The parallel axis theorem is used to find the moment of inertia about a new axis that is parallel to a known
center-of-mass axis. It is especially useful when a rigid body rotates about a pivot or hinge that does not pass
through its center of mass.
1. The theorem
If \(I_{\mathrm{cm}}\) is the moment of inertia about an axis through the center of mass, then the moment of inertia
about a new parallel axis is
\[
I_{\mathrm{new}}=I_{\mathrm{cm}}+Md^2.
\]
Here, \(M\) is the total mass of the object and \(d\) is the perpendicular distance between the two parallel axes.
2. Meaning of each term
| Symbol |
Meaning |
SI unit |
| \(I_{\mathrm{cm}}\) |
Moment of inertia about the center-of-mass axis |
\(\mathrm{kg\,m^2}\) |
| \(M\) |
Total mass of the rigid body |
\(\mathrm{kg}\) |
| \(d\) |
Perpendicular distance between the two parallel axes |
\(\mathrm{m}\) |
| \(Md^2\) |
Extra rotational inertia caused by shifting the axis |
\(\mathrm{kg\,m^2}\) |
| \(I_{\mathrm{new}}\) |
Moment of inertia about the shifted parallel axis |
\(\mathrm{kg\,m^2}\) |
3. Why the moment of inertia increases
Moment of inertia depends on how far mass is from the rotation axis:
\[
I=\int r^2\,dm.
\]
When the axis is shifted away from the center of mass, every small mass element is effectively farther from the new
axis in a systematic way. The total increase is exactly
\[
Md^2.
\]
Since \(M>0\) and \(d^2\ge0\), the shifted-axis moment of inertia is never smaller than the center-of-mass value:
\[
I_{\mathrm{new}}\ge I_{\mathrm{cm}}.
\]
4. Important condition
The theorem only applies when the two axes are parallel. If the new axis is tilted relative to the known axis, the
parallel axis theorem alone is not enough.
\[
\text{parallel axes required}
\]
5. Common center-of-mass formulas
The calculator can either use a directly entered \(I_{\mathrm{cm}}\), or compute \(I_{\mathrm{cm}}\) from common
homogeneous-object formulas.
| Object |
Center-of-mass axis |
\(I_{\mathrm{cm}}\) |
| Thin rod |
Perpendicular axis through midpoint |
\(I_{\mathrm{cm}}=\frac{1}{12}ML^2\) |
| Solid disk / solid cylinder |
Central symmetry axis |
\(I_{\mathrm{cm}}=\frac{1}{2}MR^2\) |
| Thin hoop / ring |
Central symmetry axis |
\(I_{\mathrm{cm}}=MR^2\) |
| Hollow cylinder |
Central symmetry axis |
\(I_{\mathrm{cm}}=\frac{1}{2}M(R_1^2+R_2^2)\) |
| Rectangular plate |
Perpendicular axis through center |
\(I_{\mathrm{cm}}=\frac{1}{12}M(a^2+b^2)\) |
| Solid sphere |
Any diameter |
\(I_{\mathrm{cm}}=\frac{2}{5}MR^2\) |
| Thin spherical shell |
Any diameter |
\(I_{\mathrm{cm}}=\frac{2}{3}MR^2\) |
6. Rod about one end
A classic use of the theorem is deriving the moment of inertia of a thin rod about one end. The known center-axis
formula is
\[
I_{\mathrm{cm}}=\frac{1}{12}ML^2.
\]
The end axis is parallel to the center axis and shifted by
\[
d=\frac{L}{2}.
\]
Therefore,
\[
I_{\mathrm{end}}
=
I_{\mathrm{cm}}+Md^2.
\]
\[
I_{\mathrm{end}}
=
\frac{1}{12}ML^2+M\left(\frac{L}{2}\right)^2.
\]
\[
I_{\mathrm{end}}
=
\frac{1}{12}ML^2+\frac{1}{4}ML^2
=
\frac{1}{3}ML^2.
\]
7. Worked example
Suppose a rigid body has
\[
I_{\mathrm{cm}}=0.12\ \mathrm{kg\,m^2},
\]
mass
\[
M=1.0\ \mathrm{kg},
\]
and the new axis is shifted by
\[
d=0.40\ \mathrm{m}.
\]
Apply the theorem:
\[
I_{\mathrm{new}}=I_{\mathrm{cm}}+Md^2.
\]
\[
I_{\mathrm{new}}=0.12+(1.0)(0.40)^2.
\]
\[
I_{\mathrm{new}}=0.12+0.16=0.28\ \mathrm{kg\,m^2}.
\]
Therefore,
\[
\boxed{I_{\mathrm{new}}=0.28\ \mathrm{kg\,m^2}}.
\]
Key idea: shifting the rotation axis away from the center of mass always adds \(Md^2\) to the center-of-mass
moment of inertia.
