» Download this transcript (PDF)

Last lecture, I introduced the concept of angular momentum and torque.

They're the most difficult concepts in all of 8.01.

And only when you get a lot of practice will you really get the hang of it.

You shouldn't feel bad if it takes a while.

This is very difficult.

I will spend the next five lectures exclusively on dealing with these concepts, and you will see many examples--

some intuitive, some nonintuitive and some even quite bizarre.

I'd like to briefly review the key things we discussed last time, and you see the related equations there on the blackboard.

We have a mass m, and let that mass in your frame of reference have a velocity v.

So then it clearly has a momentum p.

There may be a force acting upon that mass--

F.

And now I choose a point Q at random.

This is the position vector relative to Q.

Never forget to indicate what the origin is that you have chosen.

Then the definition of angular momentum relative to that point Q equals the cross product of the position vector with p, and that is my equation number one there.

The direction is perpendicular to the blackboard, and it will be in this case into the blackboard.

And the magnitude can be calculated provided that you take the angle into account.

You get the sign of theta because of the cross product.

The torque relative to point Q is defined as the position vector, cross F.

In this case, that would be out of the blackboard.

And, again, the magnitude can be found, but you have to take into account the angle between the position vector and the force.

The torque leads to a change in angular momentum.

You see that in equation three.

If there is no torque, then angular momentum won't be changing.

And if you have a system of objects--

not just one like we have here, but many interacting particles--

then as long as there is no external--

net external--

torque on that system as a whole, then angular momentum of the system as a whole will be conserved.

Today you will see various applications of equation four and of equation five.

Last time, we already discussed the idea of spin angular momentum, which is an intrinsic property of a rotating object.

We did an experiment with the ice-skater's delight when I was sacrificing there on this rotating turntable.

And we'll see some more examples of that during the next few lectures.

In case that you do have a rotation about an axis through the center of mass--

a stationary axis through the center of mass--

then the angular momentum is intrinsic in the sense that you don't have to specify the point about which you take the angular momentum.

You can take any point, and you always find the same, which is not true in this situation.

So that makes the intrinsic angular momentum quite unique.

The reason why it's so nonintuitive--

angular momentum--

is that the angular momentum depends on the point you choose.

And in one problem, you can sometimes pick a point about which the angular momentum changes, but in the very same problem, you can pick a point about which angular momentum doesn't change, and both solutions would be perfectly valid.

So you often have a choice, and that doesn't make it very intuitive; that doesn't make it very easy.

So let's start with an example, which I also had last time, whereby we have an object going around the Earth or around the Sun.

Let's take the Earth going around the Sun, and this is the location of the Sun, and here is the Earth.

It has a mass m, and it's going around with a velocity.

The magnitude doesn't change, but the direction does change.

The position vector relative to point C is r of C, and then we have a gravitational force F, which is pointed towards the center, and the angular velocity omega is in this direction.

Well, go to equation number one, and the angular momentum relative to point C--

L relative to point C--

the magnitude, because the direction is clear.

If it's going, seen from where you are, counterclockwise, then the direction of the angular momentum will be pointing out of the blackboard.

So I'm only interested in the magnitude.

That will be r of C times the mass times v.

And the reason why I don't worry about the cross now is because this angle is 90 degrees, so the sign of theta equals one.

Now, we may not like to leave v in there.

It's up to you.

You can always write v equals omega R, so you can also write, then, m r C squared, times omega.

Either one is fine.

And this r, if this is the radius of the circle, then this obviously becomes m R squared omega.

You could have chosen equation five, and you would immediately have said, "Aha! I have a rotation about point Q," which in this case is C, and so L... the magnitude of L of C equals the moment of inertia about point C times omega.

The moment of inertia about point C for this object is clearly mR squared, and I multiply that by omega, and you see you get exactly the same answer.

So this is equation one, but this will be equation five.

If I go to equation number two, then equation number two is telling me that if I choose point C, but only if I choose point C, that the torque relative to point C is zero--

equation number two--

because the force and the position vector make an angle of 180 degrees with each other.

Whether the object is here or here or here, that makes no difference, and so the torque relative to point C is zero, so I also know that the angular momentum is not changing relative to point C, but only relative to point C, because any other point that you would have chosen here or here or here, there would have been a torque, and the angular momentum would be changing, so there's something very special about this point C.

Angular momentum is only conserved, in this case, about point C.

Now I take another example whereby angular momentum is only conserved relative to one point but not to any other point.

I take a ruler or a rod... and the rod has mass M and length l, and C is the center of mass of that rod, but I force it to spin about point P...

and this distance is d.

Think of this as a horizontal frictionless plane, and I'm rotating it with an angular velocity.

Let's say we rotate it in this direction, force it about that point P.

I put a pin in there perpendicular to the blackboard and I rotate it.

I'd like to know what the magnitude is of the angular momentum relative to point P, and for that, I go immediately to equation five, so that tells me that it is the moment of inertia about that axis through point P times omega.

I remember the parallel axis theorem.

I know that the moment of inertia for rotation about a center of mass through this axis perpendicular to the blackboard--

I know that that one equals 1/12 Ml squared.

But I just looked it up in a table, because I don't remember that.

So that would be the moment of inertia about this axis, and then the parallel axis theorem tells me I have to add plus Md squared...

times omega.

I'm not interested in the direction of l, because that's immediately obvious.

If it's rotating clockwise, then the direction of the angular momentum would be perpendicular to the blackboard and into the blackboard.

I claim that at this point P, there must be a force acting on this ruler, and the force is in this direction.

And I can make you see that the best way by first showing you the case of a massless rod with two equal masses at both ends.

And I rotate about this axis perpendicular to the blackboard.

There is going to be a centripetal force here and a centripetal force here, and the two are equal, they cancel each other out, so there will be no force on that pivot point about which the two rotate.

However, if I had the situation such that this is my massless rod and here are the two equal masses, but now I rotate them about this point, then this centripetal force is larger than this one, so now I have asymmetry, and so now there will be a force on this pin.

The ruler will push on the pin, and action equals minus reaction--

the pin will push on the ruler.

And it is because of this same asymmetry that you have here that there will be a force from the pin on point P.

However, I don't care about that force because I'm going to take the torque relative to point P.

And when I take a torque relative to point P, any force through point P has no effect, because the position vector is zero.

But I want you to appreciate that there is a force, so if I take the torque relative to point P, I do not worry about this force.

Well, the torque relative to that point P is zero, and so it's clear that angular momentum relative to point P must be conserved.

Angular momentum relative to point P is conserved.

Take any other point--

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

## Tidak ada komentar:

Poskan Komentar