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We're now entering the part of 8.01 which is the most difficult for students and faculty alike.

We are going to enter the domain of angular momentum and torques.

It is extremely nonintuitive.

The good news, however, is that we will stay with this concept for at least four or five lectures.

Today I will introduce both torque and angular momentum.

What is angular momentum? If an object has a mass m and it has a velocity v, then clearly it has a momentum p.

That's very well defined in your reference frame, the product of m and v.

Angular momentum I can take relative to any point I choose.

I choose this point Q arbitrarily.

This now is the position vector, which I call r of Q.

Let this angle be theta.

And angular momentum relative to that point Q —

it's a vector —

is the position vector relative to that point Q cross p.

So it is r of Q cross v, and then times m.

The magnitude of the angular momentum, relative to point Q, is, of course, rmv, but then I have to take the sine of the angle theta, so let's say it is mv r sine theta and this I often call, shorthand notation, r perpendicular.

That r perpendicular is this distance, relative to point C.

What you just saw may have confused you and for good reason, because I changed my index "Q" to "C," and there is no C.

The indexes should all be Q, of course.

So this r is the length of this vector.

It is the magnitude of this vector.

So this should have a "Q." And r of Q sine theta, which I call r perpendicular, must have an index Q, and that is this part here.

This angle is 90 degrees and this here is r of Q perpendicular.

No Cs at all, only Qs —

I'm sorry for that.

The direction of the angular momentum is easy.

You know how to do a cross product.

So in this case, r cross v would be perpendicular to the blackboard and the magnitude is also easy to calculate.

Now comes the difficult problem with angular momentum.

If I chose any point on this line, say point C, then the angular momentum relative to point C is zero.

Very obvious, because the position vector, r, and the velocity vector, in this case, are in the same direction.

So theta is zero, so the sine of theta is zero.

So you immediately see that angular momentum is not an intrinsic property of a moving object, unlike momentum, which is an intrinsic property.

If you sit there in 26.100, you see an object moving with a certain velocity, it has a certain mass, you know its momentum.

What the angular momentum is depends on the point that you choose, on your point of origin.

If you had chosen this point D, then the angular momentum would even be this way, because when you put here the position vector in there you see r cross v is now coming out of the blackboard.

And this is why angular momentum is such a difficult concept.

But we will massage it in a way that it will be very useful.

Suppose I throw up an object in 26.100 and at time t equals zero, the object is here and at time t, the object is there.

So this, then, is the position vector at time t.

The object starts off with a certain velocity v and a little later, here, say, the velocity is like so.

And there is, of course, a force on it, mg, which makes this curve.

What is the angular momentum relative to point C at time zero? The angular momentum is clearly zero, because the point itself, the mass itself, is at point C.

So the position vector has no length, so it's clear that it's zero.

We are going to enter the domain of angular momentum and torques.

It is extremely nonintuitive.

The good news, however, is that we will stay with this concept for at least four or five lectures.

Today I will introduce both torque and angular momentum.

What is angular momentum? If an object has a mass m and it has a velocity v, then clearly it has a momentum p.

That's very well defined in your reference frame, the product of m and v.

Angular momentum I can take relative to any point I choose.

I choose this point Q arbitrarily.

This now is the position vector, which I call r of Q.

Let this angle be theta.

And angular momentum relative to that point Q —

it's a vector —

is the position vector relative to that point Q cross p.

So it is r of Q cross v, and then times m.

The magnitude of the angular momentum, relative to point Q, is, of course, rmv, but then I have to take the sine of the angle theta, so let's say it is mv r sine theta and this I often call, shorthand notation, r perpendicular.

That r perpendicular is this distance, relative to point C.

What you just saw may have confused you and for good reason, because I changed my index "Q" to "C," and there is no C.

The indexes should all be Q, of course.

So this r is the length of this vector.

It is the magnitude of this vector.

So this should have a "Q." And r of Q sine theta, which I call r perpendicular, must have an index Q, and that is this part here.

This angle is 90 degrees and this here is r of Q perpendicular.

No Cs at all, only Qs —

I'm sorry for that.

The direction of the angular momentum is easy.

You know how to do a cross product.

So in this case, r cross v would be perpendicular to the blackboard and the magnitude is also easy to calculate.

Now comes the difficult problem with angular momentum.

If I chose any point on this line, say point C, then the angular momentum relative to point C is zero.

Very obvious, because the position vector, r, and the velocity vector, in this case, are in the same direction.

So theta is zero, so the sine of theta is zero.

So you immediately see that angular momentum is not an intrinsic property of a moving object, unlike momentum, which is an intrinsic property.

If you sit there in 26.100, you see an object moving with a certain velocity, it has a certain mass, you know its momentum.

What the angular momentum is depends on the point that you choose, on your point of origin.

If you had chosen this point D, then the angular momentum would even be this way, because when you put here the position vector in there you see r cross v is now coming out of the blackboard.

And this is why angular momentum is such a difficult concept.

But we will massage it in a way that it will be very useful.

Suppose I throw up an object in 26.100 and at time t equals zero, the object is here and at time t, the object is there.

So this, then, is the position vector at time t.

The object starts off with a certain velocity v and a little later, here, say, the velocity is like so.

And there is, of course, a force on it, mg, which makes this curve.

What is the angular momentum relative to point C at time zero? The angular momentum is clearly zero, because the point itself, the mass itself, is at point C.

So the position vector has no length, so it's clear that it's zero.

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