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PROFESSOR: Today we're going
to talk all about moments
of inertia, and the last time
we did [? muddy ?] cards, there
were a lot of questions.
The most common question-- where
did all those terms come from?
And I'm going to do
a brief kind of intro
around that so as to facilitate
the rest of the discussions.
Some basic assumptions
here today.
We're going to
consider bodies that
rotate about their center
of mass, or fixed points.
That's the kind of
problems we're doing.
And secondly-- all
through this lecture
today, there's four or five
significant assumptions
or conditions that
are really important
to the whole discussion.
I'm going to really
try to highlight them.
And starting with two here.
The other one is we're going to
use reference frames attached
to the bodies.
That's why we did all that
work in kinematics figuring out
how to do velocities of
things that are in translating
and rotating
references frame, it's
so that we can do
these problems.
So this said, here's
a inertial frame.
Here's some rigid body.
Here's a point on
it, that's my origin.
And here is a set of coordinates
attached to the rigid body--
an xyz system-- the body
rotates and translates,
it goes with it.
And if we consider a
little piece of this body,
a little part-- I'll call
it little mass element
mi-- from here to here is some
distance y with respect to A,
then the angular momentum
of that little mass particle
is by definition of angular
momentum Ri cross Pi.
And the P is always
the linear momentum
defined in an inertial frame.
That's the definition
of angular momentum.
So you have seen this
little demo before.
So this is now my rigid
body that I'm talking about.
In this case, the axis
of rotation is this one.
And I'm just going to
define my rigid body
as essentially rotating
about this fixed point,
this is A now.
And my coordinate system is
xyz, so the rotation is omega z.
And this rigid body--
this is massless,
consists of a single point mass.
So we've done this
before, but we
tended to do this
kind of problem
using polar coordinates.
And now I really wanted to
strictly think in terms of XYZ.
The answers come out
exactly the same.
Polar coordinates also moved.
You could do this in
rhat, thetahat, z.
We're going to do it in
xyz attached to the body.
And y is going into
the board, and we'll
have no values in the y
direction for this problem.
Here's our little
mass, we'll call it m1.
It's at a position out here
x1, z1 are its coordinates,
and y is zero.
And we want to compute
its angular momentum,
h1 with respect to A. That's
our R1A cross P1 and 0.
And for this problem
that's x1 in the i hat
plus z1 in the k hat,
that's the R-- cross
P and P is mass
times velocity, m1.
And the velocity is
just omega cross R.
We've done this before.
So we need a length
on this thing.
We've done this enough
times before I'm assuming
you can make this leap with me.
The momentum is into the
board, in the y direction.
The linear momentum
is R cross with that.
And that will give us--
let me write it-- omega z.
So that's the linear momentum.
It's in the j direction,
its velocity is x1 omega z.
The radius times the rotation
rate is the velocity,
the direction is that direction.
So this is your momentum.
You multiply out
this cross product,
you get two terms-- m1 x1
squared omega z k, minus m1 x1
z1 omega z i.
And these two terms
then we would identify.
This is the angular momentum
in the k direction, angular
momentum in the i direction.
There doesn't happen to
be any in the z direction,
in general there could be.
But this is h for this
particle, particle one.
I want to particularly
emphasize this one--
oops, this is-- excuse me-- z.
And this is the
piece we call hx.
So there's a complement
of the angular momentum
that's in the z
direction, there's
a component in the x direction.
We know that if
we compute-- just
to tie it back to previous
work a little further-- d
by dt of h1A gives us
the external torques
in the system, a
vector of torques
when we compute that out.
I'm not going to calculate
it, I don't need it
for the purposes of the
rest of the discussion.
This one gives you
omega z dot term.
This one gets two
terms because you have
to take the derivative of i.
So you get three
terms which are torque
in the x, torque in the y,
torque in the z directions.
Two of them are
static, because it's
trying to bend this thing
back and out, and one of them,
the z direction one, is the
one that makes it spin faster.
So we've seen that,
that's a review.
And to make the leap from
that to rigid bodies-- Capital
H, now, a collection of
particles with respect
to its the origin
attached to the rigid body
is going to be the
summation over all
the little mass particles.
Their position vector crossed
with the linear momentum of all
those little mass particles.
Sum all that out.
Because it's a rigid
body, this is always
something of the form that
looks like omega cross R,
m omega cross R. And
I can then write this
as the summation of
[? oper ?] i of the mi,
RiA cross omega with
respect to 0, cross RiA.
That gives you the velocity, m
gives you the linear momentum,
crossed with this
one again makes it
into an angular momentum.
So the angular momentum of every
particle-- the bits of this
come out looking like
an R squared omega.
Or at least has dimensions
of R squared omega.
These can be like x1
squared, but these can also
end up being terms like x1
z1, those cross product terms.
But in general, these
things all add up.
And this is a vector,
this is a vector,
so this is a vector--
these are all vectors,
the result's a vector.
And therefore we would break
this down, this whole thing.
Once you write all that
out and sum it all up,
you're going to
get a piece of it
that we would say that
is in the i hat direction
and we call that Hx.
Another piece that's in
the j hat direction, which
we call Hy and Hz in the k hat.
That's just how
that all shakes out.
In general you get three parts.
Yeah?
AUDIENCE: Why do you have mass
multiplied by displacement
and the momentum?
PROFESSOR: Why do we have the
mass multiplied by a position
vector times the momentum?
Because that is the definition
of angular momentum.
It's R cross P, all right?
[INTERPOSING VOICES]
PROFESSOR: Ah, I
see what I've done.
Yeah you're right.
This m doesn't belong--
I got ahead of myself--
it doesn't belong here.
It pops out of this to here.
Thanks for catching that.
Absolutely right.
So every term here is
made up of things that
look like x1, z1, m1 omega x.
Or omega y, or omega z,
because this is a vector,
and it can have three
components-- a piece in the i,
a piece in the j, a
piece in the k direction.
So there's a lot
of possible terms,
and we're in the practice
of writing this out.
This H vector can
be written then
as a-- I'm going to
run out of room here.
So the Hx term-- the
piece that comes out
with all little bits
in the i direction--
when you just break it apart,
we would write it Ixx omega
x, plus Ixy omega
y, plus Ixz omega z.
All the i terms that
float out of this,
we just collect them
together-- all the ones that
are multiplied by omega x
ends up being some terms that
look like sums of mi--
x, mi z squared terms,
and there's some
y squared terms.
But we collect them
all together and we
call this constant
in front of it Ixx.
Just what floats
out of this stuff.
We break it into three pieces,
the part of the angular
momentum due to the
rotation in the x,
the part due to the
rotation in the y,
the part due to the
rotation in the z.
And you do the
same thing for Hy,
and you get in Iyz omega
x plus an Iyy omega
y plus an Iyz omega z, and
you're finally get and Hz term.
Izx omega x, Izy
omega y, Izz omega z.
That's everything that falls out
of just doing this calculation.
And where in the habit of
writing that in a matrix
notation as the
product of this thing
called the inertia matrix.
And so forth, with
this bottom term
being Izz multiplied by
omega x, omega y, omega z.
You multiply that times the top
row, you get Hx, middle row,
you get Hy, bottom
row, you get the Hz.
So where this stuff
comes from is just
from carrying out the summation
over all the mass bits.
So it basically starts with the
definition of angular momentum.
And this is just a
convenient way to write it.
So for example, the Ixz term--
the one in the upper right
there that's part of
Hx, the Ixz term--
is minus the summation of
all the little mass bits
times their location xi zi.
And this, the xz,
always matches xz.
And when you want to do this
over a continuous object,
you integrate that.
And for Izz, this is
the summation over i,
all the bits of mi xi
squared plus yi squared.
So in general this,
by summations, that's
where these terms come from.
They just come.
They've started with
this calculation.
And in general, from
that calculation.
So for our one particle
system, for this thing.
So where this is x and z,
it's one single particle.
So for our one particle system.
And there's no y1
in here because it's
a zero in this problem.
The coordinate of that little
mass particle is x1, 0, y1.
So in general, this is x1
squared plus yi squared,
but that's 0.
So it's just mx1 squared.
So that's where all
these things come from.
Now let's kind of look at--
the units of these things
are always mass
times length squared.
The diagonal terms the Ixx,
Iyy, Izz terms will all
be of the form of
something squared.
And what it is is it's
always just the distance
of the mass particles from the
axis of rotation, call that r.
It's just a summation of
this perpendicular distance
from the axis of rotation.
For Izz, the axis
of rotation is z.
And this piece here
is always the distance
squared that the mass particle
is from the axis of rotation.
Now that's the set up for today.
And when we want to get onto
the real meat of the discussion,
talking about things like
what principal axes are
and so forth.
Now we're going to move on to
the part of this conversation
about principal axes.
So a really important point.
For every rigid body,
even weird ones.
For every rigid body, there
is a coordinate system
that you can attach to this
body, an xyz set of orthogonal
coordinates that you can
fix in this body such
that you can make this
inertia matrix be diagonal.
Any weird body at
all, there is a set
of coordinates that
if you use that set,
this matrix turns
out to be diagonal.
And what that means in
dynamics terms is if you then
rotate the object about
one of those axes,
it will be dynamically balanced.
And so when you rotate a shaft--
this is an object for which I
know-- it's a
circular uniform disk,
one of the axes through the
center is a principal axis.
And if I rotate the object about
that-- this one the hole's kind
of out around so
it wobbles a bit
but if I rotate it
about that it will
make no torques about this
axis that are because it's
unbalanced.
It's essentially the dynamic
meaning of principal axis--
an axis about which you can
rotate the thing and it'll just
be smooth, no
away-from-the-axis torques.
So that's so those
are pretty important.
We want to know what
those are for rigid bodies
and how to find them.
So there's a mathematical
way to find them
and you can just read
about in the textbook.
It's sort of a-- I'm trying to
think of the mathematical term.
I forgot it.
But what I want
to teach you today
is for many, many
objects, if you just
look at their symmetries, you
can figure out by common sense
where their principal axes are.
So that's what we're
going to do next.
We started off saying that
we're talking about bodies that
are either rotating about
their centers of mass
or about some other fixed point.
So we're going to define
our principal axes
assuming we are rotating
about the center of mass.
There is an easy method known
as the parallel axis theorem
to get to any other point.
So why do we care about doing
it about the center of mass?
Why is that useful to
know about the properties
around the center for
dynamics purposes?
What kind of dynamics
problems do you
care about the center of mass?
Rotation about the
center of mass.
AUDIENCE: [INAUDIBLE].
PROFESSOR: So when I throw
this thing in the air, it's
rotating, it's translating,
and when it rotates,
what's it's rotating about?
AUDIENCE: Center of mass.
PROFESSOR: Center of mass.
So there's just lots
and lots of problem
in which in fact the rotation
is going to occur around
the center of mass.
Any time the thing is off
there and there's nothing
constraining its
rotational motion,
the rotation will be
about the center of mass.
So that's a good enough reason,
and a very practical reason
then for computing
these things or knowing
how to find these things
about the center of mass.
All right so we're now
going to do principal axes,
and we're going to teach
you some symmetry rules.
Maybe first a
little common sense.
This thing-- is this
a principal axis?
No way, right?
And we know for a fact that
it has off-diagonal terms
because I've defined
my coordinates as being
an x, y, z, like this, and this
is often at some strange angle.
So as a practical
matter, how could I
alter this object so that
this is a principal axis?
Essentially, how would
you balance this?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Take the
mass off, fix it.
[LAUGHTER]
Or?
All right.
Where do I put it?
Down here?
Ah, you want it like this.
This is steel going
into aluminum,
I have to be careful that
I don't strip the threads.
There we go.
And now test one, this
is axis of rotation,
is it dynamically in balance?
Smooth as silk.
This now is a principal axis.
Its mass distribution
is symmetric.
Where is there a plane of
symmetry for this problem,
for this object?
That's what we're going to talk
about now, planes of symmetry
and axis of symmetry.
Ah, you're saying one like back?
You're right, that's
a plane of symmetry.
Is there another
one for this one?
Like this.
So there's two
planes of symmetry.
So let's say come up
with some symmetry rules.
So this first one
is there exists--
by this I mean diagonal,
it's a diagonal matrix.
You can find a set
of orthogonal axes
such that this
matrix is diagonal.
That's what we mean when we
go to find principal axes.
And pick a board here-- rules.
And we're going to
have three of them,
so just leave room
on your paper.
We're going to build these up.
And I'm going to talk about it,
and then you add another one
and so forth.
So rule one.
AUDIENCE: [INAUDIBLE] the
inertia matrix [INAUDIBLE].
PROFESSOR: What is
the inertia matrix--
AUDIENCE: No, I
mean the second one.
PROFESSOR: This thing
with the diagonals?
This is going to have all the
off-diagonal terms are zeroes.
That's what I mean.
It's a diagonal matrix, the mass
moments of inertia and products
of inertia-- you only have
Ixx, Iyy, Izz if you properly
pick this set of orthogonal
coordinates attached
to the body-- if
you pick them right.
What it means if then you
spin that object about one
of those axes, it's in balance.
So rule one, if there it is
an axis of symmetry-- Remember
the caveat here is we're talking
about uniform density objects
or at least objects in which
the density is symmetrically
distributed.
So the geometric symmetry
and mass symmetry
mean the same thing.
What does that mean?
So axis of symmetry.
Does this have an
axis of symmetry?
Where?
AUDIENCE: It has multiple
ones down the middle
and it's also
symmetric [INAUDIBLE].
PROFESSOR: That's a
plane of symmetry,
the [? other ones you think, ?]
axis of symmetry.
AUDIENCE: Well I mean the
line right through it.
PROFESSOR: Which way?
AUDIENCE: Well any.
PROFESSOR: Nope.
AUDIENCE: Perpendicular to it.
PROFESSOR: Axis
of symmetry means
that there is a mirror
image from that point--
that point mirror image over
here is always identical.
But you think any from angle, it
doesn't matter where you start,
it's reflected through the axis.
So things that are
circularly shaped
tend to have axes of symmetry.
So this has an axis of symmetry.
Everything is reflected
exactly just across the axes,
you don't have to pick
any particular line.
So what it's saying if
it's an axis of symmetry,
then it is a principal axis,
and rotation about that
axis-- the things will be
nice perfectly in balance.
If you go through all the
hairy details of calculating
the hard way, the Ixx,
Iyy, et cetera terms--
all of the off diagonal
terms will come out zero.
And the reason is that for
every mass particle over here,
let's say there's a little
bit right here in the corner,
there is one exactly like
it on the other side.
So if this one we're trying
to bend this thing up as it
spins around, this one over
here is telling it to come back.
Yeah?
AUDIENCE: How come this is
not an axis, if you hold it--
yeah--
[INTERPOSING VOICES]
PROFESSOR: So if you do this.
So I think it's kind of just
the definition we're getting to
of an axis of symmetry.
Because in simple terms,
what the object looks
like this way is
it's a narrow object,
it is in fact a mirror
image reflection.
But it looks different
when you go to this angle.
And it looks different
when you go to this angle
so it's not an axis of symmetry,
that's a plane of symmetry.
We're going talk
about that next.
So axes of symmetry--
the thing essentially
looks the same at all
orientations about that axis.
But it is certainly
dynamically balanced.
Every mass bit here balances one
over there, it doesn't wobble.
All right.
So that's for sure true.
Now keep in mind that exists
a set of orthogonal axes.
So that means once
you find one, you
know that the other two
are perpendicular to it
and perpendicular to each other.
So for this system, when
I got the x and y here,
there has to be an x
and y perpendicular
that that are the other two.
And because it's
actually symmetric,
it doesn't matter where
you out them, you just
could say it's these two,
these two, it doesn't matter.
You just have to
decide where you're
going to put them on the
object, they're all the same.
So for this axis symmetry,
they're just two more.
They're going to be in the body
perpendicular to the first.
Now we're going to do
about g so the book uses g
to talk about the center
of mass, so I'll use g.
So the center of
mass of this thing
is right in the middle of
this and in the middle here.
So it's inside this body.
Right in the dead center of it.
So the three principal axes are
an orthogonal set, one of which
goes through it
and the other two
are embedded in it right angles.
So this is the simplest rule.
If you have an axis of symmetry,
you know right way just
about everything you need to
know about the principal axes
of the body.
Shoot.
AUDIENCE: It seems like
then there would only
be axis of symmetry in
objects that are either round
or spherical, is that correct?
PROFESSOR: Pretty much.
AUDIENCE: So it's a very
limited special case.
PROFESSOR: Yeah but there sure
appear an awful lot in machines
because they have this beautiful
property of just being balanced
in all directions.
So everything
rotating in the world
tends to have an almost
perfect axial symmetry.
Just thinking about order here.
Let's do this and then I'm going
to go do a couple of examples.
The second rule-- if you
have one plane of symmetry,
a plane of symmetry.
So this is kind of the
opposite direction,
in this you have the
least information.
Here's an object, does it
have a plane of symmetry?
AUDIENCE: Yes.
PROFESSOR: Where?
AUDIENCE: Straight
through [INAUDIBLE].
PROFESSOR: Show me.
That cut.
All right.
So I've got these little
dotted marks on this thing.
So if I slice through
that, I create
two pieces that are identical.
So that's a plane of symmetry.
There is image match across
that plane at every point.
So if I have a
plane of symmetry,
what do you think you can
say about a principal axis?
One of the principal axis?
AUDIENCE: It'll
be on that plane.
PROFESSOR: It'll
be on that plane.
That turns out to be true,
but that's the second point
I want to make.
AUDIENCE: It might
be perpendicular.
PROFESSOR: She says there might
be one perpendicular to it.
And that's the one
I was searching for,
you're right too.
There's going to
be a principal axis
that's perpendicular to
that plane of symmetry,
and since we want to define
our moments of inertia
for this to get started
with with respect
to g, where will
it pass through?
AUDIENCE: Through g.
PROFESSOR: G, kind
of by definition.
So we're going to have
it pass through g.
If there's a plane of
symmetry, then there
is a principal axis
perpendicular to it.
And we'll0 define it, for the
purposes of our discussion,
we'll just let it
pass through g.
It doesn't have to,
but that's how we're
going about this discussion.
Let it pass through the center
of mass, this point we call g.
All right, so that
means that there's
a center of mass in this thing.
And I'm just guessing
roughly where it is.
But in fact if I hung this
thing up here like this,
and let gravity find it's
natural hanging angle,
and I drew a plumb
line down here,
just drew a line that the string
would take with a plumb bob.
Then I went to some other
point and did it again
and hung a plumb bob on it drew
the line-- where they intersect
is the center of mass.
And then you know since
it's got symmetry this way
that's in the middle.
So I guesses that that's
about where it is.
And there then is
a principal axis
of this object that's
perpendicular to it,
passing through the plane--
perpendicular to the plane.
And that means now
there's two more.
But this gets a little more
difficult, and I have no clue.
There are two more, because
the principal axes always
come in an orthogonal set.
So now that I know one,
I know that there's
two more somewhere
around oriented
this way, someplace
in this plane.
Maybe one like that,
maybe one like this,
such that if you spun it
about one of those axes,
it'd be in balance.
You can see that
gets a little messy,
I can't guess where it is.
And so there are ways
to find it, one of them
would be doing an
experiment, seeing which
axis it spins nicely around.
So that's the second rule.
But even just with that
one plane of symmetry,
you get some pretty
good insight.
Now what if there are
two planes of symmetry?
Yeah?
AUDIENCE: You said that
the principal axis does not
have to pass through
the center of mass?
PROFESSOR: No, I'm saying
it doesn't have to pass
through the center of mass.
The question is?
AUDIENCE: How could
it not because it
becomes stable [INAUDIBLE].
PROFESSOR: So she's saying
it'd be unstable if it's not
passing through there.
Like if I put an axle
through this wheel,
and I spun it about
some point out here,
you don't know because I
had that shaker in here,
that thing is going
to shake like crazy.
But does it produce
unbalanced torques
about my rotation point?
It produces centrifugal force
that you'll feel like crazy,
but does it produce
a torque that you'd
have to resist with
some static torque?
What do you think?
It won't.
So there's a nuance to
this unbalanced thing,
and I was going to get to
it, probably next lecture,
but that is the when we say an
object is dynamically balanced,
we mean that it doesn't
have any unbalanced torques.
If it is statically
balanced, it's
rotating about its
center of mass.
But you can be
statically unbalanced
and let it go around this axis,
but it'll produce no torques.
It's still dynamically balanced.
And the angular momentum of
an object which is rotating--
And this we know has
a principal axis here,
I just moved it off to the side.
It's rotating about this.
It is dynamically balanced,
and if you computed
about this point now the
Ixy, Ixz off-diagonal entries
in that moment of inertia
matrix, they're all 0.
You'd get no unbalanced
torques, but you
do have an unbalanced
centrifugal force
as this thing goes around.
And we'll talk a
bit more about that.
AUDIENCE: Is that
sort of analogous
to the homework problem
we had a few weeks back
with the motorcycle wheel and
you just had mass on one side
and not on the other?
PROFESSOR: Right.
So she's asking about the
motorcycle wheel problem
where we had that little
mass that got there.
I think in the next lecture
I'm going to come back
to that problem
just so we could tie
a bow around this whole
thing and understand why it's
unbalanced, how
you can balance it,
and the difference
between static unbalance
and dynamic unbalance.
But today, we're talking
about symmetry rules.
Finally, so I was saying,
let's talk about something
that has two planes of symmetry.
This actually has three
planes of symmetry,
but we'll settle for two.
Pick a plane of symmetry
for this object.
If I pick-- OK, so she picked
the one-- slice it this way.
What about the second?
Like that.
Then a third, right?
There's even one like that.
So this has three
planes of symmetry.
But if you have two planes
of symmetry that intersect,
that are orthogonal
to one another,
what do you think you
can say about that line
of intersection?
AUDIENCE: It's the
principal axis.
PROFESSOR: It sure is.
And it's probably
right through g.
So if you have two
planes of symmetry--
Now make them orthogonal.
You can make all sorts
of symmetry rules,
and I'm just picking these
three to help you out.
This just to help you
see principal axes.
If you have two
orthogonal planes
of symmetry their intersection--
and once you know that,
then you go back to rule two.
And it tells you everything
else you need to know.
Because you have one
plane of symmetry,
you know there is a principal
axis perpendicular to it.
Well if you have two planes of
symmetry, the rule still holds.
There's one perpendicular
to each one.
The intersection, let's say,
of this plane of symmetry
and this plane of
symmetry is a line
which goes right through the
center of this thing that way.
So there's a principal
axis this way.
But since there is a
plane of symmetry here,
there must also be a principal
axis perpendicular to it.
So sure enough, three
principal axes for this thing
are through the center,
perpendicular this way,
perpendicular that way.
You instantly know.
Two planes of
symmetry-- you instantly
know where the three orthogonal
principal axes are that past
through the center of mass.
Yeah?
[?
AUDIENCE: Does this all apply
?] just like a constant mass
throughout?
PROFESSOR: Not
constant, symmetrically
distributed density.
Right so I'm choosing
my words carefully so
that I succeed in
the following--
that the planes
defining mass symmetry
will be the same as the planes
defining geometric similarity.
But you actually don't have
to have a constant density,
it just has to be distributed so
that what I just said is true.
So that the geometric
symmetries are
the same as the mass
distribution symmetries.
All right so those are my
three rules of symmetry.
You could make up others.
Those are the three
that I've made up
to help you see objects.
That object, it's
a circular disk
put on top of another object
such that their centers
of mass line up.
Where are the principal axes of
this object using those rules?
If you think you
know one, tell me.
AUDIENCE: Through the middle.
PROFESSOR: Through the
middle of both of them.
Probably, good guess.
How about another one?
Where does this thing
have planes of symmetry?
AUDIENCE: So there's a plane of
symmetry if you cut it in half.
[INAUDIBLE] cut it
in half, [INAUDIBLE].
PROFESSOR: OK, and?
AUDIENCE: The other way.
PROFESSOR: One like that,
we've got all three.
And if we're going to want it to
go through the center of mass,
then we're going to have to
find where the center of mass
is this way, but
it's about there.
So just using the
symmetry ideas,
you can right away figure out
where these principal axes be.
And that means from a
dynamic point of view,
if you spin it about
one of those axes,
it's nice and
dynamically balanced.
If you spin it off in
some other weird direction
is it necessarily
dynamically balanced
about that axis of spin?
So let me restate that question.
We know that this thing has
an axis of symmetry principal
axis through the center,
and another one this way,
and another one this way.
And if I spin it about
any one of those,
it's dynamically balanced.
But if I pick some other
strange direction for the spin,
and I spin it about that axis,
will I feel unbalanced torques
on this axle, on
the bearings having
to hold this thing in place?
Yeah, you better believe it,
this thing wobbles like crazy.
So the principal axes are
a property of the object,
they're not a property
of the angular momentum.
The angular momentum comes
then from multiplying
the mass moment of
inertia that you've
determined times the
actual rotation vector.
And you'll find out then you get
angular components of angular
momentum that are not in
the direction of spin,
and as soon as that happens,
you have unbalanced terms.
I've got to get on to something
else to help you do homework.
So for my disk, with z coming
out of the board, the Izz--
so let's say here's x, y, z
coming out of the board-- Izz,
the mass moment of
inertia about this z axis
is, from the basic definition,
the summation of the mis,
xi squared plus yi squared.
It's just that for every
little mass particle
it's the radius squared away
from the center of rotation.
That's what the x squared
plus y squared is.
And we can turn this
into an integral.
It's the integral of r
squared, that distance,
times the little mass
bit that's there.
And that's the same.
If you wanted to do the integral
as x squared plus y squared dm.
But to do this integral for a
nice circular, symmetric disk,
you can pick a little mass bit
that has thickness dr and width
rd theta.
And this angle here is d theta.
And that's a little bit of area.
That's a little dA which
has area r dr d theta.
It's just length times width.
When it's small enough,
it's a little rectangle,
and it has that area.
And it has a volume, dV--
the volume of that thing
is just the area times
the thickness of it.
So here's our disk here,
but it has some thickness,
and I'll call that h.
So the volume is
just h r dr d theta.
And the mass, dm, is
a density times dV.
So I want to
integrate this, all I
have to integrate the
integral then of r
squared dm is the integral from
0 to 2pi, 0 to r of rho dV.
Rho h-- oh, and I need
an r squared-- r squared
dV is rho h r dr d theta.
So this is 1802
integrals, right?
So is any of this a
function of theta?
No, so it's a
trivial integral, you
integrate that over
theta, you just get theta,
evaluate it 0 to 2pi.
So this is 2pi rho h, can all
come to the outside, integral 0
to R of r cubed dr.
And that ends up--
the r cubed goes to
r to the 4th over 4.
And the final result of this
one is 2pi r to the 4th over 4
rho h, and when new account
for h times pi r squared
is the volume times
rho is the mass.
This all works out to
be m r squared over 2.
So Izz-- so I needed to
do this once for you.
For simple things
integrate, Izz in this case,
you just integrate
it out, account
for all little mass bits, that
is the mass moment of inertia
with respect to the axis passing
through the center like this.
Pardon?
AUDIENCE: [INAUDIBLE]
PROFESSOR: It's this, this
is what I'm talking about.
Moving on to the last bit.
So we need to know how
to be able rotate things
about places other than
their centers of mass.
So this is a stick, I can
rotate about the center of mass,
but it's more interesting
if I rotate it
about some other point.
It makes it a pendulum
when I do it around here.
So I need to be
able to calculate
mass moments of inertia
about a point that's
not through the center of mass.
I know you've seen this
before in [? 8.01, ?]
so this is going to
be a quick reminder.
But I'll show you
where it comes from.
So here's my stick.
And it has a center of mass, and
that's where G is located here.
It has a total length l.
I'm going to give it a
thickness b, a width a.
So it's a stick.
a wide, b thick, l long.
Uniform has a center
gravity right in the middle.
And I'm going to
attach to this stick--
and this point is
kind of hard to draw.
This point is at the
center of the stick, OK?
I'm going to put my
coordinate system attached
at the center of
gravity, center of mass,
and I'm going to make it
the-- that's x prime downward,
z prime, and y prime is
then going off that way.
So this is a body
set of coordinates
at the center of mass.
x prime, y prime, z prime.
And x happens to be down.
And I want to calculate
my mass moment of inertia
with respect to a point up
here that is d, this distance.
I've moved up the
x-axis an amount d.
I'm going to set a new
coordinate system up here.
So if this was z prime,
my new z is here.
It's getting a little messy.
Maybe I'll do just a face view.
If my previously y
prime and x prime
were like that, z
coming out of the board,
now I have a new
system that is y
and x like this, z still
coming out of the board.
Now the coordinate.
So how do I calculate
mass moment of inertia?
Well I want Izz.
I probably know Iz'z'.
Iz'z' is the mass moment of
inertia about this point.
I know it's a principal
axis from all the things we
just-- that square block
is the same as this.
That's a principal axis
in the z prime direction.
I know the Izz'
with respect to G,
I want to know what with
respect to this point.
So well Izz, which
is my new location
up here, and we'll call it A.
So Izz here with respect to A
is the integral of r squared dm.
We've got to do the
same integral now.
But that's the integral of
x squared plus y squared dm.
Now I can look at this and
I can say oh, well, these
are d-- this is separated by d.
I only moved it in the x.
The ys haven't moved and
the zs didn't change.
I just moved my point
only in the x direction.
So I can now say that in
terms of my new coordinate,
it's the same as x prime plus
d, the distance from here
to a point down her,
some arbitrary mass point
xi is going to be xi' plus d.
So to do this integral
in the new coordinates,
this is going to be the integral
of x prime plus d squared
plus y.
Now y prime equals y
and z prime equals z.
Those haven't changed.
I didn't move my new
coordinate system
in the y direction
or the z direction.
So the coordinate in
the new system in y
is the same as before.
So this is just y prime squared.
And this whole thing
times d, integrated times
every little mass bit.
If I square this, I get x
prime squared, 2x'd, d squared,
plus y squared.
So this integral,
Izz with respect
to A, when you
rearrange it, looks
like x prime squared
plus y prime squared dm
and the integral of a sum
is the sum of the integrals.
So I break it into
bits here, there's
a d squared, which
is a constant, dm.
And then the last term is
plus 2d and it's x prime dm.
Just multiply this out,
rewrite it, break it apart.
Well let's do this one.
Integral of x prime squared
plus y prime squared dm.
That's something that we
already have a name for.
This is IGzz.
It's the original
mass moment of inertia
with respect to the original
coordinate system at G
in the z direction.
So it's [? IzzG, ?]
we already know that.
That's given for the object.
Plus, this is the integral
of dm over the whole extent
of the object?
Just the mass of the object.
This integral, this
is the integral
in terms of the x-coordinate.
And every mass bit from here,
if I go out here and find one,
there's an equal and
opposite one up here.
This is the definition
of the center of mass.
This integral, if I'm
at the center of mass
integrating out from it, this is
zero because of the definition
the center of mass.
And I've just proven the
parallel axis theorem.
Izz about this new point is
I about G plus Md Squared,
where d squared is
the distance I've
moved this z-axis to a
new place parallel to it.
So Izz with respect to G,
the original mass moment
of inertia for Izz is m L
squared plus a squared over 12.
And Ixx m L squared
plus b squared over 12.
And Izz-- oh, we
already know that one.
Wait a minute.
I haven't told you what that is.
That's Izz, Ixx, Iyy.
Just a little messy here.
Iyy for this problem is--
I have made a mistake.
Ixx is a squared plus b squared.
Iyy is m L squared
plus b squared.
So those are the
three-- all with respect
to G-- for this stick stick.
And I'm going to--
AUDIENCE: [INAUDIBLE]?
Is that also divided by 12?
PROFESSOR: Yeah.
That kind of sets us up where
I can pick up next time.
So let's finish by asking
ourselves the question, what
do we think about-- if
I've moved to this new put
new position, and I'm not
rotating about the center,
is this new axis-- this one
around the center before,
that one we know is
a principal axis.
If I rotate about
this new place which
I've defined the mass moment
of inertia about that place?
Is this a principal axis?
AUDIENCE: Yes.
AUDIENCE: Yes.
PROFESSOR: How many think yes?
How many think no?
OK, so lots of people not sure.
So my dynamic definition
of principal axis
is if you can rotate the
object about that axis
and produce no unbalanced
torques, it's a principal axis.
And I can do that and this thing
will just spin all day long.
Now there is a force, you could
think of a fictitious force.
There's a center
of mass out here.
As it spins around, there's
a centripetal acceleration
making it go in the circle, that
means that fictitious force is
like there is a centrifugal
force pulling out on it.
Do I feel that?
Do I have to this
resist that force
as it goes round and round?
Yes.
So that is an
unbalance of a kind
we know as a static imbalance,
but it doesn't produce torques
about my axis right
lined up on the center.
AUDIENCE: Doesn't gravity
pull on the center of mass
[INAUDIBLE].
PROFESSOR: Sure, gravity
does, but that's now
a different problem.
That's what makes this thing
act like an oscillator.
The torques of the
kind I'm talking about
is if I compute the angular
momentum of this thing
and compute dh/dt-- the time
rate of change of the angular
momentum is a torque
on the system, right?
I will get the term that
makes it spin faster,
and I will get, if
they exist, terms
that make it want to bend
this way or bend back.
It only happens if-- if I hold
this thing over here, and spin
it, get it spinning, and I
compute the angle momentum
with respect to this
point, will I get torques?
Yeah, but that's not how I--
that's a different problem.
The IG is as if I were
computing the angular momentum.
Remember I started defining
mass moment of inertia matrix
based on an angular
momentum computation at G.
So it's right there in
the center of this object,
there's no moment
arm that is causing
torques that's trying to twist
this thing about that point.
So the answer to the question
is this is a principal axis.
Yeah?
AUDIENCE: So if you take the
derivative of the angular
momentum would you get torques
that are not in that direction?
PROFESSOR: If you get torques
that aren't in that direction,
either you made a
mistake doing the math,
or you were in
error in identifying
the mass moment of inertia
matrix to begin with.
Because if you
get torques, there
must be non-zero off-diagonal
terms and in mass moment
of inertia matrix.
They are what account
for the torques.
So by this parallel axis
theorem, any other axis
you go to-- if you started
at a principal axis,
any other axis you create
is also a principal axis.
That's the movement
of just one axis.
If you do two, if you move
this way and this way,
all bets are off.
You get a difference answer.
And if you're interested in
that more complicated problem,
read that Williams
thing because he
does the complete parallel
axis, parallel planes
and comes up with a
super compact little way
of calculating them.
See you on Thursday.
