## Jumat, 27 Mei 2011

### Jaringan Fisika Asyik untuk SMP (Physics for Junior High School)

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Sumber:

Arif Kristanta

SMPN 2 Tanjungsari Gunung Kidul, Jawa Tengah, Indonesia

## Kamis, 12 Mei 2011

### Fisika untuk Universitas,

Fisika untuk Universitas

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

So far, we analyzed...

We calculated the periods of lots of oscillators: pendulums, springs, rulers, hula hoops.

We gave them a kick, moved them off equilibrium, and then they were oscillating at their own preferred frequency.

Today, I want to discuss with you what happens if I force upon a system a frequency of my own.

So we call that forced oscillations.

I can take a spring system, as we have before.

This is x equals zero, this is x, and we have the spring force, very familiar, minus kx.

But now this object here, which is mass m, I'm going to add a force to it, F zero, which is the amplitude of the force, times the cosine of omega t.

So I'm going to force it in a sinusoidal fashion with a frequency that I choose.

This frequency is not the frequency with which the system wants to oscillate.

It is the one that I choose, and I can vary that.

And the question, now, is what will the object do? Well, we have Newton's Second Law--

ma equals minus kx plus that force, F zero cosine omega t.

a is x double dot, so I get x double dot, plus-- I bring this in--

k over m times x equals F zero divided by m times the cosine omega t.

Now, the question is what is the solution to this differential equation? It's very different from what we saw before, because before, we had a zero here.

Now we have here a driving force.

It's clear that if you wait long enough that sooner or later that system will have to start oscillating at that frequency.

In the beginning, it may be a little different.

In the beginning, it may want to do its own thing, but ultimately, if I take you by your arms and I shake you back and forth, in the beginning you may object, but sooner or later, you will have to go with the frequency that I force myself upon you.

And when we reach that stage, we call that the steady state as opposed to the beginning, when things are a little bit confused, which we call the transient phase.

So in the steady state, the object somehow must have a frequency which is the same as the driver, and it has some amplitude A.

And I want to evaluate with you that amplitude A.

So this is my trial function that I'm going to put into this differential equation.

x dot equals minus A omega sine omega t.

x double dot equals minus A omega squared cosine omega t.

And so now I'm going to substitute that in here, so I'm going to get minus A omega squared cosine omega t plus k over m times A cosine omega t, and that equals F zero divided by m times the cosine of omega t.

And that must always hold.

So therefore I can divide out my cosine omega t.

I can bring the A's together, so I get A times k over m minus omega squared equals F zero divided by m.

Now, this k over m is something that we are familiar with.

If we let the system do its own thing--

we bring it away from equilibrium and we don't drive it--

then we know that omega squared, which I will give the zero, equals k over m.

This is the frequency that we have dealt with before.

This is the driving frequency--

it's very different.

And so I'm going to substitute in here for k over m omega zero squared, and so I find, then, that the amplitude of this object here at the end of the spring will be F zero divided by m divided by omega zero squared minus omega squared.

And this amplitude has very remarkable characteristics.

First of all, if I drive the system at a very low frequency so that omega is much, much smaller than omega zero, we call omega zero often the natural frequency.

It is the one that it likes.

If you have omega much, much less than omega zero, this goes--

omega zero squared is k over m--

so you get an amplitude A which is F zero divided by k.

If you go omega way above the natural frequency or, let's say, omega goes to infinity--

it becomes very, very large--

then downstairs becomes very, very large, so A goes to zero.

But now, what happens when omega is exactly omega zero? Then the system goes wacky.

Look what happens.

The downstairs becomes zero and the amplitude goes to infinity.

And that's what we call resonance.

So if we drive it at that frequency, the system goes completely berserk.

I can make a plot of A as a function of frequency.

When I say "omega," you can obviously always change to hertz, if you prefer that, because omega is two pi times F, so you can do it either in hertz or you can do it in radians per second, of course.

So if I make a plot of the amplitude versus frequency omega, then at low values-- I have here F zero divided by k--

is the amplitude.

When I hit the resonant frequency, the natural frequency of the system, it goes out of hand, it goes to infinity.

The moment that omega is larger than omega zero, notice that the amplitude becomes negative.

A negative amplitude simply means that you get all of a sudden a phase change of 180 degrees, so the object is 180 degrees out of phase with the driver.

I will not expand on that too much today, but it is negative, and so it comes up here, and then it goes here to zero for very high frequencies of omega.

So something very spectacular is going to happen at the resonant frequency of the system.

In practice, of course, the amplitude will not go to infinity, and the reason for that is that there is always friction of some kind.

There is always damping, but you get a very high amplitude but not infinitely high.

So if I make you a more realistic plot of the amplitude, and I will take now the absolute value, the magnitude, so we don't have to worry about it getting negative--

we don't have to worry about the 180-degree phase shift--

then you would get a curve that looks like this.

And here, then, if this is frequency F, then here you would get the natural frequency when things go out of hand.

And depending upon how much damping there is, this curve would look either very narrow and very spiky--

it goes very, very high, then there is very little damping--

or if there is a lot of damping in the system, it's more like this.

So the narrower this...

We call this the resonance curve.

The narrower that is, the less damping there is.

I have here a system on the air track which is an object that I can drive with a frequency that I can choose.

Here is a spring...

object mass m, and here is another spring.

It's fixed on this side, right there, and here I'm going to drive it, so I have here this variable force that you see there.

And what I want to show you now is that first I will drive it at a frequency which is way below the resonant frequency.

Then you will see an amplitude, not very large.

I will then drive it way above the resonant frequency.

Again, you will see an amplitude which is very low.

If I can go very high frequency, you will see that it almost stands still, and then I will try to hit the resonant frequency, and that will be...

It's about one hertz, the resonant frequency, if I just let this object do its own thing, just this.

Now you see the natural frequency--

but now I'm going to drive it here with a system, and we are going to pull on this spring with a frequency that we control.

So let me start.

You see here the frequency.

You see an indicator, very low, way lower than one hertz.

And when you look at the way that the system responds, if you wait long enough when the transients are died out, you will see that they go hand in hand, that the amplitude is in phase with the driver.

That's why we have plus amplitude here.

But the phase is not so important today.

See, they go hand in hand.

Very small amplitude, roughly at zero divided by k, which is the spring constant of the spring.

Now I'll go way above resonance, and this system will slip here, so don't pay attention to the arrow anymore.

Way above resonance.

You see, it starts to slip.

Look at the amplitude--

very modest, very small.

But we went over this curve, so first we probed it here and now we're probing it here.

Look, it's almost not moving at all, almost standing still, and I'm driving it at a high frequency now.

And now I'm going to find you this resonance, which is near one hertz...

which is somewhere here.

And look-- very high amplitude.

If we're not careful, then we can actually break the system.

Very high amplitude--

I'm trying to scan over it now, go a little bit off resonance.

Now I'm back on resonance.

See what a huge amplitude! Better turn it off.

So you see here the response when I drive a system.

When my system is a little bit more complicated--

for instance, if I had two masses here so I would add one here, spring constant k, spring constant k--

I could repeat this experiment, and if I did that, I would find two resonant frequencies.

And if I do it with three objects,

I would find three resonant frequencies.

If I did it with five, I would find five resonant frequencies.

And when I make, then, this curve of A amplitude as a function of frequency--

either hertz or in radians per second, whichever you prefer--

then if I had three objects there in a row, you would see something like this.

And depending upon how many of these objects you have, you get more and more resonances.

And these resonances can all be found by driving the system and searching for them.

If I go to a system whereby I have an infinite number of these masses...

We call them coupled oscillators; these oscillators are coupled through the springs.

An infinite number of coupled oscillators would be a violin string.

Here's a violin string.

And the reason why I call it "infinitely" number of oscillators is that I can think of each atom or each molecule as being driven, as being connected by springs to the neighbor.

And so it's an infinite number of coupled oscillators.

And so when I start to shake this system, I would expect a lot of resonances, and that's what I want to explore with you now.

In the case here, that the objects move in the same direction of the spring...

I call this the y direction and I call this the x direction, so the spring is in the x direction, the objects, the beats are in the x direction, and the oscillations are in the x direction.

We call those longitudinal oscillations.

There is also a way that you can have transverse oscillations, transverse...

whereby the motion is in the y direction, whereas the beats are in the x direction.

I could even do that with this system--

I could make them oscillate like this, because the springs will obviously also work if I do this with the system.

And that's the way I want this violin string or piano string to oscillate now, because that's the only meaningful way that I can make it oscillate.

And so I wonder--

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.

## Selasa, 10 Mei 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

Okay, you did have some problems with physical pendulums, and I want to talk a little bit more about physical pendulums.

Let's first look at the picture in very general terms.

I have here a solid object, which is rotating about point P about an axis vertical to the blackboard, and here at C is the center of mass.

The object has a mass M, and so there is here a force Mg, and let the separation be here b.

I'm going to offset it over an angle theta, and I'm going to oscillate it.

Clearly, there has to be a force at the pin.

If there were no force at the pin, this object would be accelerated down with acceleration g, and that's not what's going to happen.

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

Remember when we had a spring, just a one-dimensional case, we had F equals ma, and that, for the spring, became minus kx, and the minus sign indicates that it's a restoring force.

So we now get something very similar.

In rotation, force becomes torque, mass becomes moment of inertia, and acceleration becomes angular acceleration.

So now we have minus r cross F, and the minus sign indicates that it is restoring.

So if I take the torque relative to point P, then I have...

This is the position vector, which has magnitude b.

The force is Mg, and I have to multiply by the sine of theta.

So I have b times Mg times the sine of theta and that now equals minus...

I can bring the minus here--

minus the moment of inertia about point P times alpha, and alpha is the angular acceleration, which is theta double dot.

I bring them together, and I use the small-angle approximation, small angles.

Then the sine of theta is approximately theta if theta is in radians.

And so I bring this all on one side, so I get theta double dot plus bMg divided by the moment of inertia about that point P times theta--

now this is my small-angle approximation--

equals zero.

And this is a well-known equation.

It is clearly a simple harmonic oscillation in theta because this is a constant.

And so we're going to get as a solution that theta equals theta maximum--

you can call that the angular amplitude--

times cosine omega t plus phi.

This omega is the angular frequency It is a constant.

Omega here, theta dot, is the angular velocity, which is not a constant.

The two are completely different.

That is the angular frequency.

So we know that the solution to this differential equation gives me omega is the square root of this constant.

So it is bMg divided by the moment of inertia about that point P.

And so the period of oscillation is two pi divided by omega, and so that is two pi times the square root of I relative to point P divided by bMg.

And let's hang on to this for a large part of this lecture because I'm going to apply this to various geometries.

Make sure that I have it correct--

yes, I do.

This is independent of the mass of the object.

Even though you will say there is an M here, you will see that in all cases when we calculate the moment of inertia about point P that there is always a mass up here.

So the mass will disappear, as you will see very shortly.

I have four objects here, and they all have different moments of inertia.

They're all going to rotate about an axis perpendicular to the blackboard, so to speak, and we're going to massage each one of them to predict their periods.

Let's first go to the rod.

So we first do the rod.

We have the rod here.

This is point P, and here is the center of the rod.

The rod has mass M and it has length L.

So we have here Mg.

I simply go to this equation, and I want to know what the period is of this rod, of this oscillating rod.

All I have to know now is what is the moment of inertia about P.

And I know already that b in that equation equals one-half L.

So, what is the moment of inertia of oscillation about point P? I have to apply now the parallel axis theorem--

which you also had to do during the exam--

which says it is the moment of inertia of rotation about the center of mass, which, in our case, is C--

the axes have to be parallel, so there is this axis perpendicular to the blackboard, and this axis perpendicular to the blackboard--

plus the mass of the rod times the distance between P and c squared--

plus M times this distance squared--

so that is b squared, and b squared is one-quarter L squared.

What is the moment of inertia for rotation of a rod about this axis? I looked it up in a table.

I happen to remember it now, because I am lecturing 801.

Two months from now, I will have forgotten.

So I remember now that it is 1/12 ML squared plus one-quarter ML squared.

That becomes one-third ML squared.

And so the period T becomes two pi times the square root of this moment of inertia, which is the one-third ML squared, divided by bMg, and b is one-half L for this geometry.

So one-half LMg, and notice, indeed, as I anticipated, you always lose your M, you also lose one L here, and so you get two pi times the square root of two-thirds L divided by g.

So that is the period that we predict for the rod.

So let's write that under here, because we are going to compare them shortly.

So this is two pi times the square root of two-thirds L divided by g.

The rod that we have here is designed in such a way that the period is very close to one second.

That was our goal.

So T is as close as we can get it to 1.00 seconds.

And so if you substitute in this equation T equals one, you will find that the length of this rod, if it is really pivoting at the very end, should be about 37.2 centimeters.

So we did the best we can when we made this rod.

There is always an uncertainty, of course--

how you drill the holes and where you drill the holes--

so I would say the value that we actually achieved is 37.2, probably with an uncertainty of about three millimeters, so 0.3 centimeters.

That's what we have.

That is an error of one part in 370.

Let's make that...

round that off.

That's a one-percent error in the length.

Since the length is under the square root, the one-percent error becomes half a percent error, so the period that I then would predict is about 1.000 plus half a percent, so that is 0.005 seconds.

So that, then, has a one-half-percent error.

So this is my predicted period.

And so we're going to make ten oscillations of the observed oscillations.

We're going to get a number.

My reaction time is not much better than a tenth of a second.

So we're going to get a number there.

We divide this number by ten.

And so we can always calculate the period, then, and then we get a much improved error of a hundredth of a second, because we will divide this number by ten, so this also is going to be divided by ten.

And let's see how close we were able to get this to the 1.0 seconds.

So, here is the rod.

Turn this on, the timer.

We'll offset the rod, and I will start it when it stops somewhere.

Now--

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.

## Sabtu, 07 Mei 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

You see here the topics that will be on your plate on Monday.

It's clearly not possible for me in one exam to cover all these topics, so I will have to make a choice Monday.

Today I also have to make a choice.

I cannot do justice to all these topics in any depth, so I will butterfly over them and some of them I won't even touch at all.

However, what is not covered today may well be on the exam.

You will get loads of equations.

Almost every equation that I could think of will be on your exam.

There's also special tutoring this weekend you can check that on the Web--

Saturday and Sunday.

Completely inelastic collision.

We have mass m1, we have mass m2.

It's a one-dimensional problem.

This one has velocity v1, and this has velocity v2.

They collide, and after the collision they are together--

because that's what it means when the collision is completely inelastic--

and they have a velocity, v prime.

If there is no net external force on the system, momentum must be conserved.

So I now have that m1 v1 plus m2 v2 must be m1 plus m2 times v prime.

One equation with one unknown--

v prime follows immediately.

You may say, "Gee, you should really have put arrows over here." Well, in the case that it is a one-dimensional collision, you can leave the arrows off because the signs automatically take care of that.

Kinetic energy is not conserved.

Before the collision, your kinetic energy equals one-half m1 v1 squared plus one-half m2 v2 squared.

After the collision, kinetic energy equals one-half m1 plus m2 times v prime squared.

And you can easily prove that this is always less than that in case of a completely inelastic collision.

There's always kinetic energy destroyed, which then comes out in the form of heat.

Let's do now an elastic collision.

And I add the word "completely" elastic, but "elastic" itself is enough because that means that kinetic energy is conserved.

I start with the same initial condition: m1 v1, m2 v2, but now, after the collision, m1 could either go this way or this way, I don't know.

So this could be v1 prime, this could be v1 prime.

m2, however, will always go into this direction.

That's clear, because if you get hit from behind by object one, after the collision, you obviously go in this direction.

Again, we don't have to put the arrows over it, because the signs take care of it in a one-dimensional case.

This will be plus, then.

That could be... you could adopt that as your convention, and if it goes in this direction, then it will be negative.

So if you find, for v1 prime, minus something, it means it's bounced back.

So now we can apply the conservation of momentum if there is no external force on the system.

Internal force is fine.

All right, so now we have m1 v1 plus m2 v2 equals m1 v1 prime plus m2 v2 prime--

conservation of momentum.

Now we get the conservation of kinetic energy, because we know it's an elastic collision.

One-half m1 v1 squared, one-half m2 v2 squared--

that's before the collision.

After the collision, one-half m1 v1 prime squared plus one-half m2 v2 prime squared.

Two equations with two unknowns and in principle, you can solve for v1 prime and for v2 prime, except that it could be time-consuming.

And so on exams, what is normally done when you get a problem like that...

Normally you get a problem whereby either m1 is made m2 or m1 is much, much larger than m2 or m1 is much, much smaller than m2, like banging a basketball onto a Ping-Pong ball or a Ping-Pong ball onto a basketball.

I will do a very simple example whereby I will take now for you m1 equals m2, and I will call that m and I will even simplify the problem by making v2 zero, so the second object is standing still.

One-dimensional collision, one hits that object.

Very special case.

So the conservation of momentum now becomes much simpler.

m times v1--

there is no v2--

equals m v1 prime plus m v2 prime.

Conservation of kinetic energy is one-half m v1 squared equals one-half m v1 prime squared plus one-half m v2 prime squared.

Notice that now I lose my "m"s, which is very convenient.

Here I lose my one-half "m"s, even and this is very easy to solve.

If you square this equation you get something that will look very similar to this.

If you square it, you get v1 squared equals v1 prime squared plus v2 prime squared plus two v1 prime v2 prime.

And compare this equation with this equation--

they're almost identical, except for this term, so this term must be zero.

But we know that v2 prime is not zero.

It got kicked and so it'll go forward.

So what that means, then, is that v1 prime equals zero and if v1 prime equals zero, you see that v2 prime equals v1.

And this is that classic case whereby a ball hits another ball.

This one has no speed.

It hits it with a certain velocity, v1.

They have the same mass.

After the collision, this one stands still and this one takes over the speed.

You see it often with pendulums.

It is the logo on the 8.01 home page, and I showed you a demonstration here when we discussed that in lectures.

All right, let's move on, and let's do something now on torques, angular momentum, rotation and let's discuss the Atwood machine.

The Atwood machine is a clever device that allows you to measure, to a reasonable degree of accuracy, the gravitational acceleration.

Here's a pulley, and the pulley has mass m, has radius R.

It's solid, so it's a solid disk--

And there is a rope here, near massless--

we ignore the mass.

Mass m2 is here and mass m1 is here, and let's assume that m2 is larger than m1.

So this will be accelerated in this direction, this will be accelerated in this direction, and this will start to rotate with angular velocity omega, which will be a function of time.

And now the first thing we want to do

is to make up "free-body" diagrams.

"Free-body" diagrams, for this one, is easy.

m1 g down... and T1 up.

For this one, we have m2 g down and we have T2 up.

For the pulley, it's a little bit more complicated.

This is that point P.

If here is a tension T1 that's pulling down on the pulley--

so this is T1--

and this T2 is pulling down on the pulley--

so there's T2--

it has a mass, so it has weight mg.

The sum of all forces on the pulley must be zero; otherwise it would accelerate itself down, which it doesn't.

And so there has to be a force up--

I'll call it n.

And that force n has to cancel out all these three forces.

So that must be T1 plus T2 plus mg.

We will not need it any further in our calculations, but there has to be a force to hold that in place, so to speak.

So now we're going to calculate the acceleration under the condition that the rope does not slip.

That means there is friction with the pulley--

not friction here, but here.

Otherwise, the rope would slip, the rope would slip.

What it means, if there is no slip, that if the rope moves one centimeter that the wheel also turns at the circumference one centimeter.

That's what it means when there is no slip.

That means the velocity of the rope--

v of r, v of the rope--

must be omega times R of the pulley.

That's what "no slip" means.

So the acceleration--

which is the derivative of that velocity of the rope--

A, is omega dot times R, which is alpha times R.

Omega is the angular velocity and alpha is the angular acceleration.

This is a condition... it is an important condition for no slip.

So let's now start at object number one and write down Newton's second law.

I call this the positive direction for object one and I call this the positive direction for object two--

just easier for me.

So we get T1 minus m1 g must be m1 a--

one equation.

I don't know what T1 is, I don't know what a is.

Second equation, for this one.

I call this the positive direction.

m2 g minus T2 must be m2 a--

second equation.

One unknown has been added, so I need more.

Of course I need more.

I also have to think about the pulley.

The pulley... the net force on the pulley is zero.

That's why this stays in place, but it is going to rotate because this force T2 is larger than this T1.

There is a torque relative to that point P, and torque is defined as r cross F.

The torque relative to point P, the magnitude is this position vector times this force.

That is a torque in the blackboard.

What is in the blackboard, I will call positive.

The torque, due to this force, is out of the blackboard, and I will call that negative.

Since this angle is 90 degrees, I simply get that the torque relative to point P equals the radius of the cylinder--

times T2.

That's the positive part, and the negative part is the radius times T1.

Notice that this force and this force go through P, do not contribute to the torque.

And that now equals the moment of inertia about that point P, times alpha.

But since we have no slip, alpha is a divided by R.

So it's the moment of inertia about point P times a divided by R.

But since it is a rotating disk which is rotating about its center of mass...

I know the moment of inertia, I looked that up--

if you need it during your exam, you will find that in the exam--

that is one-half MR squared...

one-half MR squared times a divided by R and I lose one R, and so I find, then, that T2 minus T1 equals one-half Ma.

Notice I also lose my second R.

And so now I have a third equation and I can solve for T2, I can solve for T1 and I can solve for a.

And you can do that as well as I can.

If you find the result, you should always do a little bit of testing to make sure that your result makes sense.

And what you should do is you should say...

You should make sure that m2 g is larger than T2.

That's a must--

otherwise, it's not being accelerated down.

You should also check that T1 comes out larger than m1 g.

That's also a must.

And you should also check that T2 be larger than T1.

Otherwise, the pulley wouldn't rotate in clockwise direction.

It would also be useful, which is a trivial check, to stick in your results m1 equals m2.

That should give you that the acceleration should be zero and it should give you that T1 equals T2.

Those are obvious things and that can be done very simply.

It takes you no more than ten seconds.

And if it's... any one of these is not met, then somewhere you slipped up and it will give you an opportunity to go over the problem again.

All right, let's now do another problem--

simple harmonic oscillation of a physical pendulum.

I have here a rod--

mass M, length l--

rotating here about an axis perpendicular to the blackboard without friction.

Here is the center of the rod, and let this angle be theta.

There is a torque relative to point P.

There is also a force that goes through point P.

I'm not even interested in that force.

I know that it's here, mg.

Since I'm going to take the torque relative to point P, I don't worry about the force, but there has to be a force through point P.

Otherwise this ruler would be accelerated down with acceleration g, if this is the only force there were.

But we know that's not the case, it's going to swing.

So there has to be a force to P.

I don't want to know what it is, but there has to be one.

So the torque relative to point P--

the magnitude--

is the position vector, r of P, from here to here crossed with this force.

And so that makes it one-half l times mg times the sine of the angle, and that's... the angle is theta so that is times the sine of theta.

The cross product has the sine of the angle in it.

So this is the torque relative to point P which also must be the moment of inertia about point P times alpha.

No different from what we just had with the pulley.

So alpha equals omega dot.

It's also theta double dot.

This omega is the angular velocity--

it's d theta/dt and the derivative gives me the angular acceleration.

There has to be a minus sign here, that's important because the torque is restoring--

We were oscillating a spring, the spring force is minus kx.

The minus sign here plays exactly the same role.

So I can write down here minus I of p theta double dot.

Now, if we take a small angle approximation--

"small angle approximation"--

then the sine of theta is approximately theta, if theta is in radians.

And so I can replace now the sine of theta here by theta and so now I find, if I bring this to the other side, I get theta double dot plus one-half lMg divided by the moment of inertia about point P times theta equals zero.

Needless to say that I am happy like a clam at high tide because I see here an equation which clearly tells me that we have a simple harmonic oscillation--

theta double dot plus a constant times theta is zero.

And so we must get as a solution that theta must be some maximum angle times the cosine of omega t plus or minus phi.

This omega has nothing to do with that omega.

This is the angular frequency.

This is related to the period T of the oscillation, which is two pi divided by omega.

This is a constant.

This omega is not a constant.

It is unfortunate, in physics, that we use the same symbol.

The angular velocity is zero when the object stands still; is a maximum when the object is here.

That omega is always the same.

It's related to how long it takes to make one oscillation.

Both are called omega, both are radians per second.

Couldn't be more confusing.

All right, if we find I of P then we can solve for the frequency, angular frequency and we can solve for the period.

So let's do the... calculate the moment of inertia about point P.

In order to do that, we have to apply the Parallel-Axis Theorem because you will probably be given the moment of inertia for rotation about the axis through the center of mass parallel to this axis.

And then you will have to add the mass times the distance between these two axes squared to apply the Parallel Axis Theorem.

So this is the moment of inertia about the center of mass plus m times the distance squared.

And the distance between these two axes is one-half l, so this is one-quarter l squared.

I look up in a table what the moment of inertia is for rotation of a rod about its center--

a perpendicular rod is perpendicular to the axis--

and that is 1/12 Ml squared plus one-quarter Ml squared give me one-third Ml squared.

So I know now what I of P is, and so now I can solve for the value for omega--

the angular frequency.

I'll do that here so that we keep everything on one blackboard.

So this term here, which is omega squared...

Omega squared equals one-half lMg divided by one-third Ml squared.

I lose one l, I lose one M--

very common, always lose your "M"s in gravity--

and so this is three-halves g divided by l.

And so the period of one oscillation is two pi divided by omega--

omega is the square root of this--

so that becomes the square root of two-thirds l over g.

And that is the period of the oscillation of this ruler.

We worked on something similar in lectures and we even measured the period and we found very, very good agreement with the theoretical prediction.

You can ask, now, what is the kinetic energy of rotation of this rod, which changes with time? Kinetic energy of rotation is one-half I omega squared.

Remember, the linear kinetic energy is one-half mv squared.

The m becomes I when you go to rotation, and v becomes omega.

So the kinetic energy of rotation equals one-half I omega squared--

you'll find this equation on your exam.

This is I about that point P.

You know what theta is, as a function of time.

So omega equals d theta/dt, so you can find what omega is.

You know what I of P is, we just calculated it.

And so you know what the kinetic energy of rotation is at any moment in time.

It will change.