With our knowledge of torque...
calm down. With our knowledge of torque
and angular momentum, we can now attack rolling
objects which roll down a slope. For instance, the following...
I have here a cylinder or it could be a sphere,
for that matter, and this angle is beta. I prefer not to use alpha because that's angular
acceleration, and there is
this friction coefficient with the surface, mu, and this object
is going to roll down and you're going to get
an acceleration in this direction, a. And I will evaluate
the situation when we have pure roll. That means the object is not
skidding and is not slipping. What is pure roll? If here is an object, the
cylinder is here with radius R, and I'm going to rotate it
like this and roll it in this direction,
the center is called point Q. Once it has made a complete
rotation, if then the point Q has moved
over a distance 2pi R, then we call that pure roll. When we have pure roll,
the velocity of this point Q, and the velocity of
the circumference, if you can read that--
I'll just put a c there-- are the same. In other words, vQ is then
exactly the same as v circumference, and v
circumference is always omega R. This part always holds,
but for pure roll, this holds. You can easily imagine that
if there is no friction here, that the object could be
standing still, rotating like crazy,
but Q would not go anywhere. So then we have skidding
and we have slipping and then we don't have
the pure roll situation. If the object is skidding
or slipping, then the friction must always
be a maximum here. If the object is in pure roll, the friction could be
substantially less than the maximum
friction possible. Now I would like to calculate
with you the acceleration that a cylinder would obtain. When it pure rolls
down that slope, it has mass M, it has length l
and it has radius R. And I would like you to use
your intuition and don't be afraid
that it's wrong. I'm going to roll down
this incline two cylinders. They're both solid,
they have the same mass, they have the same length but
they're very different in radii and I'm going to have a race
between these two. Which one will reach
the bottom first? So I repeat the problem. Two cylinders, both solid,
same length, same mass, but one has a larger radius
than the other. There's going to be a race. We're going to roll them down,
pure roll. Which one will will? Win win? Will win? Who thinks that the one with
the largest radius will win? Who thinks the large with the
smallest radius will win? Who thinks there will be
no winner, no loser? Wow, your intuition is
better than mine was. We'll see how it goes. Keep in mind what your vote was and you will see it come out
very shortly. Okay, let's put all the forces
on this object that we know. This one is Mg. And we're going to decompose
that into one a longer slope, which is Mg sine beta, and one
perpendicular to the slope. We have done that
a zill... zillion times now. And this one
equals Mg cosine beta. Then there is right here
a normal force, and the magnitude of that normal
force is Mg cosine beta, so there is no acceleration
in this direction and then we have a frictional
force here for which I will write F of f. There is an angular velocity
at any moment in time, omega, which will change
with time, no doubt. And then this center point Q,
which is the center of mass, is going to get a velocity v and that v will also change
with time. And the v of the point Q
which changes with time is v of the circumference, because that is the condition
of pure roll. That equals omega R. This is always true, but this is
only true when it is pure roll. I take the time derivative. The derivative of the velocity
of that point Q is, per definition,
its acceleration, so I get a equals omega dot
times R, and that equals alpha R, alpha being the angular
acceleration. So this is the condition
for pure roll. Now I'm going to take the torque
about point Q. When I take the torque
about point Q, N has no effect because it goes
through Q, and g has no effect
because it goes through Q, so there's only one force
that adds to the torque. If this radius is R,
the magnitude is RF and the direction is
in the blackboard. But I'm only interested
in the magnitude for now, so I get R times
the frictional force. This must be I alpha, I being the moment of inertia
for rotation about this axis through point Q, times alpha,
but I can replace alpha by a/R. So I get the moment of inertia
about Q times a/R. And this is my first equation, and I have as an unknown
the frictional force, and I have as an unknown a,
and so I cannot solve for both. I need another equation. The next equation that I have
is an obvious one, that is Newton's second law:
f = MA. For the center of mass, I can consider all the mass
right here at Q. We must have f = MA. And so M times the acceleration
of that point Q, which is our goal, by the way, equals this component,
equals Mg sine beta. That is the component downhill. And minus Ff,
the frictional force, which is the component uphill, and this is
my equation number two. Now I have two equations
with two unknowns. So I can solve now. I can eliminate Ff and I will
substitute for Ff in here this quantity divided by R. And so I get Ma equals
Mg sine beta minus moment of inertia
about point Q, times a divided by R squared. And now notice that
I've eliminated F of f, and so now I can solve for a. So I'm going to get...
I bring the a's to one side. So I get a times M plus moment
of inertia divided by R squared equals Mg sine beta,
and a now we have. I multiply both sides
with R squared. I get MR squared g sine beta
upstairs, and downstairs I get MR squared
plus the moment of inertia about that point Q. This is my result, and
all I have to put in now is the moment of inertia
of rotation about that axis. I want to remind you, though, that it is only true if we have
a situation of pure roll. So we can now substitute
in there the values that we have
for a solid cylinder. If we have a solid cylinder, then the moment of inertia
about this axis through the center of mass,
which I've called Q, equals 1/2 MR squared. And if I substitute that
in here, notice that all my M's...
MR squares go away. I get 1 + 1/2, which is 1 1/2. Upside down becomes 2/3. So a = 2/3 times g
times the sine of beta. There is no M, there is no l
and there is no R. So if I have two cylinders, solid cylinders
with totally different mass, totally different radii,
totally different length and they have a race,
neither one wins. Very nonintuitive. Every time that I see it
I find it kind of amazing. Notice that everything
disappears. M, R and l disappear. So those of you who said
that if I take two cylinders with the same mass,
different radii, those of you who said that there
is no winner, there is no loser, they were correct. But even more amazing is that
even the mass you can change. You can change anything as long
as the two cylinders are solid. That's what matters. So if we take a hollow cylinder, then the moment of inertia
about this axis through the center of mass,
through Q, if this... if really most of the mass
is really at the surface, then it's very close
to MR squared, and then the acceleration-- if I substitute in here
MR squared, I get a 2 there-- equals
1/2 times g times sine beta. So this acceleration
is less than this one. So the hollow cylinder will lose in any race against a solid
cylinder regardless of mass, regardless of radius,
regardless of length. And I want to show that to you. We have a setup here and I'll
try to show that to you also on the screen there, but for those of you
who are sitting close, it's probably much better that you just look at the
demonstration right here. I have here... ooh. Uh-uh. I have here to start with a very
heavy cylinder made of brass and this one is made
of aluminum. They have very different masses,
same radii, same length. Should make no difference. There should be no winner,
there should be no loser. I'm going to start them off
at the same time. I hope you can see that there. This is... this is
the starting point. Can lower it a little. I will count down three to zero, and then you can see
that they reach the bottom almost at the same time. So very different in mass. The mass difference is
at least a factor of three. All other dimensions
are the same. Three, two, one, zero. Completely in unison. Not intuitive for me. Now I have one that has
a very small radius compared to this one. This is a sm... small
aluminum rod. Maybe you can see it here,
television. This is way more heavy,
almost 30 times heavier. Should make no difference. As long as it's solid,
should make no difference. No winner, no loser. Radii are different,
masses are different. Should make no difference. Okay? Here we start the race. Three, two, one, zero. And they hit the bottom
at the same time. But now here I have
a hollow one, and you better believe it,
that it's hollow. So now all the mass is
at the circumference, and now it takes more time. Now the acceleration as you w...
as you will see, is half times g sine beta;
in the other case it was 2/3. And you may want to think
about it tonight, why this one takes more. It has to do, of course,
with the moment of inertia, but again, it's independent
of mass, radius and length. So it's purely a matter
of geometry. This one is going to be
the loser, and this one, regardless
of mass or length, is going to be the winner. So you see them. One is hollow, one is not. This is very light;
this is very heavy. I'll put the hollow one
on your side. Three, two, one, zero. The hollow one lost
and even fell on the floor. Yeah, I find these things
always quite amazing, that nature works this way, and I'm impressed
that most of you or many of you had
the right intuition when they said it would make
no difference for the two solid cylinders. We now come to the most
nonintuitive part of all of 8.01 and arguably perhaps the most
difficult part in all of physics and that has to do
with gyroscopes. And I really urge you to pay
a lot of attention and not even
to miss ten seconds, because you're going to see some
mind-boggling demonstrations which are so incredibly
nonintuitive that unless you have followed
the steps that lead up to it, you won't have any idea
what you're looking at. It will be fun, it will be cute, but it won't do anything
for you. Imagine that you and I
go in outer space. No gravity. We're somewhere in outer space and we have
this bicycle wheel there. And I'm going to put a torque
on this bicycle wheel in this direction, so I'm going
to put my right hand towards you and my left hand away from you. And I'll do that
for a short amount of time and I'll let it go. It's obvious
what's going to happen. This wheel is going to spin like
this forever and ever and ever. I've given it a little torque. That means if there's torque, there's a change
of angular momentum. The angular momentum change
must be torque times delta t. And so I do this and let it go and it will rotate about this
axis forever and ever and ever. Simple, right? Okay. Now I'm going to torque it
in this direction. So we're in outer space,
wheel is standing still and all I do is do this
and I let it go. Then it will rotate forever
and ever and ever and ever in this direction. That's clear. Now comes
the very nonintuitive part. Now I'm going to give it
a spin in your direction and now again I'm going
to torque like this. What will now happen? You will say...
or you might say. I'm not accusing you
of anything. You might say, well, you give
the wheel a spin so this... the wheel will
probably continue to spin and you do this, so maybe
what you're going to see is that it will rotate like this
as it did before and in the same time
the wheel will be spinning. But that cannot be because if the wheel would be
spinning like this, then the angular momentum
of the spinning wheel is in this direction. And if I would give it a twist and if it would continue to spin
and it would rotate like this, then this spin angular momentum
would go around like this and that cannot be because
there's no torque on the system, because once I let go,
there is no longer any torque. So it is not possible
for the wheel to keep rotating, and as a result of this torque
that I give it, that it simply goes around. That is not possible. How is nature going to deal
with that? I'll show you that
on a view graph. It's very nonintuitive
what will happen. And we will then also... I
will also demonstrate it to you. So here is the situation precisely as I described it
to you. You are, as an observer,
in this direction. This is the direction of 26.100. So you are viewing the wheels
like this. They're spinning towards you. That's what I will do shortly
and that's what I implied now. This is my right hand
and this is my left hand. The separation between
my height... right and left hand is little "b." So the torque that I apply
is bF, is this arm, so to speak,
times this force. And the force is perpendicular
to the arm. I apply a torque for a certain
amount of time-- delta t. When I do that, I apply,
I add angular momentum in this direction. But the wheel was spinning in...
in... in this direction. You see it. And so the angular momentum
of spin of the wheel is in this direction. I add angular momentum in this
direction and you see that here. So this was originally the spin
angular momentum of the wheel. I torque for time delta t, and so I add angular momentum
like this. And then I stop. I only torque for a short amount
of time and I stop. That means after I have stopped, the angular momentum
of the system as a whole can no longer change because there's no torque
on the system, and the only way that nature now
can solve that problem is to tilt this wheel in the way
I've indicated here, and to make it spin
in this direction and it will stand still. In other words, I hold it
in my hand, the wheel-- I will get it--
I give it a spin. I hold it in my hand,
I spin it towards you... and I'm going to put
my right hand towards you and my left hand away from you. The spin angular momentum
is now in this direction, so I'm going to give
it a torque like so. That means up. And what will the wheel do? The wheel will do this. Very nonintuitive. Watch it. Isn't that strange? You wouldn't expect that. I will do it again. I'm going to torque by pushing
my hand towards you, and the wheel does something
completely unexpected. It simply tilts. If I torque
the other way around, then the torque, of course,
will make it flip like this. I'll give it a little bit more
spin angular momentum, so now I move my left hand
towards you, and my right hand towards me, and then I expect that the wheel
will do this. And that's what it does. Extremely nonintuitive. These torques applied
to a spinning wheel always do something
that you don't expect. However, there is one thing
that always helps me in terms of guiding me and that is
you can always predict that the spin angular momentum
will always move in the direction of the torque, which is this external torque
that I applied. Let's go over that again. We have here the spin angular
momentum that you saw. It was pointing
in this direction. And I applied the torque
in this direction, the vector. And what does
the spin angular momentum do? It goes in the direction
of the torque. And then when I stop
with my torque, then of course nothing
changes anymore, and so what happens is
this wheel tilts. But notice that L, the spin
angular momentum, has moved from here to here. And I was torqueing
in this direction, so it moved towards the torque. If you have digested this,
then you can test yourself now. Now we have the same wheel. I'm going to rotate it
in exactly the same direction, but now I'm not going to torque
like this, in the Z direction, or like this,
in the minus-Z direction. Now I'm going to do this. Or I'm going to do this. Try now to really concentrate
on what I just taught you. And try to give an answer
to the following question. The wheel is rotating. I hold it in my hand and I'm
going to torque it like this so that the torque factor is
in your direction. Angular momentum goes like this,
torque is like this. What will the angular momentum
vector do? Move in the direction
of... of the torque. What will the angular... what will the spin angular
momentum vector then do if the torque is
in this direction? It will do this. It's going to move
in the horizontal plane. Very nonintuitive,
but that's what it will do. And I will show that to you. I'm going to spin this wheel. I'm going to spin it with
a high angular momentum, high spin angular momentum. And then I'm going to sit
on this stool and I'm going to torque
exactly as you see on the picture on the right. I'm going to torque like this. And as long as I torque it
like this, that spin angular momentum
wants to go around in the horizontal plane. And when I torque
the other way around, it will go back
in the horizontal plane. I'm going to torque exactly
as you see on the picture there. Are you ready? I stop the torque;
nothing happens. I torque backwards;
I keep torqueing. I keep torqueing. I feel it in my hand. I really have to push. I keep torqueing. And I stop torqueing
and it stops. The angular momentum vector is chasing, so to speak,
the torque. Is that nonintuitive? Very nonintuitive? It's also dangerous sometimes. We call this motion of the
stool, and in this case, the motion of the spinning
wheel, we call that precession. So you apply a torque
to a spinning wheel. Then what you obtain
is a precession. I can show you the precession
in another way which is, in fact,
very intriguing. Suppose I have here a string,
a rope, like we have there. And I stick into that rope,
I attach to the rope this wheel, just like so. And I let it go. Well, we all know
what will happen. Pffft, clunk. It's clear. All right. But now I'm going to spin it
before I let it go. So here at the bottom, at this
point P, there is a loop. And here is the... the axis of
rotation of the bicycle wheel which is solid brass,
it's a solid piece, and I give that a length
little "r," not to be confused
with capital "R," which is the radius
of the bicycle wheel. So this is capital R. And it can rotate about this
here reasonably freely. I call that center point Q and let this be the part of
the wheel that is on your side. I'm trying to make you see it a
little bit three-dimensionally. Suppose now I give it a spin
in this direction. Omega s. "S" stands for "spin." In what direction is now
the spin angular momentum? Use your hands, your thumbs. Spinning in this direction. Yeah. Spinning this direction, angular
momentum is in this direction. That's a spin angular momentum. L spin. Well, there is a force
on this system, Mg, and that force is
in this direction. It has a mass M,
the bicycle wheel, and it has a radius,
capital "R," and this part is little "r." So relative to point P,
there is a torque and the torque is R times Mg. This is 90 degrees
so the cross product is nice. The sine of the angle is one. So the torque relative to point
P is r times M times g. In what direction
is that torque? R cross F. In what direction
is that torque? Use your hands, thumbs,
whatever you want. You think in this direction? I disagree. I disagree. R cross F is...
you must be kidding. In the blackboard. It's not out of the blackboard;
it's in the blackboard. R cross F is in the blackboard. There is a torque
in this direction. Nature, gravity
provides that torque. What will the spin angular
momentum do? It's going to move in the
direction of the torque. It's going to chase the torque. So what will it do if the
angular momentum is here? What will it do? It will do this. And as it moves, the torque
will always be perpendicular to the plane
through the string and r. You can just see that
for yourself why that is. At this very moment when angular
momentum is like this, the torque is in the blackboard
because it's r cross F. But when I'm here, this r
has changed position, and always remains perpendicular
to the wheel. So the torque
will also change direction and so this angular momentum, spin angular momentum
will keep chasing the torque and start to rotate freely. That is exactly what I was doing
when I was sitting on the stool, except that I had to apply that
torque in my hands like this. It's exactly the same direction. I had to apply it all the time, and when I stopped,
the precession stopped. Here, however, the torque
will never stop because this Mg
will always be there, and I will show that
to you shortly. You may say, "You must be crazy "because you're violating
Newton's second law, f = MA. "This object got to fall. "There's only one force
on that object. "f = MA. "How can it not? The center of mass must fall
with acceleration g." Aha. There is not just one force
on that object. What do you think is here? The tension in this cable, T,
will be exactly Mg. And so the net, the sum of all
forces on that object is zero. There is no net force on that
wheel but there is a net torque, and that's why
it's going to precess. If there had been a net force, then indeed
it would also go down, if this force
were larger than this. So nature is very clever, the way that it deals with these
rather difficult problems. Before I will show you
this demonstration by spinning this wheel and then
hanging it there in that rope, um, I want to mention that the angular frequency
of the precession, which should never be confused with the angular frequency
of spinning, is derived for you... it's only
a three- or four-minute job, on page 344 in your book. Now, I will not derive it here
but what comes out of it, that it is the torque which is the one that we have
here in this case, divided by
the spin angular momentum. That gives you the frequency
of the precession. In our case, for our bicycle
wheel, it is rMg and the spin angular momentum
of this wheel, if it is rotating with angular
velocity omega of s, would be I times omega. Remember, l is I times omega
of a spinning wheel. So I have here I
rotating about point Q-- this is the axis of rotation--
times omega of the spin. This is the spin
and this is the precession. And then the period
of precession would be 2pi divided
by omega precession. Let's take a look
at that equation and see whether that sort
of intuitively makes sense. First of all, if you increase
the torque, the upstairs, then it says that the precession
frequency will increase. That makes sense to me because the torque is persuading
the angular momentum to follow it. So the torque is persuading the
spin angular momentum to change. Well, if the torque is stronger
then it is more powerful, so you expect
that the precession frequency will be higher. However, if the spin angular
momentum is very powerful, then the spin angular momentum
says, "Sorry, torque, I'm not going to go as fast
as you want me to go." So when you increase that spin
angular momentum in the wheel, it is also intuitive that the precession frequency
will go down. As the wheel spins,
it has spin angular momentum, but as it precesses around
like this, there will also be angular
momentum in this direction because it's rotating like this. Therefore there is
a total angular momentum which is
the vectorial sum of the two. This equation will only hold as long as the spin angular
momentum is really dominating the total angular momentum and you can see that
immediately, because suppose you make
the spin angular momentum zero, that it is not spinning at all. Do you really think that the precession frequency
will be infinitely high? Of course not. So this only holds in situations where the spin angular momentum
is way, way larger than the angular momentum that
you get due to the precession. So there are restrictions. When the... when the wheel
comes to a halt, when it's no longer rotating,
you better believe it, then the thing will go clunk. There's no longer
the precession mode. For our bicycle wheel, to get a feeling for how long
the precession will take, uh, we can substitute
the numbers in there, our bicycle wheel, the... the... the rod,
the brass rod, little r has a length of 17 centimeters, and the... the radius
of the bicycle wheel is about 29 centimeters. And let me make the assumption that all the mass of the bicycle
wheel is at the circumference, which is not very accurate,
but it's close to that. I mean, there are
some spokes here, but let's assume
that everything is here, so then the moment of inertia is
MR squared. Well, if now you take
a frequency of five hertz, spin frequency, you can calculate now omega
of the spin frequency. Omega equals 2pi
times the spin frequency. And so I know now I can
substitute that in there, so I get an omega precession now equals rMg
times the moment of inertia. I assume that all the mass is
at the circumference, an approximation,
so we get MR squared and then we get omega S,
which we have here. We lose the M, and so we get rg divided by omega s
times R squared. That is the angular frequency
of the precession, and the period of the precession
is 2pi divided by omega and you find then for the period
of the precession about ten seconds. So if I gave it a spin frequency
of five hertz with these dimensions
and with this approximation that all the mass is
at the circumference, you would expect that it would
precess around very gently in about ten seconds. But I have very little control
over that frequency, so it is possible
I gave it seven hertz, it's possible
I gave it three hertz. But I will do what I can. I'll actually give it
the maximum one that I can. That is always guaranteed
success. Where is the wheel? The wheel is here. So we'll spin it up
and then we'll put it in here. Notice the way I'm spinning it. I'm holding it away from me now
and going to change it and do it differently next. And there it goes. About ten seconds. Isn't that amazing? And it rotates,
seen from below, clockwise. Now it's going this way and I'm
going to redo the experiment, changing the direction
of rotation, and then it will go
the other way around. And now the angular momentum
is rotating like this, is pointing here. Spin angular momentum
is pointing like this, torque is like this, and so the spin angular momentum
is changing that... chasing that torque. I am the spin angular momentum. I am the torque. This is the torque. It's chasing it. All right. So I have this in my right hand. That's all right. And now I will... so
when I spin it up, that's right. So let me now change
the direction. I'm turning it over and
I'm going to spin it up again. Angular momentum is now
in this direction. See, it's turning
the other way around. Angular momentum is
in this direction. Torque is now towards me. Angular momentum is chasing
the torque. I've changed the direction
of the spin angular momentum. I've not changed the direction
of the torque, and now it is rotating, as seen
from below, counterclockwise. Before,
it was rotating clockwise. If I can increase the torque by putting some weight here
on the axle, I have this... this actually extends,
in our case, and I can put some weight
on here, then I actually add
to the torque and then you will see
that it's... it goes faster. The precession frequency
goes up. So I will put some weight
on there. So let it first go around,
which was roughly ten seconds, roughly calculated, and now I'm going to put
two kilograms here at the end. And now you will see
an instantaneous increase in the precession frequency. You see it goes much faster now. I take it off and then it goes
back to its roughly ten seconds. So what I have done is
I have increased this torque but not at the expense of M, because the reason why
the M cancels is because the moment of inertia
has an M in it, but if I just hang
this object on it, that doesn't change
the moment of inertia of the spinning wheel. None of this is intuitive. None of this
is intuitive. You can do all of this
with a $5 toy gyro. And I want to show this to you. This is my toy gyro. I have it in my office. It's great fun. And this toy gyro is doing
exactly the same thing that this is doing. Let me show you
the toy gyro first. Toy gyro. Oh, yeah. Here is a toy gyro. Can you see it? Maybe I should make it
a little darker here. Can you see my toy gyro? Yeah? I'm going to spin it and then I'm going to hang it
exactly the same way that that was hanging. I'm going to spin it, for those who are
sitting close-- whoosh-- and then putting it horizontally
and hanging it in a string, and you'll see exactly
the same thing is happening. And now through friction,
of course, all this fun ultimately
comes to a halt. I have something
very special for you, or I may have something
very special for you. That depends on my helper
who is here behind the scenes. I hear him. He's there. Great. Did you get full speed? I'm going to the airport
and get a little tired and ask one of my friends
to help me. Would you please help me
and... just carry this suitcase
for me around? Pick it up, walk around
a little, make some turns. Turn the other way around,
please. (laughter) STUDENT:
What the hell? LEWIN:
What the hell, yeah. Exactly. What are you doing, man? You're behaving so strangely. Make some more turns, man. We've got to go the...
we've got to catch the plane. It doesn't quite do what you
think it will be doing, right? So in here, you've guessed it. STUDENT:
Big spinning wheel or something. LEWIN:
Spinning wheel. And when you do this,
you put a torque on it and it does exactly what
you least expect: it flips up. Isn't that fun? Yeah. You may get arrested when you go to Logan Airport
with this suitcase. Thank you very much. It's great. Spinning objects have
a stabilizing effect. If you take a bicycle wheel,
and we have one, and I put it here and I do
nothing, it will fall. No one is surprised. However, if I give it a little
spin, then it doesn't fall. Why? Because it has angular momentum. It has spin angular momentum. And so it doesn't fall. And it is not only
a bicycle wheel. Look. Nicely stable. Not only with a bicycle wheel. You take... you take a quarter
and you put a quarter like this on your desk. You bet your life it will fall. Roll it; it becomes stable. You give it spin angular
momentum, it becomes stable. Take a top. You put a top on the table,
falls over. You twist it, you give it spin,
and the top is stable. So spin angular momentum has the
property of stabilizing things. And you will see that addressed
in one of your assignments, when I want you to address that
quantitatively. This is the basic idea behind
inertial guidance systems. In inertial guidance systems,
you have a spinning wheel, at least in the days that the guidance systems
had mechanical wheels. but that spinning wheel is
mounted in such a way that you cannot put a torque
on the axis of rotation of the spinning wheel. That's the way it's mounted. We call that
three-axle-gimbaled gyros. So the moment that you put
a torque on it, the housings-- in this case, the
yellow and the black housing-- will start to rotate, and you never managed
to get that torque on the spinning wheel. You never get it
on... on this axis. And therefore,
if now you put it on your boat or you put it in a plane,
or a missile for that matter, if you can never put a torque
on the spinning wheel and if the angular momentum
for spin is in this direction, it will stay there forever
and ever, assuming that we have
no frictional losses. And if then the plane turns, the direction of the spin
angular momentum will not change, but what will happen
of course is that this yellow frame
will rotate or this black frame will rotate. And in these bearings here
are shaft encoders, and they sense that the rotation
that the outer housing makes in order to keep this thing
pointing at the same direction. And that signal is being fed
back to the automatic pilot and that keeps the plane flying in the direction
that you want to. So you use, as a reference
all the time, the spin angular momentum
of your gyro, which is now mounted
in such a way that you cannot put
a torque on it, even when the plane changes
direction, and I want to show that to you. Okay, this is the direction
of my spin angular momentum and I'm the airplane
and I'm going to fly. Look at that
spin angular momentum. It has no respect for me. It stays in the same direction
no matter how I fly. And the arrow signals
that come from the bearings of the yellow housing
and the black housing, those arrow signals are fed back
to the automatic pilot and so the plane will stay
on course. Now what I can do for you to come to a final test
on your thinking, this wheel is suspended
in such a way that there is
no gravitational torque on it like there was here. But I can put a torque on it by simply putting some weights
on the axis. And what do you think
will happen now if I put some weight here
on the axis? So the wheel is spinning, but now I'm going to put
a torque on it here. It is spinning
in this direction. Angular momentum is pointing
straight at me, away from you. I'm going to put a torque
on like this, put a little weight there. Torque will be
in this direction. What will the spin
angular momentum do? Torque is in this direction; spin angular momentum's
in this direction. Spin angular momentum
will start to chase the torque. Watch it. There it goes. The spin angular momentum
is chasing the torque. You see exactly the same thing
that I've shown you before. And if I make the torque higher, then the precession frequency
will go up. See, it stops now immediately
when I take it off. Put it back on again. Continues. Put more on it. Goes way faster. What happens now if I put
the weight on this side? So I change the direction
of the torque. If I put it on this side,
torque is now in this direction, spin angular momentum is
in this direction. It's going to reverse direction. There we go. And you see it does. Amazingly nonintuitive. If you have problems with this,
you're not alone. See you Wednesday.
This guy dots.
He was a great professor and presenter, but apparently not such a good person.
I would love to use this in traffic court to get out of a speeding ticket! After a few minutes pretty much everyone would be lost and assume you were innocent
Pure rolling motion seems to be very related to a feather or hammer being dropped in a vacuum.
Great videos. I loved that he also applied the math to one show that’s it’s true, and two, makes it a lot easier to understand. For me, I always struggled in my physics courses but did really well in applied physics.
Professor is bleeding for his demos @ 38:30
I got through a lot of my courses by watching these lectures during my computer engineering undergrad