 We have a sense from our own experience that rotation carries tremendous energy and potentially force with it, but also that energy and force are required to get things moving in the first place. So how do we relate our concepts about motion to concepts of energy and force? It is in this lecture that we'll begin to explore this question in a little bit more detail. Now in the context of rotational motion, let's revisit the older concepts of energy and force in light of this special kind of motion that combines a whole bunch of tiny linear motions of a whole bunch of points on an object that is rotating. This is a much more complicated situation than we've seen before, but as we're going to find out, we'll be able to adapt our rotational concepts back into the context of energy and force and rediscover some old friendly equations, but now recast in the form of angular quantities. The key ideas that we will explore in this section of the course are as follows. We will begin to think about energy and force concepts as they relate to extended rigid rotating bodies. These are bodies, for instance, like this saw blade pictured over on the right, where if you rotate the object through some angle, like one degree, all the points on this rigid body all rotate through one degree. We are going to find out that there are rotational analogs to linear concepts, like kinetic energy. In fact, we're going to apply kinetic energy first and see that we can develop a rotational version of the kinetic energy equation that more easily summarizes the complex motions of all the little points in a rigid rotating body, again, like this saw blade. We will come to understand how forces result in rotations, and we will explore the concept of a torque. And finally, we will connect the force and energy concepts through work, which is how they were connected in linear motion before, and we will complete a set of rotational analogs to linear energy and force concepts. So let me begin this lecture by asking you a question. Over here on the left, I have a large ventilation fan unit that can be powered by a motor and caused to spin. We'll get to the spinning part in a moment, but let me ask you this. I'm going to go ahead and throw that whole fan at you at your hand, and I'm going to throw it at your hand at a linear speed of 170 kilometers per hour. I have like a big catapult, and I can load the vent fan up into it, and I can fling it at your hand, which you'll just stick out in front of this big, big fan assembly. I'm going to get it up to a speed of 170 kilometers an hour, and I'm going to fling it straight at your hand. Is that okay? And I would hope that you probably replied, no, that is highly unsafe, or perhaps something a little less family-friendly. Why would you say that? Well, because this giant vent fan, which probably has some large mass, would also then, when accelerated up to 170 kilometers an hour, have a very large, linear kinetic energy. It's a big object, it's moving pretty fast. That's going to do some serious damage when it hits your hand. So if it strikes your hand, we know from looking at collisions that it's going to transfer some of that momentum that it has into your hand, either elastically or inelastically, but either is bad, and that collision is going to hurt a lot, a lot. And so, of course, you wouldn't agree to this. Well, fine. Let me ask you a different question then. So the vent fan has a uniform distribution of mass. There's just as much matter at the exact same distances from the center of rotation in this blade as there are in this blade, as there are in the blade you can't see here below the level of this ceiling. Okay, so if we think about the concept of center of mass and we ask for an object with a symmetric uniform distribution of material, where's the center of mass going to be? Well, it's going to be dead-centering. Dead-center is right here on the axis of rotation of the fan blade. Okay, no problem. I'm going to take the fan blade, I'm going to use this motor, and I'm going to spin the blades so that the tips of the blades are moving at 170 kilometers per hour. But the center of mass isn't going anywhere. In fact, it's the one part of this entire thing that's standing absolutely still because it's a dead-center right on the axis of rotation. Technically, it's not moving at all. So it has no linear kinetic energy. I argue that the linear kinetic energy of this fan is zero. I'm going to stick your hand in the path of the blades. Is that okay? And again, I would hope that you replied something like, heck no, that is highly unsafe. Well, why? I mean, it's true that the center of mass has no linear kinetic energy. But those blades are moving, and they are moving fast. And so there is clearly some kind of energy of motion of the rotating fan blades. That's got to be some kind of kinetic energy. And when any of those blades strikes your hand at that speed, it's going to do tremendous damage to your hand. So we obviously have to be able to quantify the kinetic energy associated with rotational motion. It's not zero. Let's see if we can figure out what that kinetic energy is. So let's consider the picture of the bicycle wheel that I showed you in the last lecture on rotational concepts, rotational quantities like angular speed, angular acceleration, and angular displacement. We can imagine that this wheel is moving as somebody is cycling over the ground. So we have frozen this wheel at a moment in time when it's technically in motion, but we're analyzing the motion over a very small period of time, an infinitesimal period of time. So if we imagine that indeed it's spinning and that we've frozen it at some very tiny little unit of time, we can also then think about what is really going on. What's moving here? Well, the bike wheel, the spokes, all of this stuff are made from a huge number of tiny little bits of mass. And those little bits of mass are distributed at various radial distances from the center of rotation, which is down here on the axis that goes through the center of this wheel. OK, so we have a wheel that's made from a huge number of bits of mass. Each of those bits of mass has its own radius from the axis from the center of rotation. We can label each distance of a bit of mass, r with a subscript i, and each bit of mass can be m with a subscript i. Now, i is a label. It labels all n, for instance, maybe Avogadro's number, of little bits of mass. Let's pick one of the little bits of mass. Here, we'll pick this little bit of mass here on the spoke, this little blue dot I've just drawn. I'm going to label that one as i equals 1. That's the first bit of mass I'm going to consider. Now, we're thinking about this wheel rotating over a very infinitesimally short period of time. So it's not in zero motion. This little point has a little bit of linear velocity in that unit of time. And so for that little blue point with its little mass, m1, we can figure out what its little speed is, v1. And from that information, m1 and v1, we could compute its linear kinetic energy. So it's going to have a linear kinetic energy, k1, which is one half m1 v1 squared. This is just the kinetic energy that we have looked at before. And again, v1 is its linear speed as it goes in circular motion over this very tiny little chunk of time that we're considering the motion of the wheel in. Let's choose a second point. I'm going to pick this red point over here, 0.2. I'm going to label it as such, 0.2, and it too will have a kinetic energy, k2, which is one half m2 v2 squared. Now, just a comment here. The linear velocity of this blue point is going to be different from the linear velocity, even in the same moment of time as this red point. Why? Because even though this is a rigid body, and even though these points are all making one revolution in the same amount of time, the points on the outer edge of the circularly rotating object have to go over a much longer distance in that unit of time. So the linear velocity of the red point is bigger than the linear velocity of the blue point, even though they make one revolution around the central axis in the same amount of time. That's just the nature of circular motion. Now, we can keep considering the kinetic energy of other points. We can pick another point, like this cyan point out here at a slightly different radius. We can calculate its kinetic energy. We can pick this purple point over here. We can pick this yellow point over here. We can pick this green point down here. We can pick all the points that make up this wheel, and for each one of them, we can write down its kinetic energy given its linear speed and its mass. Well, this activity that we're going through here, figuring out the kinetic energy of each point, this is nice, but of course we want to know the total energy of this rotating body, where all these little points make up the body. Well, if you want the total energy of a situation, add up the energies of the individual things that are happening in the situation. So we have a bunch of individual points, maybe Avogadro's number worth of them. Each of them has a kinetic energy. We have to add up all their kinetic energies. So to get the total kinetic energy, we're going to sum all the kinetic energies of all the individual points on this rigid rotating body. So the total kinetic energy of the wheel will be the sum of all the little kinetic energies of all its little pieces, and there are n of those, n could be a big number. So we can write that in a more compact notation as the sum from i equals 1 up to big n of all the little kinetic energies k labeled with the subscript i. And we can substitute in for that kinetic energy equation. Each kinetic energy is one half mi vi squared. Now this might at first seem pretty hopeless. That's a lot of terms, and even every one of them has a different speed, potentially, that we would have to then factor in and go figure out. So this might at first seem pretty hopeless. But remember, and I hinted at this a moment ago, the important fact about points on a rigid rotating body, all points on a rigid rotating body go through the same angular displacement in the same amount of time. That is, they all have the exact same angular speed omega. Angular speed is displacement in angle divided by time, and the displacement in angle in the same unit of time is the same for all points on a wheel. As long as that wheel is rigid. That's why I'm picking rigid structures here. Relative distances between points doesn't shift as the object moves. That's a more complicated situation. So in other words, what is true about every point in this is that the ratio of its speed to its radius from the center of rotation, the axis of the bike wheel, is the same. So for instance v1 over r1 is the same as v2 over r2, etc. That's the same as vn over rn, and all of those are equal to omega, the angular speed of the wheel. So if this wheel is making two rotations every second, let's say, that means it's going through 4 pi radians every second. That's its angular speed, 4 pi radians per second. So we can use this information to try to simplify this seemingly nasty sum up here. So let's go ahead and use this information to do that. So the total kinetic energy is the sum of 1 half mi vi squared. But vi over ri is the same for all points in the sum. So we can substitute for the speed of each point with the product of its angular speed and its radial location from the center. So now I have omega times ri all squared in place of vi. Now omega is exactly the same for every term in the sum, and 1 half is present in every term in the sum. So I can pull the 1 half and the omega squared, that results from squaring this quantity out, and I'm just left with this sum sandwiched in the middle. 1 half times this sum, and I'll come back to this in a moment, times omega squared. Well, this looks dangerously like 1 half times some kind of mass quantity times angular speed squared, which rings a bell. 1 half mv squared was the kinetic energy, linearly speaking, of a body moving along a straight line. Here we have a rotating body with many points that make it up. But they're all moving at the same angular speed, so we have an omega squared that's like v squared for a rotating body, and we have a 1 half out in front. So this thing here sits kind of in the seat that's reserved for inertia or mass in the kinetic energy equation. 1 half mv squared, m is inertia. This thing seems to be taking the place of inertia in a rotating situation. And in fact, we can sort of symbolically represent this equation now in a rotational context. We have this complex rotational rigid body made of a huge number of points, but we can summarize the whole kinetic energy of this rotating body in a compact equation. The kinetic energy of rotation for a rotating body is 1 half times some inertia quantity, which I'll call i for now, times angular speed squared, 1 half i omega squared, which looks very analogous to 1 half mv squared, where m was the inertial quantity in a linear motion context. Now i, this sum of the products of the masses and their positions from the center of rotation squared, this is known as the moment of inertia. It's an inertial quantity, and it summarizes how all the masses of each piece of this rotating body contributes to the overall kinetic energy of rotation through their inertia, their mass. Now moments of inertia are specific to different rotating structures. You have to know the mass and how it's distributed relative to the center of rotation of an object. So for instance, a hoop going in a circle around its center is not the same as a disc, a solid disc, going in a rotational motion around its center. The mass, even for the same mass, hoop and disc, the mass is distributed in a different way, and that has consequences for the kinetic energy of two bodies, even with the same mass that are rotating with different moments of inertia even at the same angular speed. So let's take a look at moments of inertia for a few common simple configurations involving things going in a circular motion around an axis of some kind. So for instance, let's imagine I just have a point-like mass, mass m, that's going in a circle indicated by this red path here around some axis, which is indicated in blue. In fact, thinking about the right-hand rule that tells you about angular speed and the direction of angular velocity, if you apply the right-hand rule, curl your fingers in the direction of the circular red path, your thumb indicates the direction of the angular velocity, that's where the blue arrow points. So the blue arrow is parallel to the axis of rotation and indicates the direction of the angular velocity. This is a counterclockwise rotation. It has a positive angular velocity. Now, what's the moment of inertia of this? How could we calculate the kinetic energy of this system, a point-like mass, a distance r from the axis of rotation going around the axis? Well, we have to do this sum. We have to calculate the moment of inertia of this point-like mass going in a circle. Well, the point-like mass is one point of mass m located a distance r from the center. So the moment of inertia of this point mass is the sum from i equals 1 to n of all the masses times their distances squared from the center. But there's only one term in the sum, and so we only have one thing to calculate. We have the one point with its mass m and its distance is r from the center, so mr squared. And so we would conclude that the moment of inertia of a point mass is just the mass times the distance squared from the axis of rotation, mr squared. And that's, in fact, correct. This is the moment of inertia for this situation. So if you were to be asked, what's the kinetic energy if this thing is going at some angular speed omega, it's 1 half times m times r squared times omega squared. That's the kinetic energy of rotation of this situation, of this system. Let's look at another system. I mentioned a hoop before. A hoop is a uniform distributed circle of mass m. Now, the circle, the whole thing, has a mass m, but each chunk of the circle has some little mass, and when you sum all those little chunks up, you get the total mass of the hoop. So this is depicted over here on the left. We have the hoop of mass m. It's going perhaps counterclockwise around the z-axis, so it's rotating around the z-axis. This looks a lot like a bike wheel. If we ignore the spokes of the wheel for a moment, this looks very similar to a bicycle wheel, rotating around its central axis of rotation. And in fact, you can use a hoop of mass m with most of its mass concentrated at the edge of the wheel as an analog to a bike wheel. Now, to calculate the moment of inertia for this, we actually have to do some calculus because we have mass m, but it's distributed over the circumference of this hoop. The hoop can be treated as made from a large infinite number of extremely infinitesimal, very tiny, points of mass. Each of those points has a mass dm, where dm is a differential of mass, a tiny little non-zero unit of mass, but approaching zero. Now, all those points of mass dm are located at precisely the same distance from the z-axis. Every dm is located a distance r from the center of rotation. That's what it means to be a point mass on the hoop's edge. So we have this moment of inertia of a hoop about the z-axis shown here in this picture. This is the sum from i equals 1 to n of the little masses mi times their distances r i squared. But we can convert this into an integral, a sum of infinitesimally sized mass points dm, all of them at a distance r, so we have r squared dm, and we're summing them in a calculus sense from zero to m. And if we do this integral, we find that we get mr squared evaluated between the limits of the integral, zero and m. And we also find out that the mass here, the moment of inertia for this is also just mr squared. It's the total mass of the hoop times the distance from the center squared. So this shows you how you can use calculus, for instance, to try to set up a sum over a large number of very tiny points that make up a continuous object, like a hoop or a disc, for instance. Now, here's another one. We have a rod. This is a line of mass. Think of it as a very thin rod so that it has only an extent in one dimension. It's only got physical extent along a line, but no other dimensions. It has a length, l, it has a mass, m, and it's rotating about its center. So it's going, perhaps again, counterclockwise around the blue line, which indicates its direction of its angular velocity, but also the direction of its axis of rotation. Now, to do this one also requires some calculus, but I'm going to skip it and just quote the result here. The moment of inertia of a rod depicted here on the left, rotating about its center in the sense shown, is one-twelfth times the mass of the rod times its length squared. It's a funny result, but you get it from the calculus of this particular problem. Now, compare that to taking that very same rod of mass, m, and length, l, but instead rotating it about one end. So this is, for instance, the difference between taking a baton, which is a long length of metal that you can throw into the air and get it rotating about its center. Take that same baton and instead swing it like a baseball bat from one end. It's the same object. It's got a mass, m. It's got a length, l. But it's rotating about a different axis. In one case, it's rotating about its center. That's twirling a baton. In another case, it's rotating about its end. It's more like swinging the baton like a bat. What changes? Well, more mass is now distributed further from the axis of rotation. Before, half the mass was on one side of the axis of rotation, and the other half was on the other side. But now, all the mass is on one side of the axis of rotation. So this has a very different moment of inertia. In fact, it's not one-twelfth ml squared. It's now one-third ml squared. It's much bigger than when you spin the same rod around its center. So note that for the same object, this configuration has more moment of inertia. And that means that if you get these objects, both going at the same angular speed omega around their axis of rotation, this configuration has a lot more rotational kinetic energy. So you can get very different outcomes in energy, even for the same object, but changing the axis of rotation, which changes the moment of inertia. So generally speaking, putting more mass farther from the center of rotation, the axis of rotation, means you get a bigger moment of inertia. So again, compare the rod twirled about its center versus the rod twirled about one end. The right-hand picture has much more kinetic energy for the same angular speed omega than the left-hand picture. But that also means that it takes more kinetic energy in the first place, more energy of some kind, more work, for instance, to get this thing up to the same angular speed as the one on the left. It takes less work to spin this up to an angular speed omega than it does to get this right-hand configuration up to an angular speed omega. And that's because it has a bigger moment of inertia. It is more inherent resistance to changes in its state of motion. That's what inertia is. It's the same concept that we had from way back in Newton's second law, inertia, which is mass in linear motion. Here in rotation, it is mass, but it's mass as a function of its distance from the axis of rotation, moment of inertia. It still represents inertia, and the more mass you put farther from the axis of rotation, generally speaking, the more work is required to get that object up to the same angular speed. So it works both ways. An object with a bigger moment of inertia at the same speed has more kinetic energy of rotation, but it takes more work to get it up to that speed in the first place. That is self-consistent. Now, we've been talking about spinning up objects so that they rotate about some axis, some fixed line in space, but I've totally avoided so far talking about how we get them spinning at that angular speed in the first place. I just talked about doing work, but what do I mean by work? How do I translate concepts of force and displacement and work into changes in kinetic energy? Can we recover the work kinetic energy theorem, but in a rotational sense? Can I talk about how I get something accelerating to it some distance from its axis of rotation? So we've got to get into that topic now. We've got to start thinking about forces and accelerations. How do forces cause angular accelerations? And then, forces acting over some displacement must equate to a kind of rotational work. What does that look like? So we know from Newton's second law of motion that if we want to take an object from a state of motion, so for instance, zero angular speed, that's a state of motion, no motion at all, to a different state of motion, non-zero angular speed, we have to apply a force. Force equals mass times acceleration in a linear context trying to move along straight lines. That is the statement of the relationship between force and acceleration. But we also know that forces can be applied to a body that's, say, fixed at some axis in space. If we apply a force to that somewhere along its length, we can get it rotating. So think of a wheel. If I grab the edge of a bike wheel and I push on it, I can get it to start rotating. It had no angular speed to begin with. Now it has a non-zero angular speed. I've clearly caused an angular acceleration. How does the force relate to the angular acceleration? Well, the linking concept here is when a force is applied to a body that can rotate about an axis, this generates what is known as a torque. We'll come to the formal definition of a torque in a moment, but they are essentially the rotational analog of a linear force. Now, believe it or not, we actually know a few things about these torques from our experience with doors. Doors are large, flat planes of material that can rotate about an axis, and the axis is marked by the hinges of the door. It's a vertical line on one end of the plane. And you know that if you grab the handle of a door and you push or pull, you can cause the door to start rotating about the hinges, about its axis of rotation. Opening doors is a great playground for thinking about force, rotation, and torque. So we already intuit a few things about forces and how they do or don't cause rotations. For instance, if we grab a door and we push on it or pull on it in a direction that is along the plane of the door, so toward the hinges or away from the hinges, that's a force that's parallel to the line that goes from the hinges to the edge of the door. And we know that pushing and pulling on a door like that straight along the width of the door, toward the hinges or away from the hinges, does not cause the door to open. If you can push and pull all you want and if your component of your force lies entirely along the plane of the door, the door will just sit there. It won't rotate. It neither closes nor opens. No rotation of the door results from that kind of application of force. So instead what we have to do is apply some component of our force that's perpendicular to the plane of the door, that is perpendicular to the line that connects the hinges to the other end of the door. When we do that, a rotation can be achieved. We can either open the door or close the door. Now one of the other things you may have noticed about opening and closing doors, and let's think about opening a door for a second, is that if we apply the same force closer to the hinges, it results in less rotation per unit time. It takes much longer to open the same door than if we take that same force and apply it as far as possible from the hinges themselves. The hinges mark the axis of rotation. If we pull on the door very close to the axis of rotation, it takes a lot of force to open the door in the same amount of time as if you just grab the handle which is located very far from the hinges on the other side of the plane and pull on the handle. This is why door handles are placed where they are. They are most effective. They use the least amount of force to open the door in the same amount of time if you put the door handle as far from the hinges as you possibly can physically manage on the door. So doors give us a playground to think about forces, the distance that they're applied from the axis of rotation, and the effectiveness of our ability to rotate the door as a result. That is, the effectiveness of the torque that we apply. So let's represent torques and the problem of opening a door. Let's begin by representing the door as a line. So we're looking down on the plane of the door from above. Here we have the hinges of the door at 00 in this coordinate system. I've represented that with the blue dot and the door is presently closed. It can open inward into the plane I've drawn here into this grid. So it's a door that opens one way. You have to push it to open it. I'm going to define width, which I'm denoting R, and I'm going to define a vector which is from the hinges to the place where I'm going to apply my force. That is a distance R and the vector R goes from the hinges to where I will apply the force. That is the definition of R vector. Now let's add our force. I'm going to open this door in a funny way. I'm going to push at a 45-degree angle on the handle of the door. So here's my force F applied at the handle which is at the tip of the R vector. And it's applied at a 45-degree angle. You can clearly see that depicted here. So there's my force F vector. There's R vector. That's where the handle is right at the end of the vector R. Let's think about the components of this force as I have applied it on this door. There's a component that is parallel to the length of the door either in the same direction R is pointing, as depicted here, or if I had directed my force to the left at 45-degrees pointing opposite the direction of R vector. That would also be parallel, but anti-parallel. It would point against the direction of R, but nonetheless it's a kind of parallel. Here is the component of F that is parallel to the R vector. I've drawn it kind of in purple here and I've denoted it F vector but with two parallel lines to indicate that it's a component parallel to the vector R. These forces do nothing to cause rotations. I am yanking on the hinges with this force. If I were pointing it in the other direction, I'd be pushing toward the hinges along the width of the door. That doesn't cause a door to open. That just looks foolish. What's going to cause the door to rotate is the force component that's perpendicular to R vector. So here it is. It's light blue or cyan. It's F vector with a little perpendicular symbol here from mathematics. So this means perpendicular to something and that something again is the vector R. It is only this perpendicular component that causes a rotation or in technical speak results in a torque. Torques cause rotations to occur. They create angular accelerations and thus open doors. So mathematically we could write down a formula for the torque that's represented by this picture. The torque, whatever it is, is some product of the distance from the hinges that I apply my perpendicular force and the magnitude of that perpendicular force. Well, if I think about my force F applied at an angle theta with respect to the R vector, then I can immediately use trigonometry and I know that whatever the force F is this component perpendicular to vector R is F sin theta. It's just the opposite component in this right triangle that forms the force F. So I can write my torque as R times the perpendicular component of F, which is just R F sin theta, where theta is the angle between the vector R and the force F. It's this opening angle between R and F. This might be ringing some alarm bells for you. Torque equals a magnitude of a vector times a magnitude of a different vector times the sine of the angle between them. This looks like one of the vector products and in fact it should feel familiar as a vector product because we kind of just looked at it in vectors part three, where we looked at the vector cross product. It's the magnitude of the cross product of two vectors. I have a vector a and a vector b and I take the cross product of a and b and I get the magnitude of that cross product. That is given by the magnitude of a times the magnitude of b times sine theta. That is, it's the length of a times the component of b that is perpendicular to a. Theta again is the opening angle between the two vectors and that's exactly what theta is here. It's the angle between the vector R which points horizontally and the vector F which points at 45 degrees above horizontal. So torque is also a vector. It results from a vector cross product and it's the cross product of the direction from the hinges that I apply the force and the force itself. So tau which is the symbol that's used to represent a torque is a vector and it's full vector form is given by the vector R and it's cross product with the force vector. The magnitude is RF sine theta and its direction is determined by applying the right hand rule for the vector product. So I can get that direction from the right hand rule for the vector product. Take my index finger on my right hand, point it in the direction of R vector, take my middle finger, point it in the direction of F vector and stick my thumb out perpendicular to both of those fingers. My thumb indicates the direction of the torque. The torque points up and out of the screen represent it as a circle with a dot in the middle of it. It's like a vector arrow coming straight at us out of the screen. So again this is a good chance for you to practice your vector cross product right hand rule with this situation. The magnitude of this vector is given by RF sine theta and its direction is given, for instance by using the right hand rule for the vector cross product or explicitly calculating the cross product using all the little I hat, K hat and J hat cross product rules. Now that we've seen how forces and rotation can go together, a force applied a distance from the axis of rotation generates a torque. We know that this causes an object to rotate like a door opening. So can we arrive at a simple statement about how forces generating torques then lead to angular accelerations. That was our goal in the first place. We're changing the state of motion of a door. It is accelerating in an angular sense. It gains an angular speed where it had none before. There must be a relationship between force, torque and angular acceleration. Can we figure out what it is? So let's begin by considering a simpler example than the door which has a uniform distribution of mass. Let's think about just a point mass, m that can rotate around an axis and that axis is a distance r from where the point mass is located. So in this case we apply a force to the mass and let's just apply a force perpendicular to the vector r that goes from the axis of rotation out to where the point mass is located. So we're going to apply a component of a force there that's entirely perpendicular to that r vector. That's going to begin to cause an acceleration of the mass. That mass is going to have a tangential acceleration to the circular path of motion. We've talked before when we were talking about angular accelerations. If I change the velocity of an object that's going in circular motion I'm causing an angular acceleration. The angular acceleration is related to the tangential component of the linear acceleration of that object. So I can write a equation that relates the perpendicular force and the tangential acceleration of that point mass m. So here we go. It's Newton's second law. That perpendicular component of my force, f perpendicular must be equal to the mass times the tangential linear acceleration of this point mass m. How is tangential acceleration related to angular acceleration? Angular acceleration is tangential acceleration divided by r the distance from the center of rotation. So I can substitute in for a tangential with angular acceleration alpha. That's what I get. The mass m times the angular acceleration times the distance r from the center of rotation. Now notice that if I were to multiply both sides of this equation f perpendicular and m alpha r by an additional r then I can get torque on the left hand side. So let's do that. Let's introduce torque into the above equation by multiplying both sides by the distance r. So now I have f perpendicular times r equals m alpha r squared. Because again I multiplied the right hand side by r as well. Well this is a point mass going around a center of rotation a distance r from that center. Notice I have m r squared here. And m r squared might seem a little bit familiar. It's in fact the moment of inertia of a point mass going around a center of rotation a distance r from that center. So I can substitute into this equation using the moment of inertia of this point mass. I have torque tau, the magnitude of torque on the left hand side. I have moment of inertia and angular acceleration on the right hand side. And so I get a very simple equation for all of this. Torque is moment of inertia times angular acceleration. And this is the angular equivalent of Newton's second law of motion but now no longer for linear motion but for rotational motion. So again instead of f equals m a I have tau equals i alpha. I have a rotation equivalent of force torque equal to a rotational equivalent of inertia i times the rotational equivalent of acceleration alpha. This expresses how a torque generates an angular acceleration. So now we see the link between pushing on a door its state of motion changing gaining an angular speed that must be due to an angular acceleration. How do I relate the two? I relate the two through torque and moment of inertia. Well I hinted at this earlier before if we are changing the kinetic energy of the door from a state of zero rotational kinetic energy to a non-zero state of rotational kinetic energy and if we are acting on this door with the rotational equivalent of a force, a torque a force acting at a distance from the axis of rotation then we have a relationship between a force like thing and the change in rotational kinetic energy of this body. So can we relate the change in the rotational kinetic energy of the angular analog of work? Work was in a linear sense force times displacement. What's the angular equivalent of work in a rotational context? So we know how to write the kinetic energy for a rotating body that's not the issue. We have to figure out the equation for the work in terms of forces and hopefully in terms of torques ultimately. So the work kinetic energy theorem should still hold here. A bunch of forces on an object should equal the change in the kinetic energy of the object. So whatever that work is it's got to be equal to 1 half i omega final squared minus 1 half i omega initial squared final minus initial kinetic energies. Okay well that's nice I mean that doesn't really tell us anything new work is equal to the change in kinetic energy we know how to write kinetic energy for a body that can rotate but how do we go from torque that we really care about? We have torque equals i alpha that's the Newton's second law equivalent for our rotating body. Well the definition of work hasn't really changed. Our force is being applied perpendicular to the line connecting the axis of rotation to where we apply the work and we're causing the object to go through a little displacement ds. So work is still going to be defined the way it was before. Our force and our displacement the product of those things in a vector sense adopt product. So if the force causes a change in displacement that is in the direction of motion then it's a positive work and if it causes a change in displacement opposite the direction of motion then it's a negative work it's causing something to slow down it's got to be depleting energy out of the system. Now what is s? But here s is just some little bit of path over which the force f acts. Now in a rotation it's only the perpendicular component of a force that causes a rotation to begin. And what's the path over which that force component acts? Well as a body begins to rotate and accelerate along some circular path that force perpendicular to the radial line of the path a little bit of arc length of the circular path a little bit of ds so if s is arc length ds is just a tiny differential bit of arc length it's that little differential bit of arc length over which force is acting and we can relate using angular quantities and in particular the definition of the radian we can relate changes in path over the arc length to the little bits of angle that causes paths subtent d theta and so in fact the little bit of arc length that we traverse with applying our force f is equal to the radial distance at which the force is applied r times the little bit of angle that we cause the object to rotate through d theta so again if this is something we haven't practiced yet enough so that you're comfortable with this go back to the definition of the radian a little tiny bit of arc length ds I must sweep out a little bit of angle in radians d theta and the relationship between those two has to be ds equals the radius times d theta that's where this comes from so let's again start from our definition of work, work is the integral of force times displacement and if we utilize that information that we just thought about that it's only the perpendicular component of the force and that is acting over a little bit of arc length ds then we arrive at the following equation that work is the integral of f dot ds well it's only the perpendicular component of the force that acts over that little bit of displacement ds and we can substitute ds with r d theta so we have that the perpendicular component of the force times the distance from the center of rotation the axis of rotation times the little bits of displacement d theta that's all to be integrated and thus we arrive at the relationship to torque because again we have the product of the perpendicular component of the force and the radial distance from the axis of rotation at which that force is applied that is the magnitude of the torque and so then it must be true that work is the sum from some initial angular distance to some final angular distance of the torque times d theta all I've done is substitute in here f perpendicular times r with the magnitude of the torque and I've put limits of integration between an initial angle and a final angle on this integral so work is the integral of torque times angular displacement just like in linear motion work was the product of force in linear displacement so again we see that there are analogies here between the linear world and the rotational world now finally a little bit of revisiting the work kinetic energy theorem will lead us to a revisitation of power power is the work done by a force per unit time and so we can start relating all the things that we have so far we have changes in kinetic energy we know how to write down rotational kinetic energy we now have work in terms of torque and angular displacement torque can be related to moment of inertia and angular acceleration i alpha we have all these pieces and now we can sort of come to a final concept here that will relate back to a linear concept we visited earlier and that was power so the change in kinetic energy must be equal to the integral of the torque times the displacements over which the torques are acting that's just the work kinetic energy theorem written out with change in kinetic energy on the left and the work on the right now if we just consider a teeny tiny little bit of work done by the torque tau over some little teeny change in angular location during the rotational motion then we would have this equation that dw is equal to tau d theta this is sort of just undoing the integral sign i take out the integral and i'm left with differential quantities dw equals torque times d theta now we can think about the energy per unit time that is the power that is being supplied to either drive or slow the rotation i can accelerate a wheel by making it go faster or if it's already spinning at some angular speed i can apply like a force i can put my hand on the wheel apply some friction and i can slow it down that's a deceleration so i can apply a positive torque or i can apply a negative torque i can speed the motion up i can slow the motion down i can put energy into the system by speeding it up i can take energy out of the system by slowing it down one of those is a positive power an increase in energy per unit time one of those is a negative power a decrease in energy per unit time power is just the change in work over the change in time and the in the rotational context dw is tau d theta so this can be written as tau d theta dt what is d theta dt? d theta dt is angular speed so power in a rotational context is torque times angular speed tau omega this ought to look familiar the linear equivalent of this was force times velocity power in a linear sense is force times velocity so let's take stock again of angular motion and linear motion and see the symmetries between the equations between the linear context and the angular context so let's think about linear motion in just one dimension so just along the x axis for instance we have newton's second law f equals m a the forces acting along the x direction equal the mass times the acceleration along the x direction we have the kinetic energy one half m times v squared we have work which is caused by a force acting over a displacement dx and that force can vary with distance so we might have to integrate f dx in order to get the correct work for that force and power is the change in work over the change in time but that in linear motion can be related to force times velocity at any given moment the angular quantities that are equivalent to this are newton's second law for a rotating body that's tau equals i alpha torque equals moment of inertia times angular acceleration rotational kinetic energy is one half times the moment of inertia times the angular speed squared one half i omega squared the work done on or by a rotating body is equal to the integral of the torque times the angular displacements integral of tau d theta this is analogous to the integral of f dx in a linear context and finally the power the energy supply to a body or taken out of a body per unit time is equal to the torque times the angular speed so the above are just some examples of relating linear quantities to their angular equivalence and in principle just as with angular speed and angular velocity and angular displacement you can map the linear quantities which may be more familiar at this point to you in the course onto the angular quantities all of the 1d equations of energy and force have an angular analog and to go from the linear equation to the angular equation you can change x to theta so linear position to angular displacement or angular position v to omega linear velocity to angular velocity or angular speed a to alpha so for instance linear acceleration to angular acceleration mass to moment of inertia so a moment of inertia is a distribution of mass a distance various distances from the center rotation it's more complicated than mass but it's the mass equivalent in rotation it's the inertia in rotation and finally changing force with torque f goes to tau if you do that in any linear equation you will essentially gain the angular analog for free the key ideas that we have explored in this section of the course are as follows we began to think about energy and force concepts as they related to extended rigid rotating bodies rotational motion we saw that there are rotational analogs to linear concepts like kinetic energy we started with kinetic energy we developed the rotational analog of kinetic energy in terms of the moment of inertia which is the inertial quantity and rotational motion and angular speed we've seen how forces result in rotations this has led us to the concept of a torque and we have explored that concept in the context for instance of Newton's second law relating torques and angular accelerations and we have connected the force concepts through the work concept and we've completed a set of rotational analogs to linear energy and force concepts these ideas applied to a rotating system allow us to develop a whole vocabulary and problem solving tool kit to analyze seemingly complex situations involving objects in rotational scenarios but with a fairly compact set of equations one must master the terminology one must master the application of these ideas because again we have to deal sometimes with cross products which are a bit funny compared to the kinds of vector quantities we've been dealing with so far but once you've had enough practice and you've gained some basic mastery of these ideas they become very powerful problem solving tools when thinking about the natural world