# Calculating Rotational Inertia (A Bit of Confusion)

1. Dec 4, 2013

### Lemniscates

Hey all,

Perhaps this is a bit stupid...

I'm familiar with the normal procedure of calculating rotational inertia (using integration, parallel axis theorem, etc.). However, I had a confusing thought: if the center of mass of a body is the point at which you can treat as all of the mass being concentrated there and also all the external forces, then why can't I say something like this:

If a uniform rod of length L is swinging around a pivot, and the center of mass of the rod is a distance L/2 away from the pivot, then why can't we treat that center of mass as the place where all the mass is concentrated and use the rotational inertia formula for a particle (MR^2) and subsequently get M(L/2)^2? I know it's ML^2/3, but what is the problem here?

Last edited: Dec 4, 2013
2. Dec 4, 2013

### Zag

Hello Lemniscates,

You have to be careful when you use qualitative statements such as: "the center of mass of a body is the point at which you can treat as all of the mass being concentrated there". In fact, this is only true for translational motion! If you follow the mathematical derivation of this statement, you will notice that one of the underlying assumptions is that the motion under consideration is translational. Therefore, it makes no sense and it is mathematically incorrect to apply this statement to a system undergoing rotational motion.

Best,
Zag

3. Dec 4, 2013

### Lemniscates

Ah. That's interesting. But what about problems involving physical pendulum calculations? From what I've read and have been taught, when you calculate the torque induced by gravity in a physical pendulum, you use the distance from the center of mass to the pivot as the length of the moment arm, essentially saying that all of the gravity acts at the center of mass (and is that not a use of the center of mass/external forces statement?). So how is that justified if the statement about centers of mass can only be applied to bodies in translational motion?

Last edited: Dec 4, 2013
4. Dec 4, 2013

### SteamKing

Staff Emeritus
It's like swinging a sledgehammer: if you grasp the end of the handle and swing the hammer, it takes a certain amount of effort to swing the head. However, if you turn the hammer around and swing the handle by grasping the head, it takes less effort than the first instance. The c.g. of the hammer has not changed, but its location from the axis of rotation has. Since you are not swinging the heavy head in as large a radius of arc in the second instance as in the first instance, it takes less effort.

5. Dec 5, 2013

### cjl

In effect, you somewhat can. You are correctly accounting for the moment of inertia due to the rotation of the center of mass around the pivot, but in that motion, the rod is also rotating about its center of mass. Your formulation would be correct if the rod were somehow maintaining its orientation while revolving around the pivot (I'm sure such an apparatus could be constructed). However, since the rod is rotating, you can do the following.

You can consider your situation as the superposition of two situations:
1) The rotation of the center of mass of the rod around the pivot (which, as you surmised, gives ML2/4)
2) The rotation of the rod about its own center of mass (which gives a moment of inertia of ML2/12)

If you sum these two factors, you get that the total effective moment of inertia is ML2/4 + ML2/12 = 4ML2/12 = ML2/3. This is a useful way to find the overall moment of inertia of a system, since all you need to know is the moment of inertia of the individual parts of the system about their own centers of mass, and the distance each center of mass is from the pivot of the overall system. This is known as the Parallel Axis Theorem