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- Type: Video Tutorial
- Length: 11:22
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- Posted: 07/02/2009
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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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Solving rotation dynamics problems is, in many ways, similar to solving regular old dynamics problems. Remember dynamics means F equals ma. Why does something accelerate? Because you apply a force to it. Why does an object with a fixed axis rotate? Why does its rotation rate change with time? Because you're twisting it, you're applying a torque to it. The formula equals I, it's kind of the analog of F equals ma. It looks very similar. Torque is like twisting force; it reminds me of mass, it has to do with inertia and is angular acceleration. I should really fix this formula up. First of all it's [net]. If you've got an object with an axis, you have to worry about all the external torques, so you all up all the torques acting on the body. That gives you the net torque and that will tell you then about the acceleration of the object, the angular acceleration of the object.
When you work examples, you have to think, "Do I need vector signs?" We often worried about vector signs in F equals ma, because these things have directions. They also have directions in equals I, but most of the problems that we're going to be working at first, where we have a fixed axis, it's really like a one-dimensional problem. You either have a torque in one direction or in the other. You can call it a plus or a minus, clockwise or counterclockwise. And so be aware that these are vector quantities, but, in practice, you can just think about the signs alone. But you can't forget about the signs.
Let's just work some examples. The simplest example we've seen before. If you have a wheel and you're tugging on it with a rope, so it starts to accelerate angularly. It's got a fixed axis, so it's going to have a moment of inertia I about that axis and its going to have some radius R, these are givens. And then there's going to be some tension T in the rope and presumably that tension T is pulling this end of the rope down, it's pulling this end of the rope down. equals I should tell us the story. I need to pick a convention for my signs. Let me call this direction positive for angular motion. So is R times F times the sin of the angle between them. Here's it's just R times tension. And it's a clockwise torque, so it's positive, and that's equal to I, so I'm going to get a positive . It's going to accelerate, in this sense, with a very simple formula, RT over I. Simple example, let me just make it a little bit more interesting now.
Let me hang a little mass here. Instead of pulling with a fixed T, let me hang a mass. So let's clear the board and start this problem again. And you look at this and most people's first reaction is, "I don't get it." We've got a mass m sitting down here. That means that there's going to be a gravitational force mg, and doesn't that mean tension is equal to mg? I just solved a problem, plug-in tension equals mg, I'm done. And now you've got to stop and think. That's incorrect. Supposing that the tension in this rope was mg. Look at this mass? Here's a mass m and I'm asking you to imagine what would happen if tension up was equal to mg down? This is the free body diagram for the mass. Let me get everything else out of the picture. If tension was equal to mg, the net force on that mass would be zero. And that doesn't make any sense. I've got this wheel and there's a mass hanging from it. The mass is going to start to accelerate down. It's pulling on the rope, it's going to start accelerating down as the wheel starts to spin. And if T was equal to mg, there would be no net force and it wouldn't accelerate, it would just sit there. So T is not equal to mg. This is, in fact, the right way to start the problem.
Dynamic problems, when we used F equals ma, we always started off drawing a force diagram, and you want to do that again in problems when you have rotational motion. Force diagrams are the way to get started. Here, we have two bodies. There is the wheel and it's got a tension T pulling down on one side with radius R and moment of inertia I. This is one force diagram and this is another force diagram. We have two separate bodies and we should think about each one independently. Let' think about Newton's Second Law in angular form for the wheel. It says [net] equals to I. There's a torque, which is indeed T times R, just like before, and let me call this positive, just like we did before. So there's one equation and it's got the unknown T. I don't know what T is yet and it's got an unknown . S two unknowns, I need another equation. So I think, "Great, let me go down to this free body diagram." This is not rotational motion, this is a mass, which is just moving straight up and down. In fact, it's moving straight down. So let me use Newton's Second Law in the good old-fashioned F equals ma form here. And this should be F[net]. This should be [net], but, of course, it's the only torque acting on this object. Directions, I need to pick a coordinate system. Having chosen clockwise as positive does not determine my coordinate system, so it's up to me. Let me call the down direction positive, it's completely arbitrary, and I just have to be consistent with this. So I've got to add-up all the forces. I've got mg, which is positive, and T is in the opposite direction, so mg minus tension, that's the net force. So mg minus T, and what's that equal to? That's the net force, it should equal to ma, and let's think about the sign of the acceleration. The object was going to surely accelerate down. It's not going to spontaneously climb up in the air, there's no reason for it to do that, so the acceleration should be in the positive direction, so I got all my signs right.
I have a problem, because remember my previous equation said T times R equals to I. I had two unknowns, T and . Here's that same unknown T, I've got a new unknown, a. So now I've got two equations and three unknowns. So I'm thinking, "What's going on here?" And then I remember a and are related, because, after all, there's a rope connecting the pulley to the mass. And ideal physics ropes don't stretch. So this point right here on the wheel and this point right here on the mass, those are the two ends of the ropes, they have to be moving at the same rate with the same speeds and the same accelerations. So the acceleration a, that's the acceleration of this mass, is the same as the acceleration of this little dot right here. What is that acceleration? It's the tangential component of the acceleration of the wheel. Just got to think about that for a second. Imagine drawing a little speck right at the edge of the wheel. It's accelerating tangentially with the same a as this mass is. So the connection between tangential acceleration and is a equals to R. So that's my third equation, three equations and three unknowns. It's not really so horrible as all that. I can, for instance, take this equation and solve it for tension. I don't want to know the tension; I want to know . That was what I was after.
So let me get rid of tension first. I'm going to plug it in here. I've got mg minus - let me substitute in, tension is I over R, is equal to ma. And I said I wanted to know , not a, so let me substitute in a is equal to R. And now I've got one equation. In the one unknown I was after , so let me solve for it. I've got mg is equal to times mR plus I over R. And let me solve for . It's mg divided by mR plus I over R. And I can do whatever I want to this expression to make it look a little bit better. Let me multiply numerator and denominator by one power of R. So I get mgR over mR^2 plus I. Now remember what my naïve answer was. My naïve answer, when I thought tension was just going to be mg, was going to be T times R divided by I. So I was wrong. It's a little bit less acceleration and that makes sense, because the tension is going to have to be a little bit less than mg. How do I know that? Because I know the mass is accelerating downwards. I know that gravity is winning here, so tension is less than mg, and so the angular acceleration is a little bit less than my naïve expectation.
It's a formula that is the correct answer. You could, for instance, ask yourself, "What would happen if the moment of inertia of my spinning wheel up there went to zero?" Supposing I had put all the mass of that wheel right at the hub so that it freely spins. So I goes to zero and the masses cancel, I get is g over R. What's a? The tangential acceleration of the wheel or the acceleration of the falling block was just R times , mgR^2 over mR^2 plus I. So it's R. And, in that limit, I would have R^2 over R^2 if I was equal to zero, I'd just get g, which makes sense. This wheel is basically free to spin. It's not doing a darn thing and the mass falls down with the acceleration that I would expect, g. So I think this equation is probably correct. All the signs seem to be making sense. The fact that this is a plus sign makes a little bit less than what I'd thought before. It's the right answer.
When you work equals I problems, you're always a little bit worried about signs. You've got to be careful, pick your conventions, clockwise or counterclockwise. Which are you going to call positive? Once you've done that, draw the force diagram. That's usually a useful tool in these kinds of problems. And just write down equals I, figure out the torques, solve for . It's very much like F equals ma. It's analogous to F equals ma, the same skills and ideas are what you use.
The Physics of Extended Objects
The Dynamics of Rotational Motion
Solving Problems Using Newton's Second Law for Rotational Motion Page [2 of 2]
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