## Jumat, 29 April 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

With our knowledge of torque...

calm down.

With our knowledge of torque and angular momentum, we can now attack rolling objects which roll down a slope.

For instance, the following...

I have here a cylinder or it could be a sphere, for that matter, and this angle is beta.

I prefer not to use alpha because that's angular acceleration, and there is this friction coefficient with the surface, mu, and this object is going to roll down and you're going to get an acceleration in this direction, a.

And I will evaluate the situation when we have pure roll.

That means the object is not skidding and is not slipping.

What is pure roll?

If here is an object, the cylinder is here with radius R, and I'm going to rotate it like this and roll it in this direction, the center is called point Q.

Once it has made a complete rotation, if then the point Q has moved over a distance 2pi R, then we call that pure roll.

When we have pure roll, the velocity of this point Q, and the velocity of the circumference, if you can read that-- I'll just put a c there-- are the same.

In other words, vQ is then exactly the same as v circumference, and v circumference is always omega R.

This part always holds, but for pure roll, this holds.

You can easily imagine that if there is no friction here, that the object could be standing still, rotating like crazy, but Q would not go anywhere.

So then we have skidding and we have slipping and then we don't have the pure roll situation.

If the object is skidding or slipping, then the friction must always be a maximum here.

If the object is in pure roll, the friction could be substantially less than the maximum friction possible.

Now I would like to calculate with you the acceleration that a cylinder would obtain.

When it pure rolls down that slope, it has mass M, it has length l and it has radius R.

And I would like you to use your intuition and don't be afraid that it's wrong.

I'm going to roll down this incline two cylinders.

They're both solid, they have the same mass, they have the same length but they're very different in radii and I'm going to have a race between these two.

Which one will reach the bottom first? So I repeat the problem.

Two cylinders, both solid, same length, same mass, but one has a larger radius than the other.

There's going to be a race.

We're going to roll them down, pure roll.

Which one will will? Win win? Will win? Who thinks that the one with the largest radius will win? Who thinks the large with the smallest radius will win? Who thinks there will be no winner, no loser? Wow, your intuition is better than mine was.

We'll see how it goes.

Keep in mind what your vote was and you will see it come out very shortly.

Okay, let's put all the forces on this object that we know.

This one is Mg.

And we're going to decompose that into one a longer slope, which is Mg sine beta, and one perpendicular to the slope.

We have done that a zill... zillion times now.

And this one equals Mg cosine beta.

Then there is right here a normal force, and the magnitude of that normal force is Mg cosine beta, so there is no acceleration in this direction and then we have a frictional force here for which I will write F of f.

There is an angular velocity at any moment in time, omega, which will change with time, no doubt.

And then this center point Q, which is the center of mass, is going to get a velocity v and that v will also change with time.

And the v of the point Q which changes with time is v of the circumference, because that is the condition of pure roll.

That equals omega R.

This is always true, but this is only true when it is pure roll.

I take the time derivative.

The derivative of the velocity of that point Q is, per definition, its acceleration, so I get a equals omega dot times R, and that equals alpha R, alpha being the angular acceleration.

So this is the condition for pure roll.

Now I'm going to take the torque about point Q.

When I take the torque about point Q, N has no effect because it goes through Q, and g has no effect because it goes through Q, so there's only one force that adds to the torque.

If this radius is R, the magnitude is RF and the direction is in the blackboard.

But I'm only interested in the magnitude for now, so I get R times the frictional force.

This must be I alpha, I being the moment of inertia for rotation about this axis through point Q, times alpha, but I can replace alpha by a/R.

So I get the moment of inertia about Q times a/R.

And this is my first equation, and I have as an unknown the frictional force, and I have as an unknown a, and so I cannot solve for both.

I need another equation.

The next equation that I have is an obvious one, that is Newton's second law: f = MA.

For the center of mass, I can consider all the mass right here at Q.

We must have f = MA.

And so M times the acceleration of that point Q, which is our goal, by the way, equals this component, equals Mg sine beta.

That is the component downhill.

And minus Ff, the frictional force, which is the component uphill, and this is my equation number two.

Now I have two equations with two unknowns.

So I can solve now.

I can eliminate Ff and I will substitute for Ff in here this quantity divided by R.

And so I get Ma equals Mg sine beta minus moment of inertia about point Q,

times a divided by R squared.

And now notice that I've eliminated F of f, and so now I can solve for a.

So I'm going to get...

I bring the a's to one side.

So I get a times M plus moment of inertia divided by R squared equals Mg sine beta, and a now we have.

I multiply both sides with R squared.

I get MR squared g sine beta upstairs, and downstairs I get MR squared plus the moment of inertia about that point Q.

This is my result, and all I have to put in now is the moment of inertia of rotation about that axis.

I want to remind you, though, that it is only true if we have a situation of pure roll.

So we can now substitute in there the values that we have for a solid cylinder.

through the center of mass, which I've called Q, equals 1/2 MR squared.

And if I substitute that in here, notice that all my M's...

MR squares go away.

I get 1 + 1/2, which is 1 1/2.

Upside down becomes 2/3.

So a = 2/3 times g times the sine of beta.

There is no M, there is no l and there is no R.

So if I have two cylinders, solid cylinders with totally different mass, totally different radii, totally different length and they have a race, neither one wins.

Very nonintuitive.

Every time that I see it I find it kind of amazing.

Notice that everything disappears.

M, R and l disappear.

So those of you who said that if I take two cylinders with the same mass, different radii, those of you who said that there is no winner, there is no loser, they were correct.

But even more amazing is that even the mass you can change.

You can change anything as long as the two cylinders are solid.

That's what matters.

So if we take a hollow cylinder, then the moment of inertia about this axis through the center of mass, through Q, if this...

if really most of the mass is really at the surface, then it's very close to MR squared, and then the acceleration--

if I substitute in here MR squared, I get a 2 there-- equals 1/2 times g times sine beta.

So this acceleration is less than this one.

So the hollow cylinder will lose in any race against a solid cylinder regardless of mass, regardless of radius, regardless of length.

And I want to show that to you.

We have a setup here and I'll try to show that to you also on the screen there, but for those of you who are sitting close, it's probably much better

that you just look at the demonstration right here.

I have here... ooh.

Uh-uh.

They have very different masses, same radii, same length.

Should make no difference.

There should be no winner, there should be no loser.

I'm going to start them off at the same time.

I hope you can see that there.

This is... this is the starting point.

Can lower it a little.

I will count down three to zero, and then you can see that they reach the bottom almost at the same time.

So very different in mass.

The mass difference is at least a factor of three.

All other dimensions are the same.

Three, two, one, zero.

Completely in unison.

Not intuitive for me.

Now I have one that has a very small radius compared to this one.

This is a sm... small aluminum rod.

Maybe you can see it here, television.

This is way more heavy, almost 30 times heavier.

Should make no difference.

As long as it's solid, should make no difference.

No winner, no loser.

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.

## Kamis, 28 April 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

The speed of sound is 340 meters per second--

it depends a little bit on the temperature--

When I speak to you, my sound reaches you with that speed.

I produce a certain frequency here, a certain number of oscillations per second.

They reach you, your eardrum starts to oscillate with the same frequency and you hear that tone.

I have here a tuning fork which oscillates 440 times per second.

[tuning fork produces medium-pitched tone]

Your eardrum oscillates 440 times per second--

you hear this tone.

Here I have 256 oscillations per second.

[metal rod emits lower tone]

Your eardrum is now shaking, going back and forth 256 times per second.

If you stay where you are and you don't move and I move these tuning forks, you will hear a different frequency and that's what we call Doppler effect.

If my sound source approaches you, you will hear a frequency f prime which is larger than the frequency of the tuning fork.

If it moves away from you, which I will call receding, then f prime equals lower... lower frequency.

For instance, I move to you a sound source--

I call that a transmitter--

with a speed of about one meters per second.

Transmitter is the sound transmitter.

Then if it approaches you here, you will hear f prime, which is 1.003 times f.

This three here is the one part out of 340 that you get an increase in frequency.

If I move it away from you, then f prime would be 0.997 times the frequency of the source itself.

You stay where you are.

I have here a tuning fork which generates 4,000 hertz, a very high frequency.

If I move it to you with the speed of one meter per second, which I can do, then you get an increase in pitch of 0.3%.

That makes it 4,012 hertz.

And when I move it away from you there is a decrease of 0.3%.

And you can clearly hear that difference.

I will first make you listen to the 4,000 hertz without my moving.

[tuning fork produces very high pitched tone]

Can you hear it? Very high frequency.

Is it painful, really? High frequency.

Most of you are young enough you should be able to hear 4,000 hertz.

Okay, now I am going to move it to you one meter per second and away from you.

[high tone goes up and down slightly in pitch]

Did you hear it? Once more.

[tone goes up and down quickly again]

[class laughs]

When it comes to you, it's clear that the frequency goes up, and when it moves away from you, the frequency is down.

Now imagine that I'm going to rotate the sound source around in a circle.

Now the sound that you receive, the frequency that you receive will change in a sinusoidal fashion.

If this is that circle, and this is the radius of that circle, and if you are here in the plane of the circle, then when the source comes straight to you with the velocity v-- let's say it's a uniform circular motion--

f prime will be larger than f and it will, in this case, reach a maximum.

When it is at 90 degrees relative to you--

I don't have to give it a vector notation--

f prime equals f.

When it moves away from you, f prime is smaller than f, you hear a minimum.

And when it is here again--

when the angle between the velocity and your direction is again 90 degrees-- then f prime equals f again.

And so this phenomenon is called the Doppler effect.

So if I twirl it around, you will hear a sinusoidal fluctuation in f prime.

Suppose I plot, as a function of time, f prime the way you will receive it-- you sit still, but I'm going to move the sound source around like this.

Then you will have a curve that looks something like this: some sinusoidal-cosinusoidal fluctuation of f prime.

This will be the value f produced by the sound source itself.

This will be f prime maximum and this will be f prime minimum.

If you could record this, there is an amazing number of things that you can deduce from this curve.

First of all, you can take...

You can measure f prime max divided by F, because you see this curve, so you know what f is here, you see what f prime max is, and that should allow you to retrieve immediately v velocity of the transmitter.

If that number were 1.003, then you know that the speed in the orbit was one meter per second.

So this ratio immediately gives you the transmitter velocity.

This time separation gives you immediately the period of rotation, but since two pi R-- if R is the radius divided by the velocity of the transmitter--

since that is the... oh, I can reverse it, it doesn't matter.

Two pi R divided by the time to go around is the velocity of the transmitter.

Since you know the velocity of the transmitter from this ratio since you know the period, which is this, you now also find the radius R.

So from that curve-- and keep that with you, because it's going to be important in what follows--

we can derive three things: the radius, the period of rotation and the speed of the object as I twirl it around.

I have here what we call a wind organ.

When I twirl this around, it produces a particular tone.

We will talk later about 801 why it produces a particular tone.

Sometimes you hear two tones.

I'll try to make you hear only one.

And as I swirl it around, the sound is coming...

the sound source, the transmitter is coming to you.

This, when it goes like this, it's 90-degree angle so you should not hear any Doppler shift.

When it is here, it's moved away from you and so you hear a sinusoidal change in f prime.

Try to hear that.

[wind organ producing tone that changes pitch]

Can you hear, when it's coming to you, that it's higher-pitched than when it's going away from you? Can you hear that? Just say no if you don't hear it.

Not very clear.

[wind organ again producing varying tone]

For me, it's impossible to hear because I'm standing right under it, of course.

Well, I tried.

I now want to change to electromagnetic waves.

Electromagnetic waves travel with the speed of light, which is 300,000 kilometers per second.

And if you want to treat that correctly, you would have to use special relativity.

In the case of sound, I stressed repeatedly that you in the audience should not move but that the sound source is moving.

In the case of electromagnetic radiation when you deal with the speed of light, you don't have to ask that question.

It is a meaningless question in special relativity.

To ask whether you are moving relative to me or whether I am moving relative to you, it doesn't matter.

All that matters in special relativity is the relative motion, so you can always think of yourself as standing still and make the source of electromagnetic radiation move to you, or away from you, relative to you.

All of that is electromagnetic radiation.

If the velocity of the source of electromagnetic radiation--

the transmitter--

if that is way, way smaller than the speed of light, then it is very easy to predict the change in frequency due to Doppler shift.

Let this be the transmitter which produces frequency f, and here is the receiver which receives the frequency f prime.

And let the velocity of the source of electromagnetic radiation be v--

I could put transmitter here, but we can drop that index--

and let this angle be theta.

Then this is the component in your direction--

we call that the radial component--

which is v cosine theta.

So I delete the tr.

This is just the velocity of the source relative to you at that angle.

If now we want to know what f prime is, then f prime equals f times one plus v over c times the cosine of theta.

What matters is only the radial component of the velocity.

If theta is 90 degrees, just like we had with sound, then f prime equals f.

So 90 degrees, the cosine of theta is zero, f prime equals f.

If theta is smaller than 90 degrees, then it's coming towards you, then f prime equals larger than f.

If theta equals larger than 90 degrees, it's going away from you, then f prime equals smaller than f.

You would get a similar equation for sound by replacing this c by the speed of sound.

But I want to stress that this only holds for electromagnetic radiation if v over c is much, much smaller than one.

Now, when we deal with sound, there is something mechanically oscillating.

Something is vibrating.

With electromagnetic radiation, charges are vibrating.

Electrons are vibrating, and they are vibrating with a certain frequency, and that means there is a certain period of one oscillation.

And that period of one oscillation is, of course, one over the frequency.

I can ask myself now the question, how far does electromagnetic radiation, how far does light travel in the time of one period capital T? Well, it goes with the speed of light, so in T seconds, it moves a distance cT.

And that distance we call the wavelength of electromagnetic radiation, lambda equals cT, for which you can also write c divided by F.

So this is the wavelength of the electromagnetic radiation--

the speed of light, 300,000 kilometers per second--

the period of one oscillation, say, of the electrons, and this is the frequency, which you can give in hertz.

I could give you a specific example.

I, for instance, can take a period T of two times ten to the minus 15 seconds.

That would give me a wavelength of about six times ten to the minus seven meters--

six times ten to the minus seven meters--

and that you would experience as red light.

If I make the period shorter--

say, 1.3 times ten to the minus 15 seconds--

I get a shorter wavelength.

I get four times ten to the minus seven meters, and you would experience that as blue light.

In astronomy, in optical astronomy we cannot measure the period or the frequency of optical light.

All we can measure is the wavelength.

And so if I want to use this equation, then I have to replace f prime by c divided by lambda prime and f I have to replace by c divided by lambda.

And when I do that, I get the following result.

I get lambda prime equals lambda times one minus v over c cosine theta.

If this is a plus, this is a minus.

Check that for yourself.

You have to use the small number approximation,

the Taylor expansion, namely that v over c is much, much smaller than one.

So you can see now if the object comes to you, in other words, if f prime is larger than f, if the frequency is higher, then the wavelength will be smaller.

And so let me write that down.

When the cosine of theta...

so the object is coming to you--

when the cosine of theta is larger than zero, the object is approaching you, then the wavelength lambda prime will be less than lambda.

And that has a name-- we call that blue shift.

And the reason why we call that blue shift is that if the wavelengths become shorter, it moves towards the blue end of the spectrum, because blue has a lower wavelength than red.

If cosine theta is negative, then the object is receding from you, then lambda prime is larger than lambda, and we call that red shift.

These are the terms that astronomers use all the time.

When you make a spectrum of a star--

you can do that using prisms or by other means--

and you look at the light intensity as a function of wavelength--

so here is the light intensity as a function of wavelength--

then you may expect to see some kind of a continuum.

But, in fact, what you do see is...

Superimposed on a continuum you see sometimes very sharp absorption lines--

black, missing light, called absorption lines And these absorption lines correspond to elements in the atmosphere of the star.

In fact, if you see the absorption lines, you can tell what kind of elements are present in the star.

Some are very characteristic absorption lines: some for hydrogen, some for calcium, some for silicon, some for magnesium and so on.

It's actually interesting that when you look at the spectrum of the sun--

when people did that first, when they had the means of doing that, they found absorption lines in the spectrum of the sun which could not be identified.

They had never been seen here on Earth, these lines, and so they called these lines after the sun.

The sun is Helios, and so they called it helium.

So helium was first discovered on the sun before it was later discovered on Earth by looking at the absorption lines of the solar spectrum.

If a star moves to you, then all the lines--

every single line-- will be blue-shifted.

And if the star moves away from you, all the lines will be red-shifted.

If you take an example: With lambda prime divided by lambda and you pick any one of those lines--

it doesn't matter which you pick because they will all do exactly the same.

If this were, for instance, 1.00333--

I just pick a very nice number; that means lambda prime, as you see, is larger than lambda; the wavelengths get longer, so we have a red shift--

and you substitute that in that equation, then you'll find that the velocity at which that star is moving relative to you--

that gives you immediately the answer there--

equals minus 0.00333 times the speed of light, c, and that is minus 100 kilometers per second.

And the minus, then, reminds you that the object is receding from you.

So that gives you a red shift.