### Video Transcript

In this video, we’re going to learn
about moment of inertia. We’ll learn what it is, why it
matters, and how to calculate it. To get started, imagine you’re
holding a race called “The Shapes Race.” At the top of an inclined plane,
you’ve placed a hollow ring, a solid ball, a long solid cylinder, and a hollow
spherical shell. On cue, all four of these shapes
are released at the same time and allowed to start rolling down the incline. Certain of these shapes roll
downhill more readily than others. And they pull ahead in the
race.

If you wanted to predict ahead of
time which of the shapes will win the race, it will be helpful to know something
about moment of inertia. We can begin to understand the
notion of moment of inertia by realizing that certain linear variables have
rotational analogs, and vice versa. For example, consider the linear
variable 𝑠, which stands for distance.

In the rotational world, the
variable that corresponds to 𝑠 we can call 𝜃, for an angular distance. These two variables correspond to
one another. They both describe a distance
traveled. It might be surprising to learn
that mass is also a linear variable. Here’s what we mean by that. Whenever we draw the free body
diagram of forces acting on a mass, we know that those forces move through the
center of the mass, that is, its center of mass.

Forces that act this way tend to
translate rather than rotate our mass 𝑚. So the mass in response to forces
such as these or any forces we might draw in using a free body diagram tends to move
in a line. But we’re now going to imagine a
different scenario. Instead of our forces always
originating or passing through the center of mass of our object, we’ll now consider
scenarios where the forces act along some different axis than the line through the
center of mass. Forces such as these tend to make
our mass rotate. And depending on the shape and
distribution of our mass, that rotation will be more or less difficult.

This is the notion behind moment of
inertia, often symbolized with a capital 𝐼. Instead of our forces moving
through the center of mass of an object, which will lead to its linear translation,
our forces now act outside of that center. Or even if the forces do move
through the mass’s center, we might be considering the motion of the mass relative
to an axis of rotation some distance away from its center.

We can see that moment of inertia
always has to deal with rotation. And it’s in that sense that the
moment of inertia is like a rotational mass. Say that we have a randomly shaped
mass, subject to a force that moves through the center of mass of the object. And we’ll say that this force
causes the mass to move in a circular path around an axis of rotation. Given that we have a force on a
mass that is moving in a circular path, by Newton’s second law of motion, we can
write a linear expression of this as 𝐹 equals 𝑚 times 𝑎.

Interestingly, there’s a rotational
version of Newton’s second law. And we can find that version by
considering that our mass is at a distance 𝑟 from the center of rotation. And if the vector 𝐹 and the vector
𝑟 are perpendicular to one another, we can write that 𝐹 times 𝑟 is equal to the
moment of inertia, rotational mass remember, multiplied by angular acceleration
𝛼. So not only do individual variables
have linear and rotational pairs, but so does Newton’s second law. There’s a linear and rotational
version. It’s this rotational expression of
Newton’s second law of motion that gives us a little bit of a better feel for just
what moment of inertia 𝐼 is.

As we consider this connection
between mass and moment of inertia, the question may come up: measuring mass is
straightforward enough. As long as we know the acceleration
due to gravity, we can put a body on a scale and solve for its mass that way. But what about 𝐼? What about an object’s moment of
inertia?

It turns out that an object’s
moment of inertia 𝐼 can be measured by using a simple pendulum. If we attach the shape whose moment
of inertia we want to measure to the arm of a pendulum and let it swing freely under
the influence of gravity, we can see that the back and forth motion the pendulum
makes is a rotation. We can explore just how easy or
difficult it is for the mass to make this rotation by varying the length of the
pendulum’s arm as well as the amplitude from which it’s released.

For more common shapes, an example
of which is the hollow cylinder we have here, the moment of inertia of those shapes
about particular axes has already been calculated. And we can look them up in a
table. As examples of various shapes and
their corresponding moments of inertia, the moment of inertia of a point mass is
equal to the mass of that point multiplied by the square of its distance from the
axis of rotation. A solid sphere, on the other hand,
rotating about its center, has a moment of inertia of two-fifths its mass times its
radius squared. And a solid rod rotating about one
end of the rod has a moment of inertia of one-third its mass times the length of the
rod squared.

This brings up an important point
about moment of inertia. We’ve been talking about how moment
of inertia 𝐼 and mass are similar to one another. An object’s moment of inertia 𝐼
depends not only on the object’s shape and density, like mass does, but 𝐼 also
depends on the axis of rotation. For example, if we measure the
moment of inertia of our hollow cylinder in this case, then that measured value
would depend not just on the mass of our cylinder, but also on its position or its
orientation.

If we move the cylinder so that
now, instead of its center, one of its ends is attached to the pendulum arm or if we
rotated it so it was in line with the pendulum arm, in each of these three cases, we
will probably measure a different moment of inertia, same shape different
moment. So when we do go to a table to look
up the moment of inertia of a particular shape, we’ll want to be careful to ensure
that the axis of rotation of our shape matches our scenario. Otherwise, we’ll write down an
incorrect or unmatching moment of inertia.

In addition to looking up an
object’s moment of inertia, it’s also possible to calculate it. To consider how we might do this,
let’s consider again our moment of inertia of a point mass, that is, an infinitely
small mass with mass value 𝑚 a distance 𝑟 away from an axis of rotation. Under these conditions, that point
mass’s moment of inertia has been determined to be its mass times 𝑟 squared.

Now let’s imagine that we envelop
or surround our point mass with a much larger extended mass. The point mass 𝑚 is an
infinitesimally small component of this larger mass. And we could even say that this
larger mass is simply composed of infinitely many infinitesimally small point
masses. If, for each infinitesimally small
mass element that makes up this larger mass, we calculate that mass element times
the square of its distance from the axis of rotation, then adding up all those parts
through integration, we would solve for the overall moment of inertia of this
extended mass with random shape.

Typically, when calculating the
integral to solve for an object’s moment of inertia, we express the differential
mass element 𝑑𝑚 in terms of 𝑑𝑟 in some way, depending on the shape of the mass
we want to solve for. Beyond calculating an object’s
moment of inertia from scratch, there’s one more moment of inertia tool we’ll want
to learn before getting some practice with these ideas.

This tool is known as the parallel
axis theorem. Here’s one way to think about this
theorem. Say that we have a solid sphere
that’s rotating about an axis that runs through its center. We saw earlier that the moment of
inertia of this shape rotating about its center of mass is two-fifths its mass times
its radius squared. And we noted that this result
depends on the position of the axis of rotation. If that axis were to move, then so
would the overall moment of inertia of this mass.

Let’s say, for instance, that we
move our axis of rotation from the center of the sphere to some other point. And we’ll say that that point is
separated from the center of the sphere by a distance 𝑑. Here’s what the parallel axis
theorem tells us about this situation. This theorem says that if we move
our rotation axis to another position but keep the axis parallel to its original
orientation. Then the new moment of inertia of a
rotating object is equal to the sum of the moment of inertia of the object rotating
about its center of mass plus its mass times the distance squared between the
original and final axes of rotation.

That’s a mouthful. But what it means is that we can
take advantage of the moment of inertia of an object rotating about its center of
mass when solving for its moment of inertia rotating about some other axis. This tool is very helpful in
solving for moment of inertia when our rotation axis shifts. And it’s important to know that, to
use it, our two axes must be parallel to one another. Knowing all this, let’s get some
practice with moment of inertia through an example.

A rod and a sphere are combined to
form a system. The rod’s length 𝐿 is 0.50
meters. And its mass is 2.0 kilograms. The sphere’s radius 𝑅 is 20.0
centimeters. And its mass is 1.0 kilograms. The system can either rotate about
the point 𝐴 at the opposite end of the rod to the sphere or about the point 𝐵
where the rod and the sphere connect, as shown in the diagram. Find the moment of inertia of the
system about the point 𝐴. Find the moment of inertia of the
system about the point 𝐵.

We can call these two values 𝐼 sub
𝐴 and 𝐼 sub 𝐵. Considering the information we’ve
been given that we’re told the length of the rod, its mass, the radius of the
sphere, and its mass, we can begin solving for 𝐼 sub 𝐴 by writing out this moment
of inertia as the sum of the components of this system of the rod and the
sphere. 𝐼 sub 𝐴 is equal to the moment of
inertia of the rod rotating about point 𝐴 plus the moment of inertia of the sphere
rotating about the same point.

When we consider looking up the
moments of inertia of the two parts of our system, we know we’ll be able to find the
moment of inertia of a rod rotating about its end as we have in this case. But for the moment of inertia of a
sphere, we’ll be able to find that value for a sphere rotating about its center. But in this case, the sphere does
not rotate about its center, but about the point 𝐴.

To help us out, we can recall the
parallel axis theorem. This theorem says that if we have a
mass 𝑚 with a rotation axis shifted a distance 𝑑 from the center of mass of 𝑚,
then the overall moment of inertia of this object is equal to the moment of inertia
of the object about its center of mass plus its mass times the distance 𝑑 squared
between the two parallel axes. This means that when it comes to
the moment of inertia of the sphere rotating about point 𝐴, that moment of inertia
is equal to the moment of inertia of the sphere about its center of mass plus its
mass times the distance from the axis of rotation, in our case 𝐿 plus 𝑅,
squared.

If we then go and look up in a
table the moment of inertia of a rod rotating about one of its ends, we see that
it’s one-third the mass of the rod times the length of the rod squared. Moreover, when we look up the
moment of inertia of a sphere rotating about its center, we see that value is equal
to two-fifths the sphere’s mass times its radius squared. All this means that we can rewrite
the moment of inertia of our system rotating about the point 𝐴 as one-third the
mass of the rod times its length squared plus two-fifths the mass of the sphere
times its radius squared plus, because of the parallel axis theorem, the mass of the
sphere multiplied by the length of the rod plus the radius of the sphere quantity
squared.

Since we’re given the values for
all four of these variables in our problem statement, we’re ready to plug in and
solve for 𝐼 sub 𝐴. When we plug in for all these
values, we’re careful to convert the radius of our sphere from units of centimeters
to units of meters to agree with the units in the rest of our expression. Adding these three terms together,
we find a result, to two significant figures, of 0.67 kilograms meter squared. That’s the moment of inertia of
this system rotating about point 𝐴.

Next, we want to consider the same
system, but a different rotation axis. Now our rotation axis is at the
place where the rod and the sphere join. Once again, the moment of inertia
of our system is equal to that of the rod plus that of the sphere for this
particular rotation axis. Since the rod is, once again,
rotating about one of its ends, we can again use the relationship for its moment of
inertia about such an axis. Likewise, the sphere is not
rotating about its center of mass, whose moment of inertia we know, but is rotating
about an axis a distance of its radius 𝑅 away from that center.

Our equation for 𝐼 sub 𝐵 is the
same as our equation for 𝐼 sub 𝐴 was, except for one single term. Instead of our distance 𝑑 being 𝐿
plus 𝑅, now it’s simply 𝑅, the radius of our sphere. When we plug in for these values,
again converting the radius of our sphere to units of meters, we find that 𝐼 sub
𝐵, to two significant figures, is 0.22 kilograms meter squared. That’s the moment of inertia of
this system rotating about an axis through point 𝐵.

Let’s summarize what we’ve learned
about moment of inertia. We’ve seen that an object’s moment
of inertia is equivalent to its rotational mass. Moment of inertia, symbolized by
capital 𝐼, depends on an object’s mass, shape, and the axis that the object rotates
around. Common moments of inertia can be
looked up in a table. Or they can be calculated using the
relationship the moment of inertia is equal to the integral of each infinitesimal
mass element that makes up a larger mass multiplied by each element’s distance from
the axis of rotation squared.