7.5 Inertia Intro: Parallel Axis Theorem

There are two great uses for the parallel axis theorem:

1. Finding the inertia of a complex object with multiple parts.

2. Rotating an object about an axis other than through the center of mass (y’)

To begin with, the parallel axis theorem is equal to the inertia about the center of mass (I_cm) plus the distance between the axes of rotation squared times the mass.

$$I=I_{cm}+md^2$$

Example 1:

For a disk, the distance between axes y and y’ is d and the [latex]I_{cm} = \frac{1}{2}mr^2[/latex].

$$I=I_{cm}+md^2 = \frac{1}{2}mr^2+mr^2 = \frac{3}{2}mr^2$$

You will need the table of common geometric shapes in the previous section to find the I_cm for each object.

Steps for finding the MMOI of an object

1. Determine the shape of the object (or shapes, if it a composite object).
2. Determine which axis the object is rotating about.
3. Find the center of mass for each individual shape.
4. Find the I_CM  (inertia about its center of mass) for each shape.
5. Determine the distance from the CM of the shapes to the axis of rotation.
6. Use the Parallel Axis Theorem to find the inertia for each shape.
7. Add up all these individuals inertias to find I_T.

Example 2:

Find the moment of inertia of a uniform rod with I_CM=0.05kgm², L=0.08m, and mass=0.250kg, with respect to an axis that is perpendicular to the rod and passes through at 1/4 of the length of the rod.

We know the distance (d) to be L/4 = 0.08m / 4 = 0.02m away from the z axis. Here we present the solution to the problem:

I=I_CM + md²

I=0.05 kgm² + (0.250kg)(0.02m)²

I=0.0501 kgm²

Example 3

A dumbbell consists of two .2 meter diameter spheres, each with a mass of 40 kg, attached to the ends of a .6 meter long, 20 kg slender rod. Determine the mass moment of inertia of the dumbbell about the y axis shown in the diagram.

Organize the known and unknown data in a table to complete as you go:

Find the center of mass:

• For each sphere, the r_cm is 0.3m + 1/2 radius = 0.3m + 1/2 (0.2m) = 0.4m
• For the bar, the r_cm is at 0.

Find the inertia about the center of mass for each shape separately. Use the rectangle equation:

For the sphere:

$$ I_{cm-sph}=\frac{2}{5}mr^2 =\frac{2}{5}*(40kg)*(0.1m)^2 \\ \qquad \quad =0.16kgm^2$$

For the rod,

$$I_{cm-rod}=\frac{1}{12}ml^2 = \frac{1}{12}*(20kg)*(0.6m)^2\\\qquad \quad = 0.6 kgm^2$$

Next the parallel axis theorum is needed to change the axis of rotation from the cm of the sphere to the system cm. The distance between axes of rotation is 0.3m + 1/2 radius = 0.3m + 1/2 (0.2m) = 0.4m

$$I_{o-sph}=I_{cm-sph}+md^2 = 0.16 kgm^2 + (40kg)*(0.4m)^2 \\\qquad \quad = 6.56kgm^2$$

Finally, add up the parts: the 2 spheres and the cm of the rod:

$$I_{total} = 2 * I_{o-sph}+I_{cm-rod}\\\qquad \quad = 2*(6.56 kgm^2) + (0.6 kgm^2)\\\qquad \underline{I_{total} = 13.72 kgm^2}$$

Source: mechanicsmap.psu.edu/websites/A2_moment_intergrals/parallel_axis_theorem/parallelaxistheorem.html