Physics: Torque and Angular Momentum

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Newton's Second Law; F = MA. It's really the central idea of this course. If you know about the net force on a system, you know about the change in motion. Newton didn't write down F = MA, Newton wrote down F = . It's, as we've seen, an alternative way of writing down that equation. And it's a very elegant way of writing it down. It's got physical content and meaning to write it this way. Because, for instance, if there's no net force on some object, that tells you that there's no change in the momentum.
is equal to zero means momentum is conserved and we know how useful conservation is. If you conserve anything, then you know starting from what is was in the beginning, what its going to be in the end very useful for understanding the physics of potentially very complicated problems.
This is a great way of writing F = MA. How about the angular analog? We've talked about F = MA, the angular analog would be torque equals I-alpha. Twist causes angular acceleration.
Can we write that in this form? Can we write torque equals to... what would it be? It would be the rate of change of the angular momentum, which we are calling L. And the answer is, "yes." This equation is correct. The net torque is equal to change in angular momentum with time, and it is just as useful and informative to write it down this way. A very nice formula.
Once again, if you have a complicated system with all sorts of stuff going on in it, but the net torque on that system is equals to zero, so there's no kind of outside twist, then we are going to be able to argue that there's no change in angular momentum. I'm talking about a system know. It was true with this equation as well. This equation really got the most interesting when we started talking, not about a single particle, but about a group of particles.
So, it's a nice way of writing it. I would really like to prove it, and let's begin with a proof by looking at a single object. And let me write the definition of L. L = x p for that object. And I had a , so let's just do it. Let's take the derivative of both sides with respect to time. This is a cross product and you have to know the mathematical rule. It's just like the chain rule in calculus... the derivative of the product is the derivative of the first term, cross the second, plus the first times the derivative of the second. This is just a chain rule applied to vectors. It's really correct.
And, stare at his equation for a second... we've got the rate of change in angular momentum. Look at this term, DRDT: what is that? That's velocity. Velocity vector and momentum vector point in the same direction because momentum is MV and M is just a scale, it doesn't change the direction.
So what happens when you take the cross product of two vectors that are pointing in the same direction? The sine of the angle between them is zero. So, this term vanishes, always. What's this one: R x ? Let's stop and think. , that's Newton's Second Law. That's the net force on the object. So we've got R x F. Now this term was zero, and this was R x F, its really the net force, which is Newton's Second Law. And now, you stop, we're done... R x F[net], that's the definition of torque.
So, for a single particle, we've proven that the net-torque on that particle is the change in angular momentum with time. What if we have a system of particles, because that's really what we're going to be interested in many times? So, instead of writing this formula, I would write L is the sum of L[i ]- I would just replace everything that I did here with a sum over "i" everything is "sub-i". That means I've got a complicated system, I break it up and do a bunch of into a bunch of different chunks. Each chunk obeys this equation. So when you sum this up, you get the derivative of the total angular momentum, the sum of L[i ]summed up over all of the little chunks. And this term is always zero. And, here you always get the sum of R[i] F[net i]. So you're finding the torque on each and every little piece, and you're adding that up, and that is what you mean by the net torque on the system. So it's true for a particle, its true for a system, its true for a rigid body. This formula is really important, that's why I'm going through all of this formal mathematics.
There's one thing I could do, which would make this formula a little bit useful in solving problems and understanding physical situations. The net torque on a system is kind of complicated. You are taking each and every object and figuring out the net torque on that object. Where would that net torque on each individual object come from. It could come from two places: either from outside torques, or internal torques-that means torques arising from forces between particles inside your system.
Now think about that for a second; inside your system, you've got particle A and particle B. And there's a force on A by B and there's a force on B by A, those are internal forces. And by Newton's Third Law, they're equal and opposites. So, when you add up the net internal forces, they absolutely, definitely cancel; they go to zero.
If those two forces are, like I'm showing you, on a straight line from A to B, then, remember, we're finding torques about some origin always, so here's my origin. It's a little simple geometry exercise. Try it and convince yourself that if this force and that force are equal and opposite and they ally on a line between the particles, then the torque on this one plus the torque on that one as vectors cancel and you get zero. And that's true pair by pair so, in this situation all the internal torque cancel out and so I can replace torque-net with net external torque.
So, this is a really practically useful equation. Because what it says is if you have a complex system and you add up all the external twists, then you're going to find . I mean it makes a lot of physical sense. Think of a big old wheel, and it's sitting there. And nobody is twisting on it. So, there's no external torques and there's no change in angular momentum, it just sits there. How about if it was initially spinning, so it does have some initial angular momentum? If you're not twisting on it, no friction, nobody's pushing or pulling, then is zero. Angular momentum of that big thing is going to always be constant. It will come to a halt, it will continue spinning around with the same L, pointing in the same direction forever. Even if it's in outer space, if there's no outside torques on it, it's not going to start rotating around. That's a very handy thing to know and make sure of in spacecraft.
How about if you do twist on it? What if you start pushing on one side, or pulling on the other, you're twisting it. So you're adding a net external torque. Well, the angular momentum will change; it makes sense. If you're twisting on it, it's going to start spinning up, and the angular momentum will get larger, it's going to have more rotational oomph. Or less: it depends on the direction. If you make the sum of the torques come out to be in the negative direction, you can make the be negative; you can increase angular momentum or decrease it. And the formula, if you work out the vector's carefully, will really tell you about all the signs.
Let me give you another physical example. It's an example we just saw where I dropped a ball. Here's my origin, here's the object, it's falling down. And there is an angular momentum of that object about this origin. In fact, when we worked out that problem we found that the magnitude of the angular momentum was equal to m times g times b (that was the horizontal distance between them) and times time. gt was the velocity, mgt was the momentum and this was r cross momentum. So, that was the formula.
Let's just check and see if this makes sense. Let's look at . I should put a vector sign in here but you'll remember in that problem, we worked out the vector sign. If you were looking down on the page, where you had a ball falling, r x p was into the page. If it was in the physical situation with the falling ball, r x p, which is downward, is pointing towards me. These things are always a little bit tricky to think about in three dimensions.
So, is just, derivative with respect of time, is mgc. The direction is not changing. It's always the same direction. In the real world, this-a-way. In this picture, always into the page. So the vector nature here doesn't really enter much into the story. is mgb. And what's this side? Some of the torque's external. Well, what is the external torque? There is a force of gravity, that's the external force. The direction of the force of gravity is down. And the magnitude is mg. Let me draw a little picture here. You've got a ball with a force F equals to mg down, here was my origin. What's the torque, which this force is creating? Torque is equal to r-perpendicular times F and r-perpendicular was this, b. So, it's mgb. So, indeed torque mgb is equal to .
It works, it always works. It's kind of nice to see in examples. This one is a funky example because I don't have a rigid body and I don't really have any rotation. And yet, still I'm perfectly allowed to use is equal to some of the torque externals. You can always use it in any problem. It's most convenient to use when you're thinking about systems that have some kind of rotational motion. And it's going to come into its strength, this idea and this formula, when we start thinking about complicated systems that are isolated; that have no net external torque. And when you have that situation, is zero. That's obviously going to be really nice, conservation of angular momentum is something that we'll have to get to.
The Physics of Extended Objects
Angular Momentum
Torque and Angular Momentum Page [3 of 3]