# MIT open course Physics 1 - Mechanics by Walter Lewin - Lecture 30 Simple Harmonic Oscillations of Suspended Solid Bodies

Материал готовится,

пожалуйста, возвращайтесь позднее

пожалуйста, возвращайтесь позднее

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.

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.

Загрузка...

Выбрать следующее задание

Ты добавил

Выбрать следующее задание

Ты добавил