## Selasa, 19 April 2011

### Fisika untuk Universitas

Fisika untuk Universitas

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

We have here, going back to rotating objects...

I have an object here that has a certain velocity v, and it's going around with angular velocity omega, and a little later the angle has increased by an amount theta and then the velocity is here.

We may now do something we haven't done before.

We could give this object in this circle an acceleration.

So we don't have to keep the speed constant.

Now, v equals omega R, so that equals theta dot times R.

And I can take now the first derivative of this.

Then I get a tangential acceleration, which would be omega dot times R, which is theta double dot times R, and we call theta double dot...

we call this alpha, and alpha is the angular acceleration which is in radians per second squared.

Do not confuse ever the tangential acceleration, which is along the circumference, with a centripetal acceleration.

The two are both there, of course.

This is the one that makes the speed change along the circumference.

If we compare our knowledge of the past of linear motion and we want to transfer it now to circular motion, then you can use all your equations from the past if you convert x to theta, v to omega and a to alpha.

And the well-known equations that I'm sure you remember can then all be used.

For instance, the equation x equals x zero plus v zero t plus one-half at squared simply becomes for circular motion theta equals theta zero plus omega zero t plus one-half alpha t squared--

it's that simple.

Omega zero is then the angular velocity at time t equals zero, and theta zero is the angle at time t equals zero relative to some reference point.

And the velocity was v zero plus at.

That now becomes that the velocity goes to angular velocity omega equals omega zero plus alpha t.

So there's really not much added in terms of remembering equations.

If I have a rotating disk, I can ask myself the question now which we have never done before, what kind of kinetic energy, how much kinetic energy is there in a rotating disk? We only dealt with linear motions, with one-half mv squared, but we never considered rotating objects and the energy that they contain.

So let's work on that a little.

I have here a disk, and the center of the disk is C, and this disk is rotating with angular velocity omega that could change in time, and the disk has a mass m, and the disk has a radius R.

And I want to know at this moment how much kinetic energy of rotation is stored in that disk.

I take a little mass element here, m of i, and this radius equals r of i and the kinetic energy of that element i alone equals one-half m of i times v of i squared, and v of i is this velocity--

this angle is 90 degrees.

This is v of i.

Now, v equals omega R.

That always holds for these rotating objects.

And so I prefer to write this as one-half m of i omega squared r of i squared.

The nice thing about writing it this way is that omega, the angular velocity, is the same for all points of the disk, whereas the velocity is not because the velocity of a point very close to the center is very low.

The velocity here is very high, and so by going to omega, we don't have that problem anymore.

So, what is now the kinetic energy of rotation of the disk, the entire disk? So we have to make a summation, and so that is omega squared over two times the sum of m of i r i squared over all these elements mi which each have their individual radii, r of i.

And this, now, is what we call the moment of inertia, I.

Don't confuse that with impulse; it has nothing to do with impulse.

And this is moment of inertia...

So the moment of inertia is the sum of mi ri squared.

In...

So this can also be written as one-half I, I put a C there--

you will see shortly why, because the moment of inertia depends upon which axis of rotation I choose--

times omega squared.

And when you see that equation you say, "Hey, that looks quite similar to one-half mv squared." And so I add to this list now.

If you go from linear motions to rotational motions, you should change the mass in your linear motion to the moment of inertia in your rotational motion, and then you get back to your one-half mv squared.

You can see that.

So we now have a way of calculating the kinetic energy of rotation provided that we know how to calculate the moment of inertia.

Well, the moment of inertia is a boring job.

It's no physics, it's pure math, and I'm not going to do that for you.

It's some integral, and if the object is nicely symmetric, in general you can do that.

In this case, for the disk which is rotating about an axis through the center and the axis--

that's important--

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.