The moment of inertia, I, is the rotational equivalent of mass.

For a simple object like a ball on a string being whirled in a circle, where all the mass can be considered to be the same distance away from the axis of rotation, the moment of inertia is:

For a point mass: I = mr^{2}

For something more complicated, where mass is distributed at different distances from the rotation axis, the moment of inertia is determined by integrating:

I = ∫
r^{2} dm

How do we evaluate the moment of inertia integral:

I = ∫
r^{2} dm

for a uniform rod of length L rotating about an axis passing through one end of the rod, perpendicular to the rod?

Align the rod with the x axis so it extends from 0 to L. Split the rod into little pieces of size dx. The mass of each little piece is:

dm = λ dx, where λ is the mass per unit length of the rod.

λ = M/L, where M is the rod's total mass.

The integral becomes:

I = ∫
x^{2} λ
dx, integrated from 0 to L

I = [(1/3) λ
x^{3}] with upper limit L and lower limit 0

I = (1/3) λ
L^{3}

Substituting M = λL gives:

I = (1/3) ML^{2}

for a uniform rod rotating about one end.

We could carry out such integrals for all sorts of different shapes, although many of them are inetgrals over areas or volumes instead of over lengths. It's easier to look up the result in the table on page 227 in the book.

For a given rotation axis direction, the moment of inertia will always be minimized when the axis of rotation passes through the object's center-of-mass. The moment of inertia increases as the rotation axis is moved further from the center-of-mass.

For an object of mass M, the parallel-axis theorem states:

I = I_{com} + Mh^{2}

where h is the distance from the center-of-mass to the current axis of rotation, and I_{com} is the moment of inertia for the object rotating about the axis through the center of mass that is parallel to the current axis.

The parallel-axis theorem is usually used to calculate the moment of inertia about a second axis when I_{com} is known. Let's use it to go the other way, using the moment of inertia we just calculated for a rod rotating about one end.

What is the moment of inertia I_{com} for a uniform rod of length L and mass M rotating about an axis through the center, perpendicular to the rod?

I = I_{com} + Mh^{2}

I_{com} = I - Mh^{2}

We found that the moment of inertia when the rod rotates about a parallel axis passing through the end of the rod is:

I = (1/3) ML^{2}

The distance from the end of the rod to the center is h = L/2. Therefore:

I_{com} = (1/3) ML^{2} - M(L/2)^{2}

I_{com} = (1/3) ML^{2} - (1/4) ML^{2}

I_{com} = (1/12) ML^{2}