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PROFESSOR: All right,
let's get started.
Today is all about
Lagrange method.
We will talk a lot
about what we really
mean by generalized coordinates
and generalized forces
and then do a number of
application examples.
There's a set of notes on
Stellar on the Lagrange method.
It's about 10 pages long and I
highly recommend you read them.
They're not somehow
up with the notes
associated with lecture
notes or [INAUDIBLE]
way down at the bottom.
So you have to scroll all the
way down in the Stellar website
to find them.
Our second quiz is November 8.
That's a week from next Tuesday.
OK.
Pretty much same format
as the first one.
OK.
So let's talk about how
to use Lagrange equations.
So I defined what's called
the Lagrangian last time.
t minus v. The kinetic energy
minus the potential energy
of the entire system.
Total kinetic and total
potential energy expressions.
Then we have some quantities.
qj's.
These are defined as
the generalized forces.
Generalized coordinates,
I should say.
And the capital Q sub j's
are the generalized forces.
And the Lagrange
equation says that d
by dt the time derivative of
the partial of l with respect
to the qj dots, the velocities,
minus the partial derivative
of l with respect to the
generalized displacements
equals the generalized forces.
And for a typical system,
you'll have a number of degrees
of freedom, like say three.
And if you have three
degrees of freedom,
you need three
equations of motion.
And so the j's will go from
one to three in that case.
So the j's here refer
to an [? index ?]
that gives you the number
of equations that you need.
So you do this calculation
for coordinate one,
again for coordinate two,
again for coordinate three,
and you get then three
equations of motion.
OK
So this is a little obscure.
Let's just plug in.
For l equals t minus v.
And just put it in here
and see what happens.
You get d by dt of the
partial of t with respect
to qj dot minus d by
dt of the partial of v
with respect to qj dot plus--
I'll organize it this way.
Minus the partial
of t with respect
to qj plus the partial
of v with respect to qj
equals capital Qj.
Now when we first talked
about potential energy
a few days ago, we said
that for mechanical systems,
the potential energy is not a
function of time or-- anybody
remember?
Velocity.
So if the potential
energy is not
a function of time nor velocity,
what will happen to this term?
This goes away.
So this is 0 for
mechanical systems.
If you start getting
into electrons moving
and magnetic fields, then you
start have a potential energies
involving velocities.
But for mechanical
systems, this term's 0.
And I think the bookkeeping.
So this is the form of
a Lagrange equations
that I write down when
I'm doing problems.
I don't write this.
Mathematicians like elegance.
And this comes down to this
is beautifully elegant simple
looking formula.
But I'm an engineer
and I like it to be
efficient and practical useful.
This is the practical useful
form of Lagrange equations.
So you just use what you need.
Kinetic energy here,
kinetic energy there,
potential energy there.
And I number these.
There's a lot of
bookkeeping in Lagrange.
So I call it term one, term
two, term three, and term four.
Because you have to grind
through this quite a few times.
And so when you
do, basically you
take one of the results of
1 plus 2 plus 3 equals 4.
And you do that j times to get
the equations you're after.
OK.
So now we need to
talk a little bit
about what we mean by
generalized coordinates.
qj.
What's this word
generalized mean?
Generalized just means it
doesn't have to be Cartesian.
Not necessarily
Cartesian as in xyz.
You got a lot of liberty and
how you choose coordinates.
Not necessarily Cartesian.
Not even inertial.
They do have to satisfy
certain requirements.
The coordinates, they must
be what we call independent.
They must be complete.
So it must be
independent and complete
and the system
must be holonomic.
I'll get to that in a minute.
So you need to understand what
it means to be independent,
complete, and holonomic.
So what do we mean
by independent?
So if you have a multiple
degree of freedom system
and you fix all but
one of the coordinates,
say the system can't move in
all but one of its coordinates.
That last degree
of freedom still
has to have a complete
range of motion.
So if you have a double pendulum
and you grab the first mask,
the second mask can still move.
It takes two angles to
define your double pendulum.
So independent.
When you fix all
but one coordinate,
still have a continuous
range of movement
essentially in the
free coordinate.
And that's independent.
And we'll do this
by example mostly.
And complete.
The complete really means it's
capable of locating all parts
of the system at all times.
So let's look at a system here.
It's a double pendulum.
It's a simple one just made out
of two particles and strings.
I didn't bring one today.
And I need to pick some
coordinates to describe this.
And we'll use some
Cartesian coordinates.
Here's an x and a y.
And here's particle one.
And I could choose to describe
this system xy coordinates.
And I'll specify the location in
the system with coordinates x1
and y1.
Two values to specify
the location of that.
And down here I'm going to pick
two more values, x2 and y2,
just to describe the-- so x1 is
a different coordinate from x2.
x1 is the exposition
of particle one.
x2 is the x position of
particle two. y1 and y2.
So how many coordinates do I
have to describe the system?
How many have I used?
Four, right?
How many degrees of freedom
do you think this problem has?
Two.
So there's something already
a little out of whack here.
But the point is these aren't
independent, you'll find.
You just do a test.
You'll find that these
aren't independent.
If I fix x1 and x2,
systems doesn't move.
If I say this is going to be one
and this has got to be three,
this system is now frozen.
So this system of core
coordinates is not independent.
What did we say?
Independent.
When you fix all
but one coordinate,
you still have continuous range
of movement of the final one.
I could fix only
just two of these
and I've frozen the system.
I don't even have to go to
the extent of fixing three.
I'm assuming the strings
are of fixed length.
You can't change
the string length.
So this is not a very good
choice of coordinates.
And we had a hint
that it might not be,
because it's more
than we ought to use.
We only really need two.
So and then if we choose
these angles, v1 and v2,
let's do the test with that.
Are those independent?
So those are the
coordinates of the system.
If you fix v1, is there still
free and continuous movement
of v2 of the system?
Sure.
And if you fixed v2, it means
you can require this angle stay
rigid like that, and move
v2, well, the whole system
will still move.
So v1 and v2 are a system which
satisfies the independence
requirement.
Complete.
They're both systems
that are complete.
They're both capable locating
all points at all times.
But only the pair v1
and v2 in this example
are both independent
and complete.
Now, the third requirement is
a thing called holonomicity.
And what it means
to be holonomic
is that the system, the number
of degrees of freedom required
is equal to the
number of coordinates
required to completely
describe the motion.
Now, every example we've ever
done so far in this class
satisfies that.
We picked v1 and v2 and
that's all the coordinates
that we need to completely
to describe the motion.
Let me see if I can figure
out a counter example.
I didn't write down
this definition.
So holonomic.
And if the answer to
this question is no,
you cannot use
Lagrange equations.
So let's see if I can show you
an example of a system in which
you need more coordinates than
you have degrees of freedom.
I've got a ball.
This is an xy plane.
And I'm not going to allow
it to translate in z.
And I'm not going to allow it
to rotate about the z-axis.
So those are two constraints.
So this is one rigid body.
In general how many degrees
of freedom does it have?
Six.
I'm going to constrain
it so no z motion.
Five.
No z rotation.
Four.
It's not going to
allow it to slip.
This is x and that's y.
I'm not going to allow
it to slip in the x.
So now I've got another.
Now I'm down to three.
And I'm not going to
allow it to slip in the y.
Two.
So by our calculus of how many
degrees of freedom you need,
we're down to two.
We should be able to
completely describe
the motion of this system
with two coordinates.
OK.
So I've put this piece
of tape on the top.
And it's pointing diagonally.
That way.
And I'm going to roll
this ball like this
until it shows up again.
So it's right on top,
just the way it started.
Now start off same way again.
I'm going to roll
first this way.
And then I'm going to roll
this way to the same place.
Where's the stripe?
It's in the back.
So I've gone to
the same position
but I've ended up with the ball
not in the same orientation
as it was.
I went by two different paths.
And the ball comes up over
here rather than up there
where it started.
OK
So to actually describe where
the ball is at any place
out here, having gotten
there by rolling around,
without slipping and without z
rotation, how many coordinates
do you think it'll take
to actually specify where
that stripe is at
any arbitrary place
that it's gotten
to on the plane?
Name them.
AUDIENCE: [INAUDIBLE].
PROFESSOR: So.
In order to actually
fully describe it,
you've got to say
where it is x and y
and you actually have to say
some kind of theta and phi
rotations that it's gone through
so that you know where this is.
So this system is not holonomic.
And it has to be
holonomic in order
to use Lagrange equations.
So when you go to do
Lagrange problems,
you need to test for
your coordinates.
Complete, independent,
and holonomic.
And you get pretty good at it.
So here's my Lagrange equations.
And I have itemized these four
calculations you have to do.
Call them one, two,
three, and four.
And what I'm going
to write out is just
to get you to adopt a systematic
approach to doing Lagrange.
Left hand side.
To the left hand side of
your equations of motion
is everything with t and
v. The right hand side
has these generalized forces
that you have to deal with.
And generalized forces are
the non conservative forces
in the system.
So this is going to get
a little bit cookbook,
but it's, I think, appropriate
for the moment here.
So step one.
Determine the number of degrees
of freedom that you need.
And choose your delta j's.
Not deltas, excuse me.
qj's.
Choose your coordinates.
You find the number
of degrees of freedom
and choose the coordinates
you're going to use, basically.
Verify complete,
independent, holonomic.
Three.
Compute t and v for every
rigid body in the system.
Compute your kinetic
and potential energies.
One, two, three for each qj.
So for every
coordinate you have,
you have to go through
these computations.
One, two, three, four,
for every coordinate.
And this is your left hand side.
And if you don't have any
external forces and your non
conservative external forces,
then 1 plus 2 plus 3 equals 0.
But if you have non
conservative forces,
then you have to compute
the right hand side.
So the right hand side.
So for each qj, each
generalized coordinate,
you need to find the
generalized force that
potentially goes with it.
And you do this by computing
the virtual work delta w.
I'll put the little nc
up here to remind you
these are for the non
conservative forces.
The delta w associated with the
virtual displacement delta qj.
So for every generalized
coordinate you have,
you're going to try out
this little delta of motion
in that coordinate and
figure out how much virtual
work you've done.
So delta wj is going
to be qj delta k.
So this is the thing
you're looking for.
And it's going to be a function
of all those external non
conservative forces acting
through a little virtual
displacement, a little
bit of work will be done.
Mostly I'm going to teach you
how to do this by example.
So let's quickly do a really
simple trivial system.
Our mass spring dashpot system,
single agree freedom mkb.
It's going to take
one coordinate
to describe the motion.
X happens to be Cartesian.
There'll be one
generalized coordinate.
So qj equals q1 equals
qx in this case.
It's our x-coordinate.
Actually I should
just call it x.
That's our generalized
coordinates for this problem.
Is it complete?
Yeah.
Is it independent?
Yes.
Is it holonomic?
No problem.
We need t 1/2 mx dot squared.
We need v. And we
have 1/2 kx squared
for the spring minus mgx for the
gravitational potential energy.
And now we can start.
And we have some external
non conservative forces.
What are they?
fi non conservative.
And I think I'm going
to put an excitation up
here too, some f of t.
So what are the non
conservative forces?
Pardon?
AUDIENCE: k x dot.
PROFESSOR: It's not k.
My mistake.
You're correct.
My brain is getting
ahead of my writing here.
That's normally b
and this would be k.
I'm not trying to really
mess you up there.
So would be bx dot, right.
And is there anything else?
Are there any other non
conservative forces,
things that could put energy
into or out of the system?
AUDIENCE: [INAUDIBLE].
PROFESSOR: So the damper can
certainly extract energy.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah, the force f.
That external, it might
be something that's
making it vibrate or whatever.
But it's an external force, and
it could do work on the system.
And it's not a potential.
It's not a spring
and it's not gravity.
It's coming up and
somebody's shaking
it or something like that.
So f is also non conservative.
So the non conservative
forces in this thing
are f in the i direction
and minus vx dot
in the i direction.
And we could, in our normal
approach using Newton,
we draw a free body
diagram and we identify
a bx dot on it and an f on it.
But we'd also have our kx on it.
That would be what our free
body diagram would look like.
That's a conservative force.
Oops, and we need an mg.
So we have two conservative
forces, kx and mg,
and we have two non conservative
forces, bx dot and f.
So in this case, some of
the non conservative forces
is that f in the i
direction minus bx in the x
dot in the i direction.
So let's do our calculus here.
So 1 d by dt of the
partial of the t, which
is 1/2 mx dot squared
with respect to x dot.
So that gives me the derivative
of x dot squared with respect
to x dot gives me 2 mx dot.
So this is d by dt of mx dot.
But that's mx double dot.
And as you might
expect when you're
trying to drive
equation of motion,
you're probably going to
end up with an mx double dot
in the result. And
it always comes out
of these d by dt expressions.
OK, so that's term one.
Term two in this problem.
Minus t with respect
to x in this case.
Is t a function of x?
It's 1/2 mx dot squared.
So is t a function
of displacement x?
It's a function of velocity
in the x direction,
but is it a function
of displacement?
No.
So this term is 0.
Three.
Our third term.
Partial of v with respect to x.
Well, where's v 1/2 kx squared.
The derivative of
this is kx minus mg.
And we sum those.
So we get mx double dot
plus kx minus mg equals.
And on the right
hand side this is 4.
Now we need to do four
for the right hand side.
And four is really the
summation of the fi's,
the individual forces,
dotted with dr.
These are both vectors.
dr is the movement.
Little bit of work and it's
going to be a delta quantity,
like delta x.
And f are the applied forces.
And you need to sum these up.
So this dr in general is going
to be a function of the delta
j's.
The virtual displacements in all
the possible degrees of freedom
of the system.
We do them one at a time.
This case we'll only have
one, so it's trivial.
But this could be delta
one, two, three, four.
And each one of them
might do some work
when f moves through it.
But work is f dot
the displacement.
So it's the component
of the force
in the direction of the
movement, the dot product, that
gives you this little
bit of virtual work.
OK, so in this problem, this
is going to be equal to-- we
actually have an f of t of
some function of time in the i
direction minus bx dot in the
i direction dotted width delta
x, which is our
virtual displacement
in our single
generalized coordinate.
And this whole thing is going
to be equal to Qx delta x.
So you figure out the
virtual work that's done.
So if you do this
dot product, this
is also in the i hat direction.
So i dot i, i dot i.
You just get ones.
Because the forces are
in the same direction
as the displacement.
You're going to get an ft.
f of t delta x is one of the
little bits of virtual work.
And you'll get a
minus bx dot delta x.
And that together, those
two pieces added together,
are the generalized
force times delta x.
This total here
gives you delta w non
conservative for in
this case coordinate x.
So we're trying
to solve for what
goes on the right hand side.
We need the qx.
You notice what'll
happen, it'll cancel out
the delta x is the result.
And in this case, what
you're left with is Qx
equals f of t minus bx dot.
So this is number four.
So Qx.
Delta x is the bit of
virtual work that's done.
What goes into our equation
of motion is the Qx part.
And we got it by computing
the virtual work done
by the applied external
non conservative forces
as we imagine them
going through delta x.
And we're done.
You have the complete equation
of motion for a single degree
freedom system.
You could rearrange
it a little bit.
mx double dot plus bx dot
plus kx equals mg plus f of t
if you well.
So it's the same thing you would
have gotten from using Newton.
In a trivial kind
of example, but it
helps to find each of the
steps, things that we said
were required.
OK, so we're going to go from
there to a much harder problem.
So any issues or questions
about definitions, procedure?
So we start getting into
multiple degrees of freedom.
You need is set up a
careful bookkeeping.
So I just do this myself.
The top of the page, I identify
my coordinates, write down t,
write down v. Then I
say, OK, coordinate one.
One, two, three, four.
Equation.
Then I started the
coordinate two.
Calculus for one, two, three,
and four and so forth until you
get to the end.
OK, questions?
Yeah.
AUDIENCE: On the
[INAUDIBLE] what's
that thing after
the [INAUDIBLE] It's
like an open
parentheses [INAUDIBLE].
PROFESSOR: Oh, these are
functions of the delta j's.
This dr, where it
comes from, the work
that's being done in
a virtual displacement
around a dynamic equilibrium
position for the system
is a little movement
of the system.
dr. And we express it.
It's expressed in terms
of a virtual displacement
of the generalized
coordinates of the system.
So where the dr comes
from is going to be delta.
In this case, it's only delta x.
And in the next
problem, we're going
to do the force
in the problem is
not in exactly the same
direction as the delta x's
and delta theta's and so forth.
So when you do the dot
product, only that complement
of the force that's
in the direction
of the virtual
displacement does work.
And you account for that.
So let's look into a
more difficult problem.
So the problem is this.
I tried to fix assessed
before I came to class.
I didn't really quite have
the parts and pieces I needed.
But this a piece
of steel pipe here.
It's a sleeve on the
outside of this rod.
And I've got a spring that's
on the outside connected
to this piece.
And so it can do this.
And it's also,
though, a pendulum.
So the system I really want
to look at is this system.
So this swings back and forth,
the thing slides up and down.
So this has multiple sources of
kinetic energy, multiple forms
of potential energy.
And for the purpose
of the problem,
I'm going to say that there's a
force that's always horizontal
acting on this mass pushing
this system back and forth.
Some f cosine omega
t, always horizontal.
And I want to
drive the equations
of motion of the system.
So is it a planar
motion problem?
How many rigid bodies involved?
There's two rigid bodies.
Each could have possibly
six degrees of freedom.
But when you say
it's a planar motion,
you're actually immediately
confining each rigid body
to three.
Each rigid body can move
x and y and rotate in z.
So when you [? spread ?] out
and say this is planar motion,
you've just said each
rigid body has max three.
So this is a maximum
of six possible.
Where the other
three disappeared
to is no z deflection and
no rotation in the x or y.
OK, so we have a
possible maximum six.
How many degrees of freedom
does this problem have?
How many coordinates will we
need to completely describe
the motion of the system?
So think about that.
Talk to a neighbor.
Decide on the coordinates that
we need to use for this system
while I'm drawing it.
OK.
What did you decide?
How many?
Two.
All right, what
would you recommend?
What would you choose?
Pardon?
AUDIENCE: The angle
and how far down it is.
PROFESSOR: An angle
and a deflection
of what I'm calling m2 here.
So this is m2.
The rod is m1.
And he's suggesting an
angle theta and a deflection
which I'll call x1.
And I've attached to this
bar, the rod I'm calling it,
a rotating coordinate
system x1 y1.
About point A. So A x1 y1's
my rotating coordinate system
attached to this rod.
OK.
So I'm going to locate
the position of this
by some value x1
measured from point A.
And locate the position of the
rod itself by an angle theta.
Good.
Is it complete?
So if you freeze one, do you
still have-- the complete.
[INAUDIBLE] describe the motion
at any possible position.
Those two things.
Yes.
Is it independent?
If you freeze x, can
theta still move?
If you freeze theta,
can the x still move?
OK, is it holonomic?
Right.
You need two, we
got two and they're
independent and complete.
Good.
Now the harder work starts.
So I'm going to give us the
mass of the rod, the mass moment
of inertia the rod about the
z-axis but with respect to A.
The length of the rod is l1.
The sleeve mass m2 izz
with respect to its g.
So it has a g.
There's also and I'd
better call it g2.
That's the g of the sleeve.
There's also a g1.
A center of mass for
the rod and a center
of mass for the sleeve.
Those are properties
we'll need to know
and I'll give them to you.
OK.
So we need to come
up with expressions
for potential energy
and kinetic energy.
So this problem, the potential
energy's a little messy.
Because you have
to pick references.
You have to account for
the unstretched length
of the spring.
So call l 0 is the
unstretched spring length.
We know that also.
So I propose that the
potential energy look like 1/2,
for the spring, anyway,
1/2 the amount that it
stretches in a movement x1.
The amount that
it stretches then
should be whatever
that x1 position is.
And that x1 position, and I
drew it slightly incorrectly.
I'm going to use x1 to locate
the center of mass, which
is always a good practice.
So here's the center of mass.
So my x1 goes to
the center of mass.
That's x1.
So that's the total distance.
And from that, we
need to subtract
l0, the unstretched
length of the spring.
And we need to subtract
1/2 the length of the body
because that's that
extra bit here.
So this is the amount that the
string is actually stretched
when the coordinate is x1.
And you've got to square.
And that'll be the potential
energy stored in the spring.
Then we got to do the same
thing for the potential energy.
We have two sources of
potential energy due to gravity.
And they are?
Two objects, right?
Two potential energy.
So why don't you take
a minute and tell me
the potential energy
associated with the rod.
So the rod has a center of mass.
It's a pendulum basically.
So it's the same
as all the pendulum
problems you've ever seen.
And I would recommend that we
use as our reference position
its equilibrium position
hanging straight down.
And I'll tell you
in advance, I'm
going to use the unstretched
spring position this time.
Just stay with that.
That's where it's
going to start from.
That's my reference
for potential energy.
But does the unstretched
spring position
have anything to do with the
potential energy of the rod?
No.
OK.
So its reference position is
just hanging straight down.
So figure it out.
What's the potential
energy expression
for just the rod part?
Think about that.
So I'm going to remind
you about something
about potential energy.
Potential energy, one of
the requirements about it
is the change in potential
energy from one position
to another is path independent.
So you don't actually ever have
to do the integral of minus mg
dot dr.
You don't have to
do the integral.
You just have to
account for the change
in height between
its starting position
and its some other position.
Spend a minute or
two, think about that.
Work it out.
You got a question?
OK.
Can you talk?
Talk to a neighbor,
check your ideas.
So you have a suggestion for me?
Ladies?
AUDIENCE: [INAUDIBLE].
l1/2 [INAUDIBLE].
PROFESSOR: OK.
Anybody want to make
an improvement on that?
Or they like it?
Improvement?
AUDIENCE: [INAUDIBLE].
PROFESSOR: 1 minus cosine theta.
So let's put that up and let's
figure out if we need that.
We have a bid for cosine theta
and 1 minus cosine theta.
So you need to have a potential
energy at the reference
and you need to have a potential
energy at the final point.
And the difference
between the two
is a change in
potential energy here.
So what's the reference
potential energy
is mg l1 over 2 when
it hangs straight down.
And then when it moves up
to this other position,
this is the l1 over 2
times this is a delta h.
This is the change in
height that it goes through.
So you need the 1 minus.
Do we have the sines right?
Yeah.
OK.
So now we need another term.
And I'll write this one down.
This one's a little messier.
We need a potential energy term
due to gravity for the sleeve.
And that's going to mimic this.
You're going to have
a term here plus m2g.
And it's reference,
I'm just going
to do it as a reference
amount minus the final amount.
The reference will be
at the initial location
of its center of mass, which
is l0 plus l2 over 2 minus m2 g
x1 cosine theta.
Because this one
is a little messy
because you've got this thing.
It can move up and
down the sleeve.
And if that moves, you've
lost your reference.
So you can't do this as a
concise little term like this.
You have to separate
out the reference
and then this is the final.
And the l0 plus
l2, this quantity
here is the starting height.
This x1 cosine theta is
the finishing height.
And the difference
between the two
gives you the change in
the potential energy.
So this is your potential
energy expression.
This plus this plus these.
All right.
So what about t?
We got to be able write it.
Kinetic energy is
generally easier.
Got to account for all
the parts and pieces.
So we have to chunks.
And we're going to have
rotational kinetic energy
associated with the rod,
rotational kinetic energy
associated with the sleeve.
But also some translational
kinetic energy
associated with the sleeve.
And I'll write these terms down.
Make the problem go
a little faster here.
1/2 izz about A. That's the rod.
Plus 1/2 izz for
the sleeve about g.
We'll discuss why
the difference here.
And that's theta dot squared.
Now for the kinetic energy
that comes from translation
of the center of mass.
Because I'm broken up.
Let me start over.
This system is pinned about
A. And the rod is just
simply pinned at A.
And the last lecture
I put up these different
conditions and simplifications.
You can account for a
something about a fixed pin
by computing maximum
inertia about A.
It's basically a parallel
axis theorem argument.
Times 1/2 times that
times theta dot squared.
So this gives you all
the kinetic energy in one
go with the rod.
But for the sliding mass,
because its position
is changing, you can't do that.
You have to account for the two
components of kinetic energy
separately.
This accounts for
rotation about g.
Even though g is moving.
That accounts for that energy.
Because it's only a
function of theta dot.
It's not a function
of that position x.
This term is going to account
for the kinetic energy
associated with the movement
of the center of mass.
So we need a vg2 in the
inertial frame dot vg2.
These be in vectors.
And does that get everything?
I think that does.
So vgo is it certainly
has a component that
is its speed sliding up
and down the rod, right?
And that's in the
i hat direction.
But it has another
component due to what?
Can you tell me what it is?
Its contribution to its
speed due to its rotation.
AUDIENCE: [INAUDIBLE].
PROFESSOR: It's got a theta dot.
Yep.
It needs an r, right?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
So this would be an x1 plus.
No actually, I made x1 go
right to the-- so just x1 theta
dot in one direction.
Yeah, so j hat here.
Actually that's the moving
coordinate system unit
vector in the y direction.
And so we do the dot product.
You get this times itself.
i dot i and j dot j.
This quantity here is
1/2 m2 x dot squared
plus x, this next one I guess.
x1 squared of theta dot squared.
That's the kinetic energy of
accounting for the velocity
of the center of mass.
So now we have our entire
kinetic energy expression.
So now we have how
many coordinates?
Two, right?
How many times do we
have to turn the crank
and go through the Lagrangian?
Got to go through it twice.
So let's apply Lagrange here.
And we'll just do
number one first.
So and let's see.
Which one do I have
on my paper first?
I guess we'll do
the x1 equation.
This is delta x1.
So this generalized
coordinate x1.
And we need to do
term one, which
is in d by dt of partial of
t with respect to x1 dot.
OK.
So we look at this and say,
well, is this a function?
Is this term a function of x?
Nothing.
You get nothing from there.
Is this term a function of x?
Yeah, it's down here.
We only have to take
the derivative of this.
We have to do that job.
So the derivative of
this with respect to x
dot, you get a 2x dot here.
Do you get anything from here
when you do this with respect
to x dot?
You only get a
contribution from here.
The two cancels that.
And so this should look
like m2 x1 dot but d by dt.
Do this once in two steps
here so you see what happens.
You get an m2 x1
double dot out of that.
So we've gotten the
first piece of this.
We've got a couple to go.
But you know a lot
about Newton's laws
and you know a lot about
calculating equations
of motion now using sum of
torques and all that stuff,
right?
So this is just something
moving, has a circular motion,
has translational motion.
What other
accelerations had better
appear in this
equation of motion?
And which equation
are we getting?
There's going to
be two equations
and it has physical
significance to it.
What equation does this
begin to look like?
Just physically, what movement
is being accounted for here?
Looks like translation
in the x direction.
It's this thing sliding.
It's this part of the
motion sliding up and down.
You're writing an
equation of motion and mx
double dot has units of what?
Torque?
Force.
So it's a force equation.
This is just f equals ma is
what this is going to show us.
Remember, the direct
method has to give you
the same answer as Lagrange.
So we're getting
a force equation.
It's describing mx double dot.
What other acceleration
terms do you
expect to appear in
this from what you know?
Yeah.
AUDIENCE: [INAUDIBLE].
PROFESSOR: A centripetal term.
Do you believe there ought
to be a centripetal term
in this answer?
Why?
Because it's got
circular motion involved.
For sure.
Any others?
Is there any Coriolis in this?
In this direction.
Which direction
are we working in?
Is there Coriolis acceleration
in the x direction?
By the way, these equations,
do we have any ijk's in here?
These are pure scalar equations.
No unit vectors involved.
This equation only
described motion in the x.
So will there be a Coriolis
force in this acceleration
in this problem?
Will there be an
Eulerian acceleration
in this equation of motion?
The reason I'm going
through this with you,
I want you to start developing
your own intuition about
whether or not when
you get it at the end
it's got everything
it ought to have
and doesn't have stuff
it shouldn't have.
OK.
So your forecasting, then we
better get a centripetal term.
Well, let's see what happens.
So that was number one.
Number two here is our dt
by minus the derivative
with respect to x, in this case.
So we go here.
x1 we've been calling it.
Is this a function of x?
This piece?
Nope, it's x dot.
How about this one?
Right.
Take this derivative,
you get 2x.
So this fellow is
going to give us
minus m2 x1 theta dot squared.
What's that look like?
There it is.
There's your centripetal
term you're expecting to get.
OK.
And step three is plus partial
of v with respect to x.
In this case with respect to x.
And where's our potential
energy expression?
Well, it's up here.
And where the x
dependency is in it.
There is no x in that term
and no x in that term.
But we have x's in both
of these other terms.
So when we run
through this, I'll
write down what we come up with.
We get certainly a spring term.
k x1 minus l0 minus l2 over 2.
So that's the spring piece
when you take the derivative.
The two cancels the 1/2
and the derivative of parts
inside just gives you 1.
So that's the first piece of
the potential energy expression.
And the second piece is only
going to come from here.
The derivative of this
with respect to x1
is just m2g cosine theta minus.
And you add those
bits together, you
end up with m2 x1 double
dot minus m2 x1 theta
dot squared plus k
x1 minus l0 minus l2
over 2 minus m2g cosine theta.
So those are the
three terms, 1 plus 2
plus 3, that go on
the left hand side.
And they're going to
equal my qx that I find.
I still have to find what
the generalized force is
in the x direction.
So all that's left to
do for this problem
is to find q sub x, the
generalized force that
goes on the right hand side.
So now let's draw a little
diagram here of my system.
And at the end of the sleeve.
So here's my sleeve.
I've applied this force.
This is f of t.
And maybe it's some
f not cosine omega t.
It's an oscillatory
force, external force.
Make it vibrate.
And I need to know
the virtual work done
making that force go through a
displacement in what direction?
So this equation is
the x1 equation, right?
And so the virtual displacement
I'm talking about is delta x1.
And the amount of
work that it does
is delta x1 times the
component of this force that's
in its direction.
So I'm going to take this
force and break it up
into two components.
And if this is my theta,
this is also theta.
So this will be f0 product.
And I'll leave out the
cosine omega t here.
It's a function of time.
But this side then
is cosine theta i.
No, hey, I got this wrong.
I drew this wrong, I'm sorry.
This is theta.
This is going to be sine.
This side is sine theta
in the i direction.
And this piece is f0 of
t cosine theta in the j.
So I break it up in two parts.
And the virtual work
associated with x1
is the thing I'm looking for,
qx, dotted with delta x1.
And that is f of
t here, the vector
dotted with dr, my
little displacement.
But in this case, this
then all works out
to be f0 cosine omega t.
And it has sine theta i
plus cos theta j components
dotted width delta x in the i.
So you're only going to get
i dot j gives you 0, i dot i
gets you 1.
So you're going to get
one piece out of this.
This says in the qx equals
f0 cosine omega t sine theta.
And start with you have a
delta x here and a delta x here
and that gives you the
delta virtual work.
Personally when I
do these problems,
I have to think in terms of
that little virtual deflection.
I've actually figure out
what's the virtual work done.
And then at the
end I take this out
and this is the qx
that I'm looking for.
So my final equation
of motion says,
this equals f0 cosine
omega t sine theta.
And that's your equation of
motion in the x1 direction.
So when you finish one of
these, you need to ask yourself,
does this make sense?
Does this jive with
my understanding
of Newtonian physics?
Better have a linear
acceleration term,
because that's what it's doing.
You have another acceleration
term in the same direction
due to centripetal.
A spring force for sure.
And a component of gravity
in the direction of motion,
up and down the slide,
equal to any external forces
in that direction.
So it makes pretty good sense.
OK.
Now, also another
test you can do
is does it satisfy
the laws of statics?
That's another check
you could perform.
Does this thing at static
equilibrium tell the truth?
A static equilibrium all
time derivative is 0.
So this would be
0, this would be 0.
You know its static
equilibrium hangs down,
so cosine theta is 1.
Static you don't have any
time dependent forces.
That's 0.
So the static part of this says
that k x1 minus l0 minus l2
over 2 equals m2g cosine.
And that's cosine theta
is 1, so it's m2g.
And you could figure out then
this must be k times something.
This is the x.
This is the amount of the spring
stretches, the static stretch
of the spring, so the spring
has an equal and opposite force
to the weight of the thing m2g.
So that's another check you
can do when doing the problems.
OK, I'll write up the final one.
We have one more question to go.
Got to do all the derivatives
with respect to theta.
So you take a minute to decide
how many acceleration terms
and what acceleration
terms do you
expect to see come out of this
second equation of motion.
Because now we're talking
about which motion?
Swinging motion.
And what's its direction?
In Newtonian sense, it would
have a vector direction.
It's in what we call j here.
OK, so you're about
to get the j equation.
What terms do you
expect to find in it?
Talk to your neighbors
and sort this out.
And basically tell me what
the answer's going to be.
What do you think?
What are we going to get?
AUDIENCE: We were debating
about whether or not
it was going to be like speeding
up in the theta [INAUDIBLE].
PROFESSOR: So it is a pendulum,
just a weird pendulum.
So does the theta change speed?
Sure.
When he gets up the
top of the swing at 0.
All the way down,
it's maximum speed.
So what term does that imply
that you're going to get?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Well,
maybe, maybe not.
Yeah?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Going to get an
Eulerian, which means you've
got a theta double dot term.
You're expecting a theta
double dot term to show up.
OK.
What else?
Will you get a Coriolis term?
Do you expect a Coriolis term?
Something that looks
like x dot theta dot.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah.
The thing is sliding
up and down the sleeve.
It has a non 0 value of x dot.
Any time you got
things moving radially
while something is
swinging in a circle,
you will get Coriolis forces.
It means the angular momentum
of that thing is changing
and it takes forces
to make that happen.
So here's what this
answer looks like.
That's the one term.
The two piece gives you 0.
It's not a function of
x in the three piece.
The potential energy pace
gives you m2g x1 sine theta
plus m1 g l1 over 2 sine theta.
And the fourth
piece, the q theta,
well, that's just going to
be the virtual work done.
There's a tricky
bit to this one.
Now there's virtual work,
but which direction?
So we have an f dot dr.
The only f we have is this.
What's the dr?
What direction is it?
This is the theta coordinate.
What direction does that
give you displacements?
f dot dr's a displacement,
not an angle.
To get the work done,
you got to move a force
through a distance.
So the distance, first of
all, is in what direction
when theta moves?
j.
Little j hat, right?
And now if you get
a virtual deflection
of delta theta, what's
the virtual displacement?
You had a virtual change in
[? angle ?] delta to put theta.
But is that the
virtual displacement?
What's the displacement of this
point here in the j direction,
given a virtual
displacement delta theta?
Think that out.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Can't quite hear you.
AUDIENCE: [INAUDIBLE].
PROFESSOR: x1 delta
theta will give you
the motion to displacement
at the center of mass
in that direction.
x1 comes from here to here.
So x1 delta theta will give
you a little displacement
in that direction.
But is that the
displacement we care about?
We need the displacement here.
So you're close.
So we're going to get some
force dot a displacement dr.
And that's going to be
our force, this guy,
with its i and j components.
i and j terms.
But this term out here
is x1 plus l2 over 2
to get to the end.
And it's in the j direction.
So it's a length times A.
And you need the delta.
This quantity.
And you need a delta theta.
Delta theta.
This is the term.
This is the dr for the system.
An angle, a virtual deflection
in angle times the moment arm
gives you a distance.
It's in the j hat direction
dotted with the same force
breaking the force up into
its i and j components.
It had a sine theta
i cos theta j.
So this is going to give
me a f cosine omega t
cos theta j dot j.
f0 cosine omega t cos theta
x1 plus l2 over 2 delta theta
is the delta w.
That's the work and the virtual.
The generalized force q
theta is this part of it.
So this plus this
plus this equals
that on the right hand side.
So this is part four.
And look at it.
Yeah?
AUDIENCE: [INAUDIBLE].
PROFESSOR: So f of t, I didn't
want to write it all out.
This thing breaks into
an i and a j piece,
which is written over there.
This is the sine theta
i cos theta j term.
Which I brought back
from over there.
And we dot it with the
dr that we care about,
which is this length times
that angle in the j direction.
So j dot.
We only pick up the
j piece of this.
And that gives us this
cosine theta term.
OK.
Let's look quickly.
This is a rotational thing.
It has units of is it force?
Is this a force equation?
i theta double dot
has units of what?
Torque.
This is a torque equation.
This is the total mass
moment of inertia izz
with respect to A for this
system such that the Eulerian
acceleration.
The torque it takes
to make that happen
is the sum of the mass
moment of inertia of the rod
plus the mass moment of
inertia of g plus m2 x1
squared, which looks a lot
like the parallel axis theorem.
This is izz A for
the moving mass.
There's your Coriolis term.
And here's your potential
terms and there's
your external force.
OK.
Talk more about these
things in recitation.
