T here. So we're talking about magnitudes of these vectors. T is a vector of this lamp, G is also the vector because the G is acceleration which goes down. But right now we're considering only magnitude, and it's really kind of easy. Now, as a result of this force, this object is going down with certain acceleration which we have to determine. So A is acceleration which we have to find out. Okay, fine. Exactly the same magnitude force T tension rotates the wheel. Now, we know that there is a concept of torque, right? And this torque is equal on one hand, it's equal to the force times the radius because the force is always perpendicular to the radius. It's tangential to the wheel, so my torque is equal to this. On another hand, we know that the torque is equal to moment of inertia times angular acceleration of the rotating wheel. Well, angular acceleration is actually easy to express in terms of A, because the linear acceleration of mass m, lower case m, is exactly the same as linear acceleration of the wheel, obviously, right? Because they're connected with un-stretchable and weightless thread. Now, the linear acceleration of the wheel is related to angular acceleration alpha by this formula. So instead of alpha, I can always put A divided by r. Now, what's not obvious, what it is, is i, the moment of inertia of the wheel. Now, how can I calculate the moment of inertia of a complex object like this one? Well, the best way is you have to remember, and we did address this issue in one of the lectures before, that the moment of inertia is additive. It's like mass. Mass for translational movement is additive, and if you have two separate objects of masses m1 and m2, combine them together, there will be an object of the mass m1 plus m2. So, mass is additive. The similar thing is true for moment of inertia, and we did demonstrate it in one of the previous lectures. So, I can divide this object, which is a wheel, with the mass concentrated only on the ring. I can divide it into many, many different small objects, and if I know the moment of inertia of each of these objects, I will just add them up together to get the moment of inertia of the entire wheel. Now, obviously, it's an integrating problem, and in this particular case, if I will divide this wheel into these small objects of infinitesimal linear lengths, then each one of them will have a mass differential of the mass, of an entire mass, right? And I know that it can be considered, in this case, as a point object of this particular mass, and in which case my moment of inertia is this. It's a mass times radius square. If you have just a point object on the radius r, and it's rotating, its moment of inertia is its mass times r square. Now, if I will integrate it from zero to the total mass m, obviously, I will get mass times r square. So, this is kind of easy approach, and it allows you to calculate the moment of inertia of a relatively complex object. Now, this is probably a minimally complex object, because all I have to do is just to add the same momentum of inertia infinitely number of times, infinitely small one, integrated to get an infinite number of these times, and obviously, I will get this. So, this is my i. This is the moment of inertia of the wheel. So, the entire wheel, if the mass is concentrated only on the rim, basically behaves exactly as a point object of this total mass. And it's kind of intuitive, obviously. So, I will just, obviously, use it. So, in this case, I can put this. So, if instead of i, I will substitute this, r would, one r is cancelling, so I will have mass r and a. Right? Now, if you will compare this and this, you will get that the tension is equal to m times a. Fine. Now, obviously, if I know the tension, I can substitute it here, and I will get an equation where my linear acceleration a can be very easily obtained. So, mg minus ma equals to lowercase ma. So, this goes to there. So, I have mg divided by m plus m. That's my a. Right? Ma goes here. So, it's mg divided. Okay. So, this is the final answer. So, it looks like the force of the gravity is distributed into a combined object of a combined mass to get the acceleration. But again, this is a very important property of the wheel where all the mass is concentrated within the rim. So, all the pieces, all the small object into which we can divide or break this big object, the wheel, they all have exactly the same radius. And that's what's very, very important in this case. And that's why we have such a simple formula. Now, the next is, the next problem is, I would like to calculate the moment of inertia of a little bit more complex object. But the object which I would like to consider is a disk. So, I have a flat disk of mass m and radius r. And now the mass is distributed evenly within the surface of the disk. So, let's consider, let's say, a disk is made of whatever, steel. Okay. And it's very thin, but the mass is distributed evenly within the whole surface of this disk. Now, obviously, you understand that since not all the mass is concentrated on the rim, my total moment of inertia will be smaller. It's not m times r square, as if all mass is concentrated on the rim as in the previous problem. Now, mass is evenly distributed within the whole surface. So, many different pieces of the surface into which we can basically break it, they have smaller radius and that's why smaller moment of inertia. So, this is not a total moment of inertia of the disk. So, what is it? Again, in this case, we have to resort to dividing our object into smaller objects for which we know how to calculate the moment of inertia and then add them together because the moment of inertia is additive. Okay. So, how do we do it? In this case, let me just look at this disk from the top. From the top is a circle. Okay. Now, the way how I would like to divide it is the following. I will have concentric cuts. So, I will have small rings of very small widths. Now, if I will divide my total radius into small rings, so let's consider ring at the distance r from a center. And what's the width of this wing? Obviously, not wing, ring, sorry. The width is differential of r, right? You know how integration is actually working, right? We divide it into many, many different n pieces if you wish. Each one has the widths r divided by n and then n goes to infinity and that's how sum is converted into integral. And again, I assume that all these calculus things you know. And if you don't, please take a look at the maths for teens prerequisite course using the calculus relatively freely. So, now I have a ring which is very, very thin. This is infinitely thin differential, right? Which means I can consider this ring as the one where all the mass concentrated basically on the same radius. The radius is r. So, all I have to know to calculate the moment of inertia of this ring is just have its mass and multiplied by the square of the radius. It would be exactly the same as in the previous problem. The previous problem we had a wheel where all the mass was concentrated on the ring and the total moment of inertia was mass times radius square. So, this ring is actually like the wheel in the previous problem. It's all mass is on the same distance r from, lower case r from the center and that's why its own mass times r square. That would be its moment of inertia. Now, but what's the mass? Well, that's very easy. I can just have the area of this ring and have basically the ratio of this area to the total area of the disc and that would be the part of the total m which is concentrated within this ring. So, what's the area of the ring in this case? Well, you can just cut it here and stretch it into a rectangle of the lengths 2 pi r and the widths dr, right? Again, since dr is infinitesimally small variable I can actually consider this to be a rectangle because all these little things which really make it not exactly a rectangle because the inner radius is slightly less than the outer radius. These are infinitesimal variables of the higher order obviously than dr and that's why in integration they would disappear. So, this is basically the area. So, it's 2 pi r times dr. This is the area. Now, what is the total area of the entire disc? It's pi r square. So, my ratio is 2 pi r dr divided by pi r square. So, this is the ratio of the area of the ring relative to the area of the entire disc. So, if I will multiply it by the mass of the entire disc I will have this m. That's what it is. Okay? Great. So, now, what do I have as a moment of inertia of this ring? Well, pi will cancel out. So, I will have 2 r cube, pardon me, dr divided by r square m. Right? So, this is my moment of inertia. Now, I have to integrate it by r from 0 to capital R. I have to have some of all these rings, this one, this one, this one, all together they grow up to the radius of the entire disc. And this would be the answer. That would be my moment of inertia of the disc. So, it's equal to 2 m divided by r square r constants. They go out. And I have integral from 0 to r of r cube dr, which is equal to 2 m r square. Now, indefinite integral from the r cube is r to the force divided by 4. And I have to, using the formula of Newton-Leibniz to substitute capital R and minus substitution of 0. Well, 0 will be 0, obviously. So, it would be 2 m r to the force divided by r to the 4r to the second. Right? 4 remains in r to the second, which is m r square divided by 2. And this is my moment of inertia of the solid disc. As you see, it's half of the moment of inertia if the total mass is concentrated on the rim only. Okay, that's it. This is the second one. Now, my third problem is basically a combination of the first and the second. Let me just remind you what was the first problem. The first problem was the wheel and some kind of object here, m, m. Okay, this is r. Now, my third problem is exactly the same as the first one, but instead of considering the wheel with all the mass concentrated on the rim, that was my first problem. I will have this solid disc as a wheel. And that's where I will use my second problem. Now, what difference does it make now? Well, it will be almost the same as in the previous problem. So, let me just repeat it again. But basically, you would expect that this thing, since it has a smaller moment of inertia than if the whole mass is concentrated on the rim, then it will probably make less resistance to this force and force of gravity. And that's why the acceleration would be a little bit greater, right? So, let's just make some calculations. Again, there is a tension here. And on one hand, this tension is the source of resistance to the free fall. So, gravity mg goes down, tension goes up, and the difference between between them is mass times acceleration of the going down of this particular. Now, the T, if I will multiply it by, well, in this case, I'm using capital R. By capital R is the torque, which basically forces the wheel to rotate with accelerating angular acceleration. And that would be equal to i times alpha, right? Now, I know that i is not the same as it was before. Before it was mr square. Now, it would be mr square divided by 2. And alpha is, that's an acceleration, angular acceleration, and it's equal to linear acceleration a, which is exactly the same as this one divided by R. So, I will have mr a divided by 2. Now, from here, I can have that T is equal to m a divided by 2, right? This is equal to this. And as you see, the tension is half of the tension that was in the first problem. Why? Because the moment of inertia is half, as I have calculated in the previous problem, right? So, the tension is smaller, and that's why resistance is smaller, and that's why I will have a greater acceleration. So, I will have here mg minus m a divided by 2 equals to m a. So, my a is equal to mg divided by m plus m divided by 2. So, right now, I'm having a smaller denominator. It used to be capital M, not capital M divided by 2 in the first problem. So, this one is smaller, the denominator, and that's why the acceleration will be greater. So, that's my third problem. Now, the third, the fourth problem is not related really to moment of inertia or torque or something like that, which is kind of typical for problems in rotational dynamics. Instead, it's kind of a repetition. There will be some forces, but it's more about kinetics of the rotational movement, and here is the problem. It just might have certain interest because it has very easy practical implementation. So, let's consider that you are having a thread, and it actually makes this type of movement. This is your object on the thread, and the object is circulating with the same speed. There are no linear acceleration, let's put it this way. So, what's important is that, do you remember, I will probably consider it with a pendulum. It will be very similar in some way. So, the idea of this is that I would like to connect the angle with the horizon, no, I will make it with phi. The angle was a horizon, and the speed of rotation expressed as angular speed. Now, you probably noticed that if you will just do yourself this type of motion with some kind of an object on the thread, the faster you go, the more horizontal the line on which this particular object is hanging, the more horizontal it will be, it will be higher. So, the faster you go, the higher will go, this object will go. So, I would like to make calculations. I would like to know how this angle is related to omega, and let's assume that there is a mass of this object. Now, you will see that actually it doesn't depend on the mass, but in any case. In some way, it's similar to whenever you have a free fall, the different masses are falling down with the same acceleration, so people thought that the heavier objects would really go faster, but that's not the case. Same thing here. It doesn't really depend on the mass, but let's just include it for now into the calculations. So, let's just think about what kind of forces are acting on this particular mass. Well, obviously, there is a gravity and also there is a tension on this thread, right? Threads the tension. Now, what is this particular force tension? How it's used in this particular case? Well, it has two different components. One component is vertical and another is horizontal. Now, if this angle is phi, then the vertical component is t times sine of phi, right? If this angle is phi, then this is t times sine of phi. And the horizontal component is t times cosine of phi. Now, t times sine of phi should be basically large enough to compensate the force of gravity, because our object is staying in the same horizontal plane. It doesn't fall down, it doesn't jump up, which means these two forces are exactly equal. So, mg is equal to t sine phi. Okay, now let's talk about horizontal. Now, from the kinematics of rotational movement, you know, and we did talk about this in details, that if the object rotates, then it must be something which keeps this object on the orbit and it doesn't really go tangentially to this, to a circle, to a trajectory. Now, what keeps it there? Well, obviously, there is a force, centripetal force, it's called in this case. And the result is that it doesn't really flow, it doesn't fly all the way from the circular object. This force forces this object to go back to the center. And that's actually the acceleration. So, we are instead of flying off the trajectory, we fall back to the circle. And the acceleration, centripetal acceleration is equal to v square divided by r, where v is the linear velocity. Now, linear velocity can be expressed as angular velocity and that makes it equal to r omega square. So, this is the linear acceleration. And if the mass is m, then mA, mass times acceleration should be equal to the force which is actually keeping our object on the orbit. So, this one is one equation, this one is another equation. Now, a, we can replace a with r omega square. And now we have two unknown omega and t, and t here. So, we can find out what is my omega, how omega is expressed as a function of angle phi and lengths of the, lengths of the thread, etc. The only thing is, which I did not really specify here, is radius. Now, the radius of this obviously is equal to lengths of the thread, right, times cosine, right. So, r is equal lengths of the thread times cosine of phi. So, I can substitute it here and I will have m l cosine phi is equal to and omega square, sorry, and omega square equals to t cosine phi. From this, we can find t and substituting this t into this, we will get the dependency between omega and other parameters which are given. So, mg is equal to t which is m l omega square, m l omega square times sine phi, times sine phi. And from here, we have mass cancels out, as I was saying in the beginning, it does not depend on the mass. So, we have basically the dependency between the angle equals to g divided by l omega square. So, dependency between angular speed and the angle of the thread, which this thread makes with the horizon. And obviously, the bigger, the greater, the value of omega, the smaller will be this part. And obviously, the smaller will be the sine, which means that the smaller angle will be so our thread would be more horizontal. And obviously, in case my omega is equal to infinity, which obviously cannot happen, only then my sine would be equal to zero and angle will be completely horizontal. So, basically what it says, we cannot achieve really completely horizontal position of this thread, no matter how fast we are rotating this thread. However, the faster we rotate, the higher this object will be and the more horizontal will be the thread. Okay, these are all four problems, which I would like to talk about today. I do suggest you to go to this website, to this Physics 14 proteins, and this topic, Rotational Dynamics, has these problems among other things. And just read it through or try to solve the same problems just by yourself and check if you have the same answers. So, it's just a very good exercise. And I will also complement this topic with exams for those who want to challenge yourself. That's it, thanks very much and good luck.