I would now like to show
you how to calculate
the moment of inertia of
a typical continuous body.
Let's consider a
rigid rod, very thin.
And what we want
to do is calculate
the moment of inertia of this
body about the center of mass.
Let's say the body is of length
L, and it has total mass M.
Now, recall that the
moment of inertia
about the center
of mass we defined
as an integral of dm r-squared
integrated over the body.
Now, our challenge
today is to understand
exactly what all these terms
are in this expression.
What is dm?
What is r?
And what do we mean by an
integral over the body?
So now let's do a stepwise
interpretation of each term.
The crucial thing is to
introduce coordinates systems.
So let's choose a
coordinate system.
And let's put the origin
at the center of mass.
Now, the most important
thing is that we're
going to do an integral.
So we need to introduce
the integration variable.
And that's the hardest part
of setting up intervals.
So what we want to do
is arbitrarily choose
an element dm.
So there's our elements dm.
And here's our
integration variable,
it's a distance x
from the origin.
So that's the
integration variable.
Now, there's two things
when we set up the integral.
r is equal to that
integration variable.
r is abstract in
this expression,
but in this concrete realization
it is the integration variable.
The second place the integration
variables shows up is in dm.
dm is a mass in
this small element.
But if we want to express that
in terms of our integration
variable, we have to express
it in terms of the differential
length dx.
So dm mass is equal to the
total mass per unit length.
We're assuming the
rod is uniform times
the length dx of
our small piece.
And now we've set up the
two pieces that are crucial
and all we have
to think about now
is what does an integral mean.
Well, an integral means that
we're dividing up the piece
into a bunch of small
elements and we're
adding the contribution
of each small element.
So in particular, when we
write Icm equals-- now,
we can write it as m over Ldx.
That was our dm.
And the distance of dm from the
point we're computing the axis
is x-squared.
Now, the question is what is
our integration variable doing?
Well, x is going from minus L
over 2-- that's at this end--
to x equals plus L over
2 on the other end.
So we have x minus L over
2x equals plus L over 2.
And now we've set
up the integral
for the moment of
inertia, and the rest
is just doing an integral.
Recall that the integral of dx
x-squared is x-cubed over 3.
And so this integral
then simply becomes
Icm equals m over L
x-cubed over 3-- evaluating
from the limits minus L over
2 to x equals plus L over 2.
Again when you evaluate
the limits, what
we get is m over L we have to
put in L over 2 cubed divided
by 3 minus L over 2
cubed divided by 3,
and that's 1 over 2 cubed
is an eighth-- divide
by third, that's the 24.
24 minus 24 is a 12.
And so what we get
for Icm is m over L,
12 L-cubed, or
1/12 m L-squared is
the moment of inertia
about the center of mass
of our rigid rod.
And this is a measure
of how the mass is
distributed about this axis.
