Jumat, 22 April 2011

Fisika untuk Universitas

Fisika untuk Universitas

Ditujukan untuk meningkatkan kualitas proses dan hasil perkuliahan Fisika di tingkat Universitas

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We're now entering the part of 8.01 which is the most difficult for students and faculty alike.

We are going to enter the domain of angular momentum and torques.

It is extremely nonintuitive.

The good news, however, is that we will stay with this concept for at least four or five lectures.

Today I will introduce both torque and angular momentum.

What is angular momentum? If an object has a mass m and it has a velocity v, then clearly it has a momentum p.

That's very well defined in your reference frame, the product of m and v.

Angular momentum I can take relative to any point I choose.

I choose this point Q arbitrarily.

This now is the position vector, which I call r of Q.

Let this angle be theta.

And angular momentum relative to that point Q--

it's a vector--

is the position vector relative to that point Q cross p.

So it is r of Q cross v, and then times m.

The magnitude of the angular momentum, relative to point Q, is, of course, rmv, but then I have to take the sine of the angle theta, so let's say it is mv r sine theta and this I often call, shorthand notation, r perpendicular.

That r perpendicular is this distance, relative to point C.

What you just saw may have confused you and for good reason, because I changed my index "Q" to "C," and there is no C.

The indexes should all be Q, of course.

So this r is the length of this vector.

It is the magnitude of this vector.

So this should have a "Q." And r of Q sine theta, which I call r perpendicular, must have an index Q, and that is this part here.

This angle is 90 degrees and this here is r of Q perpendicular.

No Cs at all, only Qs--

I'm sorry for that.

The direction of the angular momentum is easy.

You know how to do a cross product.

So in this case, r cross v would be perpendicular to the blackboard and the magnitude is also easy to calculate.

Now comes the difficult problem with angular momentum.

If I chose any point on this line, say point C, then the angular momentum relative to point C is zero.

Very obvious, because the position vector, r, and the velocity vector, in this case, are in the same direction.

So theta is zero, so the sine of theta is zero.

So you immediately see that angular momentum is not an intrinsic property of a moving object, unlike momentum, which is an intrinsic property.

If you sit there in 26.100, you see an object moving with a certain velocity, it has a certain mass, you know its momentum.

What the angular momentum is depends on the point that you choose, on your point of origin.

If you had chosen this point D, then the angular momentum would even be this way, because when you put here the position vector in there you see r cross v is now coming out of the blackboard.

And this is why angular momentum is such a difficult concept.

But we will massage it in a way that it will be very useful.

Suppose I throw up an object in 26.100 and at time t equals zero, the object is here and at time t, the object is there.

So this, then, is the position vector at time t.

The object starts off with a certain velocity v and a little later, here, say, the velocity is like so.

And there is, of course, a force on it, mg, which makes this curve.

What is the angular momentum relative to point C at time zero? The angular momentum is clearly zero, because the point itself, the mass itself, is at point C.

So the position vector has no length, so it's clear that it's zero.

What is the angular momentum at time t when the object is here? Well, that angular momentum is clearly not zero, because you see here position vector and you see the velocity, so clearly the angular momentum was changing.

Now you will say, "Of course it was changing--

big deal." Because angular momentum has a velocity vector in it.

And here the velocity vector is changing all the time, so, obviously, you would say the angular momentum is changing.

Well, yes, that is not a bad argument, but I will now show you a case where the velocity is changing all the time, but where angular momentum is not changing.

I choose the Earth going around the Sun.

Here's the Earth, with mass m.

At point C here is the Sun.

This is the position vector r of C and the Earth has a certain tangential velocity and the speed never changes, but the velocity does change.

So this is the position vector at a later point in time.

What now is the angular momentum of the Earth going around the Sun, relative to point C? I pick C now.

Well, that angular momentum...

If I take the magnitude of the angular momentum, because the direction is immediately obvious...

If the object is going around like this--

this is the position vector--

then the direction will be pointing out of the blackboard.

That's easy.

So I'm only worried now about the magnitude.

So the magnitude is the mass of the Earth times the magnitude of the cross product between these two vectors.

And notice the angle is 90 degrees.

So I can forget about the cross, the sine of theta is one, and so I simply get mrv, v now being the speed.

This is the case when the object is here, but when the object is here, the situation has not changed.

Again, r cross v, the magnitude, is exactly the same, because the sine of the angle hasn't changed.

And so you see here a case whereby the velocity is changing all the time but your angular momentum relative to point C is not changing.

Suppose I had chosen point Q.

Is angular momentum changing relative to point Q? You'd better believe it.

There is a time that the object will go through point Q.

Well, then the angular momentum is clearly zero because the position vector is zero.

If the object is here and you take the angular momentum relative to point Q, for sure the angular momentum is not zero.

You have a position vector and you have a velocity.

So only relative to point C--

it's a very special case now--

is angular momentum not changing.

So angular momentum is conserved in this special case, but only about point C.

And I want to address that in a little bit more general way.

I take the angular momentum and I choose a point Q, and I know that the definition is position vector relative to point Q cross p.

I take the derivative, time derivative dL/dt relative to that point Q.

It's always important that you state which point you have chosen relative to which you take the angular momentum.

That is going to be dr/dt...

excuse me--

cross p plus r of Q cross dp/dt.

This is the way that you take the time derivative of a cross product.

We calculate the angular momentum relative to point Q.

So the index has to be Q throughout the equation.

The position vector, relative to point Q.

And in this equation, you see the correct index Q here.

You see the correct index Q here, but I slipped up here and I put a "C" there.

There is no "C" in this problem, so this is also r of Q.

Sorry for that.

Ucapan Terima Kasih Kepada:

1. Para Dosen MIT di Departemen Fisika

a. Prof. Walter Lewin, Ph.D.

b. Prof. Bernd Surrow, Ph.D.

2. Para Dosen Pendidikan Fisika, FPMIPA, Universitas Pendidikan Indonesia.

Terima Kasih Semoga Bermanfaat dan mohon Maaf apabila ada kesalahan.

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