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All right...
long weekend ahead of us.

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One more lecture to go.

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If I have an object,
mass m, in gravitational field,

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gravitational force
is in this direction,

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if this is
my increasing value of y,

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then this force,
vectorially written,

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equals minus mg y roof.

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Since this is
a one-dimensional problem,

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we will often simply write
F equals minus mg.

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This minus sign is important

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because that's
the increasing value of y.

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If this level here
is y equals zero,

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then I could call this

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gravitational
potential energy zero.

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And this is y...

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Then the gravitational potential
energy here equals plus mg y.

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This is u.

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So if I make a plot of the
gravitational potential energy

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as a function of y, then
I would get a straight line.

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This is zero.

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So this equals u...
equals mg y, plus sign.

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If I'm here at point A and
I move that object to point B,

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I, Walter Lewin, move it,
I have to do positive work.

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Notice that the gravitational
potential energy increases.

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If I do positive work, the
gravity is doing negative work.

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If I go from A to some
other point-- call it B prime--

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then I do negative work.

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Notice the gravitational
potential energy goes down.

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If I do negative work, then
gravity is doing positive work.

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I could have chosen my zero
point of potential energy

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anywhere I please.

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I could have chosen it
right here

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and nothing would change

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other than that I offset
the zero point

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of my potential energy.

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But again, if I go from A to B,

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the gravitational potential
energy increases

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by exactly the same amount--

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I have to do exactly
the same work.

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So you are free to choose,
when you are near Earth,

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where you choose your zero.

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Now we take the situation

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whereby we are not
so close to the Earth.

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Here is the Earth itself.

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Of course you can also
replace that by the sun

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if you want to.

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And this is
increasing value of r.

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The distance between here
and this object m equals r.

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I now know that there is

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a gravitational force
on this object--

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Newton's Universal
Law of Gravity--

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and that gravitational force

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equals minus m M-Earth G

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divided by this r squared,
r roof

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so this is a vectorial notation.

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Since it is really
one-dimensional, we would...

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Just like we did there,
we would delete the arrow

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and we would delete
the unit vector

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in the positive r direction

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and so we would
simply write it this way.

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The gravitational potential
energy we derived last time

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equals minus m M-Earth G

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divided by r--

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and notice, here is r
and here is r squared--

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and if you plot that, then
the plot goes sort of like this.

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This is r, this is
increasing potential energy--

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all these values here
are negative--

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and you get a curve
which is sort of like this.

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This is proportional
to one over r.

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Now, of course, if the Earth
had a radius which is this big,

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then, of course, this curve does
not exist, it stops right here.

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If I move from point A to point
B, with a mass m in my hand,

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notice that the gravitational
potential energy increases.

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I have to do positive work,
there is no difference.

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If I go from A
to another point, B prime,

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which is closer to the Earth,

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notice that the gravitational
potential energy decreases.

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I do negative work.

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If I do positive work,
gravity is doing negative work.

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If I do negative work,
gravity is doing positive work.

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Right here near Earth,

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where this one-over-r curve
hits the Earth,

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that is, of course,
exactly that line.

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That dependence on y is exactly
the same as the dependence on r

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and then you can simplify
matters when you are near Earth.

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When the gravitational
acceleration doesn't change

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you get a linear relation.

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But that's only
an exceptional case

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when you don't move very far.

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The gravitational force is

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in the direction opposite
the increasing potential energy.

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Notice that when I'm here,

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the gravitational force is
in this direction.

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Increasing potential energy is
this way.

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The force is in this way.

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When I'm here,

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gravitational potential energy
increases in this way; 

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the gravitational force
is in this direction.

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When I'm here,

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the gravitational
potential energy increases

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in this direction.

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The gravitational force is
in this direction.

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When I'm here,

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the gravitational
potential energy increases

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in this direction.

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The gravitational force is
in this direction.

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The force is always
in the opposite direction

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than the increasing value
of the potential energy.

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If I release an object
at zero speed,

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it therefore will always move

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towards a lower
potential energy

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because the force will drive it
to lower potential energy.

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Now I change from gravity
to a spring.

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I have a spring
which is relaxed length l--

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I call this x equals zero--

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and I extend it
over a distance x

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and there's a mass m at the end

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and there will be
a spring force.

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And that spring force...

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F equals minus kx.

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It's a one-dimensional situation

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so I can write it without having
to worry about the arrows.

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There is no friction here.

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It's clear
that if I hold this in my hand,

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that the force Walter Lewin
equals plus kx.

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It's in the direction
of increasing x.

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If I call this point A
at x equals zero

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and I call this point B
at x equals x,

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then I can calculate the work

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that I have to do
to bring it from A to B.

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So the work
that Walter Lewin has to do

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to bring it from A to B

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is the integral
in going from A to B

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of my force, dx.

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It's a dot product,

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but since the angle
between the two is zero degrees,

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the cosine of the angle is one,

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so I can forget about the fact
that there's a dot product.

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I move it in this direction.

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So that becomes the integral

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in going from zero
to a position x

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of plus x plus kx dx,

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and that is one-half k
x squared.

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And this is what we call

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the potential energy
of the spring.

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This is potential energy.

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It means, then, that we...

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At x equals zero, we define
potential energy to be zero.

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You don't have to do that,

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but it would be ridiculous
to do it any other way.

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So in the case that we have the
near-Earth situation of gravity,

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we had a choice where we put
our zero potential energy.

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In the case that we deal
with very large distances,

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we do not have a choice anymore.

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We defined it in such a way

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that the potential energy
at infinity is zero.

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As a result of that, all
potential energies are negative.

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And here, with the spring,
you don't have a choice, either.

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You choose...

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at x equals zero, you choose
potential energy to be zero.

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So if you now make a plot

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of the potential energy
as a function of x,

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you get a parabola,

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and if you are here,
if the object is here,

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then the force is always

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in the direction opposing
the increasing potential energy.

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If you go this way,
potential energy increases.

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So it's clear that the force is
going to be in this direction.

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If you are here, the force is

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always in the direction opposing
increasing potential energy.

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Increasing potential energy
is in this direction,

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so the force is
in this direction.

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You see, it's a restoring force.

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The force is always
in the direction

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opposite to increasing
potential energy.

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If you release an object here
at zero speed,

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it will therefore go
towards lower potential energy.

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The force will drive it
to lower potential energy.

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If we know the force--
in this case, the spring force

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or in those cases,
the gravitational force--

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we were able to calculate
the potential energy.

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Now, the question is,
can we also go back?

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Suppose we knew
the potential energy.

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Can we then find
the force again?

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And the answer is yes, we can.

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Let's take the situation
of the spring first.

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We have that the potential
energy u of the spring

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equals plus one-half k x squared

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and if I take the derivative
of that versus x

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then I get plus kx,

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but the force, the spring force
itself, is minus kx,

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so this equals
minus the spring force.

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So we have
that du/dx equals minus F,

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and I put the x there because
it's a one-dimensional problem--

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it is only in the x direction.

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The minus sign is telling you

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that the force is always
pointing in the direction

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which is opposite to increasing
values of the potential energy.

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That is what
the minus sign is telling you.

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It's staring you in the face

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what I have been telling you
for the past five minutes.

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If we have
a three-dimensional situation

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that we know
the potential energy

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as a function of x, y and z,

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then we can go back
and find the forces

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as a function of x, y and z.

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It doesn't matter
whether these are springs

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or whether it is gravity

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or whether it's electric forces
or nuclear forces,

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you then find
that du/dx equals minus F of x,

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du/dy equals minus Fy

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and du/dz equals minus Fz.

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What does this mean?

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It means that if you're
in three-dimensional space,

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you move only
in the x direction.

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You keep y and z constant,
and the change equals minus Fx.

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That gives you the component
of the force in the x direction.

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You move only
in the y direction,

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you keep x and z constant,

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and then you find
the component of the force

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in the y direction.

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We call these
partial derivatives,

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so we don't give them a "d"

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but we give them
a little curled delta.

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If we go back to the situation
where we had gravity,
 

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we had there the situation
near Earth.

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We had u... was plus mg y,
so what is du/dy?

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This is
a one-dimensional situation

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so I don't have to use
the partial derivatives.

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I can simply say du/dy.

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That is plus mg, and notice

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that the gravitational force
was minus mg.

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Remember? The minus sign is
still there.

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It's still there.

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And so you see
that here, indeed,

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du/dy is minus
the gravitational force.

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Now we take the situation
that we are not near Earth--

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we have there--

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so we have u equals
minus m M-Earth G

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divided by r--

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there's only an r here--

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so du/dr...

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The derivative of one over r
isminus one over r squared.

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The minus sign eats up
this minus sign,

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so I get plus m M-Earth G
divided by r squared

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so the gravitational force...

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the gravitational force
equals minus that.

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It isminus du/dr

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and, indeed, that's
exactly what we have there--

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minus that value.

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So whenever you know

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the potential
as a function of space,

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you can always find the
three components of the forces

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in the three
orthogonal directions.

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Suppose I have
a curved surface--

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literally,
a surface here in 26.100,

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which sort of looks like this...

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something like this.

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I call this, arbitrarily,
y equals zero

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and I could call this

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u gravitational potential
energy zero, for that matter.

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So this is a function y
as a function of x

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and the curve itself
represents effectively

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the gravitational
potential energy.

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This is y and this is x.

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So the gravitational potential
energy u equals mg y,

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but y is a function of x,
so that is also u times m...

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excuse me, that is m times g
times that function of x.

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There are points here
where du/dx equals zero.

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I'll get a nice mg in here...

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Where's zero...
and where are those points?

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Those points are here, here,
here, here and here.

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If du/dx is zero,
it means that the force--

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the component of the force
in the x direction-- is zero,

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because du/dx is minus
the force in the x direction.

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So if we visit those points,
for instance here,

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then there is, of course,
gravity, mg,

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if there is an object
there in the y direction...

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in the minus y direction

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and there is a normal force
in the plus y direction

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and these two exactly cancel
each other.

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So the net result is that here,
here, here, here and there

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there is no force
on the object at all

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so the object is not going to
move, it's going to stay put.

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Well, yes,
it's going to stay put.

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However, there is
a huge difference

293
00:17:19 --> 00:17:25
between this point here
and that point there

294
00:17:22 --> 00:17:28
and you sense immediately
that difference.

295
00:17:25 --> 00:17:31
If I put a marble here,

296
00:17:27 --> 00:17:33
I will have a hell of a time
to keep the marble in place,

297
00:17:30 --> 00:17:36
because if there is a fly there
in the corner of 26.100

298
00:17:36 --> 00:17:42
which does something

299
00:17:37 --> 00:17:43
then the slightest amount
of force on this one

300
00:17:40 --> 00:17:46
and it will start to roll off.

301
00:17:42 --> 00:17:48
In fact, what will happen is

302
00:17:44 --> 00:17:50
it will go
to a lower potential energy.

303
00:17:47 --> 00:17:53
Here, however,
if this one is offset,

304
00:17:50 --> 00:17:56
then it will want to go
to a lower potential energy.

305
00:17:54 --> 00:18:00
The force is always opposing

306
00:17:56 --> 00:18:02
the direction
of increasing potential energy,

307
00:17:59 --> 00:18:05
so the force will drive it back

308
00:18:00 --> 00:18:06
and so that's why we call this
a stable equilibrium.

309
00:18:03 --> 00:18:09
It will always go back
to that point.

310
00:18:05 --> 00:18:11
And this is
an unstable equilibrium.

311
00:18:11 --> 00:18:17
We have a setup here,

312
00:18:13 --> 00:18:19
and I would like to show you
how that will work.

313
00:18:19 --> 00:18:25
So we do have something
that is a curved object.

314
00:18:24 --> 00:18:30
It's a track.

315
00:18:26 --> 00:18:32
Let me give you a little
bit better light condition.

316
00:18:30 --> 00:18:36
So you see there, there is
that object, a little ball.

317
00:18:36 --> 00:18:42
And no surprise, if I offset it
from the lowest point

318
00:18:40 --> 00:18:46
that it will be driven back
to that point-- that's trivial.

319
00:18:43 --> 00:18:49
What is less trivial is
that there is a point here

320
00:18:47 --> 00:18:53
whereby, indeed,
the net forces are zero

321
00:18:50 --> 00:18:56
and it is not easy
to achieve that,

322
00:18:53 --> 00:18:59
but I will try to put it there
so that it, indeed, stays put.

323
00:18:59 --> 00:19:05
I'm not too fortunate,
it is very difficult.

324
00:19:04 --> 00:19:10
I'm trying... no...

325
00:19:07 --> 00:19:13
Yeah! Did it.

326
00:19:09 --> 00:19:15
It's there, it's very unstable.

327
00:19:13 --> 00:19:19
I blow...
oh, it's not so unstable.

328
00:19:16 --> 00:19:22
And there it goes.

329
00:19:17 --> 00:19:23
So you see,
that's the difference

330
00:19:19 --> 00:19:25
between stable equilibrium
and unstable equilibrium.

331
00:19:21 --> 00:19:27
At the stable point,

332
00:19:23 --> 00:19:29
the second derivative of the
potential energy versus x

333
00:19:27 --> 00:19:33
is positive.

334
00:19:29 --> 00:19:35
At the unstable point, the
second derivative is negative.

335
00:19:33 --> 00:19:39

336
00:19:39 --> 00:19:45
I'm going to return now
to my spring

337
00:19:44 --> 00:19:50
and I'm going to show you

338
00:19:49 --> 00:19:55
that if you use the potential
energy of the spring alone

339
00:19:55 --> 00:20:01
that you can show that an object
that oscillates on a spring

340
00:20:00 --> 00:20:06
follows
a simple harmonic motion.

341
00:20:04 --> 00:20:10
So here...
is u as a function of x,

342
00:20:13 --> 00:20:19
and this is the parabola
that we already had

343
00:20:16 --> 00:20:22
which equals
one-half kx squared.

344
00:20:22 --> 00:20:28
Let the object be
at a position x maximum here.

345
00:20:27 --> 00:20:33
It's going to oscillate between
plus x max and minus x max.

346
00:20:37 --> 00:20:43
When it is at a random position
x, there is a force on it

347
00:20:43 --> 00:20:49
and the force is always

348
00:20:45 --> 00:20:51
in the direction opposing
the increasing potential energy

349
00:20:48 --> 00:20:54
so the force is clearly
in this direction.

350
00:20:50 --> 00:20:56
It's being driven back
to equilibrium.

351
00:20:53 --> 00:20:59
When it is there, it will have
a certain velocity.

352
00:20:56 --> 00:21:02
The velocity could either be
in this direction

353
00:20:58 --> 00:21:04
or it could be
in that direction.

354
00:21:00 --> 00:21:06
It has a certain speed

355
00:21:03 --> 00:21:09
and since spring forces
are conservative forces,

356
00:21:07 --> 00:21:13
I can now apply the conservation
of mechanical energy.

357
00:21:13 --> 00:21:19
We call this a potential well.

358
00:21:15 --> 00:21:21
The object is going to oscillate
in a potential well.

359
00:21:18 --> 00:21:24
Of course, it doesn't oscillate
like that.

360
00:21:20 --> 00:21:26
It really oscillates like this,
of course.

361
00:21:22 --> 00:21:28
It's a one-dimensional problem.

362
00:21:25 --> 00:21:31
The total energy
that I started with

363
00:21:27 --> 00:21:33
if I release it
here at zero speed

364
00:21:29 --> 00:21:35
equals one-half k x max squared.

365
00:21:34 --> 00:21:40
That is e total.

366
00:21:35 --> 00:21:41
That will always be the same

367
00:21:37 --> 00:21:43
if there is no friction
of any kind

368
00:21:40 --> 00:21:46
and I have to assume
that there is no friction.

369
00:21:43 --> 00:21:49
That must be, now,
one-half m v squared

370
00:21:48 --> 00:21:54
at a random position x,

371
00:21:50 --> 00:21:56
plus one-half k x squared,

372
00:21:53 --> 00:21:59
which is the potential energy
at position x.

373
00:21:56 --> 00:22:02
So this is the kinetic energy

374
00:21:58 --> 00:22:04
and this is
the potential energy.

375
00:22:02 --> 00:22:08
v is the first derivative
of position versus time,

376
00:22:08 --> 00:22:14
so I can write for this
an x dot.

377
00:22:11 --> 00:22:17
And now what I'm going to do,

378
00:22:14 --> 00:22:20
I'm going to rewrite
it slightly differently.

379
00:22:16 --> 00:22:22
I'll bring the x dot squared
to one side, my halfs go away

380
00:22:24 --> 00:22:30
and I divide by m,

381
00:22:27 --> 00:22:33
so I get plus k over m
times x squared,

382
00:22:33 --> 00:22:39
and then I get minus k x max
squared equals zero.

383
00:22:41 --> 00:22:47
Did I do that right?

384
00:22:42 --> 00:22:48
Yes, I divide...

385
00:22:44 --> 00:22:50
Oh, there's an m here,
and the m has to be here.

386
00:22:46 --> 00:22:52
And now what I'm going to do,

387
00:22:48 --> 00:22:54
I'm going to take the time
derivative of this equation.

388
00:22:52 --> 00:22:58
Now you will see something
remarkable falling in place.

389
00:22:55 --> 00:23:01
Just for free.

390
00:22:58 --> 00:23:04
I take the derivative
versus time.

391
00:22:59 --> 00:23:05
That gives me a two x dot, but
I have to apply the chain rule

392
00:23:04 --> 00:23:10
so I also get x double dot,
the second derivative--

393
00:23:07 --> 00:23:13
it's the acceleration--

394
00:23:09 --> 00:23:15
plus I get a two k over m
times x, with the chain rule

395
00:23:16 --> 00:23:22
gives me an x dot.

396
00:23:18 --> 00:23:24
This is a constant, that's
the total energy when I started,

397
00:23:21 --> 00:23:27
so the whole thing equals zero.

398
00:23:24 --> 00:23:30
I lose my two, I lose my x dot
because it's zero

399
00:23:29 --> 00:23:35
and what do I find?

400
00:23:30 --> 00:23:36
x double dot plus k over m
equals zero.

401
00:23:35 --> 00:23:41
And this makes my day

402
00:23:38 --> 00:23:44
because I know this is
a simple harmonic oscillation.

403
00:23:41 --> 00:23:47
You've seen
this equation before.

404
00:23:43 --> 00:23:49
We derived it
in a different way.

405
00:23:45 --> 00:23:51
We didn't use forces today.

406
00:23:47 --> 00:23:53
We only used the concept
of mechanical energy,

407
00:23:51 --> 00:23:57
which is conserved.

408
00:23:53 --> 00:23:59
We know the solution
to this equation...

409
00:23:57 --> 00:24:03
There is an x here.

410
00:23:59 --> 00:24:05
I heard someone mention the x--
thank you very much.

411
00:24:02 --> 00:24:08
The solution is:

412
00:24:04 --> 00:24:10
x equals x max times
the cosine omega t plus phi.

413
00:24:13 --> 00:24:19
This is the amplitude.

414
00:24:15 --> 00:24:21
And omega equals
the square root of k over m,

415
00:24:19 --> 00:24:25
and the period
for one oscillation

416
00:24:21 --> 00:24:27
equals two pi divided by omega.

417
00:24:26 --> 00:24:32
We were able to do this.

418
00:24:30 --> 00:24:36
We were able to apply

419
00:24:31 --> 00:24:37
the conservation
of mechanical energy

420
00:24:33 --> 00:24:39
because spring forces are
conservative forces.

421
00:24:36 --> 00:24:42
So you've seen,
in a completely different way,

422
00:24:40 --> 00:24:46
how you arrive
at the same result.

423
00:24:43 --> 00:24:49
Now I'm going to try

424
00:24:45 --> 00:24:51
something similar
to another potential well

425
00:24:50 --> 00:24:56
and that potential well
is a track

426
00:24:53 --> 00:24:59
and the track is
a perfect circle.

427
00:24:56 --> 00:25:02
And I'm going to slide down
that track an object mass m,

428
00:25:01 --> 00:25:07
and I'm going to evaluate

429
00:25:03 --> 00:25:09
the oscillation along
a perfect circular track.

430
00:25:09 --> 00:25:15
And to make it
as perfect as I can,

431
00:25:12 --> 00:25:18
I even have here
a pair of compasses...

432
00:25:17 --> 00:25:23
make it a...

433
00:25:20 --> 00:25:26
that's the track.

434
00:25:22 --> 00:25:28
And the track has a radius R,

435
00:25:28 --> 00:25:34
and at this moment in time

436
00:25:31 --> 00:25:37
the angle equals theta,
and here is the object.

437
00:25:36 --> 00:25:42
I call this x equals zero.

438
00:25:39 --> 00:25:45
That is also, of course,
where theta equals zero.

439
00:25:46 --> 00:25:52
This is increasing value of y,
and I choose this y equals zero.

440
00:25:54 --> 00:26:00
And so the gravitational
potential energy of this object

441
00:25:59 --> 00:26:05
is its own mg y, so I have
to know what this y is,

442
00:26:04 --> 00:26:10
and therefore I have to know
what this distance is.

443
00:26:09 --> 00:26:15
That's very easy.

444
00:26:12 --> 00:26:18
This one here
equals R cosine theta

445
00:26:17 --> 00:26:23
and so this one is
R minus R cosine theta.

446
00:26:23 --> 00:26:29
So the potential energy equals

447
00:26:25 --> 00:26:31
mg times R one minus cosine
theta, if I choose zero there.

448
00:26:33 --> 00:26:39
I'm free to change that

449
00:26:35 --> 00:26:41
but that's, of course,
a logical thing to do here.

450
00:26:39 --> 00:26:45
Notice if theta equals zero
and the cosine theta equals one,

451
00:26:42 --> 00:26:48
then you find u equals zero.

452
00:26:44 --> 00:26:50
That's, of course...
I have defined it.

453
00:26:47 --> 00:26:53
That's the way I defined
my y equals zero.

454
00:26:50 --> 00:26:56
So u equals zero.

455
00:26:51 --> 00:26:57
Notice that
when theta equals pi over two--

456
00:26:55 --> 00:27:01
if the object were here--

457
00:26:57 --> 00:27:03
that you find that the
potential energy u equals mg R.

458
00:27:02 --> 00:27:08
That's exactly right,
because then the distance

459
00:27:05 --> 00:27:11
between here
and the y zero is R,

460
00:27:08 --> 00:27:14
so this is the potential energy
as a function of angle theta.

461
00:27:14 --> 00:27:20
The velocity of that object
as a function of theta

462
00:27:22 --> 00:27:28
is given by R d theta/dt.

463
00:27:28 --> 00:27:34
And I can make you
see that very easily.

464
00:27:32 --> 00:27:38
Let this be the angle d theta

465
00:27:37 --> 00:27:43
so it moves in a short amount
of time over an angle d theta

466
00:27:41 --> 00:27:47
and the arc here is dS,
and the radius is R.

467
00:27:46 --> 00:27:52
The definition of theta--

468
00:27:48 --> 00:27:54
that's the definition of theta
which is in radians--

469
00:27:51 --> 00:27:57
is that dS divided by R
equals d theta.

470
00:27:57 --> 00:28:03
That's our definition
of radians.

471
00:27:59 --> 00:28:05
So I take the derivative, the
time derivative, left and right,

472
00:28:03 --> 00:28:09
so I get the dS/dt--

473
00:28:05 --> 00:28:11
which, of course, is

474
00:28:06 --> 00:28:12
the tangential velocity
along that arc--

475
00:28:10 --> 00:28:16
equals R times d theta/dt,

476
00:28:17 --> 00:28:23
for which you can write
R theta dot.

477
00:28:20 --> 00:28:26
d theta/dt... d theta/dt
is sometimes called omega,

478
00:28:24 --> 00:28:30
which is the angular velocity,

479
00:28:27 --> 00:28:33
but keep in mind
that in this case

480
00:28:30 --> 00:28:36
the angular velocity omega--

481
00:28:33 --> 00:28:39
if you want to call this omega,
which is the angular velocity--

482
00:28:36 --> 00:28:42
is changing with time.

483
00:28:38 --> 00:28:44
The angular velocity is zero
when you release it

484
00:28:40 --> 00:28:46
and is a maximum when it goes
through the lowest point.

485
00:28:46 --> 00:28:52
So I can now apply

486
00:28:48 --> 00:28:54
the conservation of mechanical
energy, because I know

487
00:28:53 --> 00:28:59
what the velocity is
at any angle of theta

488
00:28:56 --> 00:29:02
and I know what the kinetic...
what the potential energy is.

489
00:29:01 --> 00:29:07
So, let the total energy be
just the mechanical energy

490
00:29:04 --> 00:29:10
which depends on my initial
conditions wherever I start.

491
00:29:07 --> 00:29:13
Maybe it's just that I release
it here with zero speed;

492
00:29:09 --> 00:29:15
maybe I give it a little speed.

493
00:29:11 --> 00:29:17
It is a number,
it is a constant.

494
00:29:14 --> 00:29:20
So that is going to be

495
00:29:16 --> 00:29:22
one-half mv squared
at a random angle of theta.

496
00:29:20 --> 00:29:26
And that means this is v,

497
00:29:24 --> 00:29:30
so that is R squared
times theta dot squared.

498
00:29:29 --> 00:29:35
This is simply one-half
mv squared, nothing else,

499
00:29:34 --> 00:29:40
so this is the kinetic energy.

500
00:29:37 --> 00:29:43
Plus the potential energy,

501
00:29:41 --> 00:29:47
which is mg times R times
one minus cosine theta.

502
00:29:48 --> 00:29:54
And this is always the same,

503
00:29:52 --> 00:29:58
independent
of the angle of theta,

504
00:29:53 --> 00:29:59
because gravity is
a conservative force.

505
00:29:57 --> 00:30:03
So this is the conservation
of mechanical energy.

506
00:30:04 --> 00:30:10
This angle cosine theta is
really a pain in the neck,

507
00:30:09 --> 00:30:15
and therefore
what we're going to do

508
00:30:10 --> 00:30:16
is something
we have seen before--

509
00:30:13 --> 00:30:19
we are going to make a small
angle approximation...

510
00:30:18 --> 00:30:24
small angle approximation.

511
00:30:20 --> 00:30:26
And we're going to write
for cosine theta

512
00:30:24 --> 00:30:30
one minus theta squared
divided by two.

513
00:30:28 --> 00:30:34
That is a very,
very good approximation.

514
00:30:31 --> 00:30:37
That approximation isway better
than the one we did before

515
00:30:36 --> 00:30:42
when we simply said
the cosine of theta equals one.

516
00:30:39 --> 00:30:45
Remember we did that once?

517
00:30:41 --> 00:30:47
We said,
"Oh... for theta is very small.

518
00:30:44 --> 00:30:50
The cosine of theta is
about one."

519
00:30:46 --> 00:30:52
If we did that now,
we would be dead in the waters,

520
00:30:48 --> 00:30:54
because if we said
the cosine of theta is one,

521
00:30:50 --> 00:30:56
this becomes zero
and you end up with nonsense,

522
00:30:53 --> 00:30:59
because it would say

523
00:30:54 --> 00:31:00
that the mechanical energy
is changing all the time

524
00:30:57 --> 00:31:03
because this velocity is
changing all the time.

525
00:30:59 --> 00:31:05
So we cannot do that.

526
00:31:02 --> 00:31:08
We would kill ourselves
if we did that.

527
00:31:04 --> 00:31:10
The approximation is
really amazingly good.

528
00:31:08 --> 00:31:14
If I give you here
theta in radians

529
00:31:13 --> 00:31:19
and I give you here
the cosine of theta

530
00:31:15 --> 00:31:21
and here I give you one minus
theta squared over two,

531
00:31:20 --> 00:31:26
then if I take 1/60
of a radian--

532
00:31:24 --> 00:31:30
and I pick 1/60 since that is
approximately one degree.

533
00:31:30 --> 00:31:36
But I pick exactly 1/60--

534
00:31:32 --> 00:31:38
and I ask what the cosine is,
that is 0.999.

535
00:31:38 --> 00:31:44
And then I have an 861114.

536
00:31:43 --> 00:31:49
I just used my calculator.

537
00:31:45 --> 00:31:51
Then I calculate what one minus
theta squared over two is

538
00:31:49 --> 00:31:55
and I find 0.999861111.

539
00:31:56 --> 00:32:02
That isvery, very close.

540
00:31:59 --> 00:32:05
That is only... it only differs
by three parts in abillion.

541
00:32:03 --> 00:32:09
That is very close.

542
00:32:05 --> 00:32:11
That means the difference
between the two

543
00:32:07 --> 00:32:13
is only one-third
of a millionth of a percent.

544
00:32:12 --> 00:32:18
Suppose now I go
a little rougher

545
00:32:14 --> 00:32:20
and I go to one-fifth
of a radian,

546
00:32:17 --> 00:32:23
which is about 12 degrees, so
this is very roughly 12 degrees.

547
00:32:22 --> 00:32:28
Then the cosine of theta
equals 0.98007,

548
00:32:29 --> 00:32:35
and one minus theta squared
over two equals 0.98000.

549
00:32:35 --> 00:32:41
So that still is
amazingly close--

550
00:32:38 --> 00:32:44
that is, only differs
by seven parts in 100 --> 
0:32:42 --> 00:32:48
so the difference is less
than 1/100 of a percent.

551
00:32:46 --> 00:32:52
So with this in mind,
I feel comfortable to pursue

552
00:32:50 --> 00:32:56
my conservation
of mechanical energy.

553
00:32:55 --> 00:33:01
And I'm going to replace
this cosine theta

554
00:32:57 --> 00:33:03
by one minus theta squared
divided by two.

555
00:33:02 --> 00:33:08

556
00:33:05 --> 00:33:11
So I will continue here--

557
00:33:07 --> 00:33:13
the center blackboard is always
nice, you can see it best--

558
00:33:13 --> 00:33:19
and I will massage
that equation a little further

559
00:33:19 --> 00:33:25
and, of course,
you can already guess

560
00:33:22 --> 00:33:28
what I am going to do when
I massage it a little further.

561
00:33:25 --> 00:33:31
I'm going to take
the time derivative

562
00:33:27 --> 00:33:33
just as I did in the case
of the spring.

563
00:33:33 --> 00:33:39
So we are going to get
that the mechanical energy--

564
00:33:38 --> 00:33:44
which is not changing--

565
00:33:40 --> 00:33:46
equals one-half m R squared
theta dot squared plus mg R.

566
00:33:52 --> 00:33:58
Cosine squared becomes one
minus theta squared over two

567
00:33:55 --> 00:34:01
so we have a minus
times minus becomes plus

568
00:33:58 --> 00:34:04
so I get simply
theta squared over two.

569
00:34:01 --> 00:34:07
And now I take
the time derivative...

570
00:34:06 --> 00:34:12
for this becomes zero...
equals...

571
00:34:10 --> 00:34:16
Now, I get a two out of here,
which eats up this one-half,

572
00:34:14 --> 00:34:20
so I get m R squared,
then I get theta dot,

573
00:34:20 --> 00:34:26
but the chain rule gives me
theta double dot.

574
00:34:24 --> 00:34:30
Excuse me?

575
00:34:25 --> 00:34:31
Anything wrong?

576
00:34:27 --> 00:34:33
I don't think so, thank you.

577
00:34:29 --> 00:34:35
So I have to take
the derivative of this one.

578
00:34:34 --> 00:34:40
The two flips out, which eats up
this two, so I get mg R

579
00:34:39 --> 00:34:45
and then I get a theta.

580
00:34:41 --> 00:34:47
With the chain rule,
gives me a theta dot.

581
00:34:45 --> 00:34:51
I lose my m, I lose one R,
I lose my theta dot--

582
00:34:52 --> 00:34:58
I picked the wrong one; I lose 
my theta dot, not the theta--

583
00:34:59 --> 00:35:05
and what do I find?

584
00:35:01 --> 00:35:07
That theta double dot plus g
over R times theta equals zero.

585
00:35:11 --> 00:35:17
And I couldn't be happier,

586
00:35:13 --> 00:35:19
because this tells me
that the motion

587
00:35:16 --> 00:35:22
is that of a simple
harmonic oscillation.

588
00:35:20 --> 00:35:26
And the solution is x...
excuse me, not x.

589
00:35:24 --> 00:35:30
Theta equals
some maximum angle for theta.

590
00:35:28 --> 00:35:34
It's the amplitude in angle

591
00:35:31 --> 00:35:37
times the cosine
of omega t plus phi.

592
00:35:34 --> 00:35:40
This is the angle of frequency.

593
00:35:36 --> 00:35:42
It hasnothing, nothing to do
with that omega there,

594
00:35:39 --> 00:35:45
which is the angular velocity,
which is changing in time.

595
00:35:42 --> 00:35:48
This is a constant,
this is angular frequency

596
00:35:44 --> 00:35:50
and this omega equals
the square root of g over R,

597
00:35:49 --> 00:35:55
and so the period
of the oscillation is

598
00:35:51 --> 00:35:57
two pi times the square root
of R over g.

599
00:35:56 --> 00:36:02
And when you see that,
you say, "Hey!

600
00:36:02 --> 00:36:08
I have seen that before."

601
00:36:04 --> 00:36:10
Where have we seen this before?

602
00:36:06 --> 00:36:12

603
00:36:09 --> 00:36:15
Almost a carbon copy

604
00:36:11 --> 00:36:17
of something
that we have seen before.

605
00:36:13 --> 00:36:19
What is it?

606
00:36:14 --> 00:36:20
(class murmurs )

607
00:36:15 --> 00:36:21
LEWIN:
Excuse me, speak louder.

608
00:36:17 --> 00:36:23
(echoing class ):
Pendulum!

609
00:36:18 --> 00:36:24
We had a pendulum whereby we had
length l of a massless string.

610
00:36:24 --> 00:36:30
We had an object m
hanging on the end

611
00:36:26 --> 00:36:32
and what was it doing?

612
00:36:29 --> 00:36:35
It was going along a perfect arc
which is exactly identical.

613
00:36:34 --> 00:36:40
The problem is the same,
it's not a surprise,

614
00:36:36 --> 00:36:42
because now we have a surface
which is an exact, perfect arc.

615
00:36:41 --> 00:36:47
It's a circle,
we have no friction.

616
00:36:44 --> 00:36:50
We assumed
with the pendulum

617
00:36:46 --> 00:36:52
that there was
no friction either.

618
00:36:48 --> 00:36:54
So it shouldn't surprise us

619
00:36:49 --> 00:36:55
that you get exactly
the same period

620
00:36:51 --> 00:36:57
that you had
with the pendulum and...

621
00:36:53 --> 00:36:59
except that, of course

622
00:36:56 --> 00:37:02
with the pendulum,
what we called l is now R.

623
00:36:59 --> 00:37:05
Gravity is the only force
that does work

624
00:37:04 --> 00:37:10
and so it is justified

625
00:37:06 --> 00:37:12
to use the conservation
of mechanical energy

626
00:37:09 --> 00:37:15
because gravity is
a conservative force.

627
00:37:12 --> 00:37:18
We used the small angle
approximation to make it work.

628
00:37:18 --> 00:37:24
In the case of the spring,

629
00:37:20 --> 00:37:26
we had that the potential energy
was proportional to x squared

630
00:37:26 --> 00:37:32
and out came a perfect
simple harmonic oscillation,

631
00:37:29 --> 00:37:35
no approximation necessary.

632
00:37:32 --> 00:37:38
Now we forced
this potential energy...

633
00:37:37 --> 00:37:43
we forced it into being
dependent on theta squared.

634
00:37:39 --> 00:37:45
That's really what we did.

635
00:37:41 --> 00:37:47
You see, that is the term
of the potential energy

636
00:37:43 --> 00:37:49
that you have there.

637
00:37:44 --> 00:37:50
And by the approximation
of cosine theta being

638
00:37:48 --> 00:37:54
one minus theta squared
over two,

639
00:37:50 --> 00:37:56
we forced this term to become
quadratic in theta

640
00:37:53 --> 00:37:59
and therefore now,
with that approximation

641
00:37:56 --> 00:38:02
it becomes a perfect
simple harmonic oscillation.

642
00:38:02 --> 00:38:08
Now comes a key question.

643
00:38:04 --> 00:38:10
I said, "Gravity is really
the only force that does work."

644
00:38:09 --> 00:38:15
Is that true?

645
00:38:12 --> 00:38:18
There's no friction for now.

646
00:38:14 --> 00:38:20
Is that really true?

647
00:38:17 --> 00:38:23
When we had the pendulum,
it's true there is gravity.

648
00:38:22 --> 00:38:28
That's clear.

649
00:38:23 --> 00:38:29
There's a gravitational force,
which is mg,

650
00:38:27 --> 00:38:33
but there is also tension.

651
00:38:31 --> 00:38:37
We never mentioned that.

652
00:38:33 --> 00:38:39
We didn't even talk about it

653
00:38:34 --> 00:38:40
when we did the conservation
of mechanical energy.

654
00:38:37 --> 00:38:43
When the object is here,
sure, there is gravity

655
00:38:41 --> 00:38:47
and sure, there is no friction.

656
00:38:43 --> 00:38:49
So there is no force
along the arc,

657
00:38:45 --> 00:38:51
but there must be
a normal force.

658
00:38:49 --> 00:38:55
Is the tension
not doing any work?

659
00:38:52 --> 00:38:58
Is the normal force
not doing any work?

660
00:38:55 --> 00:39:01
Did we, perhaps,
forget something?

661
00:38:58 --> 00:39:04
Remember last week,
I put mylife on the line.

662
00:39:01 --> 00:39:07
I was so convinced

663
00:39:02 --> 00:39:08
that the conservation
of mechanical energy

664
00:39:04 --> 00:39:10
was going to work

665
00:39:05 --> 00:39:11
that I almost killed myself--
not quite--

666
00:39:09 --> 00:39:15
with this huge, 15½ kilogram
pendulum that I was swinging.

667
00:39:13 --> 00:39:19
I believed in the conservation
of mechanical energy

668
00:39:15 --> 00:39:21
and I overlooked the tension.

669
00:39:18 --> 00:39:24
Is it possible that the tension
does, perhaps, positive work?

670
00:39:21 --> 00:39:27
If that's the case,
I could havedied.

671
00:39:25 --> 00:39:31
What is the answer?

672
00:39:26 --> 00:39:32
Is the tension doing any work

673
00:39:28 --> 00:39:34
and in the case
of my circular track

674
00:39:32 --> 00:39:38
is, perhaps, the normal force
doing any work?

675
00:39:35 --> 00:39:41
What is the answer?

676
00:39:37 --> 00:39:43
(class murmurs )

677
00:39:38 --> 00:39:44
LEWIN:
I want to hear it loud
and clear!

678
00:39:40 --> 00:39:46
CLASS:
No.

679
00:39:41 --> 00:39:47
LEWIN:
No! Why is it no?

680
00:39:42 --> 00:39:48
Why is it not doing any work?

681
00:39:43 --> 00:39:49
Because what?

682
00:39:44 --> 00:39:50
(student answers )

683
00:39:46 --> 00:39:52
Exactly. You got it, man!

684
00:39:49 --> 00:39:55
That's it.

685
00:39:50 --> 00:39:56
The force is
always perpendicular

686
00:39:53 --> 00:39:59
to the direction of motion.

687
00:39:55 --> 00:40:01
And since work is a dot product

688
00:39:57 --> 00:40:03
between force and
the direction that it travels,

689
00:40:01 --> 00:40:07
neither the tension nor
the normal force does any work.

690
00:40:05 --> 00:40:11
So don't overlook the force,

691
00:40:07 --> 00:40:13
but do appreciate the fact
that they don't do any work.

692
00:40:11 --> 00:40:17
Great! So now I'm going to show
you a demonstration

693
00:40:17 --> 00:40:23
which I find one of the most
mind-boggling demonstrations

694
00:40:21 --> 00:40:27
that I have ever seen.

695
00:40:23 --> 00:40:29
We do have a circular track.

696
00:40:26 --> 00:40:32
You have it
right in front of you.

697
00:40:29 --> 00:40:35
That is a circle, although you
may not think it is, but it is.

698
00:40:33 --> 00:40:39
And that circle has a radius

699
00:40:37 --> 00:40:43
which, according to the
manufacturer, is 115 meters

700
00:40:43 --> 00:40:49
with an uncertainty of about...
I think it's about five meters.

701
00:40:48 --> 00:40:54
It is extremely difficult
to measure

702
00:40:50 --> 00:40:56
and even during transport,
you think it could change.

703
00:40:53 --> 00:40:59
Let me try to clean this
a little better.

704
00:40:57 --> 00:41:03
And so the radius of this... the
radius of curvature of our arc,

705
00:41:03 --> 00:41:09
which is also an air track,

706
00:41:05 --> 00:41:11
equals 115
plus or minus five meters.

707
00:41:11 --> 00:41:17
So we can calculate now

708
00:41:12 --> 00:41:18
what the period
of oscillations is.

709
00:41:15 --> 00:41:21
The whole track is
five meters long.

710
00:41:18 --> 00:41:24
So half the track is
about 2½ meters,

711
00:41:23 --> 00:41:29
so the angle theta maximum

712
00:41:26 --> 00:41:32
is approximately 2½ meters--

713
00:41:29 --> 00:41:35
which is half the length
of the track-- divided by 115

714
00:41:33 --> 00:41:39
and that is
an extremely small angle.

715
00:41:35 --> 00:41:41
That is about 1.2 degrees,

716
00:41:37 --> 00:41:43
because this is in radians
and this is in degrees.

717
00:41:40 --> 00:41:46
So the angle is very small,
so we should be able to make

718
00:41:44 --> 00:41:50
a perfect prediction
about the period.

719
00:41:47 --> 00:41:53
And I am going to do that.

720
00:41:49 --> 00:41:55
I take two pi times the square
root of R over G

721
00:41:52 --> 00:41:58
and R is 115, 115...
I divide it by G.

722
00:41:57 --> 00:42:03
I take the square root,
I multiply by two.

723
00:42:01 --> 00:42:07
I multiply by pi and I get 21.5.

724
00:42:07 --> 00:42:13
T-- and this is
a prediction...

725
00:42:13 --> 00:42:19
equals 21.5.

726
00:42:18 --> 00:42:24
The uncertainty in R
is about 4.3%.

727
00:42:23 --> 00:42:29
Since we have the square root
of R, that becomes 2.2%.

728
00:42:27 --> 00:42:33
So if I multiply that by .022,

729
00:42:30 --> 00:42:36
I get an uncertainty
of about 0.47.

730
00:42:33 --> 00:42:39
Let's call this 0.5 seconds.

731
00:42:36 --> 00:42:42
So this is a hard prediction

732
00:42:39 --> 00:42:45
what the period of an
oscillation should be--

733
00:42:43 --> 00:42:49
21.5 plus or minus
a half second.

734
00:42:47 --> 00:42:53
Now I'm going to observe it
and we're going to see

735
00:42:50 --> 00:42:56
what we're going to...
how this compares.

736
00:42:55 --> 00:43:01
I don't want to... I don't want
to oscillate it ten times.

737
00:42:58 --> 00:43:04
That will take three, four, five
minutes-- that's too long.

738
00:43:01 --> 00:43:07
It is not really necessary

739
00:43:02 --> 00:43:08
because my reaction time
is 0.1 second,

740
00:43:05 --> 00:43:11
so even if I did
only one oscillation,

741
00:43:08 --> 00:43:14
that would be enough to see

742
00:43:10 --> 00:43:16
whether it is coincident
with that...

743
00:43:12 --> 00:43:18
consistent with that number.

744
00:43:13 --> 00:43:19
However, it is such
a beautiful experiment.

745
00:43:16 --> 00:43:22
It's so much fun to see
that object go back and forth

746
00:43:19 --> 00:43:25
in 21 seconds, that I will go...

747
00:43:22 --> 00:43:28
For your pleasure
and for my own pleasure,

748
00:43:24 --> 00:43:30
I will go three oscillations.

749
00:43:25 --> 00:43:31
Not that it is necessary,
but I will do it.

750
00:43:28 --> 00:43:34
3T is going to be
something plus or minus...

751
00:43:31 --> 00:43:37
and this is my reaction time,
which is 0.1 second,

752
00:43:35 --> 00:43:41
and then we can all
divide that by three

753
00:43:38 --> 00:43:44
and then, of course,

754
00:43:39 --> 00:43:45
the error will go down
by a factor of three,

755
00:43:41 --> 00:43:47
and we will see whether this
number agrees with this one.

756
00:43:47 --> 00:43:53
All right, can you imagine

757
00:43:48 --> 00:43:54
someone making a track
like this...

758
00:43:50 --> 00:43:56
air track with a radius
of 115 meters?

759
00:43:53 --> 00:43:59
I mean, what is this?

760
00:43:54 --> 00:44:00
This may be eight meters.

761
00:43:55 --> 00:44:01
115 meters!

762
00:43:57 --> 00:44:03
That is something
like ten times higher...

763
00:44:00 --> 00:44:06
more-- 15 times higher
than this ceiling.

764
00:44:03 --> 00:44:09
Amazing that people
were able to do that.

765
00:44:05 --> 00:44:11
In fact, nowadays,
you can't even buy this anymore.

766
00:44:09 --> 00:44:15
This is probably some
50 years old, if not older.

767
00:44:14 --> 00:44:20
I have to get the air flowing
out of all these holes.

768
00:44:20 --> 00:44:26
There are many, many small holes
in here that you cannot see.

769
00:44:25 --> 00:44:31
The air is now blowing.

770
00:44:28 --> 00:44:34
And this object
is going to be put on here

771
00:44:33 --> 00:44:39
and just because of gravity,
it will go.

772
00:44:36 --> 00:44:42
That's all it is--
only gravity will do work.

773
00:44:39 --> 00:44:45
Here's the timer
and we're going to time it.

774
00:44:42 --> 00:44:48
I will start it off first

775
00:44:45 --> 00:44:51
and then when it comes back
to a stop, I will start to time

776
00:44:49 --> 00:44:55
because that's, for me,
a very sharp criterion.

777
00:44:51 --> 00:44:57
When the object comes back
and comes to a halt here,

778
00:44:55 --> 00:45:01
it's very easy for me
to start the timing.

779
00:44:59 --> 00:45:05
You may notice, as you watch,

780
00:45:01 --> 00:45:07
that some of the amplitude
will decrease

781
00:45:04 --> 00:45:10
because there is...
hold it, hold it, hold it!

782
00:45:08 --> 00:45:14
Because there is, of course,
a little bit of friction.

783
00:45:11 --> 00:45:17
It's very little,
but it is not zero.

784
00:45:15 --> 00:45:21
Enjoy this, just look at it.

785
00:45:17 --> 00:45:23
Isn't this incredible?

786
00:45:18 --> 00:45:24
It just goes simply by gravity.

787
00:45:21 --> 00:45:27
It's like a pendulum which has
a length of 115 meters.

788
00:45:26 --> 00:45:32
It's about to complete
its first oscillation.

789
00:45:29 --> 00:45:35

790
00:45:36 --> 00:45:42
It goes back...

791
00:45:39 --> 00:45:45
Actually, some of you may be
able to see the curvature.

792
00:45:42 --> 00:45:48
You can really see
that it is not straight.

793
00:45:45 --> 00:45:51
So we're coming up
to the second.

794
00:45:47 --> 00:45:53

795
00:45:53 --> 00:45:59
I better get back in position.

796
00:45:55 --> 00:46:01

797
00:46:05 --> 00:46:11
So when it stops here,

798
00:46:07 --> 00:46:13
it has made
three complete oscillations.

799
00:46:14 --> 00:46:20
Sixty-four point zero five.

800
00:46:21 --> 00:46:27
Let me turn this off.

801
00:46:27 --> 00:46:33
So 3T equals 64.05.

802
00:46:35 --> 00:46:41
I'm lazy-- 64.05,
I divide that by three.

803
00:46:40 --> 00:46:46
That is 21.35,
plus or minus .03.

804
00:46:47 --> 00:46:53
That's exactly in agreement
with the prediction,

805
00:46:49 --> 00:46:55
with the uncertainty
of the prediction.

806
00:46:55 --> 00:47:01
I have something very similar,

807
00:46:58 --> 00:47:04
and that is, again,
a curved track.

808
00:47:02 --> 00:47:08
It's not...

809
00:47:03 --> 00:47:09
Oop, I hope
I can retrieve that ball.

810
00:47:05 --> 00:47:11
It would be nice.

811
00:47:07 --> 00:47:13

812
00:47:08 --> 00:47:14
Hmm, what happened?

813
00:47:13 --> 00:47:19
Boy! You have to be...

814
00:47:17 --> 00:47:23
Gee, what's happening here?

815
00:47:19 --> 00:47:25
Oh, yeah, I got it,
got it, got it.

816
00:47:21 --> 00:47:27
Phew!

817
00:47:22 --> 00:47:28
Tricky to make a hole in here.

818
00:47:25 --> 00:47:31
This is an arc,
not unlike this one.

819
00:47:29 --> 00:47:35
There's more,
a little bit more friction

820
00:47:31 --> 00:47:37
and, in this case,
the radius is 85 centimeters.

821
00:47:36 --> 00:47:42
So we can calculate
what the maximum angle is.

822
00:47:41 --> 00:47:47
The radius is 85 centimeters

823
00:47:43 --> 00:47:49
and the arc to the edge
is about 20 centimeters.

824
00:47:48 --> 00:47:54
So we have now
a situation like this.

825
00:47:51 --> 00:47:57
R equals 85 centimeters

826
00:47:56 --> 00:48:02
and this here is
approximately 20 centimeters,

827
00:48:03 --> 00:48:09
so theta maximum is
roughly 20 divided by 85

828
00:48:09 --> 00:48:15
and that is something
like 13 degrees.

829
00:48:12 --> 00:48:18
13 degrees is
not a bad situation

830
00:48:15 --> 00:48:21
because the difference
between the cosine theta

831
00:48:18 --> 00:48:24
and one minus theta squared
over two

832
00:48:21 --> 00:48:27
is less than 1/100 of a percent,
it is that small.

833
00:48:25 --> 00:48:31
So I can make a prediction

834
00:48:27 --> 00:48:33
of the period
of this oscillation,

835
00:48:32 --> 00:48:38
predict and you can go through
exactly the same exercise.

836
00:48:36 --> 00:48:42
You take two pi times
the square root of R over d

837
00:48:42 --> 00:48:48
and you find 1.85.

838
00:48:47 --> 00:48:53
The uncertainty of this radius
is, of course, not very large

839
00:48:50 --> 00:48:56
but we are not certain
about the radius

840
00:48:52 --> 00:48:58
to about one centimeter,

841
00:48:54 --> 00:49:00
so it's 85 plus
or minus one centimeters.

842
00:48:56 --> 00:49:02
So that's about
a 1.2 percent error

843
00:49:00 --> 00:49:06
and so the error, then, in the
prediction will be 0.6 percent; 

844
00:49:04 --> 00:49:10
it's about .01 seconds.

845
00:49:07 --> 00:49:13
So I expect...
this is my prediction.

846
00:49:12 --> 00:49:18
Now, Ireally want
to challenge this .01

847
00:49:16 --> 00:49:22
and so now I'm going
to make the observations

848
00:49:19 --> 00:49:25
and surely I'm going
to do it now 10 times,

849
00:49:22 --> 00:49:28
because then the uncertainty
will be 0.1 seconds--

850
00:49:25 --> 00:49:31
that's my reaction time--

851
00:49:27 --> 00:49:33
and so I have the final period
to an accuracy of .01 seconds

852
00:49:32 --> 00:49:38
and so we can compare
these numbers directly

853
00:49:35 --> 00:49:41
and that is what I will do now.

854
00:49:37 --> 00:49:43
I have here the timer

855
00:49:39 --> 00:49:45
and I'm going to oscillate
that back and forth--

856
00:49:42 --> 00:49:48
and that would only take
20 seconds--

857
00:49:45 --> 00:49:51
zero it, we started here.

858
00:49:48 --> 00:49:54
We have great confidence
in physics, right?

859
00:49:50 --> 00:49:56
We believe in physics.

860
00:49:51 --> 00:49:57
We believe in the conservation
of mechanical energy.

861
00:49:54 --> 00:50:00
Starts... are you counting?

862
00:49:58 --> 00:50:04
Is this two? Yeah?

863
00:50:00 --> 00:50:06
Is this three? Four?

864
00:50:03 --> 00:50:09
CLASS:
Four.

865
00:50:04 --> 00:50:10
LEWIN:
I don't believe you.

866
00:50:06 --> 00:50:12
Okay, we start again.

867
00:50:09 --> 00:50:15
Now!

868
00:50:11 --> 00:50:17
One, two, three,

869
00:50:17 --> 00:50:23
four, five,

870
00:50:22 --> 00:50:28
six, seven...

871
00:50:25 --> 00:50:31
I'm getting nervous.

872
00:50:27 --> 00:50:33
Eight, nine, ten.

873
00:50:33 --> 00:50:39
Holy smoke!

874
00:50:35 --> 00:50:41
22.7 seconds!

875
00:50:39 --> 00:50:45
It should have been 18!

876
00:50:41 --> 00:50:47
22.7 seconds.

877
00:50:44 --> 00:50:50
There must be something
fundamentally wrong

878
00:50:46 --> 00:50:52
with the conservation
of mechanical energy.

879
00:50:48 --> 00:50:54
Or is there something else?

880
00:50:51 --> 00:50:57
And what is the difference
between the two experiments?

881
00:50:54 --> 00:51:00
STUDENT:
Friction.

882
00:50:55 --> 00:51:01
LEWIN:
Excuse me?

883
00:50:56 --> 00:51:02
STUDENT:
Friction.

884
00:50:58 --> 00:51:04
LEWIN:
Oh, no, the friction is so low,
that is not the reason.

885
00:51:02 --> 00:51:08
There's a huge difference.

886
00:51:05 --> 00:51:11
Think about it when you take
your shower this weekend.

887
00:51:09 --> 00:51:15
There is a huge difference

888
00:51:10 --> 00:51:16
between this object moving
and that object moving

889
00:51:14 --> 00:51:20
and when you find out,

890
00:51:16 --> 00:51:22
that is the reason why that
is way slower, not friction.

891
00:51:20 --> 00:51:26
See you next Wednesday.

892
00:51:24 --> 00:51:30

893
00:51:29 --> 00:51:35.000