AcademyRigid-Body Rotation
Academy
Parallel-Axis Theorem
Level 1 - Physics topic page in Rigid-Body Rotation.
Principle
Shifting a parallel rotation axis away from the center of mass adds \(Md^2\).
Notation
\(I\)
moment of inertia about chosen axis
\(\mathrm{kg\,m^{2}}\)
\(I_{\mathrm{cm}}\)
moment of inertia about parallel center-of-mass axis
\(\mathrm{kg\,m^{2}}\)
\(M\)
body mass
\(\mathrm{kg}\)
\(d\)
perpendicular separation between axes
\(\mathrm{m}\)
\(I_{\mathrm{tot}}\)
composite moment of inertia
\(\mathrm{kg\,m^{2}}\)
Method
Derivation 1: Separate center-of-mass distance from internal distance
The new axis must be parallel to the center-of-mass axis. Each mass element is farther from the new axis because the whole body has been shifted by \(d\).
Start from definition
\[I=\int r_\perp^2\,dm\]
Split distance
\[\vec r=\vec r_{\mathrm{cm}}+\vec d\]
Center-of-mass cross term
\[\int \vec r_{\mathrm{cm}}\,dm=\vec0\]
Parallel-axis result
\[I=I_{\mathrm{cm}}+Md^2\]
The sketch shows the only distance allowed in the theorem: the perpendicular separation between two parallel axes.
Derivation 2: Use it for composite bodies
For a composite body, move each part to the same final axis before adding the moments of inertia.
Shift each part
\[I_i=I_{i,\mathrm{cm}}+M_id_i^2\]
Add about common axis
\[I_{\mathrm{tot}}=\sum_i I_i\]
Point-mass limit
\[I=Mr^2\]
Rules
These are the compact results from the method above.
Parallel axis
\[I=I_{\mathrm{cm}}+Md^2\]
Composite body
\[I_{\mathrm{tot}}=\sum_i I_i\]
Point mass shift
\[I=Mr^2\]
Examples
Question
A rod has
\[M=2.0\,\mathrm{kg}\]
\[L=1.5\,\mathrm{m}\]
and \[I_{\mathrm{cm}}=\frac1{12}ML^2\]
Find \(I\) about one end.Answer
\[I=\frac1{12}ML^2+M\left(\frac{L}{2}\right)^2=\frac13ML^2=1.50\,\mathrm{kg\,m^2}\]
Checks
- The shifted axis must be parallel to the center-of-mass axis.
- The offset \(d\) is perpendicular axis separation.
- The theorem increases \(I\); it never subtracts \(Md^2\).
- For composite bodies, shift each part to the same final axis before adding.