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-- 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
is minus 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 is minus 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... 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. (blows) 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. And 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. Ehm... 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 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 is way 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 is very, 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 100.000 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... so 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 (SHO). 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. has nothing, 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, "He Poppelepee!"[Dutch] 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 LEWIN; 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 have died. 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? (student answers) 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, I really 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. Start... are you counting? Is this two? Yah, 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.
