The Parallel-Axis Theorem: How Moment of Inertia Changes with the Axis

The Parallel-Axis Theorem: How Moment of Inertia Changes with the Axis

The parallel-axis theorem tells you the moment of inertia about a new axis without starting the integral again. The new axis must be parallel to an axis through the center of mass. As a result, a shift in the axis always adds a simple extra term.

Moment of inertia measures how mass resists angular acceleration. Mass farther from the axis counts more. Textbooks usually list $I$I about a center-of-mass axis. Many problems use a different axis, such as a hinge, a pivot at the end of a rod, or the contact point of a rolling object. The theorem connects those two results.

If $I_{\mathrm{cm}}$Icm​ is the moment of inertia about a center-of-mass axis, the moment about a parallel axis is

$$I = I_{\mathrm{cm}} + Md^{2}$$I=Icm​+Md2

Here $M$M is the total mass and $d$d is the distance between the two axes. The added term is always positive. Therefore, $I$I about any parallel axis is larger than $I_{\mathrm{cm}}$Icm​. The center-of-mass axis gives the smallest moment of inertia for that direction.

A thin rod of length $L$L has $I_{\mathrm{cm}} = \frac{1}{12}ML^{2}$Icm​=121​ML2 for an axis perpendicular to its length. An axis through one end is parallel and sits a distance $L/2$L/2 away. The theorem gives

$$I_{\mathrm{end}} = \frac{1}{12}ML^{2} + M\left(\frac{L}{2}\right)^{2} = \frac{1}{3}ML^{2}$$Iend​=121​ML2+M(2L​)2=31​ML2

That matches the standard end-pivot result. A disk or sphere works the same way. You take the known center-of-mass value, measure $d$d, and add $Md^{2}$Md2.

Two limits matter. The axes must be parallel. A tilted axis needs a different treatment. Also, $I_{\mathrm{cm}}$Icm​ must refer to the matching direction. A rod’s formula for a perpendicular axis cannot be reused for an axis along the rod’s length.

The theorem is useful because rotation equations use $I$I about the actual axis. Torque equals $I\alpha$Iα only if $I$I matches that axis. Energy and angular momentum need the same match. Therefore, a correct $d$d often matters as much as the force.

Disclaimer: This article explains a standard mechanics result for study only. It is not a substitute for a course derivation or a graded solution. Symbols and sign conventions should follow the textbook in use.