# Question Video: Comparing the Moments of Inertia around Different Rotation Axes of an Asymmetrical Object

A rod and a sphere are combined to form a system. The rod’s length 𝐿 is 0.50 m and its mass is 2.0 km. The sphere’s radius 𝑅 is 20.0 cm and its mass is 1.0 kg. 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.

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### Video Transcript

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.