 Not all motion involves disconnected movement between X and Y and Z positions of objects. The rotors on a propeller engine for a plane function because the X motion and the Y motion is constrained. The same is true for a turbine fan blade. It's important that the motion and X and the motion and Y of the blade be correlated where this whole system would fall apart and not function. We're going to begin to explore rotational motion in more depth and develop a vocabulary to understand this. Rotational motion surrounds us in the cosmos. We would not be here if not for this kind of motion and it's important for us to take a much deeper dive into the subject and begin to adapt your ideas of force and energy and motion to this interesting but somewhat complicated form of movement. Let's begin now to describe a vocabulary in a mathematical toolkit for better understanding and labeling rotational motion. The key ideas that we will explore in this section of the course are as follows. We will transition from thinking linearly, that is, about motion only along straight lines or combinations of straight lines in three dimensions or two dimensions, to thinking rotationally and we'll come to understand why it is that these are not the same kind of motion. We will adapt our ideas about linear motion to the problem of rotational motion and specifically all of the calculus toolkit and the language of displacements that we utilized originally in order to think about linear motion will be reapplied to the problem of rotational motion. We will determine the mathematical relationships between rotational quantities and we will consider how to use vectors and essential toolkit in linear motion to describe rotation. Our experiences so far, especially in this course, have been primarily constrained to linear motion. Again, what is linear motion? It's a kind of motion where an object is confined to uncorrelated straight line movement along any of three dimensions. So we can move in a straight line along x or y or z. We can move simultaneously in a straight line motion along both x and y. For instance, we looked at this in the example of something like projectile motion. While it is true that we can relate what happens in x to what happens in y through the commonality of time, it has never been the case so far except in one case and that was uniform circular motion where motion in x dictated motion in y and motion in y dictated motion in x. That has not been something that we have dealt with rigorously so far. The problem, however, is that the universe is absolutely filled with motion where what happens in one coordinate direction constrains the motion in the other and vice versa. In rotational motion, generically, not just uniform circular motion but other kinds of rotational motion. It's a key example of exactly this phenomenon. So for instance, let's consider a particle that is executing uniform circular motion. It is doing so along a circular path with a radius of five meters. That means that the distance from the center of rotation for this particle is five meters. At some time t1, it might have a given set of coordinates. We could label them x1 and y1. And we could write those as numbers. So for instance, x1 might be three meters and y1 might be four meters. And if you check this using the Pythagorean theorem, the horizontal distance is three, the vertical distance is four, what's the length of the hypotenuse which corresponds to the length of the radius of the circular path, indeed you'll see that it's five meters. So we are, in fact, on a circle with a radius of five meters whose x coordinates are three and its y coordinate is four. Now at some later time t2, let's imagine we measure the x coordinate and we found that it's changed from three meters to one meter. However, it's doing so while maintaining a circular motion at the same radial distance. Well, that means that the y coordinate is not free to be anything it wants to be. In fact, circular motion constrains the location of y2 in space. It's not decorrelated from x. If x is now one meter and we have to maintain a radial distance of five meters from the center of the rotational motion, there are only two solutions that are allowed. And one solution, for instance, is that y2 is 4.899 meters. So for instance, for that particular solution, x2 and y2 are one meter and 4.899 meters. So in rotational motion, what happens to one coordinate, like x, affects what happens to the other coordinate, y. And this is especially the case in a rigid body, a body whose parts are not changing in orientation to one another in space. The whole thing holds together and all moves together. That doesn't mean straight line motion anymore. It was true when we considered examples in linear motion of a train moving along a track that all the atoms, when the train displaces one meter, all the atoms in the train displace one meter as well. But if that train is now a rotating object, like a space station rotating in outer space, it's true all the atoms are moving together, but their distances that they're traveling are not all the same in linear space anymore. And so we have to learn to think about this kind of more complicated situation, this rotational motion. So let's begin from something that's familiar, uniform circular motion. And let's especially go back and think about the definition of angle in our standard international system of units, and that is the radian. What is the radian? So let me remind you where the definition of the radian comes from. Radians are dimensionless measures of the degree of motion around a circle. And the definition of the radian, for instance, arises from the following considerations. I have here over on the left side of this slide a circular path shown in black, so an object might be going counterclockwise around this path. I've also shown here a little segment of the path shown with a red arrow. So let's think about the overall circular path, the black circle. The circumference of any circle of radius r is given by a well-known formula. Two times pi times the radius will give you the length of the outer edge of the circle at that radius, the so-called circumference. So let's imagine traveling along only a piece of the circumference, and that's again indicated by this red arrow here. This red arrow follows the curvature of the black circle, but it doesn't go all the way around it. It only completes part of the circle. A section of the circumference is known as an arc, and the length of that arc is the arc length s. That's the standard symbol that's used for arc length, the letter s. The question we want to answer is, what angle between this radial line at the beginning of the arc and this radial line at the end of the arc defines the length of that piece of the circumference? In other words, the phrasing here in geometry is, what angle theta, that is the opening angle between these two radial lines, does this distance s, the arc length, subtend? You can see right away that if I open up this piece, if I stretch the arc length a little further, theta gets bigger and the arc length gets longer. They have a proportional relationship to one another that we could write down and utilize to define a standard measure of angle theta. Let's do that. So let's define the relationship between s, the arc length, theta, the angle that's subtended by the arc length, and r, the radius of the circle as follows. s equals r theta. This is a nice proportional statement. Whatever s is, if theta gets bigger, s gets bigger. And we have some sense that if r also gets longer, if the radius of the circle gets longer so the circumference gets bigger, for the same angle the arc length will also get bigger. So these all have a proportional relationship to each other that very much follows this equation. So now what then does theta actually mean in all of this? Well let's consider the ratio of the arc length to the circumference of the full circle. So the arc length is a piece of the circumference of length s, and I've denoted it in red here so you can match it to this red arrow up here in the drawing. If we divide that by the circumference and then use our definition of s as being equal to r theta and c as being equal to 2 pi r, we get this nice ratio here, r theta divided by 2 pi r. And the r's cancel out and we're left with just theta over 2 pi. And we can immediately rearrange this equation and finally give a definition of theta. Now what do you notice about this equation? The number 2 and the number pi are dimensionless numbers. They don't have units of anything. It's just 2 and pi. 2 and pi are just numbers. The only thing that had dimensions in this was radius with units of meters, for instance, but it's entirely canceled out of the equation. So we have s, which has units of length, and c, which has units of length, and we have a ratio of them and the length units cancel out. So s over c is dimensionless and this thing also has to be dimensionless. So when we rewrite this equation to solve for theta, we see that the angle defined as a fraction of 2 pi is exactly the fraction of the circumference occupied by the arc length. That is, if this arc length here is 20% or 30% of the total circumference, then theta is also, say, 30% of 2 pi. Radians are dimensionless quantities. We use the term radian as in, oh, this is an angle of pi radians as if radians are a dimension of some kind, as if radians are dimensionful. But in fact, because of the way you develop the definition of the radian, radians are dimensionless numbers that represent merely fractions of a total circle. So theta is just the arc length divided by the circumference times 2 pi and it's a dimensionless quantity. There are no units on the right-hand side and so there can be no units on the left-hand side and that's something that's important to keep in mind. It's why scientists use radians when doing, for instance, mathematical developments. They're a nice dimensionless quantity. You can map them onto dimensionful quantities in a circle like degrees but they're convenient because they don't carry any dimensionality themselves. So angles in radians are very useful measures of where you've gone in a circle because they tell you how far from the right-hand horizontal line, I'll come back to this point in a moment, you've traveled around a circle. And then you can talk about relative displacements in angle. So if I start out at 30 degrees above the horizontal and I go to 60 degrees above the horizontal, I've displaced 30 degrees in angle. We begin to see that a vocabulary here about displacement and possibly even velocity and acceleration can be recovered but now from an angular perspective. So let's reconsider circular motion. So now we begin to see that when an object is moving in a circle, it's traveling a little piece of arc length which we could denote ds in a little piece of time which we could denote dt. Ds is meant to be a single thing, infinitesimal unit of arc length and dt is an infinitesimal unit of time. Ds has units of meters and dt has units of seconds. And that means that the linear speed of an object moving in circular motion at any given moment in time, freezing the motion and looking at the velocity vector of the particle going in a circle is given by our old friend, the time derivative. But it's not of x or y or z but now of the arc length. So the velocity of an object in circular motion is just the time derivative of arc length. It's ds divided by dt, an infinitesimal change in arc length over an infinitesimal change in time. This gives us the velocity tangent to the circular path. But let's remember what arc length is. It's related to your angular displacement inside the circular motion, s equals r theta. So we looked at this from our consideration of the definition of the radian and in uniform circular motion the radius of the circular path doesn't change and so we can just substitute in for the arc length. We can transform ds into something that has d theta in it. So let's do that. D equals the time rate of change of arc length but substituting in with s equals r theta and taking advantage of the fact that the radius of the circular path is not changing we get this equation, radius times the time derivative of theta, so r d theta dt. So now the linear velocity of an object in circular motion is related to the radius of the path r and the time rate of change of its angular location in space around the circle. You can think of ds again as being a change in arc length that's so small that it is effectively a tiny little straight line. You can think of the circular path as being made up of a almost infinite number of little straight line motions that curve over time and make the circle. And so this displacement is itself tangent to the circular path at any given point in time and that should remind you of the exact definition of instantaneous speed. Instantaneous speed tells you the slope of a line tangent to the path in space with respect to time. So we're just exercising our old ideas now about calculus and small changes in space and small changes in time and how we map that onto things like velocity and acceleration. Now again let's consider this equation relating the linear velocity of an object in circular motion to the changing rate of its angular position. We can then find this convenient relationship. So we take this equation, we write it down one more time, the linear speed of an object in circular motion is equal to the radius of its circular path times the time rate of change of its angle. Well now we can rearrange this and solve for this thing, the time rate of change of angle d theta dt. That's going to be equal to the linear speed divided by the radius of the circular path. And this has units of radians per second which should sound familiar and we'll use the symbol omega for this, it's not an accident. We've just defined something known as angular speed, again it has units of radians per second although radians themselves are dimensionless, it helps to remind us that this is a measure of angle. So we're talking about the radians per second that an object in uniform circular motion is actually moving and we see that we can relate linear speed v at any moment during circular motion to a rotational quantity, omega, this angular speed. Now omega as a symbol might look familiar, we've looked at oscillatory phenomena before, these are phenomena that change in time but repeat in time, and for simple harmonic motion and simple harmonic oscillations we saw that one could write down a sine or cosine driven function that contains something called the angular frequency. Well omega here actually has the exact same meaning that it did in oscillatory motion, that's the radians per second, the angular frequency of a rotational motion. So it's the fraction of a circle that you go in some unit of time, has the same meaning, fraction of a cycle that you go in some unit of time, nothing has changed here, omega really has the same meaning but now we're applying it to a physical motion, not just a conceptual description of something that changes in time for instance that might have a physical interpretation that might not. Now we've defined angular speed, there's nothing to prevent us from talking about an angular acceleration. So for instance, what if we were to suddenly start having this object increase its linear speed as it goes around the circular path, so it maintains the same path, the same radius R, but it's going faster and faster and faster and faster, it's not just changing its direction anymore, it's also changing its magnitude of its velocity, its speed. So now we can use the time derivative and go one step further and based on angular speed or angular velocity we can now define angular acceleration. This is denoted alpha, it is the time rate of change of angular speed, so it's d omega dt, the first derivative of omega with respect to time. One can plug in that omega is d theta dt and see that this is the second derivative of angular position with respect to time. This looks really familiar, right? The definition of the acceleration in terms of linear quantities was the second derivative of position with respect to time. Now we have the angular acceleration which is the second derivative of angular position with respect to time. So let's use the relationship between angular speed and the corresponding linear speed at the same moment in circular motion and we can then rewrite this in terms of linear quantities. So again we have these linear quantities like linear speed as the thing goes in a circular path but now it's speeding up, its speed is increasing and so it also has an acceleration and that acceleration is tangent to the circular path. The magnitude of the speed is increasing, the increase in the magnitude occurs perpendicular to the radial line. We call this the tangential component of the acceleration. So we have that the angular acceleration alpha is the time rate of change of the angular speed, omega. We can substitute in that omega is equal to v over r. So let's do that. So now we have the time derivative of v over r. Now r is a constant, it's not changing. So we can pull 1 over r out in front of this derivative of the velocity with respect to time, the linear velocity, linear speed. So let's do that. We have 1 over r dv dt. Well what is dv dt? It's obviously a linear acceleration. It's a time rate of change of speed. That's an acceleration but again specifically it is a change in the velocity in terms of the magnitude of the velocity and that change occurs on a tangent line to the circular path. And so we label this not just acceleration but tangential acceleration, a with a subscript t. And we see that angular acceleration alpha is given by tangential linear acceleration divided by the radius of the circle. And we see that this angular acceleration has the same relationship to linear acceleration tangent to the circular motion as does angular speed to linear speed. Again linear velocity is tangent to the circular path as well. All we have to do is divide by the radius of the circular path, the circular motion. Now again I can't emphasize this enough. We're not talking about the full acceleration of an object in uniform circular motion. This only tells us how the speed changes and that changes in a direction tangent to the circular path. That's why we emphasize that this is tangential acceleration that I've just related here to angular acceleration. But there's also centripetal acceleration that hasn't gone away. What's keeping this thing in the circular path? It's a centripetal acceleration that changes the direction of the velocity, not its magnitude. And it acts on a radial line pointing toward the center of the circular motion. That's a 90 degree angle to a tangential line. So our acceleration when we increase the speed of an object in circular motion, our acceleration has two components. The tangential component which is related to angular acceleration and the radial component which is related to centripetal acceleration which was always there in the first place because the time rate of change of the direction of the velocity vector of the object in circular motion. So if we increase the linear speed of an object in circular motion, we've actually now abandoned uniform circular motion. There's nothing uniform about this anymore. We're increasing our linear velocity as a function of time. We have now entered non-uniform or accelerated circular motion. So we now again have two accelerations. We have one tangent to the circular path, AT, which I defined a moment ago, and our old friend centripetal acceleration which points inward always toward the center of the motion. Nothing has changed here. So we actually can use the Pythagorean theorem and we can relate the tangential component and the centripetal component to the total magnitude of the acceleration. A, that magnitude is given by the square root of the sum of the squares of the tangential acceleration and the centripetal acceleration. So if I plug in with my angular quantities now, for the tangential acceleration, that's equal to alpha times r. So I get alpha squared r squared for this left term inside the square root. And remember that centripetal acceleration was given by v squared over r. So when I plug that in here, I get v to the fourth over r squared. But v is related to the angular speed, omega. And in fact, v is equal to omega r. So if I plug that relationship in, I find out finally that my total acceleration is given in terms of angular quantities as follows. The total magnitude of the acceleration A is the square root of alpha squared r squared plus omega to the fourth r squared. That's very nice, actually. That's a fairly compact formula. So if you know the angular acceleration in terms of radians per second squared, and you know the angular speed in terms of radians per second, and you know the radius of the circular path, you can very quickly solve for the magnitude of the total linear acceleration, A. So again, be careful when we are in accelerated or non-uniform circular motion. There are two components of acceleration that develop. There's the centripetal acceleration that's always there. But now you get one that's tangent to the circle, and that's related to the angular acceleration, the increase in the rate of change of theta as a function of time. Now let's go backward a little bit. We've been looking at instantaneous quantities, like the time derivative of theta or the second time derivative of theta, first time derivative of omega. We can relate this easily to our old concept of average quantities, like average velocities and average accelerations. And so just as for linear quantities, we can define average rotational speeds and accelerations. The instantaneous speeds and accelerations are just limits of those average quantities in the limit that something happens, like for instance in the limit that the time unit over which we consider the motion goes to zero. So as an example, let's consider an angular displacement, which I can write as delta theta. I start somewhere on a circular path at a location theta one on the circumference of the circular path. So this distance from the center is one radial line R. This is my starting angle. Where is this measured with respect to? Well, remember in a circle, angles are measured with respect to the horizontal line on the positive side, the side to the right of the center of the circle. So theta one is measured from the blue dotted line up to the red line. Positive angular displacements are in this sense of rotation, which is often called counterclockwise. An analog clock with an hour and a minute hand will have the hands rotating in this sense. So if we start out on the blue dashed line, you have to go down to the left, then up to the blue dashed line, then up to the right, and then down again. That's clockwise. But positive angular directions are actually counterclockwise. This is a convention that you'll want to internalize as best you can. Practice is the best way to do that. So positive angles are when you move in a counterclockwise sense away from the right-hand side of this blue dotted line that runs through the center of the circle. So this is theta one. It's a positive angle with respect to the blue dotted line. I can consider maybe a little bit of time later, some later time. I'm further along the circular path. And now I'm at some angle theta two with respect to the blue dotted line. Delta theta, then, is in a sense this displacement between where I am at time two and where I was at time one in angular space. So you can use this sort of picture to think about what delta theta looks like. And I'll come back to whether or not angular quantities like this can be represented by a true vector later on in the lecture. So we can simply define delta theta as theta two minus theta one. The angular position at the later time minus the angular position at the earlier time, that should sound familiar. Delta x was x at a later time minus x at an earlier time. Delta theta is theta at a later time minus theta at an earlier time. If this angular displacement happens in a time delta t, which is a later time t2 minus an earlier time t1, then the average angular speed is just defined as delta theta divided by delta t. And again, this should look very familiar. The average speed was defined as delta x over delta t in the horizontal direction, for instance, when we first looked at linear motion. Now almost immediately, we see how to connect the instantaneous angular speed to this average angular speed. Omega, the instantaneous angular speed, is just the limit as delta t goes to 0 of the average angular speed, which is delta theta over delta t. And that is exactly the definition of the time derivative of theta. A similar relationship holds, for instance, for alpha. Alpha is just the limit as delta t goes to 0 of delta omega over delta t. That is the first derivative with respect to time of the instantaneous angular speed. Now let's think about a rigid structure that's rotating. A bike wheel is a pretty rigid structure. It's got these spokes. They're stiff. It's got the rim, which is stiff. It's got the actual rubber tire. These things don't slide over each other. They don't bend appreciably. Everything kind of maintains its relative position to everything else in a properly working bicycle wheel. So let's think of a bicycle wheel as an archetypal rigid structure. And bicycle wheels can rotate. So this is a rotating rigid structure. And you might ask, what is the angular speed of some point, point two, maybe this red point here? If point one, this blue point here on a spoke, has an angular speed of omega. So imagine the blue point is making one revolution every half second. So it's going 2 pi every half second or 4 pi radians per second. What is the angular speed of the red point? How many radians per second is the red point traveling? So if we think carefully about such a structure, maybe take a bike wheel, mark off two points, rotate the wheel one full cycle, and then see what the two points do over that one full cycle, we will quickly draw the conclusion that if we pick any two points on such a rigid structure and then we rotate any one of them through a full circle, which means rotating the whole structure through a full circle, the other one also completes an exact same full circle of the motion. And in fact, what this means is that since the angular displacement is the same amount in the same amount of time for these two points, all points on a rigid body rotate with the same angular speed independent of their distance from the center of rotation. And this is extremely convenient. It will not be true that the linear speed of the red point and the blue point are the same, but their angular speeds will be the same. And if we know the radii, the distances from the center of rotation that each of these points lies, then we can actually figure out what the linear speed is by using the relationship between linear speed v and angular speed omega, v equals omega times r, the radial distance from the center of rotation. So this constancy of angular speed in a rigid body for two points is extremely convenient. And by the way, this same conclusion holds for angular acceleration. If the red point has a certain angular acceleration, then the blue point has the exact same angular acceleration. Again, a rigid body where the relative orientation of two points on the body never changes. That's rotating. All those points will have the same angular speeds, angular accelerations. They won't have the same linear speeds. They won't have the same linear accelerations, but you can use the constancy of angular quantities for rotating or rotating and accelerating body to solve for things that are more difficult to figure out because different points are at different distances from the center of rotation. Now what about rotational motion under constant angular acceleration? So we can imagine a situation when an object moves in circular motion, but not uniform circular motion. Again, this is accelerated circular motion, non-uniform circular motion. This is when the angular speed increases over time at a constant rate. So we're allowing for an angular acceleration, but we're going to require that that acceleration be constant in time. So we have ourselves a situation that's very similar to the one that led us to equations of motion relating linear speed, linear acceleration, linear displacement, and time. But now under consideration of angular quantities, angular speed, angular acceleration, time, and angular position, theta. So how can we describe motion in this case? And especially, can we derive equations of motion? Can we find ways to relate angular displacement, angular speed, angular acceleration, and time? And I'm saying this hintingly, of course we can. And here's how we're going to do it. Let's start from a consideration of angular acceleration. Angular acceleration, alpha, is the time derivative of angular speed, omega. So I can rearrange this equation and get set up to do an integral. So for instance, I can rewrite this equation as follows. I can move the dt to the left-hand side and I can have alpha times dt, the tiny change in time, is equal to d omega, the tiny change in angular speed. Again, alpha is constant in time. We're assuming that alpha does not change. And if that holds true, and I integrate both sides of the above equation, I get an interesting result. So what do I mean by integrating both sides of the equation? I've got dt on the left side, I've got d omega on the right. If I do an integral of the left side, it's from a minimum time to a maximum time. I'm summing up all the little pieces, alpha dt, from some minimum time to some maximum time. Over on the right, I'm summing up all the little pieces of angular velocity from some minimum angular speed to some maximum angular speed. So I'm gonna set my minimum time to be time zero, zero seconds, and I'm gonna set my maximum time to be some unknown time later, t. I'm gonna set my minimum speed to be omega zero, the initial angular speed of the body, and I'm going to have some later speed at some later time t, omega. Well, if you do these integrals out, you'll get the following equation. You'll find that alpha times time is equal to omega minus omega naught. That is, the product of the constant angular acceleration in time is equal to the difference in the angular speeds at the two times, time t and time zero. Well, my, my, my, doesn't this look eerily familiar? This looks very similar to the relationship that existed between speed and acceleration during linear motion. What was that? Acceleration along a line times time is equal to speed along a line minus the speed at the initial time zero. And in fact, if you take the angular equation, alpha t equals omega minus omega naught, and you substitute in with omega equals linear speed during rotational motion divided by the radius of the motion, and alpha equals the tangential acceleration a over r, then you'll find an equation like this. The tangential acceleration divided by r times time is equal to the speed at a later time divided by r minus the speed at the earlier time divided by r. We've immediately recovered the linear form of the equation by substituting in with the relationship between the angular and the linear quantities. Again, this is for the tangential component of the acceleration as we accelerate the angular motion of a rotating body. We can keep going. We don't have to stop here. We got one equation of motion that we kind of recognized from before. Let's see if we can get another one. Let's use the fact that omega is the time rate of change of the angular position, d theta dt. So we have that alpha t is equal to d theta dt minus omega naught. This is the velocity, the angular speed, at time zero in the problem. Well, I can rearrange this equation as well. I can get alpha and time and omega naught on one side of the equation. I can leave d theta and dt on the other side of the equation, and we can do this same trick with integrating again where we move dt to the left side and then we integrate both sides. So let's do that. So if I shuffle dt to the left side of this equation, I now have two terms. I have alpha t dt and omega naught dt. And the right-hand side, I just have one term, d theta. Well, we're gonna integrate again. We're gonna integrate the left side from time zero to a later time t, and we're gonna integrate the right side from some initial angle theta naught to some later angle theta at time t. And if you do these integrals out, you have two integrals to do, this one and this one, and then one over on the right, you'll get the following equation. So rewriting this integral I had on the previous slide here at the top, and working through this, and I encourage you to try this on your own, you will again yield a familiar equation, albeit this time in angular form. And that is that one-half times the angular acceleration times time squared, plus the initial angular speed times time is equal to the angle at a later time t minus the initial angle at time zero. And this again looks eerily like one-half at squared plus v naught t equals x minus x naught, this equation that relates initial speed, acceleration, time, and displacement along a straight line. And in fact, I encourage you to try this, you can again use the relationships between angular speed and angular acceleration, and then the linear equivalence of those quantities, plug into the angular equation and see that you get a linear equation involving the tangential component of acceleration. So let's take stock of what we've learned so far. We see that there is a very convenient mental mapping of angular motion related quantities and linear motion related quantities. So for instance, in linear motion, in one dimension only, we have the speed is the first derivative of the position with respect to time, dx dt. We have that the acceleration is the first derivative with respect to time of the speed, dv dt. And that is the second derivative with respect to time of the position. We have an equation of motion like this in one dimension, the velocity at a later time, t, is equal to the velocity at the initial time, zero, plus the acceleration times time, this assumes constant acceleration in time. We have another equation that tells us the position at a later time, t, given the initial position, the initial speed, and that constant acceleration, and of course the time that's passed, t. We have analogs in the angular world now. So in the angular world, we have angular quantities. We have the angular speed, which is the first derivative with respect to time of the angular position, theta. We have the angular acceleration, which is the first derivative with respect to time of the angular speed. That's the second derivative with respect to time of the angular position, theta. We have an equation that relates angular speed at a later time t to the angular speed at the initial time 0 and the constant angular acceleration alpha and the time. And we have similarly an equation that tells us the angular position theta at some later time t given the initial angular position, the initial angular speed, that constant acceleration, and the time that has passed since time 0. Now these are just some examples. I wanted to show you the kind of beautiful symmetry that exists between the linear world and the angular world. And there are relationships between angular quantities and their linear equivalence during circular motion or rotational motion. You have to be careful to exercise those appropriately. But in principle, all of the 1D equations of motion have an angular analog that can be written down simply by changing x to theta, v to omega, and a to alpha. Now again, if you're going to map alpha back onto acceleration in a circular or rotational motion situation, keep in mind that the acceleration, the linear acceleration you get from alpha, is only the tangential component of the total acceleration of a point in rotational motion. So I've been avoiding this pretty obviously the entire time throughout this lecture, but we used vector quantities quite successfully to describe linear motion. That's motion that has both a magnitude, like speed, but also a direction, which velocity has. It has speed, and it has direction. It has magnitude and direction. So our rotational quantities, like angular speed and angular acceleration, are these vectors? Is there angular velocity, for instance? I've avoided this, but now we're going to dig into it a little bit. So let's begin to see what can and cannot be represented cleanly as vectors in what we have so far discussed. Let's begin with angular speed. It's the most approachable of the questions. And then once we've answered this one, we'll look at angular acceleration. And then we're going to return to displacements in angle, so rotations directly, and look at that. So is there a vector that represents angular velocity omega with a vector hat over it? That would turn angular speed into the magnitude of that quantity, its length. Let's remind ourselves about vectors. Vectors point in well-defined directions in space. They have lengths that represent their magnitudes. But in rotation, the points on the rotating body are all changing direction constantly. And those directions are related to each other, what happens in x affects y, and vice versa. So we have to try to find a way to get a singular arrow that represents angular velocity. We don't want an arrow that keeps pointing all over the place all the time. Once we set an object spinning at a constant angular speed, it would be nice if there was one arrow that directly stood for the direction of the sense of rotation of this object. That way, we can reuse the rules of vector addition and subtraction quite easily. Now, in fact, there's a convention for this. We'll see why you can use vectors to represent angular velocities in a moment. But the convention for the direction that this arrow points is standard in physics. And the idea is to utilize something called a right hand rule for rotation. So set an object spinning. Maybe get it going in a counterclockwise rotational sense. Now, take the fingers on your right hand and stick them out and curl them in the sense of rotation of the object. So your fingertips should curl to point in the direction that the object is rotating. Now, this leaves your thumb sticking straight out. And where your thumb points indicates the direction of the vector part of angular velocity, omega vector. Now, as part of this convention, and I hinted at this before, positive directions, positive senses of rotation correspond to where your thumb points when your fingers are curled counterclockwise. So in this example I'm showing you, this is a positive angular velocity. My thumb points up, my fingers curl counterclockwise. This is a positive angular velocity. If I set the object rotating in the opposite sense so that now it's rotating clockwise, I have to flip my hand over to bend my fingers in the right direction of rotation. My thumb now points down and that represents a negative angular velocity. So counterclockwise rotations are positive sense rotations. Clockwise rotations are negative sense rotations. So we have a vector in space that lies along the axis of rotation of the spinning object. My thumb indicates the direction of the axle of the wheel, for instance. That's the line around which the wheel spins. So if I represent angular velocity this way with a vector that points up for a positive sense rotation, a counterclockwise rotation of the wheel, can I add velocities like vectors? If I add two velocities together, does the result match actually say doing that to the angular speed of the wheel, increasing it in a certain way? So let's look at that. Let's consider a rotating object with an angular velocity that I can write as omega vector with its magnitude omega and its direction omega hat. So omega is the angular speed, it's the number, it's the length of the vector. Omega hat is the unit vector that indicates where your thumb points if you apply the right hand rule. So in the sense of a positive rotation, counterclockwise rotation of the wheel, my thumb points up and that's where omega hat points up. It points up perpendicular to the plane of rotation. Now, double the angular speed of the wheel. Set it rotating twice as fast. We know what that means, it means that instead of going at omega, now the wheel is rotating at two omega. Is that equivalent to having added two angular velocity vectors, omega vector and omega vector together? That's what it would mean to double the velocity. It would mean adding omega vector to itself. Does that yield the correct answer mathematically? Two omega for the speed. Well, let's try adding the vector we defined above to itself and see what happens. I have omega vector plus omega vector. I write this out as the speed times the unit vector. So I have omega times omega hat plus omega times omega hat. If I group terms together that have the same unit vector in front of them, I wind up getting omega plus omega and that whole thing is in the omega hat direction. And indeed that's two omega, in the omega hat direction. So the result is a vector with twice the length of the original one pointing in the same direction and indeed adding the angular velocity to itself has resulted in a new vector with twice the angular speed of the original. That physically represents having doubled the angular speed of the object which was exactly what we knew should happen. If I double the speed of an object in an angular sense, I should go from omega to two omega and that's what we see by adding these vectors together. So it works. We can assign a direction in space to stand for the axis of rotation of the object. The object rotates around that axis with an angular speed omega and I can add vectors together to get more and more angular speeds not just in the direction I've shown but in many directions. I could have many rotations oriented along x, y and z axes for instance. Now the same argument applies to angular acceleration. One can define alpha vector as the magnitude, the angular acceleration alpha, times a unit vector alpha hat. And again the right hand rule tells you the direction of alpha hat. Curl your fingers in the direction of the change in angular speed. So if you have a positive sense rotation that's a counterclockwise rotation and you make it go faster in that direction that's a positive acceleration. Curl your fingers in the direction of the change in speed. Your thumb indicates where alpha hat points. If instead I slow that wheel down if I make it rotate less counterclockwise over time that's a negative sense angular acceleration. I'm declining the angular speed of the wheel and that gives me a negative angular acceleration. It gives us the right answer. It should slow the wheel down. The vector that's associated with both angular velocity or angular acceleration tells you where the axis of rotation lies in space. So your thumb is aligned with the line around which all the rotation is occurring, the axis of rotation. Your fingers curl in the sense of the rotation around that axis. So that's what this vector means physically. It tells you where the axis of rotation lies in space. Your fingers curling tell you the direction of the rotation or in the case of angular acceleration the direction of the increase or the decrease in the angular speed of an object. Now what about angular displacements themselves? Delta theta is theta two minus theta one. Can that be represented as a clean vector? Well it turns out that angular displacements unlike angular velocities and angular accelerations do not have a simple vector representation. Let's see why that is. Changes in angle themselves have, and this is what is known in mathematics as, non-commutative properties. What does that mean? If two activities commute, it means that it doesn't matter what order you do the activities in. For instance, if I wear rings on my left hand and on my right hand, if I put my left hand rings on first and then I put my right hand rings on second, I get the same outcome as if I put my right hand rings on first and then my left hand rings on second. I still get the same rings on the same fingers at the end of that, regardless of the order in which I did it. Let's say I'm wearing socks and I'm wearing shoes. Order matters. If I put my socks on first, then my shoes. I get a different outcome than if I put my shoes on first and then my socks. The order of the operations when it comes to socks and shoes matters. The order of the operations in terms of putting my left hand rings on my left hand and my right hand rings on my right hand, that doesn't matter. I get the same outcome there. So socks and shoes are non-commutative and the rings example I gave you was a commutative operation. Angles and specifically rotations and the order in which you do them have an order that matters. It changes the outcome and a very simple demonstration can reveal this. To demonstrate how a series of rotations depend on the order in which they're done, take a flat object, turn it 90 degrees along one axis, say pointing toward you, and then 90 degrees again along another axis, say vertically pointing toward the sky. Now take the same object and return it to its starting position and reverse the order of the 90 degree rotations, 90 degrees around an axis pointing to the sky, 90 degrees around an axis pointing toward you. And we see that even though it's the same pair of 90 degree rotations, it's not in the same orientation at the end of these two different orders. Now remember, non-commutative simply means that the order of operation matters and for angular velocities, it doesn't matter whether we add the velocities omega one plus omega two or omega two and then omega one. However, for simple changes in the angles themselves, doing one rotation around one axis and then another rotation around another axis and then a third rotation around the remaining axis, it depends on the order in which I do them as to what outcome I get, what orientation of the object I get as we saw in the simple demonstration. So for angular changes themselves, order matters, but for changes in angular velocity or angular acceleration, order doesn't matter. Vectors don't care what order you add them or subtract them in. But it's clear that changes in angle have this dependence on order and so vectors cannot simply be used to describe them. Now what does work as a footnote to this conversation? A mathematical object called a matrix is a great way to represent order dependent operations and in fact, rotations themselves in mathematics are represented by matrices of numbers. Those matrices have an order dependence in which they're applied. That's why they work great for this. But vectors simply won't fit the bill and for an introductory physics course like this, we're not gonna talk about matrices. That's usually saved for a more advanced physics course later on down the road. But matrices are the object you seek and if you care more about this, I encourage you to explore the subject a little bit further. Let's review the key ideas that we have explored in this section of the course. They are as follows. We have begun a transition from thinking linearly to thinking rotationally. Now no longer are we constrained to situations where motion and X and motion and Y don't affect each other. Now they do and we don't have to tax our brains trying to rework the old linear equations of motion into the situation. We can develop rotational concepts like angular displacements, angular speeds, angular accelerations that will describe these situations more compactly but let us recover the old mathematical frameworks like equations of motion that we had for linear velocity and linear acceleration. And as you can see, that means we've begun to adapt our ideas about linear motion to the problem of rotational motion. We've explored the mathematical relationships between rotational quantities and we've considered the slightly more advanced topic about how to use vectors to describe rotational quantities and we've seen that angular velocities can be described by vectors, angular accelerations can also be described by vectors but angular displacements themselves cannot be, they require a different kind of mathematical operation that's beyond the scope of this course, the matrix. So using this information, we begin to develop a physical toolkit for describing motion where motion in X, Y and Z gets all tangled up in one another and we have to think about higher level concepts in order to describe that motion but we'll see we can recover our old familiar notions about forces and energy from this new scenario.