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Last lecture, I introduced the concept of angular momentum and torque.

They're the most difficult concepts in all of 8.01.

And only when you get a lot of practice will you really get the hang of it.

You shouldn't feel bad if it takes a while.

This is very difficult.

I will spend the next five lectures exclusively on dealing with these concepts, and you will see many examples —

some intuitive, some nonintuitive and some even quite bizarre.

I'd like to briefly review the key things we discussed last time, and you see the related equations there on the blackboard.

We have a mass m, and let that mass in your frame of reference have a velocity v.

So then it clearly has a momentum p.

There may be a force acting upon that mass —

F.

And now I choose a point Q at random.

This is the position vector relative to Q.

Never forget to indicate what the origin is that you have chosen.

Then the definition of angular momentum relative to that point Q equals the cross product of the position vector with p, and that is my equation number one there.

The direction is perpendicular to the blackboard, and it will be in this case into the blackboard.

And the magnitude can be calculated provided that you take the angle into account.

You get the sign of theta because of the cross product.

The torque relative to point Q is defined as the position vector, cross F.

In this case, that would be out of the blackboard.

And, again, the magnitude can be found, but you have to take into account the angle between the position vector and the force.

The torque leads to a change in angular momentum.

You see that in equation three.

If there is no torque, then angular momentum won't be changing.

And if you have a system of objects —

not just one like we have here, but many interacting particles —

then as long as there is no external —

net external —

torque on that system as a whole, then angular momentum of the system as a whole will be conserved.

Today you will see various applications of equation four and of equation five.

Last time, we already discussed the idea of spin angular momentum, which is an intrinsic property of a rotating object.

We did an experiment with the ice-skater's delight when I was sacrificing there on this rotating turntable.

And we'll see some more examples of that during the next few lectures.

In case that you do have a rotation about an axis through the center of mass —

a stationary axis through the center of mass —

then the angular momentum is intrinsic in the sense that you don't have to specify the point about which you take the angular momentum.

You can take any point, and you always find the same, which is not true in this situation.

So that makes the intrinsic angular momentum quite unique.

The reason why it's so nonintuitive —

angular momentum —

is that the angular momentum depends on the point you choose.

And in one problem, you can sometimes pick a point about which the angular momentum changes, but in the very same problem, you can pick a point about which angular momentum doesn't change, and both solutions would be perfectly valid.

So you often have a choice, and that doesn't make it very intuitive; that doesn't make it very easy.

So let's start with an example, which I also had last time, whereby we have an object going around the Earth or around the Sun.

They're the most difficult concepts in all of 8.01.

And only when you get a lot of practice will you really get the hang of it.

You shouldn't feel bad if it takes a while.

This is very difficult.

I will spend the next five lectures exclusively on dealing with these concepts, and you will see many examples —

some intuitive, some nonintuitive and some even quite bizarre.

I'd like to briefly review the key things we discussed last time, and you see the related equations there on the blackboard.

We have a mass m, and let that mass in your frame of reference have a velocity v.

So then it clearly has a momentum p.

There may be a force acting upon that mass —

F.

And now I choose a point Q at random.

This is the position vector relative to Q.

Never forget to indicate what the origin is that you have chosen.

Then the definition of angular momentum relative to that point Q equals the cross product of the position vector with p, and that is my equation number one there.

The direction is perpendicular to the blackboard, and it will be in this case into the blackboard.

And the magnitude can be calculated provided that you take the angle into account.

You get the sign of theta because of the cross product.

The torque relative to point Q is defined as the position vector, cross F.

In this case, that would be out of the blackboard.

And, again, the magnitude can be found, but you have to take into account the angle between the position vector and the force.

The torque leads to a change in angular momentum.

You see that in equation three.

If there is no torque, then angular momentum won't be changing.

And if you have a system of objects —

not just one like we have here, but many interacting particles —

then as long as there is no external —

net external —

torque on that system as a whole, then angular momentum of the system as a whole will be conserved.

Today you will see various applications of equation four and of equation five.

Last time, we already discussed the idea of spin angular momentum, which is an intrinsic property of a rotating object.

We did an experiment with the ice-skater's delight when I was sacrificing there on this rotating turntable.

And we'll see some more examples of that during the next few lectures.

In case that you do have a rotation about an axis through the center of mass —

a stationary axis through the center of mass —

then the angular momentum is intrinsic in the sense that you don't have to specify the point about which you take the angular momentum.

You can take any point, and you always find the same, which is not true in this situation.

So that makes the intrinsic angular momentum quite unique.

The reason why it's so nonintuitive —

angular momentum —

is that the angular momentum depends on the point you choose.

And in one problem, you can sometimes pick a point about which the angular momentum changes, but in the very same problem, you can pick a point about which angular momentum doesn't change, and both solutions would be perfectly valid.

So you often have a choice, and that doesn't make it very intuitive; that doesn't make it very easy.

So let's start with an example, which I also had last time, whereby we have an object going around the Earth or around the Sun.

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