We know that the velocity of a particle in a rotating object is:

v = rw
Let’s take this to derive the kinetic energy of a rotating object:

K = 0.5mv^2 = 0.5m(r(omega))^2 = 0.5m(r^2)(omega^2)


The total energy is the sum of the energies of all the particles, right? And since w is the same for all particles, we have:

K[total] = 0.5(summation(mr^2))(omega^2)


It turns out that the value of the summation is the rotational equivalent of mass, and is referred to as rotational inertia or moment of inertia (I):

K[rotational] = 0.5I(omega^2)


If you are dealing with the moment of inertia for a point particle, I = mr2.

Now, for everything that rotates, there is a specific moment of inertia for that object and for that specific axis of rotation. Move the axis and the I changes. Change the shape and the I changes.

We will derive one example (which actually derives 2 common moments), show you the eight other common ones you will find, and show you how to use them in shapes derived from these ten common shapes.

But before we start the derivation, we should clarify how to find I. Remember how integrals were similar to summations? Well, since mr2 is different in different parts of a solid object, we will be integrating. We can split up an object into pieces of mass each with the same mass of Dm. Now, each piece of mass is probably at a different distance (r) from the axis of rotation. Remember how summations weren’t as accurate unless you make your rectangles really really small? Well, let’s make our masses infinitely small so that it becomes an integral of pieces of dm. Thus:

I = limit[(Delta)m goes to 0](summation(r^2(Delta)m)) = integral(r^2 dm)


Well, let’s put this integral in terms of volume, which is usually easier to work with. We are assuming that the object we are dealing with has a uniform density of r. Density is mass over volume, right? So instead, we will be concerning ourselves with the infinitely small pieces of volume (dV) that the infinitely small pieces of masses (dm) take up:

I = (rho)integral(r^2 dV)


Since density is constant, it can remain outside the integral.

Rotational Inertia of a Uniform Slender Rod

This is a sample derivation for finding the moment of inertia of a uniform slender rod. It will help set up a process you will use to derive other moments of inertia on your own. This derivation actually can be used to derive two situations, since they both fall under the same category as this one.

Rotational Inertia for Thin Rod About Axis Perpendicular to Length VRML ModelThe picture to the right shows the situation at hand (click on the illustration for a VRML model of the rod). We have a rod of uniform composition, same density everywhere, and it has a mass m and length l. We place the axis of rotation at O, a distance h from one of the ends. Simple enough, now we pick an element of volume of a short segment of length dx and cross-sectional area A, a distance xfrom O. That means:

dm = (rho)dV = (rho)Adx = (rho)Aldx/l = Mdx/l


Since we need to find the total rotational inertia of the entire rod, we need to integrate from x = –h to x = l – h:

Click to view Equation 5-11


Now I said you get two for the price of one? If the axis of rotation is at the end of the rod, h = 0, which simplifies the equation greatly. If the axis of rotation is at the center,h = l/2, which also simplifies the equation greatly. The moments of these and other shapes and axes are listed in the following section.

 

Common Moments of Inertia

Here is a table of the ten most common moments of inertia that are used:

Shape and Axis Illustration (Click for a VRML Model) Rotational Inertia
Hoop About Central Axis Rotational Inertia for Hoop About Central Axis VRML Model I = MR^2
Hoop About Any Diameter Rotational Inertia for Hoop About Any Diameter VRML Model I = 0.5MR^2
Annular Cylinder (Ring) About Central Axis Rotational Inertia for Annular Cylinder (Ring) About Central Axis VRML Model I = 0.5M(R1^2 + R2^2)
Solid Cylinder (Disk) About Central Axis Rotational Inertia for Solid Cylinder (Disk) About Central Axis VRML Model I = 0.5MR^2
Solid Cylinder (Disk) About Central Diameter Rotational Inertia for Solid Cylinder (Disk) About Central Diameter VRML Model I = (1/4)MR^2 + (1/12)ML^2
Thin Rod About Axis Through Center Perpendicular to Length Rotational Inertia for Thin Rod About Axis Through Center Perpendicular to Length VRML Model I = (1/12)ML^2
Thin Rod About Axis Through One End Perpendicular to Length Rotational Inertia for Thin Rod About Axis Through One End Perpendicular to Length VRML Model I = (1/3)ML^2
Solid Sphere About Any Diameter Rotational Inertia for Solid Sphere About Any Diameter VRML Model I = (2/5)MR^2
Thin Spherical Shell About Any Diameter Rotational Inertia for Thin Spherical Shell About Any Diameter VRML Model I = (2/3)MR^2
Slab About Perpendicular Axis Through Center Rotational Inertia for Slab About Perpendicular Axis Through Center VRML Model I = (1/12)M(a^2 + b^2)

Parallel-Axis Theorem

If you know the moment of inertia ICM of a body about an axis through its center of mass, then the moment of inertia about a new axis parallel to the first but displaced a distance h can be found through a relation called the parallel-axis theorem. The relation is stated as:

I = ICM + Mh2