### Video Transcript

A rod and a sphere are combined to
form a system. The rodโs length ๐ฟ is 0.50
meters. And its mass is 2.0 kilograms. The sphereโs radius ๐
is 20.0
centimeters. And its mass is 1.0 kilograms. The system can either rotate about
the point ๐ด at the opposite end of the rod to the sphere or about the point ๐ต
where the rod and the sphere connect, as shown in the diagram. Find the moment of inertia of the
system about the point ๐ด. Find the moment of inertia of the
system about the point ๐ต.

We can call these two values ๐ผ sub
๐ด and ๐ผ sub ๐ต. Considering the information weโve
been given that weโre told the length of the rod, its mass, the radius of the
sphere, and its mass, we can begin solving for ๐ผ sub ๐ด by writing out this moment
of inertia as the sum of the components of this system of the rod and the
sphere. ๐ผ sub ๐ด is equal to the moment of
inertia of the rod rotating about point ๐ด 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 weโll 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, weโll 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 ๐ด.

To help us out, we can recall the
parallel axis theorem. This theorem says that if we have a
mass ๐ with a rotation axis shifted a distance ๐ from the center of mass of ๐,
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 ๐ squared
between the two parallel axes. This means that when it comes to
the moment of inertia of the sphere rotating about point ๐ด, 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 ๐ฟ plus ๐
,
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
itโs 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 sphereโs mass times its radius squared. All this means that we can rewrite
the moment of inertia of our system rotating about the point ๐ด 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 weโre given the values for
all four of these variables in our problem statement, weโre ready to plug in and
solve for ๐ผ sub ๐ด. When we plug in for all these
values, weโre 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. Thatโs the moment of inertia of
this system rotating about point ๐ด.

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 ๐
away from that center.

Our equation for ๐ผ sub ๐ต is the
same as our equation for ๐ผ sub ๐ด was, except for one single term. Instead of our distance ๐ being ๐ฟ
plus ๐
, now itโs simply ๐
, 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 ๐ผ sub
๐ต, to two significant figures, is 0.22 kilograms meter squared. Thatโs the moment of inertia of
this system rotating about an axis through point ๐ต.