Selasa, 10 Mei 2011

Fisika untuk Universitas

Fisika untuk Universitas

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

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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 don't have to worry about this anymore.

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.


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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