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About this Lesson
- Type: Video Tutorial
- Length: 13:14
- Media: Video/mp4
- Use: Watch Online & Download
- Access Period: Unrestricted
- Download: MP4 (iPod compatible)
- Size: 142 MB
- Posted: 07/01/2009
This lesson is part of the following series:
Physics (147 lessons, $198.00)
Physics: Kinematics (18 lessons, $28.71)
Physics: Uniform Circular Motion & Relative Motion (3 lessons, $4.95)
This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and real-world applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for Non-Newtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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Kinematics, when you have constant acceleration, is a relatively straightforward business. You can solve for position, and velocity, and acceleration. It's projectile motion for example. What happens though when you have some very different physics or a different situation than projectile motion? For instance, suppose that you have some object, which is swinging around in a circle with a constant speed. It's a very different story, and the kinematics are going to be very different. And can we describe this?
I'm going to call this uniform circular motion, the word uniform meaning steady. The speed isn't changing. You're going around in a circle with a constant speed, uniform circular motion. It's really the physics of an awful lot of important situations. We could be talking about the moon going around the Earth, or the Earth around the Sun, a satellite around the Earth. It doesn't have to be outer space stuff. We could be talking about this little example. You could be thinking of a car running around on a circular racetrack or even the head of a golf club as you're swinging it. None of these examples are perfectly uniform circular motion. The objects might be speeding up and slowing down just a little bit or the motion might not be a perfect circle. But it's an excellent approximation to an awful lot of important physical situations.
So let's try to explain the kinematics of uniform circular motion. Kinematics means I want to tell you where it is, what's its velocity, how fast is it going, and what's its acceleration. So let's draw a picture. It's uniform circular motion so the picture is pretty easy. We've got a particle or something moving around in a circle. Imagine that it's here. So I want to describe where is it? Well this picture alone has already done that. It's here. It's at an angle theta from some fixed origin or fixed axis, and at a distance R. So that's a perfectly acceptable description of where it is. If you prefer working in a coordinate system (x, y), you put the center of your coordinate system at the center of the circle. That's always going to be the easiest , . Just look at the picture and convince yourself. So if you want, given angle and distance from the center, you can immediately figure out x and y. So the description of where you are is very simple. And then theta will be changing with time, smoothly, because it's uniform circular motion. So you can figure out what x and y are as a function of time if you want.
When people have uniform circular motion, they will often use some words. And just words that are convenient to describe this motion. One of those words is period. It's given the symbol capital T, because it's a time. It's the amount of time it takes to go once around the circle. And people also talk about frequency of this uniform circular motion. It's defined to be 1 over the period. So if you're going around let's say in two seconds, a period is two seconds, then you're going around half way each second. So your frequency would be a half a revolution per second, 1 over the period. And the units of frequency, revolutions per second, are sometimes called hertz, abbreviated Hz with a capital H. Those are just words. And they're useful for describing uniform circular motion.
How about the next step in kinematics, which is velocity. So let's think about that for a second. And the object is running around in a circle. It's here at position vector r initial. And then later it's here at position vector r final. And what's the velocity vector? Well remember velocity vector is always tangent to your path when you're looking at a picture like this. So this velocity vector is always tangent to the circle wherever you are. And what's the magnitude of this velocity vector? That's the speed. That's what you mean by speed, magnitude of velocity vector. It's constant because I'm assuming uniform circular motion. So the speed is not changing anywhere. The length of this arrow and the length of that arrow are the same everywhere. So we've really described the velocity vector as well. It's a little geometry puzzle, if you're given the angle here, to figure out what are the x and y components of the velocity vector. It's not that hard. It's just a little pair of right triangles. We'll almost never use that, so I'm not going to bother. That's the velocity vector. It's everywhere tangent. It's got a constant magnitude.
So the last issue here is acceleration. And you think that's easy. We must be done, because I'm not speeding up. Isn't acceleration the change in speed with time? If I'm not speeding up, isn't my acceleration zero? The answer is no, careful. There is an acceleration in this problem, because acceleration is not as numbers. It's a vector equation. Acceleration is defined to be the change in the velocity vector. And that is not zero. It was pointing this way, and now it's pointing in another direction. Let me slide this velocity vector over so that velocity vector initial and velocity vector final are put with their tails together so that you can really stare at the picture. And you can see that there is clearly a delta V. Even though they are not changing in length, they are changing in direction. And that means that there really is a delta V. And so there really is an acceleration. Delta theta here just represents the change in angle that we've gone through from initial to final. And in the end, of course, I'm going to want to talk about the instantaneous velocity, the instantaneous acceleration. I'm going to take the limit of delta t gets very small, and this triangle gets skinny.
So delta V over delta t, acceleration, what's the magnitude? If you think about it, just think about physics of circular motion. Imagine you're in a car driving around in a tight circle with some speed. And you know there's an acceleration going on. You can tell. And in fact, you know that the faster around the circle you go, there's more acceleration going on. It's just an intuitive sense. So if I was just trying to guess, no formulas no pictures, I might guess that acceleration grows with speed. On the other hand, if I'm going around in a tighter circle with a smaller radius, I would expect that the acceleration is bigger the smaller the radius. So this is a wild guess. And it's wrong. And I can tell it's wrong right away, because the units of velocity are meters per second. The units of radius are meters. That has units of per second. That's not the units of acceleration. Let me tell you what the answer is. We're going to prove it geometrically in just a second, but it's almost, we almost got it right. It's not V over r. It's V^2 over r. And the units work. This is meters squared per second squared upstairs, and meters downstairs. So you do get meters per second squared, which is acceleration. And the acceleration is bigger when you're going around the circle at a faster speed. And it is bigger when you are going around a smaller circle.
So the formula makes some sense to me. And let me see where it comes from. I've got to go back to my picture of this circular motion. And in fact, I immediately want to look at this picture where I've just slid the V initial vector over so they have a common point. So there's a little triangle. You know I sneaked something past you here. And I don't know if you thought about it. This angle is delta theta. And I wrote this angle as delta theta. And it is the same angle delta theta, but you have to convince yourself. It's again a little geometry puzzle. But remember that the velocity vector is tangent or perpendicular to this vector. So you can prove, it's just a little quick geometry puzzle, to convince yourself that that angle delta theta really is the same. That's the angle through which you've changed position. And this is the angle through which your velocity vector has swung.
Let me redraw this picture just so I can stare at delta r, the change in position vector. And I look at these two pictures and I say similar triangles. They are both isosceles, because these are same length. These are same length as each other. They're both speed. These are both radius. So similar triangles means that delta r, the length of this arrow, is to R, the length of that arrow, as delta V, the length of this arrow, is to the speed. When I don't put the arrow sign over it, I'm talking about magnitude. So that's a geometrical argument. Why did I do that? I'm after delta V over delta t for the acceleration. So I can get that from this equation by multiplying through by V and dividing by delta t. And I get . In the limit that delta t gets very small, delta r over delta t really is the instantaneous speed. These, remember, are magnitudes. So delta r over delta t, in the limit of delta t going to zero, just becomes or as I argued. So that's correct and it's the magnitude.
What about the direction? Once again, we go back to this picture and we stare at the direction of the velocity vector. And we can go back to our original picture. This is the delta V vector. So let me just put it back into the picture. And you know I'm going to take the limit that delta t gets really small. So these two arrows are close. But for now I'll just put it sort of at the average point. I'm talking about the average acceleration over this time interval. This is the change in velocity vector, and acceleration is delta V over delta t. It's in the same direction. It doesn't have to have the same length, because you're dividing by delta t. So here's my acceleration vector at this point. It's pointing right toward the center. It's correct. Acceleration is centripetal when you have uniform circular motion. Centripetal is a fancy word for center seeking. If you go around the circle over here, there's my acceleration vector. It's always pointing toward the center. Does that make physical sense? Am I really accelerating toward the center? You bet. Think about what would happen if I wasn't accelerating. If my acceleration was zero at this point, I would continue with no change in velocity. So I'd be traveling with constant speed along this direction. So after a time delta t I'd be over here somewhere. So which way am I accelerating if, instead of being here, I'm over there? I'm accelerating toward the center. It's always true everywhere around. And it's true when you look at instantaneous acceleration. So it's a little bit weird, and you have to think bout it. But it's really true always. You're constantly accelerating toward the center when you're moving with constant speed.
Let me show you a formula that's just handy to use in a lot of uniform circular motion problems, because often I don't tell you the velocity. What I tell you is the radius and the period. And then I ask questions. This is just something you might care about. What's the acceleration, think about a kid's merry go round. Well you know the radius, and you know how many seconds it takes to go around one. So I need the speed for . And speed is just distance traveled--in this case think about going all the way around in a circle once. You travel a total distance of . And you do that in a period T. Distance divided by time, that's going to be the speed. And so that's just a handy formula which you can use to plug in into equations. If you take a kid's merry go round with a radius of two meters, and it's maybe taking three seconds to go around once, that's sort of a reasonable number. Plug in the numbers. Find the velocity and then find . You'll find about 9 meters per second squared. Almost g, and we're used to g down. And when you're on the merry go round, you also have this acceleration toward the center that's almost as big. So kids like that. It feels interesting to be accelerating sideways at such a large rate of acceleration.
So uniform circular motion, we have described position and velocity and now acceleration vector. It's not constant acceleration. Even though the formula has a constant magnitude, the direction is always changing. So you can't use the old constant acceleration formulas. You've got to use these new ones, but they're straightforward to use and incredibly useful, because there are a lot of physical situations where you have an object moving around in a circle with constant speed.
Uniform Circular Motion
Describing Uniform Circular Motion Page [3 of 3]
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