Okay, you did have some problems
with physical pendulums, and I want to talk a little bit
more about physical pendulums. Let's first look at the picture
in very general terms. I have here a solid object,
which is rotating about point P about an axis vertical
to the blackboard, and here at C
is the center of mass. The object has a mass M,
and so there is here a force Mg, and let the separation
be here b. I'm going to offset it
over an angle theta, and I'm going to oscillate it. Clearly, there has to be
a force at the pin. If there were no force
at the pin, this object would be accelerated
down with acceleration g, and that's not
what's going to happen. But I don't care
about that force because I'm going to take
the torque about point P. Remember when we had a spring,
just a one-dimensional case, we had F equals ma, and that,
for the spring, became minus kx, and the minus sign indicates
that it's a restoring force. So we now get
something very similar. In rotation,
force becomes torque, mass becomes moment of inertia, and acceleration becomes
angular acceleration. So now we have minus r cross F, and the minus sign indicates
that it is restoring. So if I take the torque relative
to point P, then I have... This is the position vector,
which has magnitude b. The force is Mg, and I have to
multiply by the sine of theta. So I have b times Mg times
the sine of theta and that now equals minus...
I can bring the minus here-- minus the moment of inertia
about point P times alpha, and alpha is
the angular acceleration, which is theta double dot. I bring them together, and I use the small-angle
approximation, small angles. Then the sine of theta
is approximately theta if theta is in radians. And so I bring this all
on one side, so I get theta double dot
plus bMg divided by the moment of inertia
about that point P times theta-- now this is
my small-angle approximation-- equals zero. And this is
a well-known equation. It is clearly a simple
harmonic oscillation in theta because this is a constant. And so we're going to get
as a solution that theta equals
theta maximum-- you can call that
the angular amplitude-- times cosine omega t plus phi. This omega is
the angular frequency It is a constant. Omega here, theta dot,
is the angular velocity, which is not a constant. The two are
completely different. That is the angular frequency. So we know that the solution
to this differential equation gives me omega is the square
root of this constant. So it is bMg divided
by the moment of inertia about that point P. And so the period of oscillation
is two pi divided by omega, and so that is two pi times
the square root of I relative
to point P divided by bMg. And let's hang on to this
for a large part of this lecture because I'm going to apply this
to various geometries. Make sure that I have it
correct-- yes, I do. This is independent
of the mass of the object. Even though you will say
there is an M here, you will see that in all cases when we calculate the moment
of inertia about point P that there is always
a mass up here. So the mass will disappear,
as you will see very shortly. I have four objects here, and they all have different
moments of inertia. They're all going to rotate about an axis perpendicular
to the blackboard, so to speak, and we're going to massage
each one of them to predict their periods. Let's first go to the rod. So we first do the rod. We have the rod here. This is point P, and
here is the center of the rod. The rod has mass M
and it has length L. So we have here Mg. I don't have to worry
about this anymore. I simply go to this equation,
and I want to know what the period is of this rod,
of this oscillating rod. All I have to know now is what is the moment
of inertia about P. And I know already that b in
that equation equals one-half L. So, what is
the moment of inertia of oscillation about point P? I have to apply now
the parallel axis theorem-- which you also had to do
during the exam-- which says it is
the moment of inertia of rotation
about the center of mass, which, in our case, is C--
the axes have to be parallel, so there is this axis
perpendicular to the blackboard, and this axis perpendicular
to the blackboard-- plus the mass of the rod times the distance
between P and c squared-- plus M times
this distance squared-- so that is b squared, and b squared is
one-quarter L squared. What is the moment of inertia for rotation of a rod
about this axis? I looked it up in a table. I happen to remember it now,
because I am lecturing 801. Two months from now,
I will have forgotten. So I remember now that
it is 1/12 ML squared plus one-quarter ML squared. That becomes
one-third ML squared. And so the period T
becomes two pi times the square root
of this moment of inertia, which is the one-third ML
squared, divided by bMg, and b is one-half L
for this geometry. So one-half LMg, and notice,
indeed, as I anticipated, you always lose your M,
you also lose one L here, and so you get two pi times the square root of two-thirds L
divided by g. So that is the period
that we predict for the rod. So let's write that under here, because we are going
to compare them shortly. So this is two pi times the
square root of two-thirds L divided by g. The rod that we have here
is designed in such a way that the period is very close
to one second. That was our goal. So T is as close as we can get
it to 1.00 seconds. So if you substitute
in this equation T equals one, you will find
that the length of this rod, if it is really pivoting
at the very end, should be
about 37.2 centimeters. So we did the best we can
when we made this rod. There is always
an uncertainty, of course-- how you drill the holes and
where you drill the holes-- so I would say the value that
we actually achieved is 37.2, probably with an uncertainty
of about three millimeters, so 0.3 centimeters. That's what we have. That is an error
of one part in 370. Let's make that...
round that off. That's a one-percent error
in the length. Since the length is
under the square root, the one-percent error becomes
half a percent error, so the period
that I then would predict is about 1
plus half a percent, so that is 0.005 seconds. So that, then, has
a one-half-percent error. So this is my predicted period. And so we're going to make
ten oscillations of the observed oscillations. We're going to get a number. My reaction time is not much
better than a tenth of a second. So we're going to get
a number there. We divide this number by ten. And so we can always calculate
the period, then, and then we get
a much improved error of a hundredth of a second, because we will divide
this number by ten, so this also is going to be
divided by ten. And let's see how close
we were able to get this to the 1.0 seconds. So, here is the rod. Turn this on, the timer. We'll offset the rod, and I will start it
when it stops somewhere. Now-- one, two, three, four,
five, six, seven, eight, nine, ten... not bad. 9.92-- well within
the prediction. 9.92, and so this becomes 0.992
plus or minus .01, and you see that is well within
the prediction that I made. Now, all these four objects
were designed in such a way that they have exactly
the same period of one second. And now comes the question, how do the dimensions relate
to each other in order to get them
a period of one second? So let's now calculate
the period for the other three objects: for the ring, for the disk
and for the pendulum. Let's start with the pendulum. For the pendulum,
center of mass is here. Here is point P. I give the pendulum
length little l-- you see that on
the right blackboard there. So b equals l-- that's the
separation between P and C. The moment of inertia about
point P is very easy now. This has no mass-- the mass is
all here, mass capital M, so that must be
M times l squared. So I go to this equation, and so I ask, what is
the period of a pendulum? And you're not surprised
that you find two pi times the square root
of l over g. We've seen that before, but,
of course, it also comes out if you do it in this
more complicated way. So here we get two pi times
the square root of l over g. Let's now do the ring. This is the ring. Pivot about point P here. Here is the center of mass right in the middle,
in the middle of nowhere. Right here is
the center of mass. And so the distance b is
the radius of the ring, so we have to calculate the
moment of inertia about point P. Again, we have to use now
the parallel axis theorem. It is the moment of inertia
for rotation about this axis through point C,
through a center of mass. That is MR squared, if R
is the radius of the ring. All the mass is
at the circumference, all at a distance R
from the center of mass. And then we have to add
to that-- according to
the parallel axis theorem-- the mass times this distance
squared, and this distance is R, so you get MR squared. So it is 2MR squared. So what is the period
of an oscillation? I of P, that is 2MR squared,
divided by bMg-- b is R-- RMg. I lose one R, I lose an M,
and we have seen this before. I derived this
in lecture number 21. I remember everything by lecture
number, believe it or not. So the period, now, is the same as a pendulum
with length two R. This is the ring. So here we have two pi times the square root
of 2R divided by g. I'll make a comparison
very shortly. I just want to finish them all. And I now would like to do
the disk, last but not least. For the disk,
all we have to do now is calculate
the moment of inertia. Here you have the disk,
a solid disk. This is point P,
this is the center of mass, but now it's solid,
so again, b is R, the separation between c and b. And the moment of inertia
for rotation about point P is now the moment of inertia for rotation
about the center of mass, which you look up in a table-- in the case of your exam, it
was given on the front cover-- one-half MR squared. That's the moment of inertia for rotation
about the center of mass. And now we have to add
Mb squared, and b equals R, so we have to add MR squared. So we find
three-halves MR squared. So what is the period
of oscillation? Two pi times
three-halves MR squared divided by bMg-- b is R-- RMg, and that equals two pi
times the square root... I lose my M as always,
I lose an R. I get here
three-halves R divided by g. So let's write that
down here, so we have two pi times the square root
of three-halves R divided by g. So we have them all four there, and so we can now make
a meaningful comparison. We want the periods
to be the same, so we can hang on
to those numbers, so we don't need this anymore. We want the periods
to be the same. We already have established that the period of the rod
is close to the one second, so we're not going
to measure them anymore. We just want to compare them-- whether, indeed, they have
the same period of oscillation-- by making them oscillate
in unison. But we want to know what the relative ratios are
of these dimensions the way we designed it. So let's first go to the rod,
then, and make a comparison between the rod
and the pendulum. So if you look at the rod-- and we use the pendulum as our
standard, with length little l-- then you see the period will be
the same... If two-thirds capital L
is little l. So, for the rod,
two-thirds L equals little l, so capital L is 1½ l. It has to be exactly 1½ times
the length of the pendulum, so this length has to be
exactly 1½ times this length to the center of mass, which is the center
of that billiard ball. And that's what we tried,
to the best that we could. So let's now go to the ring. For the ring to have
the same period, one second, 2R has to be the same
as the length of the pendulum. So 2R equals l,
so we can put in here for L, this has to be 1½ times l, and here, this now,
which is 2R, has to be l. Very nonintuitive,
that this length here is the same as
the diameter of the ring. Not at all obvious. And so now we go for the disk. So now we require that
three-halves R, 1½ R... We want that to be l, so we want
R to be 2l divided by three. So we want the diameter 2R
to be 4l divided by three. And when you look at the disk, it's hard to see that
it is exactly four-thirds, but you can see that it is
longer than the pendulum-- it should be one-third longer--
but it is not as long as the rod because the rod is 1½ times
the length of the pendulum, and so we can now complete
that picture. And we have now
that the diameter here-- 2R-- is now
four-thirds times l. And so what we can do now,
we can play with them. We can oscillate them
simultaneously and just see how well
they track each other. There's no sense in giving
you the dimensions. The rod was 37.2 centimeters. Let me write that down,
because we calculated that. So this was 37.2 centimeters,
and I think that translates into-- for the pendulum--
24.8 centimeters. But for the others I leave it up to you
to calculate the dimensions. Yes, 25... 24.8 is correct. So timing is not useful anymore. Let's just see
how these two go together. So we offset them,
and then we let them go. And they go
pretty much in unison. If you wait long enough,
of course, you will see
there is a difference. You can never make them
exactly the same, but they track
each other nicely. We can now also use
the rod and the disk. They track each other
beautifully. They're both very close
to 1.00 seconds. And we can have
the disk versus the pendulum. And you see they track
each other very nicely. But wait long enough
and you will see that, of course, the periods
will be different. So, this is my last word
on physical pendulums, but you may see it
again on the final. Not maybe-- you can almost
count on that, I'm telling you. Okay, I want to discuss now some
other interesting oscillation-- again, simple harmonic-- and that is liquid in a u-tube,
which you see there. If I have here a tube,
which has everywhere... it's open on both sides and everywhere
the same cross-section, and I put a liquid in here,
in equilibrium, just like that, and the liquid has mass M. It has density rho. The area of the tube is A, and
the length of the liquid is l. So this is l. I'm going to offset it,
the liquid, and I wanted to see it
oscillate, and I wanted to see whether I could calculate
the period of the oscillation. The total mass of the liquid
that I have... the total mass is the volume, which is the area times
the length times rho. I'm going to offset it so that this is higher
over a distance y. So this, then, is lower
over a distance y. So this distance is also y,
same as that. So the liquid now is here,
and then I release it, and it will start to oscillate. Well, when it starts
to oscillate, there comes a time that
the liquid, the whole liquid is going to slosh back and forth
and so everywhere in the tube, the velocity at any moment
in time will be the same because the cross-section
is not changing-- it's the same everywhere. See, if there is
a certain velocity here v, then it's the same
as the velocity here, as the velocity there,
as the velocity there. And that, of course, is y dot. That's the first derivative
of that position here. I'm going to write down the conservation of total
energy, mechanical energy. I assume that there is
no energy loss, although there probably is some. Friction inside the liquid
will probably generate some heat and that will cause
some damping. You will see that
when we do the experiment. For now, I will assume
that that's not the case. So what, now, is then the total
energy of the system-- that is, the sum of the kinetic energy
plus the potential energy? And if we assume
that that's constant, we will be able to find
the period of the oscillation very shortly, as you will see. The kinetic energy
of the liquid is easy. That is one-half M times
the velocity squared, and the velocity, we agreed,
is y dot squared. Now the potential energy. I call the potential energy
here, I call that u equals zero. When the liquid is standing here
and the liquid is standing here, I call that potential energy
zero. The mass that is now above this
level here, I call that delta M, and delta M is the area
times y times rho. This is how much mass
there is here. It was taken away from here
and was put here. How much work do you have to do to take this liquid
and put it there? Well, that's the same
when you take this liquid and put it here. And when you bring this liquid
which was there here, then you have moved it up
over a distance y, and so the gravitational
potential energy increases by delta M-- this is
the amount of mass here-- times g times h, and h is y. Mgh, remember? That's the increase
in potential energy. And so I move an amount
of mass which is delta M... I move it over a distance y. I bring it here, but that makes
no difference, of course. And so this is the total energy,
and this is now a constant. So I'm going to substitute
in there the A, l and rho, so I get one-half Al rho
velocity squared plus A rho g. And I get
a y squared equals a constant, because I have a y here
and I have a y there. We've done this before-- this
is the conservation of energy, and in order to find the period of the simple
harmonic oscillation, we take the time derivative. By the way, before we do that, this is... this was delta M
and this is an A, right? Yeah. A, rho-- yeah, that's it. So we lose A, we lose rho, and we continue
with what we have. And so we're going to take the... derivative versus time
of this equation. That gives me one-half l. The two pops out,
and the two becomes a one, so I get 2y dot,
then I apply the chain rule, so I get y double dot. Here I get plus g. The two pops out, becomes 2y, and then I get the chain rule,
y dot, and that equals zero. I lose my y dot, because
I have y dot in both terms. This two eats up this two. And so I find
that y double dot plus 2g divided by l times y equals
zero, and that was my goal, because this is clearly
a simple harmonic oscillation, because this is a constant. And it'll oscillate
in the following way: y equals y max times
the cosine of omega t plus phi, and this is
the angular frequency, which is directly related
to the period. Omega, angular frequency equals
the square root of 2g over l, and so the period will be two pi times the square root
of l over 2g. So this is the period
for an oscillating liquid. Notice that it is the period
that you would have had... would have obtained
from a pendulum if the length of the pendulum
were l... l over two-- not at all obvious,
not at all intuitive. You see our setup here. I have to know what l is,
and that is not so easy. Because of this radius here,
if I measure l on the outside, it's substantially larger
than on the inside. You may not think
it's a big difference, but it's huge-- it's
a nine-centimeter difference between the outside
and the inside. If I take the average value
between the two... if I take the average,
I find 72 centimeters, and I could be off by one. If I use this number for l and I
substitute it in this equation, then I find my predicted period,
which is 1.204, and because of this error that I
have of one, that would give me an uncertainty
of about 0.01 seconds. However, before
we start measuring it-- and I will do ten oscillations to get a reasonable,
accurate result-- I want to warn you. I make a prediction-- that the period
that we will measure will probably be
larger than this, and I can think of two reasons. The first is that the damping
of this liquid will be huge. You will see
how quickly it damps. In the past, we have never
taken damping into account, and we won't do that in 801. But the damping has the effect
on making the period longer. We've always ignored that, and in most of the
demonstrations that we did-- like just now--
that was acceptable. It may not be acceptable
for the liquid. But now there is a second point
that I want you to think about. Is it correct that I take
the average length, namely the average value between the outer length
and the inner length? I don't think it is. I want you to think about
why that is not correct. Look carefully where that l comes
into my differential equation and you will probably come up
with the right answer. And I claim that the actual l
that we should have taken is a little bit larger-- I don't know how much larger,
but it's a little bit larger-- and so that will also make
the observed period become larger
than the predicted one. So I'm not too optimistic that we will go and hit this
the way we want to hit it, but that's good, because
that's where the physics lies-- that you see that
there are other factors that have to be taken
into account. I'll turn this one on. Is it on now? Is it zeroed? I'll make it completely dark
in the classroom, because you're going to see... otherwise,
you can't see the liquids. So you see the liquid now. Oh, you see
these equations, too. Okay, it's zeroed. So, let me try to give this
a swing, a large swing. It damps so enormously that I really want to get
a very large swing. That's nice. Now-- one... two... three...
four... five... six... seven... eight... nine... Ten! Ah, not bad. 12.18-- not bad. We'll get a little bit of light. 12.18. So observed... Ten T observed... is it 12.18? Okay, my reaction time is 0.1. So T observed is 1.2... let's make it two
plus or minus 0.01 seconds. Oh, that's not bad. It actually...
there is an overlap. If you add this one here,
you get one-to-one, and if you subtract it here,
you get also one-to-one. So it's not bad. I expected it
to be a little higher, but it's close enough
to be happy. Think about why l should have
been taken a little larger. Now one more
very interesting oscillation-- a torsional pendulum. There's a wire there,
2½-meter steel wire, and there's hanging something
on the bottom, which we're going to offset, and then it's going
to oscillate back and forth. That's called
a torsional pendulum. And we're going to calculate the period of oscillation
of the torsional pendulum, and they have
wonderful properties. They are in a way like a spring,
like a one-dimensional spring. Remember
the one-dimensional spring that we had a period which was
independent of the amplitude? Well, within reason, of course. If you make the amplitude
too large, then you get permanent
deformation of the spring. But you never had to make any small-angle approximation
with the spring as we had to do
with the pendulum. Here is the pendulum,
the torsional pendulum. Here's a bar,
and there is a weight here and there's a weight here-- I'll tell you more
about that later. It's hanging here
from the ceiling. It has a certain length l. This is point P. And we're going to twist it and then we are going
to let it oscillate-- this bar-- in a horizontal plane. So when you look from above,
you will see the bar here-- see point P here-- and then we can offset it
over an angle theta, and then it will oscillate
back and forth. The... The torque relative to point P
is now very similar to what we had with a spring. We have a minus sign; again, that illustrates
that it is restoring. Instead of a k now,
we have kappa, which is what we call
the torsional spring constant, and now we have an angle
which we call theta. So we generate a torque which is
proportional to the angle, very similar
to the linear spring whereby we generate a force which is proportional
to the linear displacement. Now you generate a torque which is linearly proportional
to the angle. And this is the moment of inertia
about point P times alpha, and alpha is theta double dot. So we're going to get that theta double dot
plus kappa times theta divided by I of P equals zero. Kappa, by the way,
is the torsional constant. So we have
a differential equation. It's clear that
you're going to see a... simple harmonic oscillation. This is a constant, and so you're going to get
theta equals theta maximum times the cosine
omega t plus phi. It's getting boring. This is the angular frequency,
angular frequency... And angular frequency is
the square root of kappa divided by the moment
of inertia about point P. And therefore the period-- which
is two pi divided by omega-- equals two pi times
the square root moment of inertia about point P
divided by kappa. Well... how about kappa? Kappa is a function of the cross-sectional area here
A and the length l. And it's also a function of
what kind of material you have. Whether you have steel or nylon
makes a big difference. That's very intuitive,
of course. Remember that
in an earlier lecture when we stressed a wire to
the point that it was breaking. We dealt with Young's modulus. We had a wire and we had a mass hanging
at the end of the wire. And we discussed
the vertical oscillation. We could stress it
and let it go, and then we would get
an oscillation, like a spring, and that spring constant that we
found, then, was Young's modulus times the cross-sectional area
here divided by the length. And that was kind of pleasing. The thicker the bar,
the stiffer it is; the longer the wire,
the less stiff it is. Well, there is something
very similar here, but I don't want to go
into the details of exactly
how you derive here kappa. It's a little bit
more complicated. But indeed, the same is true. If you make the wire thicker,
then kappa will go up, and if you make the wire longer,
kappa will go down. That's immediately obvious. If you have a very short rod and
you try to twist that rod... It's clamped at the top, and you
twist it and it's very short, you would need a tremendous
torque for ten degrees. If you make the wire
a hundred meters long and you want to twist it
ten degrees, it takes nothing. So you can immediately see that,
of course, the value for kappa-- the torsional constant--
is a function of the length. It will go down
when the length goes up. We have a wire here
which is 2½ meters long, and the thickness of the wire,
the diameter-- according to the manufacturer,
it's a piano string-- the thickness is
25/1,000 of an inch. And if I calculate kappa
to the best of my ability, well, I find that kappa
should be very close to four times ten to the minus
four newton-meters per radian. And so all we have to do now is to calculate the moment
of inertia of the system, and then we can predict what the period of
this pendulum is going to be-- which is not my goal. You will see, my goal
is going to be a different one. Look at the bar
and look at the wire. The wire is so thin that the moment of inertia
relative to point P of the wire is very close to zero. Remember, it is proportional
to our square. But you can forget about that. Almost all moment of inertia
is in this system. I'll blow up that system
for you-- here it is. You'll see it there,
and it has on both sides... it has 200 grams. It has 0.2 kilograms and
it has here 0.2 kilograms. And this mass is
almost negligible. And this distance here
is 30 centimeters and this distance is
30 centimeters. So to a very good approximation the moment of inertia for
rotation about that point P-- this rotation-- will be
this mass times radius squared plus this mass times
radius squared. So that will be twice,
because I have two masses, times the mass times
the radius squared, and that is about 0.036
kilogram meters squared. And so when I use that
into our equation-- so I know now what kappa is, at least I have a reasonable
idea what kappa is-- and I know what I of P is-- that's really almost exclusively
determined by that cross-bar-- I will find, then,
using that equation that the period is
very close to 60 seconds. My goal is not to prove to you
that it is close to 60 seconds. My goal is to show you that for this dimension, which
is very thin and very long, that we can make
that angle theta maximum-- this angle-- amazingly large. You're not talking about
ten degrees or 30 degrees. We can go much further. And what I want to test with you
is how far we can go. This is one
of the great things in life for you and for me--
a challenge: How far can you go
and get away with it? There comes a time that
if we make the angle too large that we permanently deform
the wire-- it will not come back
to its original position. The same happens with a spring. If you take a spring, it is true that the period of oscillation
is independent of the amplitude but only up to a point. If you go too far, that
Hooke's Law no longer holds-- that you deform it permanently--
then, of course, the period will become
a function of the extension, and the same is true here. So if we twist it too much up, then, of course, we will
permanently deform it, and then the period will not
be independent of theta maximum. Having said that, I would like
to start asking you for advice. What kind of angle in this
direction shall we start with? What do you think is reasonable without total torture
for the wire? And then we'll write down
the times. So we're not really interested
in testing the 60 seconds, but we would rather
like to compare the various angles
that we give it. So what is the first one
we will try? Any idea? (echoing class):
30 degrees? What? STUDENT:
Six pi. LEWIN:
The first try?! You are cruel! Man! The first one,
you want six pi?! You're out of your mind. (class laughs) LEWIN:
I'm willing to go
one rotation, okay? You think that's nothing--
it's peanuts for you, right? Okay, I would like to go
theta maximum of 360 degrees-- so two pi--
and measure the period. In fact, to measure the period
takes a minute, and it's not necessary. We can measure half a period. Namely, we wait
until the pendulum stops, and we measure the time
until it stops again; that's half a period. Like with the spring-- if it stops here and it stops
there, that's half a period. So we'll measure half a period. Now, I don't know what
my reaction time is going to be, It may be
another tenth of a second, because the moment that it stops is not so well defined,
you will see. I'm just guessing-- probably a little larger
than one-tenth of a second. Let's give it a shot. Let's try first 360 degrees. You see, this is black
and this is a little red, so we will rotate it
one rotation. This is back here where it was. Yeah, have you seen that?
360 degrees. Okay, now, I will first let it
go and wait until it stops. I always do that,
and then I start the time. And then when it stops again,
I'll stop the time, and then we have half a period. So let it first do
its own thing... very slowly, very gently. It should take roughly
30 seconds for half a rotation. so you'll see now that it...
it's now at equilibrium again, because we wound it up
one rotation and so back to equilibrium. And now it's going to stop
very shortly, and when it stops,
I want to start the timer. Now! Okay, so now it goes back, and
we'll wait until it stops again. That gives us half a period. Okay. Now! 28.75. 28.8. What are we going to do now? Three rotations?
Five rotations? Three. Are we in favor of three? Who is responsible for
permanent damage to the wire? Do you accept
responsibility in life? Three is a lot. Three is six pi, man, six. I can start it, because it will
take a while before it stops. Three rotations--
first we have to be sure that it is more or less
back at equilibrium, which is always a difficult
thing, because it's so slow. Yeah, close enough. Okay, three-- shall we go
clockwise or counterclockwise? It should make no difference. I went this way first;
shall we go back? Yeah? You can sleep with that? Okay? Okay. One... two... three. Six pi. The piano wire. Okay, let's go. I will start the timer
when it stops. We have some time. Six pi. If we rotate it six...
three times, then it will rotate six times
back before it stops. I hope you realize that. (student speaks up
in background) LEWIN:
Was I too late? Thank you for pointing that out. So if you rotate it three times
and then let it go, it goes first three times back,
then it's at equilibrium, and then it winds three times up
again before it stops. Notice it's going
much faster now, but the time-- that's the whole thing-- should be very close
to that number again. 28.5. Oh! Not bad. Shall we now go all the way? What do you want to do now? Break the wire or... Ten rotations? You'd love to see that, right? It will go like mad,
ten rotations. Isn't it amazing
how much faster it goes? (imitates whirring) It's still 30 seconds. I must make sure
that I have my equilibrium. This was not equilibrium. I know it's somewhere here. No, it wasn't equilibrium. I think this is it. Okay-- ten, right? Ready? One... two... three... four... five... six... seven...
eight... nine... Ten-- poor wire. We'll let it go
and we'll see what happens. When it stops,
I'll start the time. (student speaks up
in background) LEWIN:
Excuse me? (student repeats) LEWIN:
Thank you. Thank you for pointing that out. Look how fast it's going. I mean, it's really going wacko. It has to do all that
in 30 seconds. Now! So now it has to go back
to its stopping. It has to make 20 rotations now,
20 rotations in 30 seconds-- ten back to equilibrium
and then ten to come to a halt. This is going to be your Thanksgiving
farewell demonstration. Now! 29.2-- fantastic.
