Professor Ramamurti
Shankar: Okay,
so this is a new topic called
Physics or Dynamics of Rigid
Bodies.
So, a rigid body is something
that will not bend,
will not change its shape when
you apply forces to it,
and one example could be a
dime, a coin,
or a meter stick.
Of course, nobody is absolutely
rigid.
You can always bend anything,
but we'll take the
approximation that we have in
our hands, a completely rigid
body.
A technical definition of a
rigid body is that if you pick a
couple of points,
the distance between them does
not change during the motion of
the body.
So, the dynamics of rigid
bodies, in three dimensions,
is fairly complicated,
because what I want you to
imagine now is not a point mass
but some object with some shape,
like this guy.
You take this;
you throw it up in the air;
you can see it does something
fairly complicated.
In fact, we can characterize
what it does using two
expressions.
I think you can guess what they
are.
One is, if this had been a
point mass, then the point mass
will just go back and forth or
up and down.
It will do what's called
"translations."
It'll go from a place to
another place,
maybe on a straight line,
maybe on a curve.
But as it goes,
the point mass has no further
information;
you've got to tell me where it
is, and how fast it's moving.
That's the whole story.
But if you have a body like
this, it's not enough to tell me
where it is, because if I throw
this up in the air you can
imagine that as it travels along
some parabolic path one instant
the eraser may look like this,
the next instant and the eraser
may look like this;
so, we say that it is
translating and rotating.
It's a non-obvious theorem that
anything the body does from one
instant to the next instant can
be achieved by first translating
it,
then doing a slight rotation to
bring it to the configuration
you want.
In other words,
you can go from here to here by
a translation,
followed by a rotation.
Now, a translation is what
we've been studying all the
time, because for point
particles there is no notion of
rotation.
They take a point mass and
rotate it.
You don't even know it's
rotating.
It has no internal variables,
so that was the problem we've
been focusing on,
but now we're going to enlarge
our study to objects which have
a size and which have a shape,
therefore, which have an
orientation.
It's not enough to say the
football is here.
The football could be here but
its long pointy end could be up
to down or side to side.
So, what we want to do is to
break the problem into two
parts.
We want to focus on a body
which is only rotating but not
translating.
Once we've sharpened our
skills, we will then put back
the ability of the body to
translate.
So, we're going to focus
initially on a body that cannot
translate.
So, what that means is,
you take the body and you grab
one point in it;
then, you can ask,
"What can it do if I grab one
point in it?"
Well, any motion it does will
be such that that point is not
moving.
Now, in three dimensions,
if you take a football and a
certain point is held fixed,
it can gyrate in many ways.
But at any instant you can show
what it'll be doing.
It's to be doing a rotation
around an axis,
passing through the point where
you're grabbing it,
because if it rotates to an
axis, points on the axis don't
move, and the point where you're
grabbing it better lie on the
axis.
Now, that's also pretty
complicated.
You can see why I gave you a
little precautionary thing.
It's hard to visualize these
things.
So again, I want to take the
simplest possible problem;
then, make it more and more
complex.
Well, you might say the
simplest thing you've been doing
all semester long is to take
things in one dimension.
But that turns out to be too
low.
If you're in one dimension and
you've got a rigid body--By the
way, a rigid body means not only
extended but actually rigid.
For example,
if you take a cat or a snake,
a snake you can grab at one
point but the snake is not
simply doing rotations around
the one point,
right, trying to bite you,
trying to move around;
a snake is not a rigid body.
A dead snake could be a rigid
body.
A living snake -- definitely
not a rigid body.
Now, you've got to remember
that I've given you a break by
studying rigid bodies,
but real bodies can actually
not only wobble and twist and
turn, they can also vibrate and
they can move their arms around.
That's much more difficult,
so we won't go there,
and we'll try to take the
simplest rigid body.
But I said, if I take a rigid
body in one dimension,
it cannot rotate at all because
there's not room in one
dimension to rotate.
It can only go back and forth.
So, to show you there is some
new possibility,
I have to go at least to two
dimensions.
Here's one example where one
dimension's not enough.
So, I'm going to imagine rigid
bodies, which are living in the
plane of the blackboard.
Now, one way to do that is to
take a piece of metal,
just cut out a shape,
and that's a rigid body.
And of course this has a slight
thickness, because anything you
make of matter has some
thickness,
at least one atom thick,
but imagine that thickness is
negligible so it's like a piece
of metal,
a piece of foil or something,
but it's rigid,
and this is the rigid body
we're going to study.
This rigid body can move in the
plane of the blackboard and
rotate, but we're going to grab
it at one point.
So, let's pick some point here.
If I grab it at that point and
I ask you, "What can this fellow
do?"
So, imagine skewering it
through the blackboard,
you say, "Okay,
do whatever you want."
You realize that all it can do
is to rotate around an axis,
penetrating that point.
So, let's say at t = 0,
the body looked like this,
a little later it's going to
look rotated.
I have to tell you what it is
doing at a given instant.
I have to tell you how it's
oriented.
So, what we do for that is,
we draw through the point of
rotation some line.
You can pick any line you like.
Pick a line like that and we
say what angle that line makes
with a standard chosen
direction, usually the x
axis.
So, if I give you θ,
I think you'll have to think
about it and you have to agree I
have told you where the body is.
You can reconstruct everything,
right, because that point is
nailed;
it cannot move.
You take the body;
you turn it by an angle
θ.
In other words,
here is a question.
Suppose you're not in this room;
you're in another room and I
want to tell you what the rigid
body is doing.
What information do I have to
communicate to you,
without seeing the picture?
I claim, if I tell you how the
line was drawn in the body,
which is something you do once
and for all,
a ruled line,
then I say the ruled line makes
an angle θ with the
x axis.
I believe you can imagine in
another room what this body is
doing, and that's the meaning of
saying, "I've given complete
information."
Of course, that's not the
complete dynamical information,
because it also depends on how
fast it's rotating.
But to simply say the analog of
where the body is in one
dimension is this angle
θ.
So, what we are going to set
up, you'll find,
is an analogy between one
dimensional translation and two
dimensional rotations.
You'll find the analogy is very
helpful.
So, I'm going to use a piece of
the blackboard where I'll keep
drawing that analogy.
I do this because the number of
things you have to remember will
be reduced if you map this
problem mathematically to the
problem of translation in one
dimension.
In other words,
in one dimension,
when a body is moving,
x tells you where it is,
right?
I give you a number,
3, or π,
or -14;
you know where it is.
For this rigid body,
living in the plane of the
blackboard, this angle θ
tells you what the body is
doing.
So, even though the body is in
two dimensions,
you need this one angle
θ to tell you what its
orientation is.
You don't need two numbers;
a single number tells you what
the rigid body is doing.
So a little later,
it could be rotated by an angle
Δθ.
That's the analog of saying it
moved from x to x
+ Δx.
The first thing you've got to
do is, how do you want to
measure θ?
For x we know,
we have agreed.
We're going to measure in
meters θ,
the standard preference for
people at elementary status,
is to measure it in degrees.
You measure it in degrees and
when the body completes one full
revolution, you like to say it
has turned by an angle of 360
degrees.
Now, that is not the preferred
way to measure angles in more
advanced dynamics.
We're going to use something
else.
You can imagine what it is.
You know what I'm driving at.
aAnybody know where I'm going?
Students: Radians.
Professor Ramamurti
Shankar: Radians.
I'm going towards radians,
and you can ask yourself,
why would anybody think of a
radian?
What's wrong with 360 degrees?
I mean, all the textbooks--I
mean, all the novels say,
"When I was going down the
wrong path, then I did a 180."
Right?
So, everyone knows.
In fact, I saw a book by a
violinist who said,
"I was going down all the wrong
things and I did a 360."
So, you realize that 360 is not
really a way to reform your
life, but among the physicists
it struck a chord.
In physics, a 360 is really not
a trivial thing to do.
It turns out,
the particles in the world are
divided into bosons and
fermions, and half of the
particles are bosons and roughly
half are fermions.
An electron is a fermion;
the light quantum is a boson,
the photon.
For all the guys who are
photons, or bosons,
if you turn them by 360,
they come back to what they're
doing.
An electron,
it turns out,
when you turn it by 360,
all the numbers are going to
minus those numbers,
and only when you turn by 720,
you'll come back to technically
what you're doing.
So, it's quite possible this
person had in mind the behavior;
maybe she was a fermion,
but that's what it said,
"I did a 360."
I've gone through a 360 many
times, so if you use it you
better back it up with the kind
of information I gave otherwise
you're just off by a factor of
two.
Anyway, 360 is very nice.
I think the Babylonians liked
360 for a lot of reasons.
It's divisible by a whole bunch
of numbers.
It's easy to partition into
smaller and smaller pieces.
I believe, in some engineering
circles, a circle is 400 and
something else,
some other engineering degree.
So, you can decide what a
circle is, but we like radian.
So, let's ask,
"Why do we like a radian?"
For that purpose,
I want you to take a point
here, at a distance r
from the center,
and let it move to a new
location with an angle
Δθ.
You agree that if the rigid
body is rotating,
this guy will be traversing a
circle of radius r.
Because it's a rigid body,
it cannot change the distance
from this point.
So, we're going around in a
circle, and I'm asking you how
big is that segment,
that arc length ΔS that
it has traveled,
when it's rotated by an angle
Δθ.
So, one way to calculate that
is to say look,
if I did a full circle I'd have
done 2πR.
r is the distance of
this point in the rigid body
from the center.
I haven't gone a distance
2πr but I've gone
Δθ, so the fraction of
the circle I've covered is
Δθ over 360 if
θ is measured in
degrees;
so, let me indicate that in
this fashion.
So, we can write π
over180 Δθ times
r.
If you measured θ in
degrees, this is what you have
to do to convert an angular
traverse to an actual distance
along the tangent to the circle.
The linear distance you travel
is connected with the angle by
which you rotate,
by the distance r from
the center, and this number
here.
So, people say,
"Look, why don't we make life
easy by calling θ times
π /180 as a new angle."
And let's call it Δθ
times r.
So, Δθ, without any little
zero on top of it,
stands for radians from now on,
and it's related to θ,
so if you ignore the
Δ's, θ itself is
θ in degrees,
π/180.
Then, the advantage is the
distance you travel -- linear
distance you travel -- bears a
simple relation to the angle you
travel,
by the arm length r from
the center.
So, you don't have to carry
this 180 and π and
everything, you can write the
simple result.
But now, when you do this,
you are measuring it in
radians, so you've got to have
an idea how big is a radian.
Well, a full circle,
if you go a full circle;
the distance you travel should
be 2πr so a full circle
will Δθ = 2π.
So, a circle is worth
2π radians;
2π is like 6.
So, one radian is roughly 1/6
of 360, which is 60 degrees.
The real answer is close to
58-something degrees.
It looks like an odd thing to
pick.
It's odd only if you start with
360 degrees.
But if you met people from an
alien culture,
you just got to go to the
supermarket to run into these
people, those people may not be
using 360 at all;
360 is a human artifact,
nothing special about 360
that's independent of our own
culture or biases.
It may be connected to how many
fingers we have.
It may be connected to how long
a year is, and so on.
But there's nothing natural
about the number 360.
2π on the other hand,
I believe, would be discovered
in any advanced civilization.
If you haven't figured out
π, you're not an
advanced civilization.
If you haven't figured out 2,
you're way behind.
So, you know 2,
and you know π,
and that's what other cultures
and other planets and galaxies
will be using.
That's the unit we like to use.
You'll find,
over and over,
people using units which are
natural to us.
But later on,
if you want to communicate with
people in another planet,
you will have to use different
lengths.
For example,
the height of a human or the
length of Napoleon's arm,
they're not very good units,
because if you don't have a
Napoleon in your background you
don't know what you're talking
about.
But if you measure distances in
how big it is in terms of a
hydrogen atom,
hydrogen atoms all over the
universe, that's a natural
length scale to use.
This was a very popular issue
in the old days.
When they sent off a rocket or
they buried something
underground for aliens to find
out, you want to show how tall
we are;
no use giving the answer in
meters or feet or anything.
But if you can somehow relate
your size, and the size of a
hydrogen atom,
then that number will make
sense to people in all
civilizations,
because hydrogen atoms
conveniently are manufactured
and scattered all over the
universe and they have the same
size.
Okay anyway,
that's a long thing about
radians, that's why we like to
use radians.
You have to know a few popular
angles in radian measure.
I'll be using only radians most
of the time.
You've got to know a circle is
2π;
π is a half a circle,
and π/2 is a quarter of
a circle and when you are by 90
degrees that is π/2;
that's something you should
know.
Once you write it this
way--Let's get another result
that's very useful.
Take the distance it travels
along the tangents by the time
over which it does that.
This is the actual tangential
speed, or the magnitude of the
velocity in the tangential
direction.
I think you all agree with that;
that's what you and I would
mean by speed.
If at an instant,
this piece of the sample just
broke and flew off,
it'll keep flying with that
speed.
This thing, we have a symbol
for it, ω,
and ω is called the
angular velocity,
and is measured in radians per
second.
So, you can calculate angular
velocity for various things.
For example,
what's the angular velocity of
the Moon as it goes around the
Earth?
Well, the angular velocity will
be one revolution every 28 days,
right?
One revolution is 2π
radians.
At 28 days you do 28 times 24
times 3600;
it's that many radians per
second.
Or there's a chainsaw blade,
and the chainsaw is spinning at
some number of rpm -- maybe
3,000 rpm -- you can take
revolutions per minute to
revolutions per second,
but you've got to remember
every revolution is 2π
radians.
So, the angular velocity of a
blade that's spinning with
frequency f is
2πf.
You understand that?
f is the number of
complete revolutions it does.
Therefore, if you pick a point
on it, it does a traverse of
2πf radians.
So, the dictionary has got a
second member here,
which is the velocity,
which is dx/dt,
is now replaced by angular
velocity, which is
dθ/dt.
This is an analogy, by the way.
This is to help you say
ω is like what velocity
used to be in the old days when
things were moving back and
forth.
But now they're going round and
round, but even here there's a
notion of velocity,
but it's not in meters per
second, it's measured in radians
per second.
Now, here is something you guys
have got to keep very clear in
your head.
I can only mention it to you
but I cannot do it for you.
This ω is an analogous
thing to ordinary velocity but
it's also an actual velocity
here,
the good old velocity,
that's the tangential velocity
of points which are part of a
rotating rigid body.
That is not part of my
dictionary but it's a very
useful result to know.
The tangential velocity is the
angular velocity times the
distance of that point from the
center of the rotation.
So the rigid body as a whole
has a single angular velocity.
That's what it means to be
rigid.
The whole thing is rotating.
You cannot say the angular
velocity is so and so here and
so and so there;
it's the property of the entire
body.
With a linear velocity,
it's not the same everywhere;
points near here in the same
time go that far,
and points near here in the
same time go that far.
Therefore, the linear velocity
increases from the center.
The angular velocity is
constant, and a rigid body
cannot have two angular
velocities.
It can have only one angular
velocity.
Okay, then, what we do is we
say okay, let's take a problem
where the angular velocity is
itself changing.
Fine, that's the analog of
saying there is a rate of change
of velocity, and here we have a
quantity called α,
which is the rate of change of
angular velocity,
or the second rate of change of
the angle,
and that's denoted by α.
So, if you took one of these
rotating circular saws,
and it was spinning at a
certain speed,
and due to friction,
let us say, it was slowing
down, or when you turn on the
motor it's speeding up,
suppose you just turned down
the saw motor and it's speeding
up in its rotation,
you've got to be very clear
that the acceleration of
parts--This is the saw blade
okay?
I'm trying to draw for you one
of these saw blades and it's
spinning.
If you pick a point,
say on the circumference;
you remember long back we
learned, even if it's going at a
constant speed,
a constant angular speed,
it has a tangential velocity,
which is ω times
r.
And because of that,
it'll have an acceleration
towards the center,
centripetal acceleration,
which is v^(2)/r and you
can write it as ω^(2)r.
That's one acceleration it has
even if it's spinning at a
steady rate.
But I'm saying it can also have
a tangential acceleration,
if on top of everything else,
the angular speed itself is
changing.
Well, let's take our time
digesting that.
This is not too hard but I want
you guys to understand.
If it's spinning at a steady
ω, we have learned
anything going in a circle has
an old-fashioned linear
acceleration towards the center,
which is v^(2)/r.
But we have realized v in the
tangential direction is
ωr, so if you put that
in -- if you want you can call
it v tangential -- that
gives me ω^(2)r,
that's just the v^(2)/r.
But if you put the brakes on
this by speeding it up or
slowing down it'll have in
addition, an acceleration in
this direction.
So, the acceleration of a point
on the edge or anywhere else can
have two components.
The tangential one will be
there only if you speed up or
slow down the rotating disk,
but the radial one will always
be there as long as it's
rotating.
Okay, once you've got this
dictionary, you can have a whole
bunch of analogous quantities.
For example,
if you focus on problems where
the body has a constant angle of
acceleration α,
what can you say?
Well, you can already say that
θ would be the
θ_0 +
ω_0 + ½ αt^(2).
This is because θ is
behaving just like x.
If x had a constant
acceleration a,
we all knew x was equal
to the starting x +
initial velocity times time plus
½ at^(2).
Similarly, the angle by which
it rotates will be the initial
angle.
The initial angular velocity
times t plus something
involving t^(2).
This is only if α is a
constant, just like I hope you
guys know this is true only if
a is a constant.
All right, now I'm going to
pursue the analogy and ask what
is the kinetic energy of a rigid
body, when it's rotating?
Yes?
Student: Could you
please explain [inaudible]
Professor Ramamurti
Shankar: Here?
Student: No,
the acceleration with
α.
Professor Ramamurti
Shankar: Here?
Student: Right next to
the saw blade.
Professor Ramamurti
Shankar: Oh here?
Student: Up.
Professor Ramamurti
Shankar: This one?
Student: Yeah.
Professor Ramamurti
Shankar: Okay,
I'm saying, if you have a
rotating disk,
take any point on the rotating
disk, even if the disk is going
at constant angular velocity,
it has a tangential velocity,
and the direction of the
velocity is constantly changing.
This is the oldest thing we
learned, so it has an
acceleration towards the center.
But it's not an angular
acceleration;
it just a standard acceleration
towards the center of anything
going in a circle.
That's always going to be there.
But on top of it,
if you speed up or slow down
this rotation rate of the disk,
you'll also have a tangential
acceleration,
which is α times
r.
Not clear yet, or?
That's the acceleration lay
people would know.
If you're on a rotating
platform and somebody sped up
the rotation rate,
you will notice you're going
faster.
That's the acceleration in the
tangential direction.
The acceleration towards the
center is a more sophisticated
one we have learned,
and it's due to the fact that
velocity's a vector and not a
scalar.
So, if the direction is
changing there is an
acceleration we associate with
that, and it's as real as any
other acceleration;
it needs a force to make it
happen.
Maybe the better way to ask you
is this.
Suppose this little speck here
is you, okay?
You happen to be stuck on a
rotating chainsaw blade,
and you're clinging to it for
dear life.
What force do you have to apply
to stay onto that rotating
blade?
This may not happen to you and
me but if you're in organized
crime this is not a very unusual
situation.
You can find yourself
intermingled with all kinds of
machinery, and you want to know
what do I do to stay on.
Well, Newton tells you the
answer.
Newton says,
if you want to be on this
rotating chainsaw blade,
your acceleration towards the
center has to be paid for,
and if the blade is picking up
speed, that's got to be paid for
too.
That's the net acceleration;
that, times your mass will be
the force that has to be applied
to you.
So, you will cling onto it with
that force, and that force will
be applied in turn by the third
law by the disk on you.
For example,
suppose it's a top view of a
rotating platform.
You're just standing there and
the platform is rotating.
You can ask,
"What frictional force do I
need to stay on the platform?"
Well, you need the force to
provide you with ma,
and I'm telling you that your
a has two parts.
Even if the merry-go-round is
spinning at a steady rate,
you will need to have a
frictional force directed
towards the center to bend you
into a circle.
Now, if on top of it,
the merry-go-round starts
increasing its speed,
or decreasing its speed.
We'll also need a tangential
component of force to provide
the tangential acceleration.
Okay, now this is kinematics of
rotation at constant angular
acceleration,
which is identical to this one
mathematically,
because mathematically,
if you know
d^(2)θ/dt^(2) is
α,
the answer is just like
d^(2)x/dt^(2) = a.
You change the symbols,
you get the same answer.
So, you can imagine a variety
of problems I don't feel like
doing.
I think you guys can easily do
with that.
I'd rather focus on more
difficult ones.
For example,
a chainsaw blade,
spinning at 3,600 rpm,
is decelerated at some rate
α.
How long does it take to come
to rest?
Well, for that you will write
the second formula,
ω = ω_0 +
αt.
That's the analog of v =
v_0 + at.
And you start with some
ω_0,
you put on the brakes;
α it's a negative
number, you see at what time
this ω goes to 0;
that's when it'll stop.
Another result which I will use
today in fact is the following:
ω^(2) =
ω_0^(2)
+2α(θ -
θ_0).
That is the analog of v^(2)
= v_0^(2) +
2a(x - x_0).
This says that if the body has
an acceleration α,
then it's going to pick up
speed,
of course, and the final speed
squared is related to the
initial angular velocity squared
plus 2 times the acceleration
times the angle rotated.
So, this is the analogy,
so we don't have to reinvent
all of these things.
They just come from the fact
that things which are
mathematically the same have the
same mathematical results.
They obey the same equation;
they have the same results.
All right, now let's find the
kinetic energy of a rigid body.
It's got mass,
and if it's spinning,
all the little atoms making up
the body are moving,
and they've got their own ½
mv^(2).
And we want to ask,
"What is the total ½
mv^(2) summed over all
the particles?"
For that purpose,
it's convenient to take the
following simpler rigid body.
We take a rigid body by taking
a mass m_1 at a
distance r_1,
connected to some central hub
with a massless rod,
and you take another mass,
and you take a third mass.
This is m_2,
at a distance
r_2,
there's m_3,
at a distance
r_3,
and so on.
You can make a rigid body by
modeling it in this simple
fashion.
If you are a little more
abstract, you can take a
continuous rigid body and do the
same thing;
but let's start with the simple
thing.
Here's a rigid body,
three rigid rods with masses
m_1,
m_2,
m_3,
and it's nailed here and there
is a skewer going through the
board,
and the things can only rotate
around that.
By the way, I should tell you
the convention for ω.
In one dimensional motion,
velocity is positive,
you're moving to the right,
and negative you're moving to
the left.
With angular velocity,
we've got to have a convention
on what's positive and what's
negative.
This thing which is spinning
can only spin clockwise or
counter-clockwise.
It turns out to be that
ω is positive for
counter-clockwise.
That just happens to be the
convention people use.
In fact, you notice the way I
drew the rigid body when I did
Δθ, I had rotating
counter-clockwise.
That's the preferred direction
for measuring the angle
θ from the x
axis.
Okay now, let's take this rigid
body.
It is rotating with a common
angular velocity.
I hope you all understand
that's the meaning of being
rigid.
So, everybody's got the same
angular velocity ω.
So, what's the kinetic energy
of this object?
Well, it's the kinetic energy
of each mass summed over all the
masses i,
m_iv
_i^(2).
Now, what's the velocity of
each object?
If you take this body,
m_1,
its velocity is necessarily
perpendicular to the line
joining it to the point of
rotation.
This one is moving this way;
it cannot move in the line
joining it, because if it is
it's not a rigid body.
For a rigid body,
all distances are maintained.
So, you don't come near the
center and you don't go far from
the center.
So, all you can do is rotate
around the center.
So, the velocity of this guy,
v_1 =
ωr_1.
The velocity of this guy,
v_2,
is the same
ωr_2,
and so on.
So, you can then add it all up.
And what do you get?
You get ½ sum over i
m_ir
_i^(2)ω^(2).
And you have to do the
summation over all the bodies.
In my example,
the summation goes from 1 to 3.
You may have 1 to infinity,
and 1 to a million;
it doesn't matter.
That's the kinetic energy.
You guys follow this?
This is not a new kinetic
energy;
this is what we know to be
kinetic energy,
just ½ mv^(2) for every
piece that makes up the rigid
body.
For convenience,
I've taken the rigid body to be
made up of a discreet set of
masses.
But now, if you notice,
as you go from 1 to 3,
m_1 and
m_2 and
m_3 are three
different numbers.
We don't know what the masses
are.
r_1 and
r_2 and
r_3 are three
different numbers.
We don't know;
each one could be at a
different distance from the
origin.
But the ω doesn't have
a subscript i;
there's nothing called
ω for this one and
ω for that one.
There's only one ω for
everything.
So, you can pull it out of the
sum and write it this way.
Then, the answer we write as
identical to ½ Iω^(2),
and we give this I a
name;
I is called the moment
of inertia.
So, continuing with the
analogy, this board is going to
be used for analogy,
we used to have K = ½
mv^(2).
Yet, we can have K = ½
Iω^(2) and I is
called the moment of inertia.
So, the moment of inertia is
determined not only by the
masses that make up the body but
how far they are from the
center.
If all the masses just fell on
top of the center,
the body would have no moment
of inertia.
It'll weigh the same;
the moment of inertia would
vanish.
And likewise,
if the mass is spread out the
moment of inertia is more.
For example,
if I'm standing around here,
and you come along and decide
to spin me,
I'm standing like this,
you try to spin me,
I stick my hands out,
my moment of inertia has
changed because the sum of my
mass now is further away from
the axis of rotation.
So, the moment of inertia
depends on the arrangement of
the masses.
We're taking a rigid body;
it's not like ice dancers who
do this.
We're talking about a rigid
body;
therefore, the moment of
inertia is constant.
But it requires a calculation
to find the moment of inertia.
And here's another important
thing.
If I decide to rotate the body
about a new location,
say around this point,
the moment of inertia will
change,
because all the distances
r will change.
The moment of inertia--If
someone says,
"Here are the masses,
here's where they are,
please find me the moment of
inertia," you will say,
"I cannot do it."
You will say,
"I cannot do it until you tell
me the point around which you
plan to rotate the body.
Unless that is given I cannot
tell you what the moment of
inertia is."
Therefore, the moment of
inertia is with respect to a
point, and there's nothing
called the mass with the respect
to the point.
The mass is just the mass.
The moment of inertia depends
on the point around which you're
computing the moment of inertia;
it's a variable.
It's also worth knowing that if
the body's being rotated about
mass m_3,
then m_3
doesn't contribute to the moment
of inertia;
it's out.
Anything sitting on the axis
doesn't contribute.
Things which are further away
contribute more in proportion to
their mass, in proportion to the
square of the distance from the
axis.
So, this is my dictionary here.
So, who is playing the role of
mass?
The mass is played by moment of
inertia.
It's got units of kilogram,
meters squared,
and no one has given a name for
that.
For example,
Newton times meter is called a
joule.
You can ask,
well this is kilogram times
meter squared,
maybe it's named after
somebody.
So far, it's not been named
after anybody,
it is just called Newton meter
squared.
So, this is going to be the
analog of mass in our world of
rotations.
You remember,
in the world of rotation
[correction: should have said
"translation"],
we define something called
momentum, which is mass times
velocity.
Here, we are going to call
something called angular
momentum is going to be I
times ω.
This L is called the
angular momentum.
One extremely important point
that you guys should notice is
that all the concepts I'm using,
like mass and energy,
they are the same old concepts
that we learned earlier on.
The fact that it's a rotating
body doesn't change anything,
okay?
Kinetic energy still means the
same;
this K,
though it looks so different,
is in fact the ½ mv^(2)
of every part of the body,
summed over the bodies.
All right, so this L is
called angular momentum.
It'll take a while to get used
to this, but right now it's the
product of the moment of inertia
and the angle of velocity.
So, now we are looking for the
most important equation in
mechanics, which is F =
ma.
That's going to be equal to
some mystery object.
Now, you should be able to
guess what goes on the
right-hand side,
okay?
You should know by now,
by analogy, it can be written
as dL/dt,
if you like,
the rate of change of momentum,
or it can be written as
I times α;
it doesn't matter.
I hope you understand that if
you take rate of change of
momentum, you will get
ma,
because when you take the
derivative, m is not
changing.
Likewise here,
if you take dL/dt,
I is not changing,
ω is changing to give
you the dω/dt which is
α.
I is not changing
because as the body rotates of
course particles in the body are
moving around,
but their distance from the
point of rotation is not
changing.
Even though the bodies are
moving to new locations,
the r^(2) is the same,
because they're just rotating.
That's why in the time
derivative you don't have to
take the time derivative of the
moment of inertia.
But we don't know who's going
to play the role on the
left-hand side,
okay?
So, we're going to find that.
We're going to find out the
analog of that.
So, I'm going to tell you there
are many ways to find this out.
It's a lot more subtle than
generally appreciated,
on what the correct law is.
You can hear the law stated
quite often, but some of the
reasoning behind it is
incorrect.
So, let me tell you one way to
get to the law.
I hope you know what I'm
looking at.
I'm trying to find out what is
the entity that is possible for
the rate of change of angular
momentum,
or what's the analog of the
left-hand side for ma.
I know the right-hand side is
Iα.
By analogy, who's on the
left-hand side?
First of all,
it's pretty clear that if you
want the angular momentum of the
body to change,
ω has to change.
You can see if something's
spinning around at some fixed
ω, you want to change
it, you've got to push or pull
or do something.
So obviously,
you're going to apply a force,
but the thing that comes on the
left-hand side is not simply the
force but something else and
we'll find out what the
something is.
And that's done by the
following device.
I'm going to use the good old
Work Energy Theorem that says:
when the force acts on a body,
the work that it does,
which is the force times the
distance it travels,
is equal to the change in the
kinetic energy.
That is the standard Work
Energy Theorem that we
established.
So, if you took a rigid body
like this, and let's apply to
the rigid body,
I'm just going to take a
simpler case of a rigid body
with these two masses,
m_2 and
m_1;
it doesn't matter.
I apply a force now,
F, a perpendicular to
that body, at the distance
r from the point of
rotation.
And let it that turn the body
during that time;
let's say the body moves by an
amount Δθ.
Maybe I should draw you a
better picture for this because
this is very,
very important.
So, here is the body;
I just took the simplest one,
with two masses,
m_1,
m_2.
I'm just going to push it here
and I decide to push it in this
direction, and it's at a
distance r from the point
of rotation.
And let it turn by a small
angle and come to this position,
and this angle is Δθ.
What is the work done by this
force?
I'm going to write it down but
I want you to think about it.
The work done by the force is
the force multiplied by the
distance over which it has
moved.
And basically,
I'm asking you what is that
distance it has moved,
and if you remember anything at
all from what I did earlier,
it's approximately this arc
length, which we have seen,
is r times Δθ.
In other words,
if you apply a force F
and you rotate the body by
Δθ, the distance you
actually move is rΔθ;
so, this is force times
distance.
Now, that's got to be equal to
the change in kinetic energy.
It's got to be equal to the
change in kinetic energy.
Kinetic energy,
I remind you,
is ½ Iω^(2).
Therefore, the change in
kinetic energy is going to be ½
I(ω^(2) -
ω_0^(2)).
By the way, these are two very
neighboring instants of time.
If in the neighboring instants
of time, during that brief
period we may take the body to
have a constant acceleration;
therefore, α is
constant.
So, this formula you can also
calculate when α is
constant, period.
In a real body,
when it's rotating,
α need not be constant,
but for a small enough
interval,
as small as you wish,
we may apply the formula that
works when α is a
constant.
In that case,
you will write it as ½ times
2α times the rotation
angle.
Right?
Look at this formula here.
I'm referring to this result,
ω final squared
-ω initial squared is 2
times α times the change
in angle.
So, I want you to then equate
this work done to this change in
kinetic energy.
I want you to equate this guy
to this guy, and we get our
great result that F times
r = I times α.
Iα is ma;
that's what I'm looking for.
So, this fellow here,
Fr, is the analog of
force, and it's called the
torque.
It's very important to notice
that the torque due to this
force is the value of the force
times the distance from the
point of rotation,
and it's only the external
force that I'm talking about.
There are also internal forces
in a rod.
Every part of the rod is being
dragged along by another part of
the rod, but that analysis in
terms of forces can be done but
it's very tricky and
complicated.
It's much easier to look at the
energy where the energy changes
only due to external forces.
If you only look at forces
you've got to be careful.
This pivot point is also
applying a force on the rod,
but it's not moving and not
doing any work.
So, the only external force is
the one I'm applying and is
responsible for the change in
the kinetic energy.
There are other forces inside
the rod.
In other words,
this mass is going to pick up
speed, not because I'm directly
pushing it.
I'm probably pushing this part
of the rod and that's pushing
the part next to it,
and it finally pushes this one.
So, I'm not directly applying
the force law to this mass.
I'm saying, whatever is
happening internally,
I don't care;
the external force times
distance is a change in the
kinetic energy of an object.
Yes?
Student: But how are you
saying the angular [inaudible]
Professor Ramamurti
Shankar: No,
I said this quantity,
the change in kinetic energy,
is the final kinetic energy,
½ times ω^(2) minus
the initial kinetic energy;
but final/initial referred to
when the rod was there and the
rod had rotated by a small
amount.
Student: Right.
Professor Ramamurti
Shankar: Okay.
Student: So,
do you say it's 2α of
the [inaudible]
Professor Ramamurti
Shankar: Right.
Student: You said there
needs to be something about
velocity [inaudible]
Professor Ramamurti
Shankar: Here.
No, I used this equation,
the change, the ω^(2)
at the end, minus the
ω^(2) at the beginning,
is 2 times the acceleration and
the angle of acceleration times
θ - θnot is the
change in θ over that
time,
right.
In other words,
that formula can be used for
even finite periods of time
provided α is constant.
We are applying it for a tiny
period of time,
small enough for us to believe
that the acceleration during
that tiny period is some number
α.
So, this ω^(2) -
ω_0^(2) is that
really if you want the change in
ω^(2),
and that's proportional to
α times the change in
angle.
So, α is not a constant;
α can vary from instant
to instant.
But at that one instant,
depending on the forces,
there'll be some α,
and this thing called torque is
I times α.
Now, I think it's very clear
that if I had several forces
acting on the body,
then the torque will be
F_ir
_i,
and there's just one final
caveat to this thing,
one final qualification,
which is that,
in practice,
the forces that are used to
rotate a body may not always be
perpendicular.
You can apply a force in some
other direction,
like that.
Now, you've got to realize that
torque is the ability to change
the rotational state of a body,
and what we are realizing is
that it depends on the force you
apply and depends on how far
away the force is from the point
of rotation.
So, if you have a door,
you're trying to open a door,
here is the door,
here are the hinges,
and you want to open this door
by applying a force.
The question is,
"Where do you put the
doorknob?"
Okay?
You suddenly invented the door,
you thought about hinges,
but you haven't quite figured
out where to put the doorknob.
You're going to ask yourself,
"Well, I've complete freedom,
maybe it'll look nice if I put
it there right next to the
hinges,"
then you realize that there are
parts of the universe,
like Moron Land,
where if you go,
all the hinges,
all the doorknobs are here and
they're saying,
"You know, we're applying a lot
of force,
we're not getting anywhere.
What's wrong?"
Well, what we're realizing is
that whereas force was
everything in linear motion,
force is not everything in
rotational motion.
If you want to get your money's
worth, you've got to take the
doorknob as far as you can and
put it near the end.
Okay, then, you figure that out.
Then you come along and you
say, "Okay, I've put the
doorknob at the right place,
open the door," then you're
applying enormous force this
way.
And again frustrated,
again you're getting no
results, you say,
"I'm going as far as I can from
the hinge,"
and it dawns on you that if you
really want to get something
going, you should really take
the line of separation between
where it's rotating and where
you're applying the force,
and apply a force perpendicular
to it.
Any part parallel to it is not
doing anything,
unless you kind of rip it off
the hinges.
You don't want to do that.
So, here this force has got a
useless part,
and got a useful part,
and the useless part,
which is trying to pull the rod
off the hinges,
is going to be balanced by a
force from the hinges,
because the rod is a rigid rod,
it's not going to move.
But the one perpendicular to
the rod can, in fact,
turn the rod without affecting
the rigidity.
So, we need an extra factor
here, which is sin
θ_i,
where θ_i is
defined as follows.
If you're at a point
r_i,
and you apply a force F
that way,
θ is the angle between
the line joining the point of
application of the force,
and the direction of the force.
You are here and you're
separated by this vector and as
the angle between that
separation vector and the force,
that angle θ can vary
from force to force.
But this force
F_i,
the angle could be
θ_i.
That's the final formula for
the torque, okay?
So, we have built up all the
pieces, and this is the final
answer.
This is the definition of
torque, and with this torque we
can now complete the dictionary
and say what we want.
On the left-hand side is the
torque, where the torque is
defined as
F_ir_i
sin θ_i.
F sin θ is just a
component of the force
perpendicular to the separation
r.
That's what F sin θ
does.
I hope you can see that.
If you've got a force acting
this way, F cos θ is
this part, which is no good,
and F sin θ is that
part, which is good for
rotations.
Also, if you want,
in terms of work done,
you want the force times the
distance traveled.
The distance traveled is only
in this direction.
There is no distance traveled
in this direction.
So, the distance traveled is
F dot product with the
direction in which you move,
and that applies only to the
force perpendicular to the
separation, not to the part
that's parallel.
Anyway, I hope we understand
intuitively why this is the guy
that's responsible for rotating
objects.
You want to rotate something
around some axis,
you've got a certain amount of
force,
you're best off applying the
force as far as you can from the
point of rotation.
In other words,
suppose we want the rear end of
this table here,
keeping this end fixed,
but I want to turn it so it's
this way.
Then I'm saying,
I think we all know,
we want to start pushing here.
Also, we don't want to apply a
force this way or that way,
we want to apply a force
perpendicular.
If, for some reason,
we are tied to a rope and can
only pull in that direction,
then the component of the force
this way is to be discounted,
no use, this is the part that
does the turning and that's the
part that also does the work.
Any force perpendicular to the
motion is not going to do any
work.
Any force perpendicular to the
displacement is not going to do
any work.
This rigid body can only be
later in that position,
so the displacement is the
tangential direction,
and only the force in the
tangential direction contributes
to the work.
And in my work energy argument
that's what comes here.
This is the reason why you come
up with this definition of
torque.
Okay, so we've got most of the
pieces for this.
I will write one more thing by
analogy.
The work done by a force that
displaces a body by an amount
dx is this.
The work done by a torque will
be τ times -- I think
you can guess what I want to put
there -- times dθ or
Δθ.
When a torque rotates a body by
an angle Δθ,
then the work done is τ
times Δθ.
You can see that even here you
can write the work done as
F times r times
Δθ;
but that is the torque,
that's the Δθ.
So, it's very useful to know.
So, if you guys are going to
reason by analogy,
I would ask you to know the
counterparts,
only then you can do the
problems.
So, let's get used to this
formula here,
ΔW = τΔθ.
I'll show you one example.
Suppose I have a rod hanging
from the ceiling.
Well, let's say it's massless
and there is a mass m at
the bottom and it's got some
length L;
it's a pendulum.
So, I'm going to take this
pendulum and I'm going to bring
it to this position,
by an angle
θ_0,
which unfortunately
θ_0 stands for
the final angle.
I want to know how much work I
do;
that's what we're going to
calculate.
So, we take some intermediate
position when we are here.
What is the force acting on
this guy?
This rod is massless;
the force on this is mg.
If that angle is θ,
you can see that angle is the
same θ,
so mg cos θ acts this
way, and mg sin θ acts
that way.
To hold the pendulum from
falling when the angle is
θ, I have to apply a
tangential force in this
direction of size mg sin
θ.
I hope you guys can see that.
The mg cos θ is going
to be provided by the rod.
Because the rod is a rigid
body, it won't let the mass move
in this direction.
It'll provide whatever force it
takes to keep the mass moving
radially.
In the tangential direction,
to keep it from sliding back to
where it began I have to apply a
force mg sin θ.
So, the torque that I apply is
mg sin θ times L.
So, I did all that by taking
the force, finding the component
of the force perpendicular to
the thing, and balancing it with
a countering force.
You can equally say the torque
that I have to apply is really
mg, which is the force,
R, which happens to be
L here,
and sine of the angle between
them.
No matter how we write it,
that's the torque.
Let's find the work done by
that torque when I rotate it
from the starting angle to some
final angle.
That'll be mgL sin θ
dθ from 0 to some final
angle θ_0.
I'm just saying,
the work done is τΔθ summed
over all the angles;
that means integrate.
So, you've got to go back to
your calculus thing and you've
got to remember that the
integral of sin θ is - cos
θ,
and you want to start from
θ = 0 to θ equal
to some final angle
θ_0.
And if you crank it out,
you will find it's mgL
times 1 - cos θ.
So, that's the work done to
take a pendulum which is
horizontal, which is vertical
and turn it to an angle
θ_0.
I'm going to do a cross-check
on this to tell you that it
works out from another point of
view.
This mass was down here,
now it's climbed up there.
So, you can ask,
"What happened to the work I
did?"
Going back to the Work Energy
Theorem, let's take this mass
here.
Forces were acting on it.
One was the rod,
but the force of the rod was
always perpendicular to the
motion of this bob here,
so it couldn't do any work.
The only work was done by me.
What happened to the work I did?
The work I did has to be
changing the total energy of the
body, but the bob was not moving
before,
and it's not moving after,
because I give just the right
force to cancel gravity;
I didn't give it any speed.
So, it must be that the
potential energy of this bob
when it's here should explain
the work I did.
That's exactly what it is,
because let me rewrite this as
[mg]
- mgL + mgL cos θ.
But maybe we can see that it is
mg + mgL.
This one, well,
I think it's best to write it
as L - L cos θ,
and you guys can see what's
going on.
L is this distance here,
L cos θ is that
distance, and L - L cos
θ is the increase in the
height of this bob.
This height here is in fact
L - L cos θ,
because if that is L and
that is θ,
this is L cos θ till
that point.
So, this is just mg
times h,
and that's where your work went
in.
So, what's the point of this
exercise?
This exercise is to get you
used to the notion of
calculating work done,
not as force times distance,
but as torque times angle.
And here's the problem where
the torque itself is changing
with the angle.
The torque is itself a function
of θ,
so you have to do an integral
and that's how you get the work.
Okay, so now,
we've got all the machinery we
need, got all the machinery we
need to do kinematics or
dynamics of rigid bodies in this
analogy.
So, let me summarize what it is.
You've got a rigid body;
it's got a moment of inertia;
it's got an angular velocity.
The product of the two is
called the angular momentum,
L, and to change it
you've got to apply a torque.
To find the torque,
find all the external forces
acting on the body,
multiply each force by the
distance from the point of
rotation,
but the component of each force
perpendicular to the line of
separation times the distance of
separation,
that's the Fr sin θ,
and add it up.
Again, when you do the torques,
you've got to give it a sign.
If a torque tends to rotate a
body counter-clockwise,
we will take it as positive,
and if the torque tends to
rotate a body clockwise,
we'll take it as negative.
So, it's possible that in a
given body, there could be
torques trying to move it in
different directions.
So, for example,
in the simplest rigid body made
up of, say, two masses,
you could be pushing this way
and I could be pushing that way.
We're all trying to rotate them
in opposite directions.
The torque you apply would be
your force times that distance;
this is called positive because
you're going counter-clockwise.
The torque that I'm applying
would be my force times the
smaller distance,
and that'll be considered
negative because it's clockwise.
And you'll add them up;
you'll get a net torque.
That net torque will cause an
acceleration of this
dumbbell-like object.
What acceleration will it cause?
It'll be my force times my
distance, sin θ happens to be
1, minus your force,
times the distance over which
you're doing it,
times the moment of inertia.
The moment of inertia for the
simple problem is
m_1r
_1^(2) +
m_2r2^(2),
if this is m_1
and this is
r_1,
and this is
m_2 and that
distance is
r_2.
So, this is the simplest
problem you could do in rigid
body dynamics.
Take a couple of forces,
work out the torques,
divide it by moment of inertia
and get the angle of
acceleration.
If the torque is a constant,
it's hard to maintain that.
What do I have to do to keep
the torque constant?
I've got to start running
around with this as it rotates,
and you could imagine doing
that;
then, the torque will be
constant, αa will be
constant, α constant is
like a constant,
and you can do the whole bunch
of problems you guys did before.
Okay, now there is one
technical obstacle you've got to
overcome if you want to do rigid
body dynamics,
and that is to know how to
compute the moment of inertia
for all kinds of objects.
If I give you 37 masses,
each with a distance r
from the point of rotation,
it's a trivial thing.
Do mr^(2) for each one
and add it.
But here is the kind of objects
you may be given,
not a discreet set of masses
but a continuous blob.
Just like in the case of center
of mass, when you had a bunch of
masses, you took
m_1x_1 +
m_2x_2 +
m_3x_3
and so on.
But if you had a rod which is
continuous, you have to do an
integral.
So you will have to do an
integral here also.
So, let's take certain rigid
bodies and try to find the
moment of inertia,
which are rigid bodies which
are continuous.
So, here is a ring of radius
r and mass M.
It's not a disk,
in spite of what it looks like.
It is a ring and I want to find
its moment of inertia.
So, you have to think about
what you will do.
So, you think for a second,
and I'll tell you the way I
will do that is to partition it
into tiny pieces,
each of which is small enough
for me to consider a point.
This tiny piece has a mass
Δm_i,
and i goes from 1,2,
3,4 to whatever,
10 million, till I go around
the circle.
The moment of inertia would
be--I use Δm_i
to tell you it's a tiny mass for
this section.
But what is the r^(2)?
If I'm planning to rotate it
around that point,
I plan to rotate around that
point;
it is the same r^(2) for
everybody.
So, you can pull it out of the
summation and you just get
Mr^(2).
So, this is a very easy problem.
This is as easy as this
problem, but you clearly had
only one mass and all of it was
at distance r from the
point of rotation,
it's Mr^(2).
Here, you got the mass spread
out, but luckily spread it out
in such a way that every part of
it is the same distance r
from the center,
so it's very easy to do the
summation.
Pull out the r^(2),
do the sum over mass and get
the total mass.
Then we're going to do--Any
questions about this?
Everything is going to be built
on understanding this.
So, for the ring,
this is the moment of inertia,
which is Mr^(2).
But now, I want to do the
following.
I want to take a disk now,
of radius r;
I've got to do the sum of
Mr^(2) for every mass in
the problem.
So, here is where you have to
organize your thinking.
I plan to do the moment of
inertia through the center.
First of all,
if I didn't tell you the moment
of inertia through where you
cannot even start;
I tell you I want it through
the center.
To want it through the center
we have the ability then to
think of this as made up of a
whole bunch of concentric rings,
each one at radius r and
of thickness dr.
If you can find the moment of
inertia of this tiny shaded
region, I sum over all the
little washers,
the annuli that I have,
to fill up the whole disk.
So, what's the moment of
inertia of this annulus?
Well, I already told you.
If the annulus has a mass
M, then it's
Mr^(2) is the moment of
inertia of that annulus.
So, the question is,
"What is the mass of the shaded
region?"
Clearly, the r^(2) we
use is the r at which it
is sitting.
But what's the mass I should
use?
Well, for the mass I should
use, I argue that if the total
mass is that [M]
that times [should have said
"over"]
πR^(2) is the mass per
unit area.
Then, I need the area of this
shaded region.
That's the tricky part.
Have you got that?
Yes?
Student: 2πrdr.
Professor Ramamurti
Shankar: Right,
so let's ask why it's
2πrdr.
If you take this annulus,
and you take a pair of scissors
and you cut it out,
and you open it out,
it's going to look like a long
rectangle whose thickness will
be dr;
its length will be 2πr.
So, 2πrdr is the--that
says contribution,
here is the dr.
Now, you've got to add it up
over everything.
Adding it up is what we do by
doing the integral.
Now, I don't want to do this
integral;
you guys can at least guess
what's going to happen.
There's going to be no
π in the answer;
π is gone.
You've got rr^(2) is
r^(3),
r^(4)/4 and that'll,
if you combine it with that,
you get this MR^(2)/2.
So, the moment of inertia of a
disk is MR^(2)/2.
That's because if I came and
told you moment of--suppose I
made a calculation with a
mistake somewhere,
and I said it's MR^(2),
will you know I'm wrong?
And why will you argue that it
cannot be the right answer?
Yes?
Student: Well,
it's like the ring but the
radii are smaller.
Professor Ramamurti
Shankar: Right.
In other words,
his answer, the correct answer
is if it is MR^(2) it
means the entire mass is at a
distance R^(2).
But we know some of the mass is
a lot closer to the center.
In fact, some of it is right at
the center, so you cannot get
the same R^(2) as a
contribution from all the
pieces.
Some will be at 0 distance;
some will be at the full
distance.
So, whenever you do the moment
of inertia for a disk,
it's got to look like
MR^(2) times a number
which is definitely less than 1.
It turns out it's half;
if we got a third or a fourth
you wouldn't know it's obviously
wrong.
But if I got 1 in front,
or worse, if I got 2 times
MR^(2),
then you know I've made a
mistake.
Okay, now, the last of the
objects I want to look at today,
is a rod, I want to take--This
is actually the example that's
done quite often as the first
example but I will do it now.
So, here is a rod, okay?
It has length L,
mass M,
and I say, what's the moment of
inertia?
Do you know what you could do?
What's the first question you
will ask when I say,
"What's the moment of inertia?
Student: Around what
point?
Professor Ramamurti
Shankar: Around what point,
right?
Till I tell you that,
you cannot begin.
So, I will say,
"Let's find the moment of
inertia around this end."
So, what you should imagine in
that case is--You stick a nail
through here and the guy rotates
like that on a big circle with
that as the center.
If you stuck the nail through
this midpoint,
it's a different story,
and the answer is different.
So here, what I do is I take a
distance x.
I take a sliver of thickness
dx.
That sliver is small enough for
me to consider there's a point
mass.
Therefore, the moment of
inertia of that tiny portion
will look like the square of the
distance times the mass of the
tiny portion.
The mass of the tiny portion
looks like this.
Let me see;
that's the mass per unit
length, that's the length,
that's the mass of the object.
Then I have to do the integral
from 0 to L.
Well, x^(2) will give me
x^(3)/3.
That'll give me
ML^(2)/3,
and that's the moment of
inertia of the rod.
Again, it looks like
ML^(2),
it's got to have a number in
front of it that's less than 1,
because if you're measuring
from here the furthest you can
be is L;
a lot of the guys had a lot
closer.
So, the weighted average of the
square of the distance cannot be
all L^(2);
it's got to be something less
than L^(2).
The fact that it's 1/3 is of
course what you do all the work
for.
Student: You need an
L too.
Professor Ramamurti
Shankar: Pardon me?
Student: An L too.
Professor Ramamurti
Shankar: You forgot the
L here?
Student: Yeah.
Professor Ramamurti
Shankar: Also,
the dimension for the moment of
inertia should be in the end
mass times length squared.
Yes?
You're not happy with that?
Student: No, I'm not.
Professor Ramamurti
Shankar: You agree,
okay.
Look, sometimes I do get these
things wrong,
so you should always try to
keep an eye on this.
The last thing I want to do is
to find the--Well,
when I say last thing I want to
do,
I don't mean I'm reluctant to
do it, I'm just saying I'm near
the end of the day.
The final thing that I will now
calculate is the moment of
inertia around the center.
Let's see what's going to
happen.
It's the same integral,
x^(2),
dx/L,
M/L, but the range of
integration for x will
now go from -L/2 to +
L/2.
Then, I simplify my life by
writing it as 2M/L times
x^(2)dx from 0 to
L/2.
This is something I encourage
you to do all the time.
If an integral is an even
function of x,
namely, -x contributes
as much as +x,
then you can take half the
reason of integration and double
the answer.
You can do it only if it's an
even function,
so if x and--yes?
Student: How come it's
still over L if you put
x in with the other one?
Professor Ramamurti
Shankar: Oh,
I'm sorry, you're absolutely
right.
The way to do that would be
here.
See, you keep trying,
you will catch me.
That's right, that's correct.
Okay, so let's do this now,
2M/L,
okay, now I'm really anxious,
because this L^(3),
it's the X^(3)/3,
but the thing that's being
cubed is L/2,
and that's giving me
ML^(2)/12.
So, the moment of inertia is
not a fixed number.
It got a lot smaller when I
took it around the center of
mass.
When I took it around the left
edge, I got ML^(2)/3;
I went to the midpoint,
I got ML^(2)/12.
I think it's pretty clear that
I'm going from here I reduce the
answer.
So, when I keep going,
it will get even smaller.
That's where your common sense
will tell you that if you got
some answer from there you're
going to get the same answer
from here and you hit the best
possible moment of inertia you
can.
You should be able to argue
intuitively, and if you do the
calculation, it'll come out to
be the same.
Now, the thing to check if you
do this, the moment of inertia
around one end is the moment of
inertia around the center less
the mass times (L/2)^(2).
This is something you can
actually verify.
Just take the numbers I gave
you and you'll find that this
answer plus M times
(L/2)^(2) will be equal
to that answer.
And this happens to be a very
general theorem I'll talk about
next time.
If you know the moment of
inertia through the center of
mass, you're done,
okay?
If I want the moment of inertia
through any other point,
I take the moment of inertia
through the center of mass and
add to it the total mass of the
object and the square of the
distance by which you made me
move my axis,
which here happens to be
L/2.
I will of course prove this;
I don't want to give you
anything without proof.
We can actually,
in our class actually,
satisfy ourselves that this is
true.
Not only is it true for a rod,
it's even true for a disk.
It's true for any object in two
dimensions, that if you've got
the moment of inertia of this
creature,
through the center of mass,
and you want it through that
point, just take the answer
through the center of mass and
add to it Md^(2) where
d is that distance.
That'll be proven next time.
