KAHIBARO
Table of Contents
Shifting the Axis of Rotation
The moment of inertia depends not only on how much mass an object has, but also on which axis it rotates about. If the axis changes, the moment of inertia usually changes too. The parallel-axis theorem gives a simple way to find the moment of inertia about a new axis when that new axis is parallel to an axis through the center of mass.
This theorem is especially useful because moments of inertia are often known for axes through the center of mass, and many real problems involve axes located somewhere else.
The Main Idea
Suppose an object has moment of inertia $I_{\mathrm{cm}}$ about an axis that passes through its center of mass. Now consider another axis that is parallel to the first one and is a distance $d$ away. The moment of inertia about the new axis is
$$
I = I_{\mathrm{cm}} + Md^2
$$
where $M$ is the total mass of the object.
This means that moving the axis away from the center of mass always increases the moment of inertia.
For two parallel axes separated by distance $d$,
$$
I = I_{\mathrm{cm}} + Md^2
$$
Here, $I_{\mathrm{cm}}$ is the moment of inertia about the axis through the center of mass, $M$ is the total mass, and $d$ is the perpendicular distance between the axes.
Why the Moment of Inertia Increases
Moment of inertia measures how far the mass is distributed from the axis of rotation. When the axis is shifted away from the center of mass, the particles of the object are, on average, farther from the new axis. Because moment of inertia involves the square of distance, even a modest shift can produce a noticeable increase.
The center of mass axis gives the smallest moment of inertia among all axes parallel to it.
Among all axes parallel to each other, the axis through the center of mass has the smallest moment of inertia.
Simple Geometric Picture
Imagine a thin rod. If it rotates about its center, the mass is spread on both sides of the axis in a balanced way. If it rotates about one end, almost all of the mass is farther from the axis, so the moment of inertia is larger.
A Short Derivation
Take the center of mass axis as the reference axis. Let a small mass element $m_i$ be at distance $r_i$ from that axis. Now shift to a parallel axis a distance $d$ away. The distance of the same mass element from the new axis becomes
$$
r_i'
$$
and when the geometry is worked out, the total moment of inertia becomes
$$
I = \sum m_i {r_i'}^2
$$
Expanding this result gives three parts. One part is the original center of mass moment of inertia, one part is proportional to $Md^2$, and one mixed term appears. That mixed term becomes zero because the axis through the center of mass is used. So the final result is
$$
I = I_{\mathrm{cm}} + Md^2
$$
The important point is that the theorem works because the original axis passes through the center of mass.
The parallel-axis theorem must start from an axis through the center of mass. If the known axis is not a center of mass axis, the formula cannot be used directly in this form.
Common Example, Rod About One End
For a uniform thin rod of length $L$, the moment of inertia about its center is
$$
I_{\mathrm{cm}} = \frac{1}{12}ML^2
$$
The axis through one end is parallel to the center axis and is a distance
$$
d = \frac{L}{2}
$$
away. Applying the theorem,
$$
I = I_{\mathrm{cm}} + Md^2
= \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2
$$
$$
I = \frac{1}{12}ML^2 + \frac{1}{4}ML^2
= \frac{1}{3}ML^2
$$
So the moment of inertia of a uniform rod about one end is
$$
I = \frac{1}{3}ML^2
$$
Common Example, Disk About a Tangent Axis
For a solid disk of radius $R$, the moment of inertia about an axis through its center and perpendicular to the disk is
$$
I_{\mathrm{cm}} = \frac{1}{2}MR^2
$$
A tangent axis perpendicular to the disk is parallel to this center axis and is a distance
$$
d = R
$$
from it. Then
$$
I = I_{\mathrm{cm}} + MR^2
= \frac{1}{2}MR^2 + MR^2
= \frac{3}{2}MR^2
$$
Summary Table
| Object | Known center of mass axis | $I_{\mathrm{cm}}$ | Shift distance $d$ | New moment of inertia |
|---|---|---|---|---|
| Thin rod, length $L$ | Through center, perpendicular to rod | $\frac{1}{12}ML^2$ | $\frac{L}{2}$ | $\frac{1}{3}ML^2$ |
| Solid disk, radius $R$ | Through center, perpendicular to disk | $\frac{1}{2}MR^2$ | $R$ | $\frac{3}{2}MR^2$ |
| Hoop, radius $R$ | Through center, perpendicular to hoop | $MR^2$ | $R$ | $2MR^2$ |
When to Use It
Use the parallel-axis theorem when you know the moment of inertia about a center of mass axis and need the moment of inertia about another axis that is parallel to it.
Do not use it if the axes are not parallel. Do not use it if the distance $d$ is measured incorrectly. The distance must be the perpendicular separation between the two axes.
Checklist for using the theorem correctly:
$$
I = I_{\mathrm{cm}} + Md^2
$$
Only use it when the new axis is parallel to the center of mass axis, and $d$ is the perpendicular distance between the axes.
Physical Interpretation
The term $I_{\mathrm{cm}}$ describes the spread of mass around the center of mass axis. The extra term $Md^2$ appears because shifting the whole object by distance $d$ makes all of its mass effectively farther from the new axis.
This added term depends only on total mass and shift distance. It does not depend on the shape of the object. The shape information is already contained in $I_{\mathrm{cm}}$.
Final Insight
The parallel-axis theorem is a shortcut that saves a full integration. Instead of recalculating the moment of inertia from the beginning for every new parallel axis, you start from the center of mass result and add $Md^2$.
That simple correction makes it one of the most useful tools in rotational motion.