All right... long weekend ahead of us.
One more lecture to go.
If I have an object, mass m, in gravitational field,
gravitational force is in this direction,
if this is my increasing value of y,
then this force, vectorially written,
equals minus mg y roof.
Since this is a one-dimensional problem,
we will often simply write F equals minus mg.
This minus sign is important
because that's the increasing value of y.
If this level here is y equals zero,
then I could call this
gravitational potential energy zero.
And this is y...
Then the gravitational potential energy here equals plus mg y.
This is u.
So if I make a plot of the gravitational potential energy
as a function of y, then I would get a straight line.
This is zero.
So this equals u... equals mg y, plus sign.
If I'm here at point A and I move that object to point B,
I, Walter Lewin, move it, I have to do positive work.
Notice that the gravitational potential energy increases.
If I do positive work, the gravity is doing negative work.
If I go from A to some other point-- call it B prime--
then I do negative work.
Notice the gravitational potential energy goes down.
If I do negative work, then gravity is doing positive work.
I could have chosen my zero point of potential energy
anywhere I please.
I could have chosen it right here
and nothing would change
other than that I offset the zero point
of my potential energy.
But again, if I go from A to B,
the gravitational potential energy increases
by exactly the same amount--
I have to do exactly the same work.
So you are free to choose, when you are near Earth,
where you choose your zero.
Now we take the situation
whereby we are not so close to the Earth.
Here is the Earth itself.
Of course you can also replace that by the sun
if you want to.
And this is increasing value of r.
The distance between here and this object m equals r.
I now know that there is
a gravitational force on this object--
Newton's Universal Law of Gravity--
and that gravitational force
equals minus m M-Earth G
divided by this r squared, r roof
so this is a vectorial notation.
Since it is really one-dimensional, we would...
Just like we did there, we would delete the arrow
and we would delete the unit vector
in the positive r direction
and so we would simply write it this way.
The gravitational potential energy we derived last time
equals minus m M-Earth G
divided by r--
and notice, here is r and here is r squared--
and if you plot that, then the plot goes sort of like this.
This is r, this is increasing potential energy--
all these values here are negative--
and you get a curve which is sort of like this.
This is proportional to one over r.
Now, of course, if the Earth had a radius which is this big,
then, of course, this curve does not exist, it stops right here.
If I move from point A to point B, with a mass m in my hand,
notice that the gravitational potential energy increases.
I have to do positive work, there is no difference.
If I go from A to another point, B prime,
which is closer to the Earth,
notice that the gravitational potential energy decreases.
I do negative work.
If I do positive work, gravity is doing negative work.
If I do negative work, gravity is doing positive work.
Right here near Earth,
where this one-over-r curve hits the Earth,
that is, of course, exactly that line.
That dependence on y is exactly the same as the dependence on r
and then you can simplify matters when you are near Earth.
When the gravitational acceleration doesn't change
you get a linear relation.
But that's only an exceptional case
when you don't move very far.
The gravitational force is
in the direction opposite the increasing potential energy.
Notice that when I'm here,
the gravitational force is in this direction.
Increasing potential energy is this way.
The force is in this way.
When I'm here,
gravitational potential energy increases in this way;
the gravitational force is in this direction.
When I'm here,
the gravitational potential energy increases
in this direction.
The gravitational force is in this direction.
When I'm here,
the gravitational potential energy increases
in this direction.
The gravitational force is in this direction.
The force is always in the opposite direction
than the increasing value of the potential energy.
If I release an object at zero speed,
it therefore will always move
towards a lower potential energy
because the force will drive it to lower potential energy.
Now I change from gravity to a spring.
I have a spring which is relaxed length l--
I call this x equals zero--
and I extend it over a distance x
and there's a mass m at the end
and there will be a spring force.
And that spring force...
F equals minus kx.
It's a one-dimensional situation
so I can write it without having to worry about the arrows.
There is no friction here.
It's clear that if I hold this in my hand,
that the force Walter Lewin equals plus kx.
It's in the direction of increasing x.
If I call this point A at x equals zero
and I call this point B at x equals x,
then I can calculate the work
that I have to do to bring it from A to B.
So the work that Walter Lewin has to do
to bring it from A to B
is the integral in going from A to B
of my force, dx.
It's a dot product,
but since the angle between the two is zero degrees,
the cosine of the angle is one,
so I can forget about the fact that there's a dot product.
I move it in this direction.
So that becomes the integral
in going from zero to a position x
of plus x plus kx dx,
and that is one-half k x squared.
And this is what we call
the potential energy of the spring.
This is potential energy.
It means, then, that we...
At x equals zero, we define potential energy to be zero.
You don't have to do that,
but it would be ridiculous to do it any other way.
So in the case that we have the near-Earth situation of gravity,
we had a choice where we put our zero potential energy.
In the case that we deal with very large distances,
we do not have a choice anymore.
We defined it in such a way
that the potential energy at infinity is zero.
As a result of that, all potential energies are negative.
And here, with the spring, you don't have a choice, either.
You choose...
at x equals zero, you choose potential energy to be zero.
So if you now make a plot
of the potential energy as a function of x,
you get a parabola,
and if you are here, if the object is here,
then the force is always
in the direction opposing the increasing potential energy.
If you go this way, potential energy increases.
So it's clear that the force is going to be in this direction.
If you are here, the force is
always in the direction opposing increasing potential energy.
Increasing potential energy is in this direction,
so the force is in this direction.
You see, it's a restoring force.
The force is always in the direction
opposite to increasing potential energy.
If you release an object here at zero speed,
it will therefore go towards lower potential energy.
The force will drive it to lower potential energy.
If we know the force-- in this case, the spring force
or in those cases, the gravitational force--
we were able to calculate the potential energy.
Now, the question is, can we also go back?
Suppose we knew the potential energy.
Can we then find the force again?
And the answer is yes, we can.
Let's take the situation of the spring first.
We have that the potential energy u of the spring
equals plus one-half k x squared
and if I take the derivative of that versus x
then I get plus kx,
but the force, the spring force itself, is minus kx,
so this equals minus the spring force.
So we have that du/dx equals minus F,
and I put the x there because it's a one-dimensional problem--
it is only in the x direction.
The minus sign is telling you
that the force is always pointing in the direction
which is opposite to increasing values of the potential energy.
That is what the minus sign is telling you.
It's staring you in the face
what I have been telling you for the past five minutes.
If we have a three-dimensional situation
that we know the potential energy
as a function of x, y and z,
then we can go back and find the forces
as a function of x, y and z.
It doesn't matter whether these are springs
or whether it is gravity
or whether it's electric forces or nuclear forces,
you then find that du/dx equals minus F of x,
du/dy equals minus Fy
and du/dz equals minus Fz.
What does this mean?
It means that if you're in three-dimensional space,
you move only in the x direction.
You keep y and z constant, and the change equals minus Fx.
That gives you the component of the force in the x direction.
You move only in the y direction,
you keep x and z constant,
and then you find the component of the force
in the y direction.
We call these partial derivatives,
so we don't give them a "d"
but we give them a little curled delta.
If we go back to the situation where we had gravity,
we had there the situation near Earth.
We had u... was plus mg y, so what is du/dy?
This is a one-dimensional situation
so I don't have to use the partial derivatives.
I can simply say du/dy.
That is plus mg, and notice
that the gravitational force was minus mg.
Remember? The minus sign is still there.
It's still there.
And so you see that here, indeed,
du/dy is minus the gravitational force.
Now we take the situation that we are not near Earth--
we have there--
so we have u equals minus m M-Earth G
divided by r--
there's only an r here--
so du/dr...
The derivative of one over r isminus one over r squared.
The minus sign eats up this minus sign,
so I get plus m M-Earth G divided by r squared
so the gravitational force...
the gravitational force equals minus that.
It isminus du/dr
and, indeed, that's exactly what we have there--
minus that value.
So whenever you know
the potential as a function of space,
you can always find the three components of the forces
in the three orthogonal directions.
Suppose I have a curved surface--
literally, a surface here in 26.100,
which sort of looks like this...
something like this.
I call this, arbitrarily, y equals zero
and I could call this
u gravitational potential energy zero, for that matter.
So this is a function y as a function of x
and the curve itself represents effectively
the gravitational potential energy.
This is y and this is x.
So the gravitational potential energy u equals mg y,
but y is a function of x, so that is also u times m...
excuse me, that is m times g times that function of x.
There are points here where du/dx equals zero.
I'll get a nice mg in here...
Where's zero... and where are those points?
Those points are here, here, here, here and here.
If du/dx is zero, it means that the force--
the component of the force in the x direction-- is zero,
because du/dx is minus the force in the x direction.
So if we visit those points, for instance here,
then there is, of course, gravity, mg,
if there is an object there in the y direction...
in the minus y direction
and there is a normal force in the plus y direction
and these two exactly cancel each other.
So the net result is that here, here, here, here and there
there is no force on the object at all
so the object is not going to move, it's going to stay put.
Well, yes, it's going to stay put.
However, there is a huge difference
between this point here and that point there
and you sense immediately that difference.
If I put a marble here,
I will have a hell of a time to keep the marble in place,
because if there is a fly there in the corner of 26.100
which does something
then the slightest amount of force on this one
and it will start to roll off.
In fact, what will happen is
it will go to a lower potential energy.
Here, however, if this one is offset,
then it will want to go to a lower potential energy.
The force is always opposing
the direction of increasing potential energy,
so the force will drive it back
and so that's why we call this a stable equilibrium.
It will always go back to that point.
And this is an unstable equilibrium.
We have a setup here,
and I would like to show you how that will work.
So we do have something that is a curved object.
It's a track.
Let me give you a little bit better light condition.
So you see there, there is that object, a little ball.
And no surprise, if I offset it from the lowest point
that it will be driven back to that point-- that's trivial.
What is less trivial is that there is a point here
whereby, indeed, the net forces are zero
and it is not easy to achieve that,
but I will try to put it there so that it, indeed, stays put.
I'm not too fortunate, it is very difficult.
I'm trying... no...
Yeah! Did it.
It's there, it's very unstable.
I blow... oh, it's not so unstable.
And there it goes.
So you see, that's the difference
between stable equilibrium and unstable equilibrium.
At the stable point,
the second derivative of the potential energy versus x
is positive.
At the unstable point, the second derivative is negative.
I'm going to return now to my spring
and I'm going to show you
that if you use the potential energy of the spring alone
that you can show that an object that oscillates on a spring
follows a simple harmonic motion.
So here... is u as a function of x,
and this is the parabola that we already had
which equals one-half kx squared.
Let the object be at a position x maximum here.
It's going to oscillate between plus x max and minus x max.
When it is at a random position x, there is a force on it
and the force is always
in the direction opposing the increasing potential energy
so the force is clearly in this direction.
It's being driven back to equilibrium.
When it is there, it will have a certain velocity.
The velocity could either be in this direction
or it could be in that direction.
It has a certain speed
and since spring forces are conservative forces,
I can now apply the conservation of mechanical energy.
We call this a potential well.
The object is going to oscillate in a potential well.
Of course, it doesn't oscillate like that.
It really oscillates like this, of course.
It's a one-dimensional problem.
The total energy that I started with
if I release it here at zero speed
equals one-half k x max squared.
That is e total.
That will always be the same
if there is no friction of any kind
and I have to assume that there is no friction.
That must be, now, one-half m v squared
at a random position x,
plus one-half k x squared,
which is the potential energy at position x.
So this is the kinetic energy
and this is the potential energy.
v is the first derivative of position versus time,
so I can write for this an x dot.
And now what I'm going to do,
I'm going to rewrite it slightly differently.
I'll bring the x dot squared to one side, my halfs go away
and I divide by m,
so I get plus k over m times x squared,
and then I get minus k x max squared equals zero.
Did I do that right?
Yes, I divide...
Oh, there's an m here, and the m has to be here.
And now what I'm going to do,
I'm going to take the time derivative of this equation.
Now you will see something remarkable falling in place.
Just for free.
I take the derivative versus time.
That gives me a two x dot, but I have to apply the chain rule
so I also get x double dot, the second derivative--
it's the acceleration--
plus I get a two k over m times x, with the chain rule
gives me an x dot.
This is a constant, that's the total energy when I started,
so the whole thing equals zero.
I lose my two, I lose my x dot because it's zero
and what do I find?
x double dot plus k over m equals zero.
And this makes my day
because I know this is a simple harmonic oscillation.
You've seen this equation before.
We derived it in a different way.
We didn't use forces today.
We only used the concept of mechanical energy,
which is conserved.
We know the solution to this equation...
There is an x here.
I heard someone mention the x-- thank you very much.
The solution is:
x equals x max times the cosine omega t plus phi.
This is the amplitude.
And omega equals the square root of k over m,
and the period for one oscillation
equals two pi divided by omega.
We were able to do this.
We were able to apply
the conservation of mechanical energy
because spring forces are conservative forces.
So you've seen, in a completely different way,
how you arrive at the same result.
Now I'm going to try
something similar to another potential well
and that potential well is a track
and the track is a perfect circle.
And I'm going to slide down that track an object mass m,
and I'm going to evaluate
the oscillation along a perfect circular track.
And to make it as perfect as I can,
I even have here a pair of compasses...
make it a...
that's the track.
And the track has a radius R,
and at this moment in time
the angle equals theta, and here is the object.
I call this x equals zero.
That is also, of course, where theta equals zero.
This is increasing value of y, and I choose this y equals zero.
And so the gravitational potential energy of this object
is its own mg y, so I have to know what this y is,
and therefore I have to know what this distance is.
That's very easy.
This one here equals R cosine theta
and so this one is R minus R cosine theta.
So the potential energy equals
mg times R one minus cosine theta, if I choose zero there.
I'm free to change that
but that's, of course, a logical thing to do here.
Notice if theta equals zero and the cosine theta equals one,
then you find u equals zero.
That's, of course... I have defined it.
That's the way I defined my y equals zero.
So u equals zero.
Notice that when theta equals pi over two--
if the object were here--
that you find that the potential energy u equals mg R.
That's exactly right, because then the distance
between here and the y zero is R,
so this is the potential energy as a function of angle theta.
The velocity of that object as a function of theta
is given by R d theta/dt.
And I can make you see that very easily.
Let this be the angle d theta
so it moves in a short amount of time over an angle d theta
and the arc here is dS, and the radius is R.
The definition of theta--
that's the definition of theta which is in radians--
is that dS divided by R equals d theta.
That's our definition of radians.
So I take the derivative, the time derivative, left and right,
so I get the dS/dt--
which, of course, is
the tangential velocity along that arc--
equals R times d theta/dt,
for which you can write R theta dot.
d theta/dt... d theta/dt is sometimes called omega,
which is the angular velocity,
but keep in mind that in this case
the angular velocity omega--
if you want to call this omega, which is the angular velocity--
is changing with time.
The angular velocity is zero when you release it
and is a maximum when it goes through the lowest point.
So I can now apply
the conservation of mechanical energy, because I know
what the velocity is at any angle of theta
and I know what the kinetic... what the potential energy is.
So, let the total energy be just the mechanical energy
which depends on my initial conditions wherever I start.
Maybe it's just that I release it here with zero speed;
maybe I give it a little speed.
It is a number, it is a constant.
So that is going to be
one-half mv squared at a random angle of theta.
And that means this is v,
so that is R squared times theta dot squared.
This is simply one-half mv squared, nothing else,
so this is the kinetic energy.
Plus the potential energy,
which is mg times R times one minus cosine theta.
And this is always the same,
independent of the angle of theta,
because gravity is a conservative force.
So this is the conservation of mechanical energy.
This angle cosine theta is really a pain in the neck,
and therefore what we're going to do
is something we have seen before--
we are going to make a small angle approximation...
small angle approximation.
And we're going to write for cosine theta
one minus theta squared divided by two.
That is a very, very good approximation.
That approximation isway better than the one we did before
when we simply said the cosine of theta equals one.
Remember we did that once?
We said, "Oh... for theta is very small.
The cosine of theta is about one."
If we did that now, we would be dead in the waters,
because if we said the cosine of theta is one,
this becomes zero and you end up with nonsense,
because it would say
that the mechanical energy is changing all the time
because this velocity is changing all the time.
So we cannot do that.
We would kill ourselves if we did that.
The approximation is really amazingly good.
If I give you here theta in radians
and I give you here the cosine of theta
and here I give you one minus theta squared over two,
then if I take 1/60 of a radian--
and I pick 1/60 since that is approximately one degree.
But I pick exactly 1/60--
and I ask what the cosine is, that is 0.999.
And then I have an 861114.
I just used my calculator.
Then I calculate what one minus theta squared over two is
and I find 0.999861111.
That isvery, very close.
That is only... it only differs by three parts in abillion.
That is very close.
That means the difference between the two
is only one-third of a millionth of a percent.
Suppose now I go a little rougher
and I go to one-fifth of a radian,
which is about 12 degrees, so this is very roughly 12 degrees.
Then the cosine of theta equals 0.98007,
and one minus theta squared over two equals 0.98000.
So that still is amazingly close--
that is, only differs by seven parts in 100 -->
so the difference is less than 1/100 of a percent.
So with this in mind, I feel comfortable to pursue
my conservation of mechanical energy.
And I'm going to replace this cosine theta
by one minus theta squared divided by two.
So I will continue here--
the center blackboard is always nice, you can see it best--
and I will massage that equation a little further
and, of course, you can already guess
what I am going to do when I massage it a little further.
I'm going to take the time derivative
just as I did in the case of the spring.
So we are going to get that the mechanical energy--
which is not changing--
equals one-half m R squared theta dot squared plus mg R.
Cosine squared becomes one minus theta squared over two
so we have a minus times minus becomes plus
so I get simply theta squared over two.
And now I take the time derivative...
for this becomes zero... equals...
Now, I get a two out of here, which eats up this one-half,
so I get m R squared, then I get theta dot,
but the chain rule gives me theta double dot.
Excuse me?
Anything wrong?
I don't think so, thank you.
So I have to take the derivative of this one.
The two flips out, which eats up this two, so I get mg R
and then I get a theta.
With the chain rule, gives me a theta dot.
I lose my m, I lose one R, I lose my theta dot--
I picked the wrong one; I lose my theta dot, not the theta--
and what do I find?
That theta double dot plus g over R times theta equals zero.
And I couldn't be happier,
because this tells me that the motion
is that of a simple harmonic oscillation.
And the solution is x... excuse me, not x.
Theta equals some maximum angle for theta.
It's the amplitude in angle
times the cosine of omega t plus phi.
This is the angle of frequency.
It hasnothing, nothing to do with that omega there,
which is the angular velocity, which is changing in time.
This is a constant, this is angular frequency
and this omega equals the square root of g over R,
and so the period of the oscillation is
two pi times the square root of R over g.
And when you see that, you say, "Hey!
I have seen that before."
Where have we seen this before?
Almost a carbon copy
of something that we have seen before.
What is it?
(class murmurs )
LEWIN: Excuse me, speak louder.
(echoing class ): Pendulum!
We had a pendulum whereby we had length l of a massless string.
We had an object m hanging on the end
and what was it doing?
It was going along a perfect arc which is exactly identical.
The problem is the same, it's not a surprise,
because now we have a surface which is an exact, perfect arc.
It's a circle, we have no friction.
We assumed with the pendulum
that there was no friction either.
So it shouldn't surprise us
that you get exactly the same period
that you had with the pendulum and...
except that, of course
with the pendulum, what we called l is now R.
Gravity is the only force that does work
and so it is justified
to use the conservation of mechanical energy
because gravity is a conservative force.
We used the small angle approximation to make it work.
In the case of the spring,
we had that the potential energy was proportional to x squared
and out came a perfect simple harmonic oscillation,
no approximation necessary.
Now we forced this potential energy...
we forced it into being dependent on theta squared.
That's really what we did.
You see, that is the term of the potential energy
that you have there.
And by the approximation of cosine theta being
one minus theta squared over two,
we forced this term to become quadratic in theta
and therefore now, with that approximation
it becomes a perfect simple harmonic oscillation.
Now comes a key question.
I said, "Gravity is really the only force that does work."
Is that true?
There's no friction for now.
Is that really true?
When we had the pendulum, it's true there is gravity.
That's clear.
There's a gravitational force, which is mg,
but there is also tension.
We never mentioned that.
We didn't even talk about it
when we did the conservation of mechanical energy.
When the object is here, sure, there is gravity
and sure, there is no friction.
So there is no force along the arc,
but there must be a normal force.
Is the tension not doing any work?
Is the normal force not doing any work?
Did we, perhaps, forget something?
Remember last week, I put mylife on the line.
I was so convinced
that the conservation of mechanical energy
was going to work
that I almost killed myself-- not quite--
with this huge, 15½ kilogram pendulum that I was swinging.
I believed in the conservation of mechanical energy
and I overlooked the tension.
Is it possible that the tension does, perhaps, positive work?
If that's the case, I could havedied.
What is the answer?
Is the tension doing any work
and in the case of my circular track
is, perhaps, the normal force doing any work?
What is the answer?
(class murmurs )
LEWIN: I want to hear it loud and clear!
CLASS: No.
LEWIN: No! Why is it no?
Why is it not doing any work?
Because what?
(student answers )
Exactly. You got it, man!
That's it.
The force is always perpendicular
to the direction of motion.
And since work is a dot product
between force and the direction that it travels,
neither the tension nor the normal force does any work.
So don't overlook the force,
but do appreciate the fact that they don't do any work.
Great! So now I'm going to show you a demonstration
which I find one of the most mind-boggling demonstrations
that I have ever seen.
We do have a circular track.
You have it right in front of you.
That is a circle, although you may not think it is, but it is.
And that circle has a radius
which, according to the manufacturer, is 115 meters
with an uncertainty of about... I think it's about five meters.
It is extremely difficult to measure
and even during transport, you think it could change.
Let me try to clean this a little better.
And so the radius of this... the radius of curvature of our arc,
which is also an air track,
equals 115 plus or minus five meters.
So we can calculate now
what the period of oscillations is.
The whole track is five meters long.
So half the track is about 2½ meters,
so the angle theta maximum
is approximately 2½ meters--
which is half the length of the track-- divided by 115
and that is an extremely small angle.
That is about 1.2 degrees,
because this is in radians and this is in degrees.
So the angle is very small, so we should be able to make
a perfect prediction about the period.
And I am going to do that.
I take two pi times the square root of R over G
and R is 115, 115... I divide it by G.
I take the square root, I multiply by two.
I multiply by pi and I get 21.5.
T-- and this is a prediction...
equals 21.5.
The uncertainty in R is about 4.3%.
Since we have the square root of R, that becomes 2.2%.
So if I multiply that by .022,
I get an uncertainty of about 0.47.
Let's call this 0.5 seconds.
So this is a hard prediction
what the period of an oscillation should be--
21.5 plus or minus a half second.
Now I'm going to observe it and we're going to see
what we're going to... how this compares.
I don't want to... I don't want to oscillate it ten times.
That will take three, four, five minutes-- that's too long.
It is not really necessary
because my reaction time is 0.1 second,
so even if I did only one oscillation,
that would be enough to see
whether it is coincident with that...
consistent with that number.
However, it is such a beautiful experiment.
It's so much fun to see that object go back and forth
in 21 seconds, that I will go...
For your pleasure and for my own pleasure,
I will go three oscillations.
Not that it is necessary, but I will do it.
3T is going to be something plus or minus...
and this is my reaction time, which is 0.1 second,
and then we can all divide that by three
and then, of course,
the error will go down by a factor of three,
and we will see whether this number agrees with this one.
All right, can you imagine
someone making a track like this...
air track with a radius of 115 meters?
I mean, what is this?
This may be eight meters.
115 meters!
That is something like ten times higher...
more-- 15 times higher than this ceiling.
Amazing that people were able to do that.
In fact, nowadays, you can't even buy this anymore.
This is probably some 50 years old, if not older.
I have to get the air flowing out of all these holes.
There are many, many small holes in here that you cannot see.
The air is now blowing.
And this object is going to be put on here
and just because of gravity, it will go.
That's all it is-- only gravity will do work.
Here's the timer and we're going to time it.
I will start it off first
and then when it comes back to a stop, I will start to time
because that's, for me, a very sharp criterion.
When the object comes back and comes to a halt here,
it's very easy for me to start the timing.
You may notice, as you watch,
that some of the amplitude will decrease
because there is... hold it, hold it, hold it!
Because there is, of course, a little bit of friction.
It's very little, but it is not zero.
Enjoy this, just look at it.
Isn't this incredible?
It just goes simply by gravity.
It's like a pendulum which has a length of 115 meters.
It's about to complete its first oscillation.
It goes back...
Actually, some of you may be able to see the curvature.
You can really see that it is not straight.
So we're coming up to the second.
I better get back in position.
So when it stops here,
it has made three complete oscillations.
Sixty-four point zero five.
Let me turn this off.
So 3T equals 64.05.
I'm lazy-- 64.05, I divide that by three.
That is 21.35, plus or minus .03.
That's exactly in agreement with the prediction,
with the uncertainty of the prediction.
I have something very similar,
and that is, again, a curved track.
It's not...
Oop, I hope I can retrieve that ball.
It would be nice.
Hmm, what happened?
Boy! You have to be...
Gee, what's happening here?
Oh, yeah, I got it, got it, got it.
Phew!
Tricky to make a hole in here.
This is an arc, not unlike this one.
There's more, a little bit more friction
and, in this case, the radius is 85 centimeters.
So we can calculate what the maximum angle is.
The radius is 85 centimeters
and the arc to the edge is about 20 centimeters.
So we have now a situation like this.
R equals 85 centimeters
and this here is approximately 20 centimeters,
so theta maximum is roughly 20 divided by 85
and that is something like 13 degrees.
13 degrees is not a bad situation
because the difference between the cosine theta
and one minus theta squared over two
is less than 1/100 of a percent, it is that small.
So I can make a prediction
of the period of this oscillation,
predict and you can go through exactly the same exercise.
You take two pi times the square root of R over d
and you find 1.85.
The uncertainty of this radius is, of course, not very large
but we are not certain about the radius
to about one centimeter,
so it's 85 plus or minus one centimeters.
So that's about a 1.2 percent error
and so the error, then, in the prediction will be 0.6 percent;
it's about .01 seconds.
So I expect... this is my prediction.
Now, Ireally want to challenge this .01
and so now I'm going to make the observations
and surely I'm going to do it now 10 times,
because then the uncertainty will be 0.1 seconds--
that's my reaction time--
and so I have the final period to an accuracy of .01 seconds
and so we can compare these numbers directly
and that is what I will do now.
I have here the timer
and I'm going to oscillate that back and forth--
and that would only take 20 seconds--
zero it, we started here.
We have great confidence in physics, right?
We believe in physics.
We believe in the conservation of mechanical energy.
Starts... are you counting?
Is this two? Yeah?
Is this three? Four?
CLASS: Four.
LEWIN: I don't believe you.
Okay, we start again.
Now!
One, two, three,
four, five,
six, seven...
I'm getting nervous.
Eight, nine, ten.
Holy smoke!
22.7 seconds!
It should have been 18!
22.7 seconds.
There must be something fundamentally wrong
with the conservation of mechanical energy.
Or is there something else?
And what is the difference between the two experiments?
STUDENT: Friction.
LEWIN: Excuse me?
STUDENT: Friction.
LEWIN: Oh, no, the friction is so low, that is not the reason.
There's a huge difference.
Think about it when you take your shower this weekend.
There is a huge difference
between this object moving and that object moving
and when you find out,
that is the reason why that is way slower, not friction.
See you next Wednesday.