# MIT open course Physics 1 - Mechanics by Walter Lewin - Lecture 19 Rotating Rigid Bodies, Inertia, and Axis Theorems

Материал готовится,

пожалуйста, возвращайтесь позднее

пожалуйста, возвращайтесь позднее

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.

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.

Загрузка...

Выбрать следующее задание

Ты добавил

Выбрать следующее задание

Ты добавил