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PROFESSOR: So today we're going
to do two things in particular.
One is finish off
with the discussion
of this device, a shaker.
This, by the way, this
is a commercial thing.
And out of the catalog,
this is the littlest one.
This is a 50 pound shaker.
At full speed it actually
puts out 50 pounds.
All it is is masses inside
going around and around.
AUDIENCE: What is its
commercial purpose?
PROFESSOR: Ah, what's
its commercial purpose?
Well, the big ones that are
maybe 100 pounds of moving mass
are they bolt them to the
floor in nuclear power plants
and test them.
Shake the buildings to
represent earthquake kind
of loads and things like that.
And the smaller ones,
you can buy these for.
This kind is actually
if you're running
an operation like
in a flour mill
and you've got particulate
stuff trying to get
it to slide through chutes.
Does stuff slide down
chutes easier if the things
are vibrating a little bit?
Have you ever
banged on something
to get stuff to come loose?
You can just stick one of these
on the side and let it run.
Nothing sticks.
So there's lots and lots of
purposes for shakers like this.
So we were in the process of
analyzing how one of these
works.
And I want to finish that.
And then part two today is
we're going to-- we've only
really talked about angular
momentum with respect
to particles,
individual particles.
And even in your
physics classes you
did things with mass
moment of inertia.
And so we're going to
make the connection
today between particles,
mass moment of inertia,
unbalanced shakers.
It all comes together in the
second part of today's lecture.
So the problem we are analyzing,
literally that little shaker,
can be modeled.
Well, in that one
particular application
we can find that
thing on rollers.
This is the problem
we are discussing.
It has inside of it an
unbalanced rotating mass
with an arm that's E long.
It's called the
eccentricity in the trade.
And it has some mass m.
And this body that it's in,
we'll call it mass of the body,
mb.
And this thing's going
around and round.
So this is some angle theta
which is described as omega t.
And they're constant
rotation rate devices.
So theta dot equals omega
and that's a constant.
That's how they're
basically designed.
And we label this
point A. Over here we
have a inertial
coordinate system, xy.
This point a, this point we've
called b in our analysis.
And we set out to
find the equation
of motion of this thing
in the x direction.
Has no movement in the y.
It's confined in the y.
It puts out lots of
force in the y direction.
You really have to restrain
it to keep it from moving.
But it doesn't move
in the y direction,
but it will move in the x.
OK.
So we came to the
conclusion that we
could write for the
main body the summation
of the external forces on mb.
It's mb times its acceleration.
And it's acceleration
is completely
defined by this coordinate.
And if we draw a free
body diagram of this mass,
you're going to have a
normal force I'll call n
in the y direction upwards.
You're going to have
its weight downwards.
And you're going to
have some force exerted
on it through this shaft that
comes from the little mass.
So we're accounting for
everything the little mass,
all its influence
on this big block
by the forces that are passed
through that rod, which
is hinged at the center.
OK.
And I'm going to call that f mb.
OK, the force from
that little link.
Now, that happens to be equal
to minus the forces on the mass
that this rod exerts.
It must exert some
force on the mass
to make it go around and around.
And because of
Newton's third law,
those two forces have
to be equal and opposite
because they're operating
on the same massless
shaft for the purpose
of this example.
OK, in order find that
equation of motion,
the sum of the
external forces, these
are both in the y direction.
So we just need to find
the horizontal component
of this force and we'll be
able to complete that equation.
So the point of
the exercise here
is just is to find this
horizontal component.
So to do that, let's move on to
thinking about the little mass,
small mass, and what its
free body diagram looks like.
So viewed from the
here's our rod.
Here's the small mass.
This is a side view.
So there's the point
it's rotating about A.
But I want to draw a free
body diagram of this rod.
The rod puts a
force on this mass,
which will have a
vertical component, fm.
And I'll just call it y.
And it'll put a force that's
in horizontal component fm x.
And I'm drawing
them both positive,
because I don't know
which direction they act.
And if the answer turns
out to be positive,
then I guessed right.
If it's negative, it means
it's going the other way.
And what other forces
that are acting on this?
Well, there is certainly
is an mg downwards
acting on that mass.
OK.
And there's no forces in
and out of the page on it.
And this is operating
in the plane.
So this is a planar
motion problem.
And we note that in here r dot
equals r double dot equals 0.
This thing doesn't
change in length at all.
It's just going round
and round fixed length.
So we can write, then,
that the summation
of the external forces
on this little mass
had better equal its mass times
the acceleration of point A
with respect to
the inertial frame.
And whoops.
Not A, but what?
B. The acceleration
of this point.
This is B.
We need to figure out what
the acceleration of that point
is in the inertial frame.
But we've done enough
of these problems,
so this should be pretty easy.
This is the mass
times the acceleration
of point A with respect
to O plus the mass
times the acceleration of B
with respect to A. B and A.
These are all vectors
until I break them down
into their x and y components.
So what's the acceleration
of A with respect
to O in the coordinate systems
that we have written here?
So that's just kind of
our generic representation
of acceleration, right?
But we've chosen some
coordinates here.
Specifically have
a coordinate that
describes the motion of
the main mass, right?
What is that?
So what's the
acceleration of point A?
x double dot.
So we know that this then
is m x double dot and plus.
Now, it's easiest to
describe this in terms
of cylindrical coordinates.
And we can then write
that, well, then this
must be a mass times the
terms in the r hat direction.
r hat.
And then terms over here
in the theta hat direction.
Theta double dot plus
2r dot theta dot.
Now, which of these are 0?
Does that arm change length?
No.
So this is 0.
Is the angular
acceleration constant?
So this is 0.
The arm doesn't change length.
The Coriolis is 0.
So there's no Coriolis
force, no [INAUDIBLE] force,
no radial acceleration,
only a single term.
Just the centrifugal.
So this becomes a pretty
simple expression.
So the summation of the external
forces on our little mass,
then, we can write as mx double
dot in the i hat direction.
I'm going to break it into
its vector components here.
Minus m.
And I know that r here equals e.
That's the eccentricity.
I'm going to start
using these terms.
Minus me omega squared.
r hat.
But I'm going to break that r.
Goes round and round.
I need to break it into
x and y components,
but we've done that
many times before.
That looks like a cosine
omega t in the i hat
direction plus a sine omega
t in the j hat direction.
So as this thing goes
around and around,
it has a cosine term
and a sine term.
And this is in the x
direction, this is in the y.
So we're really interested
in the equation of motion
on small mass m in
the x direction.
So we just need to pull out
the x components from this.
So we have an m x double
dot i hat minus me
omega squared cosine
omega t i hat.
And we can drop the i hats
now because we just have
one single component equation.
And this is this
quantity I called fm.
And fm x then is the x term in
my little free body diagram.
And the force that it
exerts on the main mass
is in the x direction.
So this is the
force that the rod
places on that little
mass in the x direction.
What's the force
that the rod places
on the big mass in
the x direction?
Minus that.
So this is minus f mb
in the x direction.
That's what we're after.
We need that force so we can go
back now and we'll finish out
are the equation
of motion that we
were after for the main mass.
This up here.
We need to sum the external
forces to get that,
to fill out that expression.
But while I'm here,
just to have it,
the summation of the forces
on the small mass in the y
direction.
Look at our free body diagram.
It has a minus mg.
And it then has this term, minus
me omega squared sine omega t.
So just for completeness,
we have also the y component
of the force that the rod
places on the small mass.
And minus this amount is what
it places on the main mass
that it's connected to.
So now let's go back to
our equation up here,
the summation of the
forces on the main body.
In the x direction.
This is going to
be the main body.
x double dot.
And it's now the x
direction forces.
There's only x
component of this force.
And that's what we
have right here.
It's minus that.
And that's our
equation of motion.
We can rearrange it a little bit
and it remarkably simplifies,
actually.
You end up, if you collect
the motion terms involving
x on the left hand side
equals an external excitation
on the right hand side.
And I've been kind of following
the commentaries in mb.
Little confusion
about some questions.
When you're asked to find
an equation of motion,
is that the same
thing as meaning
solve the equation of motion?
No, asking find the equation
of motion means get this far.
Now, if I wanted to know a
solution for this, pretty
trivial in this case, it's going
to look like cosine omega t,
but then I'd say solve
that equation of motion.
OK, now let's see.
We also know-- let's
just finish this--
that the summation of the forces
on this main body in the y
direction must be 0
because it can't move.
No acceleration.
And from the free
body diagram for that,
we can write that this
is n and y minus mmbg
minus mg from the
little mass plus me
omega squared sine omega t.
That's the other phase of this.
And the interesting thing here,
then, is to solve for the force
that it takes to hold
this thing in place.
So you get mb plus m
times g plus or minus.
All right, yeah.
All right, so what's that?
So just kind of step back
and look at these things
and say what's it telling us.
So first of all, just
to keep this thing
from moving up and down,
there's a force on it
that has to support its weight.
And it's the combined
weight of whatever's
inside that container.
The weight of the rotating mass
and the weight of the object.
They have to be supported
by a normal force, which
this is a constant term.
Weight down, normal force up.
And around that constant
force is an oscillating force.
me omega squared sine omega t.
e omega squared you
should recognize
as a centripetal acceleration.
Mass times
acceleration to force.
And because it goes
round and round, when
it's like this it's
pulling up and when
it's like this it's pulling
down and when it's like this is,
it's only going to the sides.
So sine omega t for the
vertical parts, cosine omega
t for the horizontal.
And that's actually all
there is to the shake.
That's all there
is to the shakers.
The rotating mass inside.
Now, in the homework, from
the second homework where
you had this thing, this
ball running around inside,
where I posed the
question in a way
I didn't really quite intend.
But I asked here's the track.
And you had this roller
going around inside.
And I asked to find the normal
force that the track exerts
on the roller.
So it's an unknown.
And there must also be a
tangential force on this thing.
And there's also going to
be this thing certainly has
weight mg.
And so that's the complete
free body diagram.
Now, let's if this is
frictionless, which it won't
be in reality, but for
the purposes of analysis,
let's say it's frictionless,
it's only a normal force.
Where does this tangential
force come from?
Why's it there in this problem?
There's a key piece of
information you're told,
and that is that the angular
acceleration of this thing
is constant.
It's constant
speed going around.
If you had a ball rolling
around there at constant speed,
would it go constant
if you just pushed
it and it started rolling?
It would slowdown going up and
it would speed up coming down.
Why?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Gravity, right?
So there must be something
that has to overcome gravity
going uphill and holding
it back coming downhill.
So the way these
things actually work
is they've got ports pushing
compressed air in here.
And this is driven
around by compressed air
and there's a
pressure difference
between this side and
that side and that
generates the necessary
tangential force to make
the thing go around and around.
But they're really easy to make.
You can imagine very
few moving parts.
Just hook up a compressed
air hose to that
and it's just pushing
the ball around inside.
You get the same outcome.
On this ball, on
this roller, if there
is a-- the problem we
just solved is we found fm
in the y and fm in the x.
And this problem said yeah, but
why can't we get the same thing
but have those coordinates
be f normal and f tangential?
And sure, that's just
a coordinate rotation.
So what can you say
about these forces?
Well, one thing you could say
is fn squared plus ft squared
had better be equal to fmx
squared plus fmy squared,
right?
And then just like
converting from polar
to Cartesian coordinates,
you can do these conversions.
And you could find
out, for example,
that fn is-- keep my
notation consistent
here. fn will be fm in the x
cosine omega t plus fm in the y
sine omega t.
And so there's the answer.
This is what you're asked
for in that problem set.
OK.
So all you need to
know about shakers.
If you're ever confronted
with something like this,
what's the magnitude of the
force that the shaker puts out?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Little louder?
AUDIENCE: Is it
mr omega squared?
PROFESSOR: mr omega
squared, but substitute
for r the actual eccentricity.
It's whatever that
mass in length
out there that's
spinning around.
me omega squared is the
magnitude of the force
and it's going to
oscillate up and down
and it's going to have
gravity that it adds to.
But the important part
is me omega squared
is the magnitude of the force.
OK, now we're going to
move on to the next topic.
The next topic is mass
moments of inertia.
And it has a strong
connection to these.
And I'm going to use
this kind of analysis
as the transition to talking
about moments of inertia.
Moments of inertia and
products of inertia.
So any final questions
about this before we go on?
Yeah?
AUDIENCE: Your summation
of [INAUDIBLE],
why did you not include mg?
PROFESSOR: Why didn't I include?
AUDIENCE: Mg.
PROFESSOR: Mg.
in the.
AUDIENCE: First summation.
[INAUDIBLE]
PROFESSOR: Oh.
Yeah, you're right.
And where's my
free body diagram?
Has it on it, right?
Just didn't get it down into
the-- and what direction's
it in?
Because then we
did get it back in.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Back in the last
line for the y component.
Now, does it appear in this
one, this equation at all?
It has absolutely
nothing to do with it.
Gravity is in the
j hat direction.
This is a force
equation in i hat.
But it does appear in that
normal tangential expression
when you go look at the
solution for that problem,
because it has compounds in
both of the i and j directions.
And so it'll show up.
Gravity will show up
in this expression.
Right through this term.
Yeah?
AUDIENCE: In the bottom
equation on that middle board,
you have my minus
mbg minus mg plus.
I don't understand where
that last plus came from.
Because in your
equation on the left
you're using the force
of little f, correct?
And you have two
negatives there.
PROFESSOR: This is 0.
I left the n where it
was and moved everything
to the other side.
So that plus becomes a minus.
AUDIENCE: No, from the board to
the left to the middle board.
PROFESSOR: OK.
AUDIENCE: So down.
So you have the
summation of the force
on the little mass as negative
mg minus m-- yep, that one.
And from what I can
understand, you just
moved that force over
to the large force,
but you [INAUDIBLE], correct?
PROFESSOR: It should
be minus this thing.
The summation here.
This force is minus
the little mass force.
So that ought to become
a plus and a plus, right?
And so if I do that carefully.
To this one is OK.
But this one appears to
have a sign problem, right?
But these two terms
have got to be the same.
And so I've got a
mistake somewhere.
And rather than spend 10
minutes fixing it on the fly,
I'll take note of that.
This should be OK.
AUDIENCE: Yeah, intuitively
makes sense too,
I just don't
understand [INAUDIBLE].
PROFESSOR: Ah, wait a second.
No, I'm not going to
try to fix it right now.
I made a slip in
my notes somewhere.
But I will repair that.
Yeah?
AUDIENCE: Why do
we need mg at all?
Because doesn't this force
the angular acceleration
is constant?
Or the angular velocity
is constant, right?
So the centrifugal
acceleration is
going to be constant, which
means that the part that's
driven by the motor is
going to be changing
to account for gravity.
So isn't gravity
taking into account
that we have a constant
force or [INAUDIBLE]?
PROFESSOR: Yeah, you're
asking if gravity is not
taken into account
somehow by that rotating.
The gravitational force that
is on the main mass that
comes from the
little mass certainly
has to pass through the rod.
It's got to be contained in the
forces in the connecting rod.
So it's definitely there.
But the force that causes
the centripetal acceleration
of that rotating
mass is completely
independent of gravity.
With or without gravity,
it takes a particular force
to make that thing travel
in a circular path.
And that's m minus mr
theta dot squared always.
Yep?
AUDIENCE: So doesn't that mean
that on your first expression
on that board, there
should be no mb?
PROFESSOR: On which expression?
AUDIENCE: That one.
PROFESSOR: This one.
OK, this is the total
forces on the little mass.
AUDIENCE: [INAUDIBLE].
PROFESSOR: We need
to back up to here.
The total forces are mass
times the acceleration
of the main body it's
connected to plus the mass
times the acceleration
of B with respect to A.
So we have to have that term.
And we then go
into our four terms
here and find there's
only one left.
So that's the force exerted
on the small mass by the rod.
And that is positive mx double
dot minus mr theta dot squared.
So we sum the forces
on that little mass.
It has got to be
equal to-- ah, I
know where we made the mistake.
So we've just
discovered our mistake.
This has got to be able to
mass times acceleration.
And what are the forces?
The summation the forces
is mass times acceleration.
So the acceleration
is this plus this.
But the sum of the forces.
The problem here is I've
used a notation where
this is very similar
looking to the forces
that I've noted here.
So this is the actual force
in the y direction, j hat,
plus the actual force in the
x direction, i hat, minus mgi.
So when I solve for
the i component,
I'm going to get the i
pieces of that plus mgi.
I mean, excuse me, j component.
Should it be like that?
The j component will have
this piece times sine omega
t with a minus.
And you move the mgj to that
side and it becomes a plus.
All right.
That makes sense.
The rod has to hold up the
weight of that little mass,
right?
The weight's down.
But the rod has to push up
on it in the y direction.
So the force the rod
puts on the little mass
has got to be equal to the
weight of the small mass
minus this me omega
squared term, which
is the force necessary to create
the centripetal acceleration.
OK, so we've got this now fixed.
This term is OK.
And the minus that force is
then the force on the main body.
So minus.
Plus.
Now I've got to figure
out what I did wrong here.
You're doing what I said
I wasn't going to do.
We're on the fly trying
to figure out where the.
AUDIENCE: [INAUDIBLE].
PROFESSOR: OK.
You're happy now.
Good.
OK.
All right, we're
going to move on.
AUDIENCE: [INAUDIBLE].
PROFESSOR: What
about the summation?
AUDIENCE: It's not
really a summation.
It's just the force
that arm is [INAUDIBLE].
PROFESSOR: Right.
Fair enough.
Yep.
This is just minus f.
This is on the little mass.
This is the force on the
little mass from the free body
diagram.
Right.
And that helps.
And that's different from
the summation one here.
OK.
All right.
I think we've got it sorted out.
Now, I've put on
the Stellar website
under readings a little
one page thing called
"Moments of Inertia."
It's two pages of information
taken from the Williams
textbook on dynamics.
And it's going to
show some of what
I'm going to put on the
board, and especially
the detailed stuff you
don't have to copy.
OK, we're going to come up with
some expressions for angular
momentum in terms of
particles and their positions.
And this is now the
subject of mass and moments
of inertia and
products of inertia.
And I'm going to put some of
these equations on the board
and you don't have
to copy them all.
All these expressions
become the definitions
of mass moments of inertia
and products of inertia.
And if you just drop
down one last little bit,
we come up with an expression
for angular momentum.
Three vector components look
like ixx omega x plus ixy omega
y and so forth.
These compounds in terms of
particle masses and positions
are defined in these
final equations.
So I'm going to tell you
what I'm going to tell you.
We're going to
make the transition
from dealing with particles and
angular momentum of particles
to angular momentum
of rigid bodies.
OK?
And in my own experience
this is something
that is generally done badly.
And I'm going to
try to do it well.
I'm going to try to give you an
intuitive understanding of why
we have these diagonal terms
called the moments of inertia
and what they're
useful for and why
these off diagonal terms call
products of inertia turn up
and what they actually mean.
When I was taught
the stuff, I never
got a gut feeling
for why or what
the off diagonal terms meant.
You don't know it, but
we've been using them.
And then I'll tell
you the answer.
The answer is that when we
have a problem like well,
the motorcycle problem
were talking about is this.
Basically here's the motorcycle
wheel spinning around
and round.
And it has these two masses.
This is set up B.
So one little mass
was off to the side
of the rim a bit.
And the other mass
was off to the side.
On the picture it
looked like this.
Here's the axle,
motorcycle, and forks
would be coming down like this.
And these two little masses.
Equal distance but opposite
sides from one another.
And if this spins, it
puts a heck of a wobble
into this thing.
And this puts a moment
about this point.
It tries to make this
thing rock back and forth
as it's spinning.
It's really hard to hold.
You hold the axle there.
And you got to do
it so you don't
get hit by the-- there you go.
Now tell me if
you feel a moment.
It's really hard to keep
that thing straight, right?
Well that's what it's trying
to do to that motorcycle wheel.
OK.
For this problem, those
off diagonal terms,
those products of
inertia are not 0.
The product of inertia
terms cause these things
called dynamic imbalances.
It causes there to be angular--
makes the angular momentum
terms instead of
the angular momentum
being aligned with the axis of
rotation, the rotation vector,
it's pointed off
in this direction.
Anytime the angular momentum
vector and the rotation vector
are not aligned, you
have off diagonal terms
and you will have
dynamic imbalance.
So there's a
physical consequence
of those off diagonal terms.
And they explain the
dynamic imbalance.
So let's see if we can't
make some headway on that.
So you've seen the rotor.
Let's look at two cases.
One that looks like that, which
I just had set up a second ago.
One that looks like that.
Call this A, B.
And in both cases,
the rotation is around
the vertical axis
and it's constant at omega.
And I just mean these to
be two different cases.
I'll make it lowercase
so I don't confuse it
with my coordinate
system notation.
This is going to be point A
in both of these problems.
And it's going to be the
origin of a coordinate system.
So if you cause this to
spin, these both have--
did I write these
masses as m over 2?
For a moment, let's just think
of these as being equal masses.
If you do this problem, do you
think this one will wobble?
No, it's perfectly balanced.
And it'll just spin
nice and smoothly.
It has angular momentum around
the z-axis, the omega axis.
It has angular momentum
in that direction,
certainly, when calculated.
This one has same mass, same
distance away from this axis,
but now one up and one down.
This one wobbles.
But this one has a component
of angular momentum
in this direction, which is
exactly equal to this one.
But this also has a component
that's in this direction.
And we're going take a
look and see what that is.
So we're going to do
this problem here.
We're going to analyze B. This
case B. And here's the goal.
The goal is to show
you that the angular
momentum of this system
with respect to this point
can be written as a matrix
with constants in it,
which you can call the mass
moment of inertia matrix.
Times the vector components
of the rotation rate.
Now, this problem,
the z-axis will be
upwards and will only have one
component one non 0 component.
But in general, we want to be
able to express the angular
momentum as a product
of this inertia matrix.
And these are the
inertias we'll find out
with respect to A. Times the
vector of angular velocities.
We've got to be very careful
about some definitions.
So we're going to do
this specific problem,
but we're going to use methods
that are completely general.
So I want to describe
the general problem.
Here is a inertial
coordinate system fixed.
Here's a body out here in space.
And it is rotating about
some point A. So point one
and the rotation
vector, the angular
rotation and some omega.
And it's just in some direction.
And that omega is with respect.
We always in these
angular momentum problems
define rotation rate with
respect to an inertia
coordinate system.
Now, this point A.
So first carefully
define A is a fixed point.
So is that an inertial point?
Yeah.
You can do Newton's
laws from this point
just as well as you
could any fixed point
in this inertial reference
frame is an inertial point
and you can use Newton's laws.
So this is a fixed point.
I'm defining it that way.
This body is rotating about that
point with this angular rate.
But attached to the body
is a coordinate system
that rotates with the body.
So this would be
some a xyz coordinate
system attached to the body.
So it's like this
problem where I've
got a coordinate system
attached to my wheel.
There's x, here's y,
z coming out of it.
And in a really
simple case, it's
rotating around the z-axis.
But I can make it rotate
around some other axis.
I pushed a nail
through here and I'm
trying to hold it constant here.
And now it's rotating about
a different axis, right?
Same rotation rate,
but it doesn't
have to be lined up
in any pretty way.
If I make that thing rotate
around that other axis,
it looks weird, but
we can define it.
And that's what we're
talking about here.
So this body is rotating
around, has some rotation
rate with respect to a reference
frame attached to the body.
So A xyz is a frame that can't--
going to make this go up.
Come on.
This is attached to the body.
And I've drawn them at
kind of funny angles here,
just to emphasize that
they're not necessarily
lined up with these.
And it's going to rotate.
OK.
Omega.
Just to emphasize.
It's always in the
inertial frame.
The last point may be
confusing to start with.
Omega measured with
respect to O can
be expressed in terms of
the axyz unit vectors.
We're going to do that.
It turns out it vastly
simplifies the problem
to express the
rotation in the unit
vectors of the frame
attached to the body.
Remember, that
frame is still fit.
Its origin A is at a fixed
point in the inertial frame.
So it's just the system's
going around and inside
of that system you
have a rotation
and you can break it
down into xyz components.
Just a vector and you can
express it in those components.
That's all we're saying here.
Now I want to do the
motorcycle problem.
I'm going to just
turn it on its side.
And the reason I'm going to
do this specific example,
the hope here is to
actually now give you
a physical feeling
for what's going on.
We've done a lot of
illustrations of it.
And you know that it
produces imbalances.
So here's my z-axis
and my rotation rate.
Omega with respect to O
is some omega in the k hat
direction in the fixed frame.
And in this case, it's
going to be simpler
than the general case,
so that we can do it
in a reasonable length of time.
So actually here's my rod.
Here's my point A. This is
my coordinate system axyz.
So this is now
attached to the body.
My rigid body is a massless
rod with two masses on it.
And this distance,
this is the x.
Going that way will be a y,
which we have little use of.
There's nothing happening
in that direction.
So this distance
here I'll call x1.
This distance here is z1.
Over here, this is z2 and x2.
Now we're going to
make this problem.
We'll substitute a number.
So this is symmetric.
So x2 is going to be
minus x1 and so forth.
But we want to keep them
separate for the moment
so you see what happens
to different terms.
OK, so that defines a problem.
So the coordinates.
And we'll call this mass m1.
I'll keep this a little
general for a moment.
And this is m2.
So m1 is at the coordinates
x1, i, 0, and z1 k.
And m2 is at x2, i, 0, and z2 k.
Just points in a plane.
And I want now to
compute the-- I
want to find the angular
momentum of this object
with respect to point a.
Remember we compute angular
momentum in respect to points.
So I'm going to do it
with respect to point A.
And that's going to be
the sum of the angular
momentum of mass 1
with respect to A,
plus the angular momentum
of mass 2 with respect to A.
So the angular momentum of any
particle i with respect to A
is r cross p.
r cross the linear momentum.
So it's r i with
respect to A cross p i.
Now the p, this is the
momentum of the particle.
That's always with respect
to what kind of frame.
When you compute
angular momentum.
Must be the inertial
frame, right?
So technically to start with,
just remind you of that,
we'd say oh.
But we've already
said our A is a fixed
point in an inertial frame.
So it's OK to write r i
with respect to A cross,
in this case, p i with respect
to A. They're the same thing.
These two things are
exactly the same thing.
The momentum measured
at any two fixed
points in an inertial
frame is the same.
Doesn't matter where
you're measuring it from.
OK.
And we know that p
i with respect to A
now, we'll call it, is the
mass i times the velocity of i
with respect to A. That's
just ordinary linear momentum.
So I need an expression for
the velocity of i with respect
to A. Any point.
So these are fixed now.
These are fixed length things.
The velocity of a moving
point is just the derivative
of the position vector.
But you have this equation
some people call a transport
equation.
So the length of this
thing's not changing any,
so it's just going to
have one term in it.
So what's a velocity?
In vector notation, omega cross.
All right.
Right.
And this could also.
All right, these are vectors.
And because I can say that, then
I can say hi with respect to A
is mi riA cross
omega with respect
to O cross ri with
respect to A. OK.
All vectors.
So any rigid body.
So here's the link now.
Here's the jump from points
particles to rigid bodies.
Any rigid body is made up of
the whole mess of particles,
connected rigidly together.
No relative motion.
But a whole mass of particles.
So I can compute the total
momentum of a rigid body
as the summation over all
the little particles in it.
mi.
riA cross omega with
respect to O cross riA.
Just sum them all up.
And when you have
continuous bodies,
these summations
turn into integrals.
So you'll find definitions
for like there's
a mass moment of inertia
about this axis of this wheel.
It's mr squared over 2.
And it comes from
the-- and that's
the number that you have
to multiply by omega
to get the angular momentum.
So it comes from summing up
all these little particles
in this thing is the total
momentum, angular momentum,
of the object.
All right, let's do that.
We're going to do that
for our too little masses
here and see what kind
of things result. Oops.
I want to get my h
with respect to A
is the sum of h1 with respect
to A plus h2 with respect to A.
And I'm just going
to use that formula.
So it's m1.
So if I were just work out that
little vector products there.
m1.
here's riA.
It's x1 i times z1k cross
omega zk cross x1i plus z1k.
And then I have a second
term, the m2 term.
x2i plus z2k omega
zk x2i plus z2k.
So just a lot of
little vector terms.
That is that expression for
our two little particles.
With their specific positions
at x1 and z1 and x2 and z2.
So if I multiply all
that out, then I'll
get the following result.
An h with respect to A here.
It's m1 x1 squared omega
zk minus m1 x1 z1 omega
z in the i hat direction
plus an m2 x2 squared omega
zk minus m2 x2 z2 omega
z in the i direction.
So this is the angular
momentum of particle one.
This is angular momentum
of a particle two.
And I'm going to
do a special case.
And the special case
I'm going to let
m1 equal m2 equal m over 2.
So they'll do sum to m.
And x1 equals minus x2
and z1 equals minus z2.
So they're nice and
symmetrically opposite
like drawn in the picture.
That I'm making an equal
masses in equal distances
on either side of the origin.
And that's going to make this
thing simplify quite a bit.
This is of the form.
This angular momentum
vector is of the form
has three vector components.
In this particular
case, this one's 0.
And we call the first component,
this one here will be hx.
And this one here is
clearly hz, the component
in the z direction.
And if we draw,
here's our system.
Here's our coordinate system.
The coordinate system
attached to the body.
It has a z component of angular
momentum positive upwards.
And it has an x component of
angular momentum in the minus
direction like that.
When you add them
together, you get that.
So this is h with respect to
A. This is hz, this is hx.
Now, we found this before.
We didn't talk anything about
moments of inertia, anything.
We just deal in particles
earlier as we did problems.
We found out that when you
have this kind of unbalance,
the direction of the
angular momentum vector
is not in the same direction
as the rotation vector.
In this case, the rotation
doesn't make a zk.
It's like that.
The vector is going around it.
Angular momentum.
Now, in general
you would write hx.
General case.
And this is what you can pull
off, this little two sheet
handout that you
can download and you
don't have to copy everything.
This is going to look
like an ixx omega
x plus ixy omega y
plus ixz omega z.
So if we look at
that and we look
at this, this particular case
the hx term is this, right?
So this is the general
expression for hx.
And in this
particular case, that
will look like minus
m x1 z1 omega z.
And this is the piece
that's in the i direction.
That's why we call it hx.
And this is then ixz omega z.
So this piece here
is what we call ixz.
It's where it comes from.
And we can write it.
So this is our particular case.
Get this result. And we
find there's h in the h.
y is 0.
And hz is mx1 squared
omega z omega z.
And that's got to be of
the form izz omega z.
Now, how do you remember
what the subscripts mean?
ixz means this is
the h component
and this is the omega
component it's multiplied by.
So ixz is the product
of inertia for hx.
It's related to rotation in
the z component rotation.
That's what the subscripts mean.
Maybe I'll do this.
So in general, if you know
what these constants are
for your rigid body and you
know your rotation rate,
you instantly know
your angular momentum.
These things, the products
in moments of inertia,
are basically cataloged--
you'll find them
in the back of your
textbook-- for all sorts
of different objects.
So I know that if you have z in
this direction and this thing's
rotating around the z,
hz is the total mass
of the system times
the radius squared
divided by 2. mr squared over
2 would be izz for this object.
And for all sorts of objects.
These are just cataloged values.
And then there's ways of
moving the axes, called
parallel axis theorems that
you've probably run into,
that allows you then to
construct these values from one
known point to moving the
point to someplace else
and having it move around that.
So these values are
tabulated, calculated,
with respect to the
centers of mass.
And if you want to have
the mass moments of inertia
with respect to any
other point, then you
will use something which we call
a parallel axis theorem, which
we'll get to in due course.
Pretty good on timing here.
A note about textbooks.
Textbook conventions.
This I matrix.
In some they write it
ixx ixy ixz and so forth.
ix.
No, iyx.
iyy.
iyz.
cx.
Some write it like that.
And others write it with all
of these with minus signs
on the off diagonal terms.
So Hibbler uses the minus signs.
Williams does not.
So the diagonal terms
are always positive.
Yeah?
AUDIENCE: [INAUDIBLE].
PROFESSOR: All the off
diagonals are negative.
So this is positive.
Positive, positive,
positive and then
negative, negative,
negative, negative, negative.
Now there are actually
negative-- they'll
be negative-- the numbers will
pop up negative and so forth.
It's just that in the notation,
some authors have adopted
putting the minus signs here.
Others have embedded
them in the value itself.
So Williams' notation,
he would say that ixz
is minus m x1 z1 for this body.
Hibbler would say it's plus
and he'd put the minus sign
in the notation.
So just beware of that.
Because all your
life you're going
to run into people saying
the product of inertia
of this thing is and you got to
know which way they define it.
All right.
Compute torques.
You just take time to
[INAUDIBLE] angular momentum.
And we'll do that as a
last little step next time.
But you've got the
essence of the movement
from talking about
particles to how we're going
to talk about rigid bodies.
So you have muddy cards.
You have two or three minutes.
Write down what was
tough for you here.
Write down what wasn't.
And see you next Tuesday.
Oh, I must say, so this
stuff about-- the mass moment
of inertia matrix.
That stuff is not on the exam.
But knowing about particles and
particle moments of inertia is.
