12.4.1 The Parallel-Axis Theorem

The moment of inertia depends on the rotation axis. Suppose you need to know the moment of inertia for rotation about the off-center axis in FIGURE 12.15. You can find this quite easily if you know the moment of inertia for rotation around a parallel axis through the center of mass.

The figure illustrates a thick oval with an off-center axis.

FIGURE 12.15 Full Alternative Text

If the axis of interest is distance d from a parallel axis through the center of mass, the moment of inertia is

I = I_{cm} + M d^{2}

(12.18)

Equation 12.18 is called the parallel-axis theorem. We’ll give a proof for the one-dimensional object shown in FIGURE 12.16.

The figure illustrates the method to prove the parallel axis theorem.

FIGURE 12.16 Full Alternative Text

The x-axis has its origin at the rotation axis, and the $x ′ -axis$ has its origin at the center of mass. You can see that the coordinates of $d m$ along these two axes are related by $x = x' + d$ . By definition, the moment of inertia about the rotation axis is

I = \int x^{2} d m = \int {(x' + d)}^{2} d m = \int {(x')}^{2} d m + 2 d \int x' d m + d^{2} \int d m

(12.19)

The first of the three integrals on the right, by definition, is the moment of inertia $I_{cm}$ about the center of mass. The third is simply $M d^{2}$ because adding up (integrating) all the $d m$ gives the total mass M.

If you refer back to Equations 12.5, the definition of the center of mass, you’ll see that the middle integral on the right is equal to $M x^{'}_{cm}$ . But $x^{'}_{cm} = 0$ because we specifically chose the $x'$ -axis to have its origin at the center of mass. Thus the second integral is zero and we end up with Equation 12.18. The proof in two dimensions is similar.

EXAMPLE 12.7 The moment of inertia of a thin rod

Find the moment of inertia of a thin rod with mass M and length L about an axis one-third of the length from one end.

SOLVE From Table 12.2 we know the moment of inertia about the center of mass is $\frac{1}{12} M L^{2}$ . The center of mass is at the center of the rod. An axis $\frac{1}{3} L$ from one end is $d = \frac{1}{6} L$ from the center of mass. Using the parallel-axis theorem, we have

I = I_{cm} + M d^{2} = \frac{1}{12} M L^{2} + M {(\frac{1}{6} L)}^{2} = \frac{1}{9} M L^{2}

12.4.1 The Parallel-Axis Theorem

FIGURE 12.15 An off-center axis.

FIGURE 12.16 Proving the parallel-axis theorem.

EXAMPLE 12.7 The moment of inertia of a thin rod