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About this Lesson
- Type: Video Tutorial
- Length: 12:19
- Media: Video/mp4
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- Posted: 07/01/2009
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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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A parachutist in free-fall, it's a clear visual image. To a physicist, it's a lousy use of the word free-fall, because a parachutist who's falling feels lots of friction, air friction. To a physicist, free-fall means motion under the influence only of gravity. So we've talked about free-fall in one dimension. What about free-fall in two dimensions? People use another name for that, because when you're in two dimensions and you fire something up in the air, it'll follow a projectile motion. You can think of launching a rocket. Now while the engines are going, that's not free-fall. There's a rocket force pushing on it. But with an ICBM, once the engines cut off, the rocket will follow projectile motion, free-fall in two dimensions. There are lots of examples. You can toss a ball. You can jump the high jump. You can take a gun and fire it at the cameraman.
Free-fall, projectile motion, can we describe it? Can we understand what's going on? The answer is yes. We're all set. Free-fall means the only force acting on you is gravity. What we know is that under the influence of gravity, all objects accelerate straight down with an acceleration of 9.8 meters per second squared. It's constant acceleration. And we've solved constant acceleration problems in 1D, how about in two dimensions? In two dimensions, you have to ask yourself is there any complication? Does the fact that you're moving sideways and up and down change any of the story? And the wonderful answer is no. We're all set up to solve free-fall in two dimensions.
Here are the equations that we've been using in one dimension . I stuck a little x on all the terms. In one dimension, we didn't need to bother, because there was only one velocity. It was in the x direction. It was understood. Now we're talking about a projectile that's moving through two dimensions. It's got x motion and y motion. These are the equations for the x components. It's the exact same equations as we had before, the same equation for position as a function of time, the same equation for V final in terms of displacement in the x direction. It's wonderful. In fact, it's even better than this. If we're talking about projectile motion, what's a[x]? What does that mean? It's the component of the acceleration of gravity. That's an arrow that points straight down. What's the component of such an arrow in this direction? Nothing, it's pure y. It's got zero components in the x direction. So for free-fall, this term vanishes, . It's nice and simple. And it tells you something. Maybe it's obvious. Maybe it's not. If you throw some object, it's got an initial velocity vector, which might be pointing up at some funny angle. That initial velocity has an x piece and a y piece. And what this equation tells me is that the x piece never changes with time. If you start off with some sideways motion, you will continue to have that sideways motion for all time if you're in free-fall. This equation, well this a term also disappears. And it looks like very simple motion. In fact, that's just essentially uniform motion in the x direction, . This equation, well if acceleration is gone, nothing new. V final equals V initial, we already knew that. So the x motion of projectiles is as simple as can be.
How about the y motion? Well, , it's the same equation as we had in one dimension. This is the y component of the vector equation V vector equals V initial vector plus a vector delta t. So do we know A[y]? Sure, it's constant acceleration of gravity. It's 9.8 meters per second squared down. Now you have to worry about your coordinate system. If you've decided to call down positive, this would be the number +9.8 meters per second squared. If you've decided to call up positive, then this would be a - 9.8 meters per second squared.
So these equations are a little bit more complicated, but it's still constant acceleration. And the point, if I tell you the initial position in velocity, what do I have to tell you when I tell you an initial velocity in 2D? I either have to tell you the length and the angle, the magnitude and the angle. Or equivalently, I tell you V[i] in the x direction and V[i] in the y direction. I tell you the initial conditions. These equations tell you the final, V final x, V final y. We had x final, now we've got y final. We know everything there is to know about the path and motion of a projectile.
Let's talk about some specific examples. Suppose that you were to take two objects and drop one of them. And the other one you give a little sideways whack. So one of them is going to be go flying off exactly horizontally, while the other starts to drop. What's going to happen? Let's just watch. What we've seen is x and y motion can each be thought of independently. Falling objects satisfy the equations of motion in one dimension for uniform constant acceleration. Whether the ball is originally horizontally moving or not, it's got no up or down component of velocity. So both objects start off with zero y component of velocity, and therefore their vertical motion is identical. They both fall with the same acceleration. At any given moment in time, they're at the same height. How about in the x direction? The one that was just dropped had no x motion to start with, so it never gets any. The one that was projected sideways began with some x component, and that just continues. So it marches across uniformly to the left. It's going faster and faster downward, and so it follows a parabolic arc.
Think about a similar example. You're in an airplane cruising along, and you've got some payload and you drop it. You're moving this way and you drop the payload. How is the payload going to move? If you aren't thinking too carefully, this is a common intuition, the airplane continues along. Acceleration is down. Doesn't the object fall straight down? Doesn't the object land directly below where the airplane was when it dropped it? No it does not, because the equation of motion said . X motion has nothing to do with the y motion. If you're in an airplane, you're moving to the right at 150 miles an hour. And you release something. At the moment of release, it was in the airplane. The object was moving along with the airplane at 150 miles an hour. And once you let it go, there is nothing going on in the x direction. There's no acceleration in the x direction. Of course, remember I'm talking about no friction. And so as the airplane moves along, the object that drops is in free-fall. It will get farther and farther below, going faster and faster downward. But its horizontal motion will be the same as the airplane's. And it will always be under the airplane.
If I fire a gun, and I drop a bullet at the same time, which is going to hit the ground first, the object which I dropped, or the bullet that I fired at some huge speed. If I fire horizontally, what we've just seen is kind of counter intuitive, but that's what the equations of constant acceleration tell us. The bullet fired horizontally and the bullet dropped, they both hit the ground at the same time.
Let's look at one kind of numerical example of this. It's a numerical example that's very similar to the examples that we've been talking about. Here's Thelma and Louise heading toward the cliff. And at the moment they reach the cliff edge, they're moving at 50 meters per second. They're moving dead horizontal. I guess that's a bad adjective to use in this case. What's going to happen? Any object, which starts off horizontally, is going to follow this parabolic path. And there are all sorts of quantitative questions, which you might want to answer. A little bit, maybe, sick in this particular context, but how long is it going to take? How much life have they got left? Where are they going to land? How fast is the car going to be going? That's an important one. Are they going to survive the crash? So we can figure all of this out. We've got the equations of motion.
When you first start working problems in two dimensions, it can be a little intimidating, because we've got three equations in the x direction, and three equations in the y direction. That's six equations, and you think how am I going to figure out which equation to use? But it's usually just a question of take a look at what question you're being asked, what you're given, and it'll be clear. For example, how long before they hit the ground? That's a perfectly legitimately question. So think for a second. What information am I given, and what am I asked for? I want time. So I've got to look for an equation that involves time. And "when they hit the ground," those are words for what? It means that y is equal to zero. In order to solve this problem, I need one more piece of information. It depends on how high the cliff is. If it's a one-foot cliff, no sweat. If it's a 500-meter cliff, then clearly the time it takes for them to fall down is going to be a lot longer.
If I've told you the height of the cliff, and I asked you for the time when they hit, I would hunt immediately for the equation that says y final equals y initial, because I know y final and I know y initial, plus V initial yt. What's V initial y? It's not 50. That's V initial x. V initial y is zero. This arrow has no up component and no down component. So that term just disappears. In fact I've got only one term, , and I know what a is. There is only one unknown in that equation. I can solve for time.
And once I know time, if I wanted to know how far they'd go, what would I do? I would go look at the x equations, because I'm asking an x question. If I was really being careful, which I really should be, I would pick a coordinate system. You should always pick a coordinate system when you start one of these problems. If I'm calling y up, and this was zero, and this was 500, y initial would be plus 500. Y final would be zero. And a would be in the y direction, -9.8 meters per second squared, of course it's zero in the x direction. So when I'm looking for the range, in the x direction I would just use x final, that's this, equals x initial plus V initial in the x direction. That's 50 meters per second times time. I already got the time. And I'm all set. Any question I can ask, those equations are all you need.
Investigating Motion in Two Dimensions
A First Look at Projectile Motion Page [2 of 2]
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