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Last lecture,
I introduced the concept

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of angular momentum and torque.

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

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And only when you get
a lot of practice

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will you really get
the hang of it.

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

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This is very difficult.

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I will spend the
next five lectures exclusively

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on dealing with these concepts,
and you will see many examples--

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some intuitive,
some nonintuitive

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and some even quite bizarre.

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I'd like to briefly review

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the key things we discussed
last time,

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and you see
the related equations

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there on the blackboard.

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We have a mass m,

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

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So then it clearly has
a momentum p.

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There may be a force acting upon
that mass-- F.

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And now I choose a point Q
at random.

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This is the position vector
relative to Q.

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Never forget to indicate
what the origin is

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that you have chosen.

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Then the definition
of angular momentum

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relative to that point Q

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equals the cross product
of the position vector with p,

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and that is my equation
number one there.

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The direction is perpendicular
to the blackboard,

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and it will be in this case
into the blackboard.

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And the magnitude
can be calculated

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provided that you take
the angle into account.

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

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The torque relative to point Q

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is defined as the position
vector, cross F.

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

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And, again, the magnitude
can be found,

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but you have to take
into account the angle

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between the position vector
and the force.

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The torque leads to a change
in angular momentum.

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You see that in equation three.

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If there is no torque,

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then angular momentum
won't be changing.

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And if you have a system
of objects--

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

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then as long as there is no
external-- net external-- torque

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on that system as a whole,

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then angular momentum
of the system as a whole

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will be conserved.

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Today you will see various
applications of equation four

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and of equation five.

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Last time, we already discussed

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the idea of spin
angular momentum,

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which is an intrinsic property
of a rotating object.

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We did an experiment with
the ice-skater's delight

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when I was sacrificing there
on this rotating turntable.

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And we'll see some
more examples of that

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during the next few lectures.

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In case that you do have
a rotation about an axis

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through the center of mass--

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a stationary axis
through the center of mass--

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then the angular momentum
is intrinsic

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in the sense that you don't
have to specify the point

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about which you take
the angular momentum.

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You can take any point,
and you always find the same,

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which is not true
in this situation.

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So that makes the intrinsic
angular momentum quite unique.

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The reason why
it's so nonintuitive--

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angular momentum--

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

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And in one problem, you can
sometimes pick a point

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about which the angular
momentum changes,

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but in the very same problem,
you can pick a point

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about which angular
momentum doesn't change,

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and both solutions
would be perfectly valid.

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So you often have a choice,

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and that doesn't make
it very intuitive; 

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that doesn't make it very easy.

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So let's start with an example,
which I also had last time,

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whereby we have an object
going around the Earth

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or around the Sun.

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Let's take the Earth
going around the Sun,

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and this is the location of
the Sun, and here is the Earth.

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It has a mass m, and it's
going around with a velocity.

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The magnitude doesn't change,
but the direction does change.

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The position vector relative
to point C is r of C,

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and then we have
a gravitational force F,

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which is pointed
towards the center,

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and the angular velocity omega
is in this direction.

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Well, go to equation number one,

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and the angular momentum
relative to point C--

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L relative to point C--

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the magnitude, because
the direction is clear.

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If it's going, seen from where
you are, counterclockwise,

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then the direction
of the angular momentum

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will be pointing
out of the blackboard.

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So I'm only interested
in the magnitude.

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That will be r of C
times the mass times v.

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And the reason why I don't
worry about the cross now

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is because this angle
is 90 degrees,

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so the sign of theta equals one.

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Now, we may not like
to leave v in there.

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It's up to you.

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You can always write
v equals omega R,

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so you can also write, then,
m r C squared, times omega.

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Either one is fine.

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And this r, if this is
the radius of the circle,

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then this obviously becomes
m R squared omega.

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You could have chosen
equation five,

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and you would immediately
have said,

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"Aha! I have a rotation
about point Q,"

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which in this case is C,

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and so L... the magnitude
of L of C

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equals the moment of inertia
about point C times omega.

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The moment of inertia
about point C for this object

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is clearly mR squared,

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and I multiply that by omega,

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and you see you get exactly
the same answer.

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So this is equation one,
but this will be equation five.

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If I go to equation number two,

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then equation number two
is telling me

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that if I choose point C,
but only if I choose point C,

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that the torque relative
to point C is zero--

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equation number two--

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because the force and the
position vector make an angle

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of 180 degrees with each other.

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Whether the object
is here or here or here,

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that makes no difference,

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and so the torque relative
to point C is zero,

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so I also know that the angular
momentum is not changing

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relative to point C,
butonly relative to point C,

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because any other point
that you would have chosen

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here or here or here, there
would have been a torque,

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and the angular momentum
would be changing,

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so there's something very
special about this point C.

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Angular momentum is only
conserved, in this case,

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about point C.

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Now I take another example

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whereby angular momentum is only
conserved relative to one point

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but not to any other point.

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I take a ruler or a rod... and
the rod has mass M and length l,

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and C is the center of mass
of that rod,

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but I force it to spin
about point P...

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and this distance is d.

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Think of this as a horizontal
frictionless plane,

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and I'm rotating it
with an angular velocity.

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Let's say we rotate it
in this direction,

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force it about that point P.

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I put a pin in there
perpendicular to the blackboard

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and I rotate it.

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I'd like to know
what the magnitude is

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of the angular momentum
relative to point P,

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and for that, I go immediately
to equation five,

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so that tells me that it is
the moment of inertia

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about that axis through
point P times omega.

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I remember
the parallel axis theorem.

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I know that the moment
of inertia

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for rotation
about a center of mass

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through this axis perpendicular
to the blackboard--

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I know that that one equals
1/12 Ml squared.

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But I just looked it up
in a table,

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because I don't remember that.

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So that would be the moment
of inertia about this axis,

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and then the parallel axis
theorem tells me

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I have to add plus Md squared...
times omega.

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I'm not interested
in the direction of l,

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because that's
immediately obvious.

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If it's rotating clockwise,

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then the direction
of the angular momentum

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would be perpendicular
to the blackboard

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and into the blackboard.

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I claim that at this point P,

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there must be a force
acting on this ruler,

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and the force is
in this direction.

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And I can make you
see that the best way

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by first showing you the case
of a massless rod

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with two equal masses
at both ends.

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And I rotate about this axis
perpendicular to the blackboard.

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There is going to be
a centripetal force here

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and a centripetal force here,
and the two are equal,

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they cancel each other out,
so there will be no force

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on that pivot point
about which the two rotate.

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However, if I had the situation

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such that this is
my massless rod

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and here are
the two equal masses,

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but now I rotate them
aboutthis point,

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then this centripetal force
is larger than this one,

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so now I have asymmetry,

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and so now there will be
a force on this pin.

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The ruler will push on the pin,

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and action equals
minus reaction--

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the pin will push on the ruler.

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And it is because of this same
asymmetry that you have here

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that there will be a force
from the pin on point P.

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However, I don't care
about that force

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because I'm going to take
the torque relative to point P.

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And when I take a torque
relative to point P,

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any force through point P
has no effect,

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because the position vector
is zero.

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But I want you to appreciate
that there is a force,

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so if I take the torque
relative to point P,

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I do not worry about this force.

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Well, the torque relative
to that point P is zero,

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and so it's clear that angular
momentum relative to point P

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must be conserved.

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Angular momentum relative
to point P is conserved.

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Take any other point--

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it doesn't matter
which one you take;

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take this point Q,
take this point here,

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take this point here--

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and angular momentum
is not conserved.

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You immediately see that
if I take this point Q here,

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that this is
the position vector,

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and you see that r cross F
is not zero,

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so there is a torque
relative to point Q.

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Angular momentum
is not conserved.

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Only this point--
that point is very special.

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I now take the same ruler,
but I'm going to make it rotate

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about the center of mass.

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So here is now the same ruler,

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and now I'm going to rotate
about the center of mass.

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Rotation-- same direction.

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I have a stationary axis
in 26.100,

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and it's rotating about there.

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Center of mass--
C is my center of mass.

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Now there is no force on this
pin because of the symmetry,

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which I just explained,

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so the pin is not pushing
on the rod either,

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so there is no force at all if
this is a frictionless object.

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So if there is no force at all,

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then the torque relative
toany point must be zero--

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not only relative to this point,
but also relative to this point

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and relative to this point.

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Because if the force is zero,

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then any cross product
of r and F is zero.

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So now you see the
special case that I alluded to

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in equation number six.

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Now you have the case

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that the angular momentum
of this rotating object

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is the same, no matter
which point you choose.

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We call that
the spin angular momentum.

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It is an intrinsic property
of a spinning object,

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and it is the same relative
to any point that you choose.

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And if you want to know
how large it is,

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well, the magnitude
of the angular momentum--

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I calculated about C,
which is the center of mass--

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is simply equation number six,

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is the moment of inertia
about that point C,

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which is the center
of mass,

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times omega
about the center of mass,

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so CM-- center of mass--
and C are the same point.

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What is the moment
of inertia of rotation

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about the center of mass?

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That is my 1/12 Ml squared.

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So now I have 1/12 Ml squared
times omega.

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That is now the intrinsic
spin angular momentum,

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and it is the same for
any point that you choose.

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Even if I choose a point
here in space,

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you can prove that it is still
the same angular momentum.

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Now we're going to look
at other applications.

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There are a huge number
of applications,

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and in the next four
or five lectures,

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we will go through many of them.

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In some cases, you will say,

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"Yeah, yeah, that's intuitive; 
that's obvious."

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In some cases, you will say,
"Hmm, not so intuitive."

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And in some cases,
you will fall off your chair--

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it is completely bizarre.

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It is so nonintuitive that
you don't even believe it.

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You won't believe it
until I actually show you

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with a demonstration that
that's the way nature works.

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But that comes later, not today.

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On assignment number seven,

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I give you a problem--
problem 7-9--

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in which I have a ruler or a rod

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on a frictionless
horizontal table,

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and the ruler has a mass l...
a mass M and it has a length l,

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and here is the center
of mass at C.

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And I hit that ruler--
I give it an impulse--

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perpendicular
to the direction of the ruler.

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Here, I give it an impulse.

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A force acts upon it
for a certain amount of time.

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And let this distance be...

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I think I call it "d"
in your problem.

286
00:15:24 --> 00:15:30
And it's on
a frictionless table.

287
00:15:28 --> 00:15:34
I do this-- bang--
very short hit.

288
00:15:33 --> 00:15:39
And now the question is,
what will this object do?

289
00:15:36 --> 00:15:42
What does your instinct
tell you?

290
00:15:40 --> 00:15:46
You will say, "Well, for sure,

291
00:15:42 --> 00:15:48
it will move
in this direction."

292
00:15:43 --> 00:15:49
Remember it's frictionless, so
whatever happens after the hit

293
00:15:46 --> 00:15:52
happens forever and ever
and ever.

294
00:15:48 --> 00:15:54
It will never stop.

295
00:15:49 --> 00:15:55
So you all say,

296
00:15:50 --> 00:15:56
"Well, it's going to move
in this direction."

297
00:15:52 --> 00:15:58
That's very vague,
but that's true.

298
00:15:53 --> 00:15:59
But how about rotation?

299
00:15:55 --> 00:16:01
Will it rotate?

300
00:15:57 --> 00:16:03
And if it rotates,
about which point?

301
00:16:00 --> 00:16:06
Well, you may say,

302
00:16:01 --> 00:16:07
"Well, maybe where... the point
of rotation may depend

303
00:16:04 --> 00:16:10
on where I hit it."

304
00:16:06 --> 00:16:12
That would be a reasonable
intuition, but it's not true.

305
00:16:10 --> 00:16:16
It must rotate
about the center of mass.

306
00:16:12 --> 00:16:18
It cannot rotate
about any other point.

307
00:16:15 --> 00:16:21
Suppose it rotated
about this point,

308
00:16:17 --> 00:16:23
just for sake of argument.

309
00:16:18 --> 00:16:24
So after the hit,
it's rotated about this point.

310
00:16:22 --> 00:16:28
That would mean that the center
of mass would then do this.

311
00:16:26 --> 00:16:32
And that's not allowed, because
the center of mass behaves

312
00:16:29 --> 00:16:35
like a point in which all
the mass is concentrated.

313
00:16:34 --> 00:16:40
F equals Ma holds
for the center of mass.

314
00:16:36 --> 00:16:42
And so after the hit,
a single point--

315
00:16:39 --> 00:16:45
because the center of mass can
always be considered to be

316
00:16:42 --> 00:16:48
a single point with
all the mass in it.

317
00:16:44 --> 00:16:50
That point can never do this.

318
00:16:45 --> 00:16:51
That would be absurd.

319
00:16:46 --> 00:16:52
So from that reasoning alone,
you must conclude

320
00:16:50 --> 00:16:56
that the only way that nature
can digest that impulse

321
00:16:57 --> 00:17:03
is by giving the center of mass
a certain velocity

322
00:17:04 --> 00:17:10
in this direction,
which will never change.

323
00:17:08 --> 00:17:14
In addition, it will give it
an angular velocity

324
00:17:12 --> 00:17:18
about the center of mass.

325
00:17:14 --> 00:17:20
And if there is no friction,
that will never change.

326
00:17:16 --> 00:17:22
So a little later in time,

327
00:17:20 --> 00:17:26
the center of mass will still
be on this line,

328
00:17:24 --> 00:17:30
and it may have rotated one
or two or three rotations.

329
00:17:27 --> 00:17:33
That depends... that depends
on how large I is

330
00:17:30 --> 00:17:36
and that depends
on this distance d,

331
00:17:32 --> 00:17:38
but let's assume
that it is now like so,

332
00:17:37 --> 00:17:43
and then it is happily rotating

333
00:17:39 --> 00:17:45
with that angular velocity,
omega C,

334
00:17:42 --> 00:17:48
and continues all the time

335
00:17:44 --> 00:17:50
with that velocity
v center of mass.

336
00:17:47 --> 00:17:53
And in that problem, I'm asking
you to calculate the velocity

337
00:17:50 --> 00:17:56
of the center of mass

338
00:17:52 --> 00:17:58
and to calculate
the angular velocity

339
00:17:54 --> 00:18:00
about the center of mass.

340
00:17:56 --> 00:18:02
I'll help you a little bit,

341
00:17:59 --> 00:18:05
and I hope that PIVoT
will help you a little more.

342
00:18:02 --> 00:18:08
For the center of mass,
what must always hold...

343
00:18:06 --> 00:18:12
The center of mass acts
like a point source,

344
00:18:09 --> 00:18:15
so it must always hold that F
equals Ma of the center of mass.

345
00:18:14 --> 00:18:20
That is really one
of the major characteristics

346
00:18:16 --> 00:18:22
of the center of mass.

347
00:18:18 --> 00:18:24
So if we look at magnitude--

348
00:18:22 --> 00:18:28
F delta t equals M times
a center of mass delta T--

349
00:18:27 --> 00:18:33
the force acts for a certain
amount of time,

350
00:18:31 --> 00:18:37
and that's the impulse.

351
00:18:33 --> 00:18:39
But if-- before I hit--
if this velocity is zero,

352
00:18:39 --> 00:18:45
then a delta t is obviously

353
00:18:42 --> 00:18:48
the velocity of the center
of mass afterwards.

354
00:18:44 --> 00:18:50
v equals at, right,
and v was zero to start with.

355
00:18:49 --> 00:18:55
So you see now immediately
that I-- that is the impulse--

356
00:18:54 --> 00:19:00
equals M times the velocity
of the center of mass,

357
00:18:58 --> 00:19:04
and so the velocity
of the center of mass

358
00:19:02 --> 00:19:08
equals I divided by M.

359
00:19:04 --> 00:19:10
And so I have solved for you
the first part of that problem.

360
00:19:09 --> 00:19:15
And what is remarkable--

361
00:19:11 --> 00:19:17
that it's independent of b...
or d, sorry.

362
00:19:13 --> 00:19:19
We called that d.

363
00:19:14 --> 00:19:20
It's completely independent
of d.

364
00:19:17 --> 00:19:23
Whether you hit here
with the very same impulse

365
00:19:20 --> 00:19:26
or hit there makes
no difference.

366
00:19:22 --> 00:19:28
The center of mass behaves like
a point source-- F equals Ma--

367
00:19:26 --> 00:19:32
and the velocity of the center
of mass only depends on I

368
00:19:31 --> 00:19:37
and on the mass of the object.

369
00:19:32 --> 00:19:38
The larger I,
the larger the velocity.

370
00:19:34 --> 00:19:40
That's, of course,
quite intuitive.

371
00:19:36 --> 00:19:42
And the larger the mass,
the lower the velocity.

372
00:19:40 --> 00:19:46
But now comes the hardest part,

373
00:19:42 --> 00:19:48
and I will leave you
largely with that hardest part

374
00:19:45 --> 00:19:51
but give you a few clues,
because now I want you

375
00:19:49 --> 00:19:55
to calculate omega
about the center of mass.

376
00:19:53 --> 00:19:59
And now you have to choose
an origin,

377
00:19:54 --> 00:20:00
because you're dealing
with torques,

378
00:19:56 --> 00:20:02
you're dealing with changing
angular momentum,

379
00:20:00 --> 00:20:06
and so you would probably
choose C as your origin.

380
00:20:04 --> 00:20:10
And you will say,

381
00:20:05 --> 00:20:11
"Okay, let's think about
what the angular momentum is

382
00:20:09 --> 00:20:15
"relative to point C before the
collision, after the collision--

383
00:20:14 --> 00:20:20
"if you call it a collision.

384
00:20:15 --> 00:20:21
What is the torque relative
to point C?"

385
00:20:20 --> 00:20:26
Well, the torque relative
to point C,

386
00:20:23 --> 00:20:29
there is here the position
vector r of C,

387
00:20:29 --> 00:20:35
and the torque is r cross F,

388
00:20:31 --> 00:20:37
and this F acts for
a certain amount of time,

389
00:20:34 --> 00:20:40
so this certainly is not zero.

390
00:20:37 --> 00:20:43
You can clearly see that there
is a torque relative to point C,

391
00:20:41 --> 00:20:47
and if there is a torque
relative to point C,

392
00:20:43 --> 00:20:49
the angular momentum relative
to point C must be changing.

393
00:20:46 --> 00:20:52
Yeah, you bet your
life it's changing,

394
00:20:49 --> 00:20:55
because the angular momentum
before the hit is zero,

395
00:20:55 --> 00:21:01
butafter the hit,

396
00:20:56 --> 00:21:02
the angular momentum is
I about the center of mass

397
00:21:01 --> 00:21:07
times omega, center of mass.

398
00:21:03 --> 00:21:09
You can see that if it rotates
about the center of mass,

399
00:21:06 --> 00:21:12
it has an angular momentum

400
00:21:08 --> 00:21:14
of rotation about the center
of mass which is I omega.

401
00:21:12 --> 00:21:18
Before, it was zero; 
afterwards, it's this.

402
00:21:14 --> 00:21:20
So clearly there must be a
torque about the center of mass.

403
00:21:18 --> 00:21:24
And so now, all you
would have to do

404
00:21:21 --> 00:21:27
is to somehow relate that change
in angular momentum

405
00:21:25 --> 00:21:31
with this distance d and with I.

406
00:21:28 --> 00:21:34
And you will see that if d
becomes larger,

407
00:21:31 --> 00:21:37
that indeed the omega
of center of mass--

408
00:21:35 --> 00:21:41
the angular velocity of our
center of mass-- will increase.

409
00:21:38 --> 00:21:44
It's very sensitive.

410
00:21:40 --> 00:21:46
That's also intuitive because
if you hit right in the middle,

411
00:21:43 --> 00:21:49
you didn't expect it
to rotate at all.

412
00:21:45 --> 00:21:51
You would just expect
this object to translate.

413
00:21:49 --> 00:21:55
And so clearly, the velocity
of the center of mass

414
00:21:53 --> 00:21:59
is insensitive of where you hit,

415
00:21:55 --> 00:22:01
but the angular velocity
around the center of mass

416
00:21:58 --> 00:22:04
is very sensitive,
as I will show you shortly.

417
00:22:00 --> 00:22:06
If you hit here, you don't
expect any rotation.

418
00:22:05 --> 00:22:11
Some of you may say, "I don't
like to take C as my origin.

419
00:22:09 --> 00:22:15
"I'm going to pick another point
as my origin.

420
00:22:11 --> 00:22:17
"I'm going to take this point P
as my origin

421
00:22:15 --> 00:22:21
or any other point
on this line."

422
00:22:20 --> 00:22:26
Angular momentum must be
conserved about that point,

423
00:22:24 --> 00:22:30
because there is no torque
relative to that point.

424
00:22:28 --> 00:22:34
Because any position vector
from this point to here

425
00:22:32 --> 00:22:38
is in the same direction as
the force, so r cross F is zero.

426
00:22:37 --> 00:22:43
So now we have a point--
an infinite number of points--

427
00:22:41 --> 00:22:47
all these points here on this
line, the torque is zero.

428
00:22:46 --> 00:22:52
So the torque relative to point
P is zero,

429
00:22:50 --> 00:22:56
so there is angular momentum L
relative to point P

430
00:22:54 --> 00:23:00
is conserved.

431
00:22:58 --> 00:23:04
It is zero before,
and so it's zero after the hit.

432
00:23:04 --> 00:23:10
Isn't that interesting?

433
00:23:05 --> 00:23:11
And you can solve that problem

434
00:23:08 --> 00:23:14
and find exactly the same answer
for the angular velocity

435
00:23:11 --> 00:23:17
about the center of mass
by choosing point P.

436
00:23:13 --> 00:23:19
It's just a matter of taste.

437
00:23:15 --> 00:23:21
Most of you would
probably pick C,

438
00:23:17 --> 00:23:23
but if you pick point P,

439
00:23:20 --> 00:23:26
the problem may even be
a little easier.

440
00:23:23 --> 00:23:29
Both should work,

441
00:23:24 --> 00:23:30
and I will make sure
that the solutions to both

442
00:23:26 --> 00:23:32
will appear
on the final solutions.

443
00:23:30 --> 00:23:36
So you will see that it can be
done in two different ways.

444
00:23:34 --> 00:23:40
I have here a ruler with a...
which I can hit.

445
00:23:41 --> 00:23:47

446
00:23:48 --> 00:23:54
Here's this ruler.

447
00:23:50 --> 00:23:56
This is the center of mass, and
this is just a reference line.

448
00:23:55 --> 00:24:01
What I will try is to hit it

449
00:23:57 --> 00:24:03
in a direction parallel
to the reference line.

450
00:24:00 --> 00:24:06
If I succeed, then it
may start to rotate,

451
00:24:04 --> 00:24:10
but the center of mass
should stay on that line,

452
00:24:06 --> 00:24:12
depending upon where I hit.

453
00:24:08 --> 00:24:14
However, if I don't hit it
exactly parallel to this line,

454
00:24:11 --> 00:24:17
it will not stay on this line--
the center of mass--

455
00:24:13 --> 00:24:19
so don't start laughing then.

456
00:24:15 --> 00:24:21
It simply means that my angle is
not perfectly in this direction.

457
00:24:18 --> 00:24:24
If I hit like this,

458
00:24:20 --> 00:24:26
then obviously it will go
in this direction.

459
00:24:22 --> 00:24:28
So I will try to hit it first

460
00:24:26 --> 00:24:32
along the direction
of this line.

461
00:24:28 --> 00:24:34
I will hit it right in the
middle, so I expect no rotation.

462
00:24:32 --> 00:24:38
I only expect that the
center of mass will move.

463
00:24:35 --> 00:24:41
Now, there is lots
of friction here,

464
00:24:36 --> 00:24:42
so the thing will come
to a grinding halt.

465
00:24:38 --> 00:24:44
It will not go on
forever and ever.

466
00:24:40 --> 00:24:46
So let's try this.

467
00:24:41 --> 00:24:47
And you see that's fantastic.

468
00:24:44 --> 00:24:50
It moved, it translates,
stays on the line,

469
00:24:46 --> 00:24:52
but it hardly rotates at all

470
00:24:48 --> 00:24:54
because I hit it very close
to the center of mass.

471
00:24:51 --> 00:24:57
Now I'm going to hit here,
far below the center of mass,

472
00:24:55 --> 00:25:01
so now I expect this point
to stay on the line

473
00:24:57 --> 00:25:03
if I'm lucky enough that I hit
it in the right direction,

474
00:25:01 --> 00:25:07
and then, of course, it will
start to rotate like this.

475
00:25:03 --> 00:25:09
And it will come to a halt
because of the friction.

476
00:25:07 --> 00:25:13
So let's try that.

477
00:25:10 --> 00:25:16
Not bad.

478
00:25:12 --> 00:25:18
You see it stays very close
to the center, to that line.

479
00:25:15 --> 00:25:21
That means my hit was almost
perfectly in this direction,

480
00:25:17 --> 00:25:23
and you saw it rotate.

481
00:25:19 --> 00:25:25
I'll do it once more.

482
00:25:24 --> 00:25:30
Oh, look at that.

483
00:25:25 --> 00:25:31
It goes much farther.

484
00:25:26 --> 00:25:32
It rotated over
a much larger angle.

485
00:25:28 --> 00:25:34
The center of mass indeed
is very close to this line,

486
00:25:30 --> 00:25:36
so I was lucky, and my direction
in which I hit it

487
00:25:34 --> 00:25:40
was quite close to being
parallel to that black line.

488
00:25:40 --> 00:25:46

489
00:25:45 --> 00:25:51
Now comes another application
of the... use of torques.

490
00:25:55 --> 00:26:01
And that has to do with
simple harmonic oscillations.

491
00:26:01 --> 00:26:07
We were able to calculate

492
00:26:03 --> 00:26:09
the period of oscillation
of a pendulum.

493
00:26:06 --> 00:26:12
That was relatively easy.

494
00:26:07 --> 00:26:13
Most of you have even learned
how to do that in high school--

495
00:26:10 --> 00:26:16
a string, no mass, an object at
the end, and you give it a kick,

496
00:26:13 --> 00:26:19
and it goes simple harmonic
if the angle is not too large.

497
00:26:19 --> 00:26:25
But now suppose you have
a ruler like this,

498
00:26:22 --> 00:26:28
and you have here
a little pinhole,

499
00:26:24 --> 00:26:30
and you're going
to oscillate it like this.

500
00:26:28 --> 00:26:34
What now is the period
of oscillation?

501
00:26:31 --> 00:26:37
That looks like
a mathematical headache.

502
00:26:35 --> 00:26:41
And yet, with the knowledge
that we have now,

503
00:26:37 --> 00:26:43
this can be solved quite easily,

504
00:26:40 --> 00:26:46
and we can make a quite
accurate prediction

505
00:26:42 --> 00:26:48
about the period of oscillation
of this rather complex system.

506
00:26:48 --> 00:26:54
Even a system as bizarre
as this hula hoop,

507
00:26:52 --> 00:26:58
which oscillates about that pin,

508
00:26:54 --> 00:27:00
can now be calculated
very easily

509
00:26:57 --> 00:27:03
with the knowledge that we have,

510
00:26:59 --> 00:27:05
or at least that you will
have very shortly

511
00:27:02 --> 00:27:08
if you spend the time
and you study this.

512
00:27:06 --> 00:27:12
So let's first do the...

513
00:27:09 --> 00:27:15
We first take the ruler.

514
00:27:13 --> 00:27:19
This is a ruler.

515
00:27:16 --> 00:27:22
The ruler has mass M
and length l.

516
00:27:20 --> 00:27:26
This is the center of mass
of the ruler, and I make...

517
00:27:24 --> 00:27:30
I force it to rotate about point
P, put a pin through there,

518
00:27:31 --> 00:27:37
nearly frictionless,

519
00:27:33 --> 00:27:39
and the separation between
the center of mass and P,

520
00:27:38 --> 00:27:44
I call that now "b",
as in "bridge."

521
00:27:41 --> 00:27:47

522
00:27:46 --> 00:27:52
This angle is theta.

523
00:27:48 --> 00:27:54
It rotates about the pin
perpendicular to the blackboard.

524
00:27:53 --> 00:27:59

525
00:27:58 --> 00:28:04
For sure, there will be forces
through that point P.

526
00:28:02 --> 00:28:08
I don't care about that

527
00:28:04 --> 00:28:10
because now I'm not going
to negotiate with you

528
00:28:06 --> 00:28:12
about which point I'm
going to take the torque.

529
00:28:09 --> 00:28:15
I will insist on taking torques

530
00:28:11 --> 00:28:17
exclusively
through that point P,

531
00:28:13 --> 00:28:19
so that I get rid of any forces
that go through that pin--

532
00:28:17 --> 00:28:23
any forces that the pin
exerts on the object.

533
00:28:20 --> 00:28:26
Gone-- don't care about it.

534
00:28:22 --> 00:28:28
All I worry about now is this Mg
and this position vector rP.

535
00:28:34 --> 00:28:40
That determines the torque.

536
00:28:37 --> 00:28:43
And the magnitude
of that torque--

537
00:28:40 --> 00:28:46
and I'm sure that torque
is going to change in time--

538
00:28:43 --> 00:28:49
equals Mg times this distance b,

539
00:28:48 --> 00:28:54
but I have to take the sine
of this angle theta

540
00:28:50 --> 00:28:56
because remember, it is a cross
product between r and F.

541
00:28:56 --> 00:29:02
And the cross product has
the sine of the angle in it.

542
00:28:59 --> 00:29:05
So the magnitude of that

543
00:29:01 --> 00:29:07
is going to be Mg times b
times the sine of theta.

544
00:29:08 --> 00:29:14
This is r cross F--
the magnitude of r cross F--

545
00:29:12 --> 00:29:18
this being F, this being r.

546
00:29:15 --> 00:29:21

547
00:29:19 --> 00:29:25
I go to equation number four,

548
00:29:23 --> 00:29:29
and I say, aha, that must be
equal to the moment of inertia

549
00:29:28 --> 00:29:34
about point P-- about the axis
through the perpendicular

550
00:29:32 --> 00:29:38
to the blackboard
through point P--

551
00:29:34 --> 00:29:40
times alpha, which is
the angular acceleration,

552
00:29:40 --> 00:29:46
and, of course,
this may change with time

553
00:29:43 --> 00:29:49
and theta will change with time.

554
00:29:46 --> 00:29:52
However, there is
one important thing.

555
00:29:49 --> 00:29:55
This torque is
a restoring torque.

556
00:29:52 --> 00:29:58
It wants to drive it
back to equilibrium.

557
00:29:55 --> 00:30:01
And therefore, we need
a minus sign here

558
00:29:58 --> 00:30:04
for the very same reason
that when we did the spring

559
00:30:01 --> 00:30:07
that the force was minus kX
and not plus kX.

560
00:30:06 --> 00:30:12
Now I say,
aha, small angle approximation.

561
00:30:10 --> 00:30:16
By the way, alpha equals
omega dot--

562
00:30:16 --> 00:30:22
first derivative
of the angular velocity.

563
00:30:19 --> 00:30:25
Therefore, it also equals
theta double dot.

564
00:30:21 --> 00:30:27
I also wrote that there.

565
00:30:23 --> 00:30:29
And theta is this angle.

566
00:30:25 --> 00:30:31
This is the change of this angle
that you see there--

567
00:30:28 --> 00:30:34
the second derivative.

568
00:30:30 --> 00:30:36
I say, aha, small angle
approximation,

569
00:30:32 --> 00:30:38
then the sine of theta
is theta--

570
00:30:34 --> 00:30:40
we've done that before, provided
the angle is in radians.

571
00:30:38 --> 00:30:44
And so I'm going to get

572
00:30:39 --> 00:30:45
that Mg b theta plus I
relative to P

573
00:30:48 --> 00:30:54
times theta double dot
equals zero.

574
00:30:53 --> 00:30:59
And now I already begin
to feel it in my bloodstream.

575
00:30:57 --> 00:31:03
I already begin to smell

576
00:30:59 --> 00:31:05
the simple harmonic oscillation,
and you, too.

577
00:31:03 --> 00:31:09
Let's bring the theta double dot
naked out here

578
00:31:08 --> 00:31:14
plus Mg b divided by I of P
times theta equals zero,

579
00:31:17 --> 00:31:23
and this, for sure, is a simple
harmonic oscillation in theta.

580
00:31:24 --> 00:31:30
And the solution must be

581
00:31:26 --> 00:31:32
that theta in time must
be some maximum angle

582
00:31:31 --> 00:31:37
times the cosine
of omega t plus phi.

583
00:31:37 --> 00:31:43
This omega hasnothing
to do with that omega.

584
00:31:41 --> 00:31:47
This is angular velocity, and
this is angular frequency.

585
00:31:48 --> 00:31:54
This will never change.

586
00:31:50 --> 00:31:56
That's related to the
period of oscillation.

587
00:31:53 --> 00:31:59
This is changing all the time.

588
00:31:55 --> 00:32:01
d theta/dt
is changing all the time.

589
00:31:58 --> 00:32:04
When the object is here,

590
00:31:59 --> 00:32:05
d theta/dt has
the largest value.

591
00:32:01 --> 00:32:07
When the object stands still,
d theta/dt is zero.

592
00:32:05 --> 00:32:11
So the angular velocity
is changing with time,

593
00:32:08 --> 00:32:14
but this omega is a constant,

594
00:32:10 --> 00:32:16
and it is very awkward
in physics

595
00:32:12 --> 00:32:18
that we use the same symbol
in one problem

596
00:32:14 --> 00:32:20
and have totally
different meaning.

597
00:32:16 --> 00:32:22
In any case, we know that this
omega, the angular frequency

598
00:32:20 --> 00:32:26
equals the square root
of Mg b...

599
00:32:26 --> 00:32:32
divided by I of P, and so
the period of an oscillation,

600
00:32:30 --> 00:32:36
which is 2pi divided
by omega has to be this.

601
00:32:38 --> 00:32:44
This, by the way, is independent
of the mass of the object.

602
00:32:42 --> 00:32:48
You may say, "Yeah, but
there is a mass downstairs."

603
00:32:45 --> 00:32:51
Yeah, yeah, but there
is also a mass upstairs,

604
00:32:47 --> 00:32:53
because the moment of inertia
about point P has a mass in it.

605
00:32:52 --> 00:32:58
The moment of inertia about
point P we find immediately

606
00:32:55 --> 00:33:01
by using the parallel
axis theorem.

607
00:32:58 --> 00:33:04
If it were rotating about the
center of mass, about this axis,

608
00:33:02 --> 00:33:08
then it would be
1/12 Ml squared,

609
00:33:06 --> 00:33:12
and so we have
to add Mb squared.

610
00:33:10 --> 00:33:16
We did something similar there.

611
00:33:11 --> 00:33:17
So this is 1/12 Ml squared

612
00:33:17 --> 00:33:23
plus M times b squared--
parallel axis theorem--

613
00:33:21 --> 00:33:27
and so now we find
that the period

614
00:33:24 --> 00:33:30
is 2pi times
the square root of...

615
00:33:28 --> 00:33:34
I lose my M, so I get
1/12 l squared plus b squared,

616
00:33:35 --> 00:33:41
and I get here gb.

617
00:33:38 --> 00:33:44
And that is a very simple result
when you come to think of it.

618
00:33:41 --> 00:33:47
I mean, this would have been
a complete headache

619
00:33:43 --> 00:33:49
to do it in any other way.

620
00:33:45 --> 00:33:51
And this always
baffles me so much.

621
00:33:48 --> 00:33:54
Some people... some of you think
that physics is difficult,

622
00:33:50 --> 00:33:56
but I always think of it
the other way around.

623
00:33:53 --> 00:33:59
I always think of that...

624
00:33:54 --> 00:34:00
physics is there to make
very difficult things easy.

625
00:33:58 --> 00:34:04
Look at it.

626
00:33:59 --> 00:34:05
This is an incredibly
complex system

627
00:34:01 --> 00:34:07
that is going to rotate
about a pin,

628
00:34:03 --> 00:34:09
which is offset from the center,

629
00:34:05 --> 00:34:11
and here is a very
straightforward prediction

630
00:34:07 --> 00:34:13
about the period.

631
00:34:09 --> 00:34:15
In our case,
a b equals 40 centimeters,

632
00:34:15 --> 00:34:21
and it is an exact meter stick.

633
00:34:17 --> 00:34:23
So l equals 1.00,

634
00:34:22 --> 00:34:28
and b equals, oh,
it's about 40 centimeters,

635
00:34:27 --> 00:34:33
but we could be off by something
like maybe two millimeters.

636
00:34:32 --> 00:34:38
It's not so easy to get
that whole in precisely.

637
00:34:37 --> 00:34:43
So it's 40 centimeters
plus or minus 0.2 centimeters.

638
00:34:43 --> 00:34:49
So that is an uncertainty
of about half a percent.

639
00:34:46 --> 00:34:52
I have upstairs l square
and b square

640
00:34:49 --> 00:34:55
and downstairs b,

641
00:34:51 --> 00:34:57
and that makes the error
analysis a bit complicated.

642
00:34:55 --> 00:35:01
If I stick in the numbers,

643
00:34:57 --> 00:35:03
then I find a period which will
be very close to 1.565 seconds.

644
00:35:04 --> 00:35:10
That's taking the exact numbers
that I gave you for l,

645
00:35:09 --> 00:35:15
and I take for g 9.8,
and I take the values for b.

646
00:35:13 --> 00:35:19
Since I prefer to be
on the conservative side,

647
00:35:16 --> 00:35:22
I will allow for an uncertainty
of about 0.01 seconds,

648
00:35:20 --> 00:35:26
and I would like to make
a measurement and show you...

649
00:35:24 --> 00:35:30
yeah, and show you what we get.

650
00:35:29 --> 00:35:35
So we have here the ruler.

651
00:35:34 --> 00:35:40
We're going to oscillate it...
small angle.

652
00:35:38 --> 00:35:44
Well, we can be
a little bit rougher,

653
00:35:40 --> 00:35:46
make it a little larger.

654
00:35:43 --> 00:35:49
Let's stop it first,

655
00:35:45 --> 00:35:51
and I will start it
when it stops at my side.

656
00:35:47 --> 00:35:53
That is always
the easiest to do.

657
00:35:50 --> 00:35:56
Now-- one, two,

658
00:35:54 --> 00:36:00
three, four,

659
00:35:57 --> 00:36:03
five, six,

660
00:36:00 --> 00:36:06
seven, eight,

661
00:36:03 --> 00:36:09
nine, ten.

662
00:36:07 --> 00:36:13
Look-- on the button,
way within the error.

663
00:36:12 --> 00:36:18
Amazing, isn't it, that
such complex systems

664
00:36:15 --> 00:36:21
can be so easily dealt
with in physics?

665
00:36:19 --> 00:36:25
It is 15... what is it-- .71,

666
00:36:22 --> 00:36:28
and my reaction time
is always .1 second,

667
00:36:26 --> 00:36:32
so the agreement is
absolutely fantastic.

668
00:36:31 --> 00:36:37
Now I will show you, once you
have done this with your ruler,

669
00:36:37 --> 00:36:43
you can now apply this knowledge

670
00:36:40 --> 00:36:46
on a way more
interesting system,

671
00:36:44 --> 00:36:50
and that is the, um... the hoop.

672
00:36:50 --> 00:36:56
We'll take the hula hoop,

673
00:36:54 --> 00:37:00
and we're going to do the
same thing for the hula hoop.

674
00:36:56 --> 00:37:02

675
00:37:01 --> 00:37:07
And the system, the approach,
will be exactly identical.

676
00:37:06 --> 00:37:12
We have a pin
through the hula hoop.

677
00:37:09 --> 00:37:15
This is the hula hoop,
this is point P,

678
00:37:15 --> 00:37:21
this is the center of mass--
I call it "C".

679
00:37:19 --> 00:37:25
It's now like this.

680
00:37:21 --> 00:37:27
A little later, the center
of mass will be here,

681
00:37:25 --> 00:37:31
and this angle will be theta.

682
00:37:27 --> 00:37:33
The mass of the hula hoop is M,
and it has a radius R.

683
00:37:33 --> 00:37:39
I know there will be forces
through that point P,

684
00:37:36 --> 00:37:42
but I couldn't care less
about them.

685
00:37:38 --> 00:37:44
All I worry about is this Mg
and this position vector rP.

686
00:37:45 --> 00:37:51
And the whole thing
goes parallel.

687
00:37:48 --> 00:37:54
I'm going to get

688
00:37:49 --> 00:37:55
that the torque about
that point P equals Mg...

689
00:37:56 --> 00:38:02
and I'll fill in for this
immediately the radius R...

690
00:38:01 --> 00:38:07
times the sine of theta.

691
00:38:05 --> 00:38:11
And that must be equal

692
00:38:06 --> 00:38:12
minus the moment of inertia
about point P

693
00:38:10 --> 00:38:16
times theta double dot.

694
00:38:14 --> 00:38:20
What is the moment
of inertia about point P?

695
00:38:17 --> 00:38:23
Well, it is the moment
of inertia of rotation

696
00:38:20 --> 00:38:26
about the center of mass
plus MR squared--

697
00:38:23 --> 00:38:29
that's the parallel
axis theorem--

698
00:38:25 --> 00:38:31
and the moment of inertia
for rotation about this axis

699
00:38:28 --> 00:38:34
is clearly MR squared, because
all the mass is distributed

700
00:38:31 --> 00:38:37
at the circumference
at distance R,

701
00:38:34 --> 00:38:40
so it is the summation
of all the masses

702
00:38:37 --> 00:38:43
times the radius squared.

703
00:38:38 --> 00:38:44
So this is going to be
MR squared plus MR squared,

704
00:38:45 --> 00:38:51
so that is 2MR squared.

705
00:38:49 --> 00:38:55
So we bring this to one side,

706
00:38:52 --> 00:38:58
so we're going to get
theta double dot plus MgR.

707
00:39:01 --> 00:39:07
We make this theta--
small angle approximation--

708
00:39:04 --> 00:39:10
divided by IP equals zero.

709
00:39:07 --> 00:39:13
Simple harmonic oscillation
in theta

710
00:39:10 --> 00:39:16
with exactly that same solution,
but now we get as a period,

711
00:39:14 --> 00:39:20
we get 2pi times the square root
of IP divided by MgR.

712
00:39:24 --> 00:39:30
But IP is 2MR squared.

713
00:39:29 --> 00:39:35
You see IP divided by Mg, not b,

714
00:39:32 --> 00:39:38
but now we have MgR...
divided by MgR,

715
00:39:39 --> 00:39:45
and that is 2pi times
the square root of 2R/g.

716
00:39:46 --> 00:39:52
And that is a cute result,
because if I had a pendulum

717
00:39:52 --> 00:39:58
and the pendulum had a length l
which is 2R

718
00:39:59 --> 00:40:05
and I put all the mass
at the end of the pendulum

719
00:40:02 --> 00:40:08
and the string has no mass,

720
00:40:04 --> 00:40:10
I would have found
exactly the same result.

721
00:40:06 --> 00:40:12
Remember, the period
of a pendulum

722
00:40:08 --> 00:40:14
is 2pi times the square root
of l/g.

723
00:40:11 --> 00:40:17
Notice again there is
no mass in there.

724
00:40:15 --> 00:40:21
It's only a matter of geometry.

725
00:40:16 --> 00:40:22
Geometry is the only thing
that determines

726
00:40:18 --> 00:40:24
the period of oscillation.

727
00:40:20 --> 00:40:26
That's always the case.

728
00:40:21 --> 00:40:27
With a pendulum--
a simple pendulum--

729
00:40:23 --> 00:40:29
with a mass at the end

730
00:40:24 --> 00:40:30
is also the period is
independent of the mass,

731
00:40:27 --> 00:40:33
not with springs,
but always with gravity.

732
00:40:32 --> 00:40:38
So the first thing I want
to show you is

733
00:40:34 --> 00:40:40
that if I have here
an apple on a string,

734
00:40:40 --> 00:40:46
and to the center of the apple

735
00:40:42 --> 00:40:48
is about the same to here,
to this pin,

736
00:40:45 --> 00:40:51
is about the same distance
as the diameter of the...

737
00:40:48 --> 00:40:54
of the hula hoop,

738
00:40:50 --> 00:40:56
that they probably have
very closely the same period.

739
00:40:56 --> 00:41:02
You see, they track
each other quite well; 

740
00:40:58 --> 00:41:04
not precisely,

741
00:41:00 --> 00:41:06
because I was not able to get
the length exactly right,

742
00:41:03 --> 00:41:09
but very closely.

743
00:41:05 --> 00:41:11
So already you see that
indeed that result makes sense,

744
00:41:08 --> 00:41:14
but now I want to go
one step further.

745
00:41:11 --> 00:41:17
I want to do
a quantitative measurement

746
00:41:14 --> 00:41:20
because I know the radius
of this hula hoop.

747
00:41:19 --> 00:41:25
The radius, I think,
is 40 centimeters,

748
00:41:22 --> 00:41:28
and we know that to an accuracy
of about half a centimeter.

749
00:41:28 --> 00:41:34
So R is 40.0
plus or minus 0.5 centimeters.

750
00:41:35 --> 00:41:41
You will say, "Boy, can you
not measure that any better?

751
00:41:38 --> 00:41:44
"I mean, half a centimeter--

752
00:41:40 --> 00:41:46
it's almost yea big,
half a centimeter."

753
00:41:42 --> 00:41:48
Well, the reason is,
it's not a perfect circle.

754
00:41:44 --> 00:41:50
So if you measure the diameter
at various places,

755
00:41:47 --> 00:41:53
you don't always get the same,

756
00:41:48 --> 00:41:54
and so that's why I allow for
half a centimeter uncertainty.

757
00:41:51 --> 00:41:57
So that's a one percent
uncertainty in R, one percent.
 

758
00:41:56 --> 00:42:02
Under the square root,
that becomes half a percent.

759
00:42:00 --> 00:42:06
And so the prediction
then would be,

760
00:42:05 --> 00:42:11
if I put in the numbers
here that we have...

761
00:42:07 --> 00:42:13
is 1.795,

762
00:42:12 --> 00:42:18
and that would become
half a percent uncertainty,

763
00:42:16 --> 00:42:22
half of 1.79 is about one,

764
00:42:18 --> 00:42:24
so let's make it easy--
.01 seconds.

765
00:42:22 --> 00:42:28
That is my prediction, and I'm
going to make ten oscillations,

766
00:42:28 --> 00:42:34
which gives me
an uncertainty

767
00:42:30 --> 00:42:36
of 0.1 seconds
in ten oscillations,

768
00:42:33 --> 00:42:39
so that gives me an uncertainty
of .01 in one oscillation.

769
00:42:40 --> 00:42:46
Imagine that you come home
at Thanksgiving,

770
00:42:44 --> 00:42:50
and you show this to
your parents, and you say,

771
00:42:47 --> 00:42:53
"Oh, by the way, Dad and Mom,

772
00:42:49 --> 00:42:55
would you be able to calculate
the period of this oscillation?"

773
00:42:53 --> 00:42:59
I mean, they would turn pale,
green, purple,

774
00:42:57 --> 00:43:03
and you take out your pocket,

775
00:42:59 --> 00:43:05
and you just do it
like that-- simple.

776
00:43:03 --> 00:43:09
And not only that,
but you can do a demonstration.

777
00:43:06 --> 00:43:12
You can show it to them.

778
00:43:07 --> 00:43:13
They'll be proud of you.

779
00:43:09 --> 00:43:15
(students laugh )

780
00:43:10 --> 00:43:16
Okay, let's try it.

781
00:43:15 --> 00:43:21
There we go.

782
00:43:16 --> 00:43:22
Small angle-- this is about
five degrees, seven degrees.

783
00:43:20 --> 00:43:26
It doesn't make
that much difference

784
00:43:21 --> 00:43:27
as long as you don't
get much above ten.

785
00:43:23 --> 00:43:29
I don't like the wiggle.

786
00:43:27 --> 00:43:33
I'll start it
when it stops here.

787
00:43:31 --> 00:43:37
Now!

788
00:43:32 --> 00:43:38
One, two,

789
00:43:35 --> 00:43:41
three, four,

790
00:43:39 --> 00:43:45
five, six,

791
00:43:43 --> 00:43:49
seven, eight,

792
00:43:46 --> 00:43:52
nine, ten.

793
00:43:50 --> 00:43:56
Holy smoke!

794
00:43:52 --> 00:43:58
Look at that.

795
00:43:54 --> 00:44:00
18.07... 18.07.

796
00:44:00 --> 00:44:06
If I divide that by ten,
then I get 1.807.

797
00:44:05 --> 00:44:11
Let's call it one
plus or minus 0.01.

798
00:44:11 --> 00:44:17
And if you take this error
into account and this...

799
00:44:14 --> 00:44:20
well, you can make that a 07...
pretty impressive.

800
00:44:20 --> 00:44:26
We have five minutes left,

801
00:44:21 --> 00:44:27
and I want to use
these five minutes...

802
00:44:23 --> 00:44:29
First, I'm going to ask
your forgiveness

803
00:44:25 --> 00:44:31
for what I'm going to do
the next five minutes

804
00:44:27 --> 00:44:33
because by now,
you may be lost already

805
00:44:30 --> 00:44:36
and I'm going to make it
a little worse

806
00:44:32 --> 00:44:38
by challenging you a little.

807
00:44:36 --> 00:44:42
And this challenge makes
life really interesting.

808
00:44:43 --> 00:44:49
If I spin an object-- a top--

809
00:44:46 --> 00:44:52
then clearly, there will always
be external friction

810
00:44:50 --> 00:44:56
on that top, and the
top will come to a halt.

811
00:44:53 --> 00:44:59
It's obvious.

812
00:44:55 --> 00:45:01
We've all seen it.

813
00:44:56 --> 00:45:02
You turn a top, you wait
a while, and it stops.

814
00:44:59 --> 00:45:05
And if it doesn't stop

815
00:45:00 --> 00:45:06
like it didn't when
I gave the exam review,

816
00:45:03 --> 00:45:09
then something stinks.

817
00:45:05 --> 00:45:11
Then you're being cheated,
and you were cheated.

818
00:45:07 --> 00:45:13
There was something going on
in that little black box

819
00:45:10 --> 00:45:16
which kept it going, of course.

820
00:45:12 --> 00:45:18
It can't keep going forever.

821
00:45:15 --> 00:45:21
So whenever you spin something
in the laboratory here,

822
00:45:18 --> 00:45:24
different from outer space,

823
00:45:20 --> 00:45:26
there are always external
frictional torques on it

824
00:45:22 --> 00:45:28
and it will come to a halt.

825
00:45:25 --> 00:45:31
It is inconceivable

826
00:45:27 --> 00:45:33
that all by itself
it would come to a halt

827
00:45:29 --> 00:45:35
and then turn
the other way around.

828
00:45:31 --> 00:45:37
That would be absurd.

829
00:45:33 --> 00:45:39
It couldn't possibly happen

830
00:45:34 --> 00:45:40
because the moment
it comes to a stop,

831
00:45:36 --> 00:45:42
all the external torques go
away, so it will just sit there.

832
00:45:41 --> 00:45:47
Imagine that you slide an object
on the table-- whoosh-- a plate,

833
00:45:45 --> 00:45:51
and it comes to a stop
because of friction

834
00:45:47 --> 00:45:53
and then it comes back at you--

835
00:45:49 --> 00:45:55
it would be equally
absurd, right?

836
00:45:51 --> 00:45:57
Unless you attached a spring
to it, of course.

837
00:45:53 --> 00:45:59
There are no other
external forces on it.

838
00:45:55 --> 00:46:01
You shove it on the table,

839
00:45:57 --> 00:46:03
friction takes out
the kinetic energy,

840
00:45:59 --> 00:46:05
it comes to a halt,
and it sits there.

841
00:46:01 --> 00:46:07
It couldn't possibly
come back at you--

842
00:46:03 --> 00:46:09
would be a violation
of basic laws of physics.

843
00:46:06 --> 00:46:12
So if you rotate...
if you spin a top,

844
00:46:09 --> 00:46:15
then it has to come to a halt,

845
00:46:11 --> 00:46:17
and it couldn't possibly rotate,
come to a halt

846
00:46:15 --> 00:46:21
and reverse its direction
of rotation.

847
00:46:18 --> 00:46:24
And as long as we agree to that,
then you will pass this course.

848
00:46:23 --> 00:46:29
Now, you already smell
a rat, don't you?

849
00:46:30 --> 00:46:36
You already smell a rat.

850
00:46:33 --> 00:46:39
Okay, I have here
a piece of plastic,

851
00:46:38 --> 00:46:44
a nice piece of plastic.

852
00:46:40 --> 00:46:46
You can come and examine it.

853
00:46:42 --> 00:46:48
And this plastic
behaves quite nicely.

854
00:46:45 --> 00:46:51
I will give it a spin.

855
00:46:47 --> 00:46:53

856
00:46:49 --> 00:46:55
Friction, external torque,

857
00:46:52 --> 00:46:58
friction will bring it
to a halt, and it stops.

858
00:46:56 --> 00:47:02
No problem, right?

859
00:46:57 --> 00:47:03
Inconceivable
that after the stop,

860
00:46:59 --> 00:47:05
it would reverse
the direction of motion,

861
00:47:03 --> 00:47:09
which it did just now.

862
00:47:07 --> 00:47:13
You may not believe your eyes.

863
00:47:08 --> 00:47:14
In fact, when I saw this first,
I did not believe my eyes.

864
00:47:12 --> 00:47:18
When I saw it again,
I felt sick in the stomach

865
00:47:15 --> 00:47:21
and couldn't sleep all night.

866
00:47:18 --> 00:47:24
I mean, look at this.

867
00:47:20 --> 00:47:26
I'm rotating it,
it comes to a halt.

868
00:47:22 --> 00:47:28
Friction brings it to a halt,
and it reverses direction.

869
00:47:26 --> 00:47:32
Don't have sleepless nights
about it,

870
00:47:29 --> 00:47:35
but give it some thought.

871
00:47:31 --> 00:47:37
Is this a complete violation

872
00:47:32 --> 00:47:38
of the conservation of angular
momentum of some kind?

873
00:47:35 --> 00:47:41
Think about it,
and see you on Wednesday.

874
00:47:39 --> 00:47:45

875
00:47:46 --> 00:47:52.000