 Things moving in circles power the world around us, from the whirring of electric generators to the gentle swing of wind turbine blades in the breeze. Things moving in circles can also be a sort of delight to us. Think of a ferris wheel moving around at an amusement park. And of course, things moving in circles are also a source of incredible beauty and regularity in the cosmos, the moon orbiting the Earth every 28 days. The moon is sort of a cosmic tennis ball tethered to us by some unseen influence whirring around the Earth. And we will begin to investigate this kind of motion here from the simplest examples through the mathematics and into the consequences of circular motion and ceasing circular motion. Let's begin to explore the mathematical description of circular motion so that we can come to a deeper understanding of the interplay between speed and velocity and position, time and acceleration. The key ideas that we're going to encounter in this section of the course are the following. We're going to come to recognize that some motion occurs not along straight lines, but rather along circular paths that require at least two dimensions to describe. We've already seen some examples of this. We will consider this kind of motion when the object in circular motion has a constant speed along the circular path. So the object is moving on some kind of circular shape, and if you were to ask at any moment what is its speed, you would be told its speed is the same as the last time you asked. Now as you'll see, something is changing and we're going to come to understand that circular motion is a kind of motion where the speed may remain the same, at least in uniform circular motion, but where the direction of the velocity, not its magnitude, but its direction is constantly changing. We're going to come to understand the nature of the kind of acceleration that results in this form of motion and its magnitude and its direction and how that relates to the magnitude and direction of the velocity itself. Let's begin by thinking about circular motion and how you can make it happen. Again, we've seen examples of this, but let's take a really simple example. Take a coin. I've depicted a ridiculously large coin in the picture on the right here. Put it flat on a nice level even surface. Now give it a push. What kind of motion does it execute? So let's go ahead and let's do this. Go ahead and try that over and over. The coin may not exactly go the way you flick your finger. Your finger may not be as straight and flicking as you think it is. It may veer off to the left. It may veer off to the right. It might be going straight ahead the way you thought you were flicking your finger in the first place. But when it starts moving, it moves along a straight line until the force of friction, that is the contact that it makes with the surface you push it on, stops it. Now consider this. Can you make that coin go in a circular path just by pushing on it once? Go ahead and try that over and over again. Just push on it once and let it go. Does it go in a circle? You should find if your surface is level, so gravity doesn't get to play a role in this, that you just can't do it. And really what you're going to find out by trial and error is that if you want to make an object go in a circle, you're going to have to provide some kind of force or influence that constantly changes the direction of the object's motion, even as it tries to move along a straight line. These simple experiments reveal something about the character of motion in the world, that objects in motion tend to remain going in the motion along the direction they were traveling unless they were acted upon by some other influence that alters that direction or alters the speed of the motion. The coin slows down because of friction, but its direction of motion is entirely determined by the way you flick it with your finger and then your finger stops making contact and your coin will continue on the path determined by the angle of the flick and so forth, but you just can't make it go in a circle unless you constantly influence the coin in some way. And that's an interesting observation about the natural world. Why does the Moon continue to circle the Earth even though nothing seems to be touching it? These are questions that we will grapple with as we go forward in the course, certainly the 28-day orbit of the Moon around the Earth must be accomplished by some influence, some force that keeps the Moon, even if it's moving in a constant speed, forever changing its direction so that it remains in a circular path around our own planet. So let's begin to think about circular motion and to think about and describe and to come to understand circular motion, we really have to combine everything that we've encountered in this course so far in one synthesis exercise. This is why this is such an interesting topic. It forces you to combine position and time and velocity and acceleration and vectors and even calculus into one synthesis exercise to come to grasp what seems like such a simple thing. It's really actually a little bit complicated under the hood, but I promise you that you have the toolkit at this point to pick apart the how of what's going on in circular motion. So let's consider an object that is moving in circles. For instance, a ball tied to a string swung above your head in a plane that is parallel to the ground above your head. I've depicted that in the right-hand graph. So I've got a coordinate system I've made up. It's as if we're looking down on the ball being spun over your head so we can see the perfect circular path that the ball is making over some period of time. Time is not represented here. We are merely getting the x-coordinate and the y-coordinate in this plane. And the blue circular arrow that closes on itself is a depiction of the direction of motion, which in this case is counterclockwise. It's going from the right hemisphere to the left hemisphere and back in this plane. And it's centered at 3. And I'll come back to the point of the center of motion in a little bit. It's centered at 3 along the y-axis, centered at 0 along x. Now let's consider a single point along the motion curve. For instance, this might be the ball that's moving in the circular path. The ball is effectively only at one place. We can think of it as a point. The location and space, that point is constantly changing as it moves around the circle. So we could freeze the ball at some instant of time and just consider a very tiny slice of time after that and consider the motion of the ball along the circular path in that tiny slice. What would that small motion reveal? What would the velocity of the ball be doing? So think about that for just a moment. If we froze the ball at this moment at just a slice of time, a microsecond, a picosecond of time, drift so we could watch the movie of the ball moving just edge forward slightly, what would that tell us about its direction of motion at the moment ago where we froze the motion? And what we would learn by thinking about this, by playing with this kind of at the experimental level would be that if we were to draw the velocity vector of the ball in that moment, that instantaneous velocity vector, it would be a arrow that is tangent to the circular motion path. And I would hope that that would be a bit of a reminder to something that came up earlier when we were talking about motion paths and derivatives, calculus. The motion, the velocity vector, is the tangent line to the curve. That's what the derivative is. The derivative of x with respect to t tells you the direction of the tangent line at that point x, at that moment t. So it tells you how you're going to get to the next point in the curve. It's pointing in the sense of motion, and that's what I've depicted over here on the right in the graph. So we have the ball represented in red, we have the instantaneous velocity vector whose direction I've indicated with a unit vector v hat, I don't have to draw the whole vector, and it points tangent to the circle at the point where the ball was when we froze the motion. But if we let time lapse forward now, sometime later the ball has moved, it's still moving at the same speed, maybe it was moving at a quarter of a meter per second or a tenth of a meter per second, something like that, that hasn't changed. But something has changed, and the something that has changed is the direction. If we now freeze the ball at the top of this circular path, the ball, if we again let a little bit of time move and think about, okay, where does the ball move? Well, it moves not from down to up as it did on the right-hand side of the curve, but at the top of the curve it moves from right to left, just a little bit. And so that gives us a sense of the velocity vector at that point, and it points to the left, indicating that this is the direction that the ball is going at that moment. It's not pointing up, it's pointing left. And we can consider again the ball at a later point, some other time t, down here for instance. And in here, again, the velocity vector is tangent to the motion path, and it would point in the sense that I have drawn it here. So while the speed may not be changing, the direction of this velocity vector certainly is. Something is changing. So let's take stock of what we have observed. An object in uniform circular motion whose speed is not changing, nonetheless experiences a change in direction. There is a change. The change, if we go back and take a look at that graph, what's the tip of the arrow that represents the velocity vector at any moment doing? What is that tip of the arrow doing? It's tilting toward the center in each moment of time, and it's this tilting toward the center as if something were constantly nudging the vector, hey, move back toward the center of the circular path. That's what keeps it on the circle. And this kind of acceleration is a specific kind of acceleration, given its own name, centripetal acceleration. And centripetal simply means center seeking. So whatever it is, whatever influence, whatever force, keeps that motion on a circular path, it does so in a way that makes the nose of the velocity vector seek the center of that circular motion. We would like to describe this because we would like to make concrete predictions about the velocity, its direction, acceleration, and how those things relate to each other. And we can find the equation that does this that describes centripetal acceleration by considering a very tiny moment in the motion of this circular path and an infinitesimal change in that motion over an equally infinitesimal moment of time. You've heard this before. This is calculus. So we've got it all here. We've got space. We've got time. We've got speed. We've got velocity with the speed, which is the magnitude and the direction. So we've got vectors. And now we're going to employ the toolkit of calculus to come to a deeper mathematical understanding to move at a constant speed in a circular path. And to do this, we have to keep in mind the definition of acceleration. And by that, of course, I mean instantaneous acceleration. Although I do not say that when I say acceleration instead of average acceleration, I always mean instantaneous. And the equation for that, vector A, is given by the average velocity in some extremely small piece of time, sending the slice of time over which you calculate the average to zero. And this is defined as the first derivative with respect to time of the velocity vector. And you can go ahead and use the full limit formula for this. Of course, this is the correct way to mathematically calculate a derivative. But if you're comfortable at this point with the fact that this beast can be summarized notationally with this seemingly innocent-looking little operator, dv dt, then you should feel free to go ahead and look up the first derivative of anything that we encounter. But as you'll see here when we get to the point of actually assessing what the derivatives are, we already have those definitions in our pocket and we don't have to really do any hard work looking them up. Let's get to that. So let's go back to our picture here. We've got our circular motion. That motion path is always going to be represented by the blue circling arrow that closes on itself. We've got the object, the ball. It's going to be the point that we're considering the motion at. And I've selected this point in the lower left quadrant of the circular motion. The center of the motion is somewhere right about here. And notice that this is displaced from the origin of the coordinate system. I did this on purpose to move you a little bit outside your comfort zone at this point. You can put the circle at the center of the coordinate system, but it turns out it doesn't matter. And it's assessment of what's going on in motion as you will come to learn more and more has aspects that are absolutely invariant to your choice of coordinate system origin and things like that. It's just that some calculations are easier under certain circumstances than others. But as you'll see here, this one's not so bad. However, it is a bit of a tour de force of everything we've seen so far. So let's get moving. Now that object, that ball, is associated with it at this moment on the path. And it's got components. It's got a component that is horizontal. So that is that points only along the x-axis. And we can label that vx vector. So here's v vector, the full velocity vector. It's tangent to the motion curve. It's horizontal component. It's x component, of course, always points along the x-axis. And it's y component, which in this case points down because the vector is moving to the right down at this moment will always point along the y-axis and only along the y-axis. Now, as the ball moves through the circle, of course, it's x component length and it's y component length are ever changing. This is what nudges the nose of the velocity vector back toward the center of the circle and keeps the ball on the circular motion path. And we can represent this, of course, mathematically using the equations for components of vectors we've written before. The velocity vector is equal to its x component plus its y component, and these are each vectors. And this is the simplest form we could write any vector in that is two-dimensional. Now, I can further break the components down into a piece representing their magnitude, although there may be a plus or a minus sign hidden in here. And as you'll see in this exercise, I've decided to go ahead and let the minus sign kind of leak in via the vx and vy. And we've got the unit vector that represents being on the x-axis and pointing along that direction, being on the y-axis and pointing along that direction, j hat. So that's it. That's our equation for the velocity. This is an equation we could write down for anything that is moving. It doesn't tell us very much right now, but as you'll see, with a little trigonometry, it will. Now, we can also recognize that the location in space relative to the center of the motion, which is the sort of key stable point in all of this, this point is not moving. The ball might be moving, but in an ideal situation, the center of motion is fixed in space. It has some coordinates. I'm going to think of the center of motion as my origin for this problem. So relative to that point, the center of motion, the ball has an x-coordinate, which is just a distance along the x-axis from the vertical line that intersects the center to where the ball is, so that's labeled here x. And it's got a y-coordinate. And so to get from the center of the circle to the point on the circle, the motion path of the ball, I have to go down and I have to go left at this moment in time. If I want to go from the center of the circle out here, so think about this as a displacement vector that gets me out to the ball, I have to go down and I have to go left. And just like with any vector system, so here's the radial vector, the line that points from the center of motion to where the ball is on the circular motion path, that's the radius of the circle, I'm going to call that r. And we've just formed a right triangle, which you can do with any vector. Any vector has two components, if it's a two-dimensional vector, an x-component and a y-component, and they always form a right triangle whose hypotenuse is the length of the original vector. So I've got a y-component, I've got an x-component, and I've got a hypotenuse represented by the radius of the circle. And so it must be that this beast can be described by the Pythagorean theorem. So let me go back and add some stuff in here. So I've now put the y-component back. We're going to need that in a moment. As I said, the Pythagorean theorem should give me the relationship at any time between x, y, and r. But there are some other things lurking inside of this distance-right triangle with r, x, and y as its sides. There are angles. So let me go ahead and label those. I'm going to take the circle away to make this a little bit easier to see, otherwise this picture gets a little bit cluttered very fast. So we've got one angle that we always have in a right triangle, and that is the right angle. It's the angle between the y-side and the x-side. There are two other angles in a right triangle, and we don't necessarily know what their values are. We do know some things mathematically about them, but we don't know specifically right now without more information what exactly those angles are. So I'm going to label them. They're variables. I don't know their values. And I'm going to label this angle up here at the apex of the right triangle. He's the way I've drawn it as the Greek letter theta. That's a lower case letter in the Greek alphabet, theta. I'm going to label down here on the lower left. I'm going to label with the lower case Greek letter phi. Again, another letter from the Greek alphabet. So we have a 90 degree, a right angle here. We have a theta up here and phi down here. Now we don't know what theta is and we don't know what phi is, but we can start relating theta and phi to other things we do know. For instance, we know that the sum of all angles in a triangle is 180 degrees. If you prefer radians, that's pi radians. I prefer radians, but I put degrees down here to give you some measure of comfort in case radians are still a little uncomfortable to you. So we know that the sum of all of these three angles must be 180 degrees. Well, what are the angles? Well, one of them is 90 degrees, or pi over 2 radians. One of them is theta and one of them is phi. So I can already write an equation that says 90 degrees plus theta plus phi must equal 180. That has to be true in any triangle. Three interior angles must add up to 180 degrees. And this is awesome because this allows me then to rewrite this equation and relate phi and theta. And if we rewrite this equation in one way, we'll find that phi is equal to 90 degrees minus theta. But we can flip that equation around and also solve for theta in terms of phi. Theta is 90 minus phi. So we've already got two interesting equations out of this and we're going to see why these are useful in a moment. Now continuing on with this right triangle picture, again, I've kept the Pythagorean theorem up here to remind you of the right triangle nature of what's going on here with vector component composition. And I've rewritten these angular relationships here for you as well. Now, remember that since the velocity vector, v down here, is tangent to the circular path, that tells us what the angle is between the radial vector, which points from the center out to the ball and the velocity vector, which points tangent to the circle at that point. That angle must also be 90 degrees. The radius of a circle, the radial line of a circle, is always 90 degrees to the tangent of the point on the circle that it intersects. So if you imagine extending the radial line out through a point on the circle, go ahead and emphasize that point with a pen or pencil on your own hand drawn circle. Draw a tangent line there that only touches the circle at one point. Even if your drawing's not so good, even if that doesn't look like a 90 degree angle, mathematically, geometrically, trigonometrically, that angle is always exactly 90 degrees. And this is a very handy piece of information because we can use this to figure out what the heck the angles are down here that we would need to decompose our vector v into its components, v y and v x. So with that information in mind that this is 90 degrees here, we can begin to label other bits of information. If the angle between r and v is 90 degrees and this piece of the 90 degree angle is taken up by phi, then this remaining piece can be solved for. This piece here is just 90 degrees minus phi. This little angle interior between the horizontal component of the radial vector and v itself is 90 degrees minus phi. And given our relationship up here that 90 degrees minus phi is theta, we can go ahead and replace this with theta. Let's keep going. Note here that because this is an entirely vertical component, v y, and this is an entirely horizontal component, x, the distance along the x axis to go from the center of the circle to the point on the circular motion path, this is another 90 degree angle. And so again, we have a situation where this whole thing is 90 degrees, we've already said this is theta, so this must be 90 minus theta, and we're good. We've now labeled the angle between v y and v. We've actually also labeled the angle between v x and v. I haven't written v x here, v x is the horizontal component, but it's absolutely parallel to the x component of the radial vector, and so it makes the same angle theta with respect to v. And this is great because now we suddenly have a whole bunch of trigonometric information relating sides and angles and hypotenuses within a right triangle. Remember your sines and cosines and how those relate the sides to the hypotenuse of a right triangle. So again, you could realize, oh hey, this is phi, but it might be more convenient to keep this as 90 minus theta. I did this just to remind you that we could use these relations up here again to do some substitutions. So let's put the motion path back. Let's take the radial vector r away, let's take its x component and its y component away. Let's put the components of the velocity vector back on the graph, v x and v y. And I've left theta and phi here labeling the angles between what is now, as you can see, the horizontal component of the velocity vector and phi, or 90 minus theta, which is the angle between the vertical component of the velocity vector and v. Now we know a couple of things. We've just learned the angles between the components of velocity and the full velocity vector. Check that off. But now we also know a couple of other things thanks to trigonometry. We know that the cosine of phi, which is the length of the adjacent side over the hypotenuse, is given by v y over v. And we know that the cosine of theta, which again is given by the adjacent side over the hypotenuse, is equal to v x over v. And we know one more thing. If you go back and review your trigonometric identities or look them up on Wikipedia, which is what I did, you can remind yourself that the cosine of phi, which can be written as the cosine of 90 degrees minus theta, has an identity to the sine function. The cosine of 90 minus any angle is the sine of that angle. So we can write everything up here in terms of theta. We don't need phi and theta. We can just eliminate phi in terms of theta and have some nice, simple trigonometric formulas for v x, v y, v, and sines and cosines. And here we go. We've got that now. So employing that identity, v y will be related to v in the sine of theta, v x will be related to v in the cosine of theta. Now this is where I'm going to start sticking direction information in. I know that v y points down. So I know that it must have a negative sense to it. So I'm going to put a minus sign in front of v times sine theta when I solve for v y. v x points in the positive x direction. So I'm going to leave that as a positive quantity, v cosine theta. Okay, well, this is great. But now we've got a new variable to deal with. Now we've got theta. What the heck's theta doing is a function of a time. I have no idea. And you shouldn't either. So let's see if we can eliminate this. Let's see if we can eliminate theta in terms of something that we've been dealing with more commonly. And those are coordinates, x and y. Now remember your right triangle. If you go back and take a look at the r, x, y right triangle, that's also got theta and phi in it. And that means it's also got sines and cosines that relate y and x to r. And if you take a look at the directional information, you know, that to get to from the center of motion of the circle to the point where the ball is on the circular motion path, you have to go down and left. Or negative y direction, negative x direction. You would find that the components of the radial vector, r vector, would be negative r cosine theta and negative r sine theta. Which then allows us to write the following. That cosine of theta is negative y over r and sine of theta is negative x over r. And bam, I can eliminate theta entirely from my vy and vx equations in terms of y, x and r coordinates. That seems a lot nicer. So let's do that. So now, if I do that substitution, I have vy equals two minus signs cancel out because they multiply each other. Positive vy over r and then vx is negative vx over r. Here I pick up a net minus sign in this component. Alright, so finally in terms of x and y and r and the speed v, we can finally write the velocity in terms of all of these things. And here's the velocity vector written in terms of x, y, v and r. And as you can see, I've got this formula here. So I've still got my i-hats and my j-hats but now I've picked up y's, x's, r's and there's my overall speed v just sitting there now. So remember this is uniform circular motion. So v never changes. So to get acceleration from velocity, what do we do? We know the speed isn't changing because this is uniform circular motion. So something's changing. Let's figure out what it is. And to figure that out, we have to take the first derivative of velocity with respect to time. And so, again, to remind you, that's going to be written in the following way. I showed you this a few slides ago. And one important thing to understand as we're about to step forward here is something I haven't really emphasized before. And that has to do with the behavior of the time derivative, the behavior of any derivative really. Imagine a generic equation that you're going to take the derivative of and it's got some constants in it and I'm going to label those a and b. And I'm going to do a time derivative. So they are constant in time. They do not depend on time. If I ask you what is the value of a at zero seconds, you might say, well, it's five. And if I ask you, well, great, what's the value of a at ten seconds? You'd say, well, it's five because it's a constant. It doesn't change. Constants are special when it comes to the time derivative. The time derivative answers the following question. How is the thing I am taking the derivative of changing with time, the thing with which I'm taking the derivative? A constant, the answer is it's not. So if you have only a constant, the time derivative of that is zero. It's not changing. And the derivative with respect to time, the first derivative with respect to time, answers simply the question, how is this thing changing with time? And if you only have a constant and you take the derivative of it, the answer is not at all, zero. It's not changing in time. If you have a constant multiplying something that does change in time, well, you still have something that's changing in time, but that constant isn't, and you can move the constant out of the way of the derivative. It's one of the special rules that you will work up to in first semester calculus. You'll prove it, or at least you'll show how to derive it. I'm just going to use it. So I've written that kind of out down here. If I have this fancy formula for the derivative, the limit is delta t goes to zero of the function. Well, the function, we have to do the change in the function over the change in time. And let's say the function now, part of it that actually does change with time, I'm writing as f of t. But the constants a and b, which are dividing one another out in front, they're just multiplying f of t. And so the rule is that if you have a bunch of constants doing stuff that multiply the thing that actually changes with time, you are free to take those constants and move them out to the left of the derivative operation. So that's what I've done over here. I've taken a over b, a over b is still a constant, a is a constant, b is a constant, the ratio doesn't change. So I've taken this liberty, this little mathematical freedom that I have, and I've moved the multiplication activity of a over b out in front of the whole derivative. So there is no dependence on the order of multiplication, whether you do the multiplication, then the derivative, or do the derivative, and then the multiplication when you have constants that are doing the multiplication. There is a penalty if the thing that's multiplying another thing is time dependent. If they're both time dependent, you have to be careful. But a and b are constants, as I defined them, and it doesn't matter. So I've moved a over b out in front, now I just have the limit of delta f of t over delta t, and this is just the derivative of f of t with respect to time. So this is the final formula that I'm going to arrive at here. You just move the action of the multiplication of these constants in time to later. You do the derivative first, that's much easier. Now you've simplified the function. You don't have to worry about these stupid constants anymore. And then once you've got done with the derivative, you can multiply by over b. So that's just a little mathematical aside. I didn't prove anything. You would learn that in first semester calculus, but we're going to use it. So let's write out the equation that represents the acceleration. We're going to see one more thing that happens with time derivatives here. One more cool mathematical feature of time derivatives that you would learn in first semester calculus. We're just going to use it here. So we're going to take the time derivative of the velocity vector that we've written down to get the acceleration. Acceleration is the first derivative with respect to time of the velocity vector. That's the definition of acceleration. Well, we've got to write out v vector. So here I've got the derivative operation and it's acting on this thing in big parentheses. And that thing is the velocity vector from a couple of slides ago. Negative v y over r i hat plus v x over r j hat. Going one step further, how can I simplify this? Well, the derivative acting on a sum like this can be distributed. So I can have the derivative of this first thing plus the derivative of this second thing. That's another feature of the derivative. It can be distributed kind of like multiplying this whole thing in parentheses by a constant. You can distribute the multiplication to each piece of the sum. You can distribute the derivative operation to each piece. Of course, you still have to act accordingly and responsibility on each piece with the derivative. So I'm going to write it out just the next step. So I can do the derivative of the first piece plus the derivative of the second piece. Okay. Before we go further with this, before we start trying to pull constants out of this and then finally figure out what we're taking the actual derivative of, let's answer a question. Let's take stock of things. What in these big parentheses is constant in time and what is not? Okay. So I want you to think about this. Of course, you can kind of already see the answers here, but I just want you to think about this. If you let time step forward a tiny bit from the picture that we've been looking at, will a specific quantity change? Will x change? Will y change? Will r change? Will v change? Will i hat change? Will j hat change? Which of these are constant? Which of them can you confidently say no? I know this thing will not change or confidently say yes, I know that this thing will change. Think about that a moment. Come up with your own answers and then advance the video. Well, the constants are the radius of the circle, the speed of the object, v, i hat and j hat. Why those four? Well, this is uniform circular motion. The radius of the circular path is fixed. It never changes. So even if the ball moves along the circumference of the circular path, r hasn't changed. So r is constant. The speed never changes because by definition this is uniform circular motion. So v never changes either. That's the definition of uniform motion. The unit vectors i hat and j hat, they always point in their assigned directions. i hat always points along the x direction, j hat points along the y direction. They never deviate from that. So you might have been tempted to think ah well i hat and j hat have to change when the ball moves, but what's really changing are its coordinates. Not the definition of the unit vectors. The unit vectors constantly and always point along their assigned axes, x and y for i hat and j hat respectively. So all four of these do not change in this problem in time. But what is variable? What varies with time? Well certainly if you let the ball move along the circumference of the circular path by letting time slip a bit, its x coordinate is different and its y coordinate is different. It is not in the same x and y position it was in a moment ago. You have to update those numbers. So those things vary in time and if we're going to have a time dependent piece of this above equation, y is one of them and x is the other. v, r, i hat and j hat do not change. So they are our a's and b's from the previous slide. v, i hat, j hat and r those are our constants of time. We can pull them out in front of the derivative operation and multiply them later after we've taken the derivative. So let's go ahead and do that. I have rewritten my acceleration equation with the constants pulled out in front. So I've got negative v over r, i hat those are all constant. Now we've got this time derivative of y with respect to t. And then I've got plus v over r, j hat and now I've got this time derivative of x. So now we have to answer the following question. Can we go further with this? What are these time derivatives? What physically is the time derivative of the x position? And what does it mean physically to take the time derivative and hopefully now with some experience relating speed, velocity, position, time all this stuff to one another you have a fairly ready notion of what this would be. Why these are the x and y components of the velocity vector. dx dt is the velocity in the x direction and dy dt is the velocity in the y direction. And together these components form the full velocity vector v. So if we substitute that into the acceleration equation now acceleration is equal to negative v over r, i hat, v, y plus v over r, j hat, v, x. It's interesting to note that because of the way that the trigonometric functions relate to r, x and y that for the acceleration the y component of the velocity plays a role in the x component and the x component of velocity plays a component in the y component of acceleration. Just an interesting observation and that has implications physically for what happens next. So then we can take one step further. We've got this v over r that multiplies both terms. Let's pull it out in front and group the components together. So we've got v over r out in front and then we've got negative v, y, i hat plus v, x, j hat. And I'm not a big fan of having v over r, but I'm going to write them back again in terms of these trigonometric relations. I'm going to substitute for v, y and v, x in terms of the speed v and cosine and sine of theta. And the reason I do that is that I can then pull an additional v that multiplies all of this stuff inside the parentheses out in front one more time. Okay, well we have arrived essentially at an answer. The question was how do we describe this intripetal acceleration that accompanies this velocity whose direction is constantly being nudged toward the center of motion? Well, that was our velocity vector here. We had velocity vector v is v sine theta i hat minus v cosine theta j hat. And this is the acceleration that now accompanies graphically the components I've shown here. So the intripetal acceleration is v over r times the quantity v cosine theta i hat plus v sine theta j hat. Notice the sines and cosines have flipped. The cosine now goes with the x component of the acceleration and the sine goes with the y component of the acceleration. Note also that while the velocity for this moment in time points down and to the right the acceleration vector which you can write out points up and to the right. It points toward the center of the circle. It is truly center seeking. The acceleration which is a change of direction but not the magnitude of the velocity, not the speed is something that points directly at the center of the circular motion path. And this explains why it is that this object remains on this path. The acceleration vector is a vector that is at always a right angle to the velocity. It makes a 90 degree angle with respect to the velocity at any point on the circle and it dutifully points in toward the center of the circle. Calculus, vectors, position, time, velocity these have all come together to tell us the answer to the question. Indeed centripetal acceleration is truly center seeking. Now it's got components of course it's got a horizontal component which happens to coincide with vx in this case but it's got a vertical component that points exactly in the opposite direction of vy and that's what flips the direction of the acceleration toward the center of the motion. It's the sum of these two things here that net give us the acceleration that points toward the center. We've actually got one more thing we can do here because the components of any vector form the sides of a right triangle and the hypotenuse of that right triangle is the length of the vector itself. We have centripetal acceleration and it must be true that its components squared and summed must be equal to the square of the length of the acceleration vector itself. Let's go ahead and write that out. We've got the x component which is v over r, v cosine theta and we've got the y component which is v over r, v sine theta and if I square the x component and I square the y component and I've just written that out here and then I sum those together of this centripetal acceleration I get this equation and then I go one step further and I pull out like terms and multiply that times the stuff that I still can't simplify which is cosine squared plus sine squared. Think back to your trigonometry. This is one of the beautiful identities in trigonometry. The cosine squared of an angle plus the sine squared of an angle is always equal to 1. It's one of the nice relationships between sine and cosine. Square the cosine of the angle, square the sine of the angle, Adam, you always get 1. I can substitute this thing in parentheses with the number 1 and that leaves me with just v to the fourth over r squared times 1 which is just v to the fourth over r squared and if I take the square root of this I can get back to the centripetal acceleration and I find out that the magnitude of the centripetal acceleration is simply given by the speed squared over the r, the radius of the circular path. That's a nice relationship. If I know v and I know r I can get a centripetal for free. If I know a centripetal and I know r I can figure out the speed for free. This is an awesome relationship. It allows us to predict so many things about an object that's going in circular motion around some point. So let's take stock of everything that we've learned about uniform circular motion. An object maintaining a constant speed while moving in a circle of fixed radius uniform circular motion. Now, such objects are accelerating but it's not the magnitude of the velocity that is the speed that's changing. Rather, it's the direction of the velocity that is changing. So the speed remains constant but the direction changes over and over and over again as some influence nudges the velocity vector toward the center of motion, center seeking or centripetal acceleration. That's how we describe that change and the magnitude of that centripetal acceleration is given by this deceptively simple looking equation v squared over r. The centripetal acceleration always points in toward the center of the motion, that is the center of the circle, the middle of the circle. And finally let's look at one more aspect of this and that is what happens when the influence maintaining an object in circular motion stops acting. It is important to note that if that happens for whatever force or influence or action that's keeping the object in the circular motion ceases to act then at the very moment that the force stops, this centripetal acceleration also stops. What does that mean for the motion of the object? Does the object keep whirling in a circle? Well, that's what people thought at one time if you do the experiment and we're going to do the experiment in a moment, that is not what happens. In fact what does happen observationally in the natural world is that the object suddenly freed from its influence that keeps it going in a circle will simply continue along a path at least initially that's given by whatever its velocity vector v is at the moment it's released. So if you think about again the whirling ball overhead and you're looking down on it and you're keeping the ball whirling over your head because you've got it tied to a string that's affixed to your palm of your hand and gripped by your fingers, what if you cut the string? Well, if you cut the string at a particular moment remember the velocity vector is pointing tangent to the circle, it's pointing at a right angle to the string which is the radius of your circle. So when you cut the string the ball should continue roughly straight along the trajectory aimed on, tangent to the circle 90 degree angle to your string at that moment. It will cease circular motion and it will continue on that path unless acted upon by another force. Now when we do this experiment as you'll see in a moment, we can have a ball on a string, we can arrange a situation where the string can be cut so that it is no longer able to influence the motion of the ball. Unfortunately on the surface of the earth there's always at least one other force acting and that is gravity. So yes, the ball's trajectory after that will not be perfectly straight, it will fall toward the center of the earth but it will continue in the component that is parallel to the surface of the earth along the straight line that it was aimed on at the moment the string was cut. So let's contrive this experiment and let's observe the results. I have here a tennis ball which is affixed to a very thin piece of string. A piece of string that could be easily cut by scissors or some other sharp blade. What I'm going to do is I'm going to begin to move the ball over my head in a circular motion simply by whirling the ball and tugging on the string a little bit to keep the string pulling the ball back. Now I have a razor blade on a stand pointing up into the air such that when the ball swings past my right shoulder it can cut the string. You can see here it really is a blade, it's clamped in and it is pretty sharp. I begin by swinging the ball in such a way that the string is above the blade and once I'm confident that I've got this moving in a pretty good plane above my head I'm going to begin to slowly lower the string. Lower my arm a little bit so that the string swings right through the blade. Here we go, we're approaching the moment watch. Now let's take a look at that in slow motion one more time and you can see the string is swinging around I've now lowered it, the blade is going to make contact with the string and then sure enough it cuts the string and the ball doesn't continue moving in a circle, it continues straight along the velocity vector it had at the moment that the blade separated the string into two pieces. I can even show you here that the string has in fact been cut into two pieces by holding the now frayed and damaged end up to the camera. So this is an example of having something that's moving in circular motion under an influence, in this case a thin piece of string but then suddenly removing the influence and observing what happens to the ball. We can do this again with a view from the side whirling a ball from the side. Of course in this case it'll be under the influence of gravity and projectile motion once it leaves but watch carefully as I'm whirling the tennis ball on the string watch what happens when I let go of it and we'll try to look at this very slowly so that you can see the direction that the tennis ball is moving on just as I release and it continues along that direction gravity of course begins to accelerate it down toward the center of the earth as soon as the string is no longer influencing the ball and so it will move on a projectile motion curve, a parabola after it's released but nonetheless for that moment in time just as it's freed it's only moving in a straight line determined by its velocity vector just before I release the string. So we can see here the transition from circular motion to linear motion or projectile motion by removing the influence that was once holding the object in circular motion. Let's do a little experiment using some simple graphical tools that would be available on any modern computer. I'm going to use a video playing program to snap frames from a video and save them to disk as images I'm going to use a power point like program to load those images and then draw arrows on it and I'm going to use those arrows to make predictions about what I would expect for instance the motion of the tennis ball to do when I release the ball from my hand by letting go of the string so to begin here is the still frame of me at the very moment that my hand really stops making contact with the string and now the ball is no longer held by the influence of my hand along the string in circular motion. It is this moment that the velocity vector of the tennis ball will determine its ultimate fate. What's neat about this is that even though I've extracted a still frame from a slow motion video, the tennis ball appears as a green blur and the length of the blur is a little bit like taking the derivative of the position of the ball at a very tiny moment in time. You know, perhaps one hundredth or one two hundredth of a second. This blurring motion is a stretched line in space that indicates at sort of that instant of time that one two hundredth or so of a second in time what the sense of motion of the ball was where its velocity vector was pointing at the moment I released it and we can use a PowerPoint like program. I'm using LibreOffice Impress which is a free open source presentation creating software framework and I'm going to take the image and overlay a little red arrow after darkening the image so that we can better see the contrast between the ball and the wall behind it. I'm going to overlay a little red arrow whose tail begins at the bottom of the blurring tennis ball and ends at the top of the blurring tennis ball and kind of splits this blur in half so that it roughly points along the axis of travel of the tennis ball. To extend the red arrow out into space right up to roughly the ceiling in this frame of video this is my prediction. I've made a prediction using the sense of motion of the tennis ball at the moment it's released and knowing that once an object is released from circular motion because the influence no longer acts upon it that it will move along a straight line determined by its last velocity vector while it was in circular motion I can make a prediction that this red line indicates the approximate direction of travel of the ball. Certainly this tells me it won't go in a circle anymore. Now of course we know that the tennis ball is in projectile motion once it is released from the prison of being attached to the string attached to my hand and so gravity is going to nudge the ball down a little bit and in fact if we skip to a later frame where the tennis ball is about to strike the ceiling indeed we see the tennis ball is a little bit below where the prediction was but that's okay we know gravity is pulling the ball down toward the floor the whole time it's traveling even though it's moving quite fast nonetheless this is still a pretty good prediction even ignoring gravity which is what I did to draw this arrow we did okay and the ball is moving fast enough that the acceleration of gravity over the short period of time the ball is flying free through the air is not that great. We can even run the video and overlay the arrow and watch as the path first begins under the arrow and gently as gravity nudges the ball down in the vertical direction it deviates a little bit from the red arrow but nonetheless this is a pretty good prediction considering the fact that I neglected gravity and we know about projectile motion we know that this ball is supposed to fall under the influence of gravity so it's not a surprise that it winds up a little bit below the head of the arrow pointing at the ceiling. So we have here a real laboratory with just video and still images and some presentation and drawing software that we can use to play around predictions of the natural world and see whether our notions of the gross outcomes of a physical situation will be correct and that's pretty powerful. Let's review the key ideas that we have seen in this section of the course. We've seen that some motion occurs not along straight lines but rather along circular paths that require at least two dimensions in order to be described. We've focused on an example with an x and a y component and considered the motion when the object in circular motion has a constant speed along the circular path. We've come to understand the nature of the acceleration. It is a center seeking acceleration that doesn't change the magnitude of the velocity but does constantly alter its direction and so we have been able to come to a mathematical description of this kind of acceleration, this centripetal acceleration using the tools of vectors, calculus, space, time and velocity.