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About this Lesson
- Type: Video Tutorial
- Length: 12:06
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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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Supposing you have a rigid body, which has got a fixed rotation axis, like a bicycle wheel, and I apply an external force to it. So I start pushing down like so. You know what happens, it starts to accelerate in an angular sense. It's interesting to think about Newton's Second Law, F equals ma. I'm applying a force, how come the whole wheel doesn't start accelerating? Well, it's because it's mounted on some rigid axis, and so there must be some other force acting right here at the axis, which is causing the acceleration of the center of mass to be zero. But still, I'm applying an external force and its causing angular acceleration. And the way that it's causing an acceleration is because I'm applying a torque. I'm not applying a force straight towards the center, I'm applying a force that generates a torque. I'm generating a torque - remember is equal to r perpendicular times F, it's force times moment arm. And if you have a torque, you know you're going to get an angular acceleration. I know that the harder I push, the more angular acceleration I'm going to get, and the question is how exactly are these related? Is torque equal to ? Is it proportional?
I'm thinking back to F equals ma. Let me just look at magnitudes here. I know that when you apply a force to a mass, it will accelerate and F equals ma. And these formulas, they look so similar. Torque is like the angular analog of force. is the angular analog of a. So what belongs here? You might guess m. Torque equals m, and that's wrong, because if you increase the mass of this wheel by putting lots of mass right at the hub, it's just as easy to turn. It's not so easy to move sideways, but it's just as easy to turn. If I put a mass right out here at the edge, that's going to have a much bigger effect than if I put the mass towards the center. In fact, the correct angular analog of mass is I, the moment of inertia. Remember moment of inertia, you can think of it as the sum of the mass of each individual chunk times the r^2 of those little individual chunks. The moment of inertia is called moment of inertia for good reason. Remember mass was inertia, it told you how resistive something is to moving, and moment of inertia is telling you the resistance to rotating. The bigger the moment of inertia, the less angular acceleration you get for a given torque.
This is the correct formula; it's Newton's Second Law in the angular version. Torque is really a vector and so is , so I could put arrows over everything if I wanted to. Not everything - moment of inertia is not a vector. And I usually don't bother, because I'm dealing right now with an object with a fixed rotation axis when I write down this equation. And if the axis is fixed, I can think about sort of like a one-dimensional problem. It's either clockwise or counterclockwise. The torque is either clockwise or counterclockwise. So I can just think of these as being plus or minus rather than vectors. At some point, we'll want to talk about what are the consequences of the vector nature of rotational motion, but not for now.
So where does this formula come from? It sure looks analogous. Let me not prove it to you in complete generality, but let's think of my favorite example, which is the world's simplest complex object. So here's my rotation axis, it's fixed, and here's my little mass m. And that m is sort of connected by a rigid rod, so all it can do is run around in a circle. I could apply any force I like, but it's a rigid object, it can only go in a circle. So supposing that I supply some funky old force, like this. That's my external force and it's at a radius r, and I ask myself, "What's going to happen?" Well, this force has two components; it's got a tangential component and a nontangential component, a radial component. And this component is pulling on the rod, but it's a rigid object, so there's going to have to be some compensating force at the center and it's not going to do anything. It's this one that does all the interesting stuff when you've got a rigid rotating object with a fixed axis. It's the tangential force that causes some angular acceleration. In fact, F[Tan] is equal to ma[Tan], that's just Newton's Second Law written for this point mass in component form, where I'm breaking up my components. Rather than the usual x and y, I can break them up into any coordinate system I want. And now I say, "Ah, tangential acceleration of a point on a rigid rotating object is rotated to angular acceleration." Tangential acceleration is just radius times . That's one of our connections between linear and angular variables. So tangential force is m times r times .
This is the tangential force, what's the torque that I'm applying? So now we've got to stop and think. Torque is defined to be r times F times sin . And you can write that in various ways, and one way is to write it as F[Tan] times r. In this case, this is F[Tan] is the perpendicular component of F. It's the component of F perpendicular to r, so F[Tan] times r is the torque. And so, if I took this equation and multiplied it through by r, I would get torque on the left side and I'd have mr^2 on the right side. So I would have torque on the left side and mr^2 on the right side. For this little simple situation, where I just have one mass m, this is the moment of inertia. Moment of inertia is supposed to be the sum of mr^2, but the sum is only over this one chunk, so I really do have I. So this little proof worked for this little simple system, and now what happens if I start adding more m's, m[i]'s? Well, I would have the sum of the forces and I could just sum both sides and I would get the sum of the torques equals to the sum of m[i]r[i]^2 times . is going to be the same for everybody, because it's a rigid body. Everybody has to rotate together, that's my starting assumption here, that I have a rigid body with a fixed axis. And I would get the net torque is equal to I times . I would be the sum of mr^2. That proof is a little bit of a cheat. You have to think a little bit about the fact that there could be internal forces between these objects, because it's a rigid body. And you have to convince yourself that the sums of the torques from those internal forces will cancel out. So I leave that as a slightly tricky exercise for the viewer, but the result is correct. Torque is equal to I, and let's just work a simple example.
Here's a rigid object. Let me apply a force. You could imagine I'm tying a string to it and pulling, or I'm just giving it some fixed force. It's going to start to accelerate. And if I keep applying a fixed force, it's going to go faster and faster. And let's work out the angular acceleration rate. So here's a nice picture of my wheel and I'm applying a force. Let me pull a rope down on this side and let me pull with a constant tension T. So what's going to happen? I need to know something about the wheel. I need to know its moment of inertia, and I need to know the radius of the wheel and I need to know the tension that I'm applying. These are the givens that I have to start with. Newton's Second Law in angular form says torque equals I. I'm looking for . I want to know how rapidly does it start to angularly accelerate? It's got no linear acceleration, because there's this axle that's pushing up and a rope that's pushing down. It's a fixed wheel, but it will spin. And the torque here, what's the torque? Well, torque is R times F times the sin of the angle between them. R is just a just exactly the radius of the wheel. F, in this case, is T, the tension in the rope. The tension in the rope is downwards, and so torque is just R times F and they're 90 degrees apart, so that's sin is 1, so that's the whole torque. And that's equal to I. So that's really all you need. is R times F divided by moment of inertia. So if you know all these numbers - whoops, F is the tension T. And if you're given these numbers, you can immediately calculate . The larger the radius, the larger the angular acceleration, except remember I has some hidden radial dependence inside of it. Let's write that down. I'm afraid I can't write it down in general, because different objects have different moment of inertia about their center. If this was just a uniform disk, the moment of inertia would be MR^2. And if it was a hoop, like this bicycle wheel is roughly a hoop, it's got a fairly thin, heavy outside and almost no mass anywhere on the inside, so this would be MR^2. The formula would depend on the details of what exactly my circular object was like. So you can't write down any one of these formulas as being correct, it just depends on the details of the moment of inertia. In any case, what you can see is that there's an R on this side and an R^2 on this side, so, in fact, generally speaking, the angular acceleration will decrease for larger objects. And that makes some sense to me. Larger objects tend to have more moment of inertia. They tend to be more resistive to acceleration; they have more inertia. Similarly, the more massive the wheel, the smaller is going to be, the more resistance they have to rotating.
So the central formula for dynamics of circular motion is torque, and let me now write it as [net] equals I. You don't have to have just one torque in a problem. For instance, in this problem what would happen if I was pulling on both sides? So I'm pulling on the backside and I'm pulling on the front side with equal and opposite tensions. Is the net torque equal to 2Tr? The answer is certainly not. This torque is in the clockwise direction. This torque is in the counterclockwise direction. r F is into the page on this side. r F is out of the page, so whether you think about it as plus and minus or as vectors, there's zero net torque if I'm pulling with equal tensions. And so, you've got to think about the net torque. Clearly, if you pull on both sides, you won't get any angular acceleration.
If you want to calculate angular accelerations, you use equals I. You need to compute I for your specific problem. Working out torque is always a bit of a pain. You've got to look at the geometry and think about it, but the formula is r times F times the sin of the angle between the, or if you prefer r F, and you can figure out the motion. This is dynamics of circular motion.
The Physics of Extended Objects
The Dynamics of Rotational Motion
Newton's Second Law for Rotational Motion Page [2 of 2]
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