AcademyRigid-Body Rotation
Academy
Calculating Moments of Inertia
Level 1 - Physics topic page in Rigid-Body Rotation.
Principle
Moment of inertia is mass weighted by squared distance from the rotation axis.
Notation
\(I\)
moment of inertia
\(\mathrm{kg\,m^{2}}\)
\(r_\perp\)
perpendicular distance to axis
\(\mathrm{m}\)
\(dm\)
small mass element
\(\mathrm{kg}\)
\(\lambda\)
linear mass density
\(\mathrm{kg\,m^{-1}}\)
\(\rho\)
volume mass density
\(\mathrm{kg\,m^{-3}}\)
Method
Derivation 1: Build the definition from point masses
Each mass element contributes according to how far it is from the axis. The square means distant mass matters strongly.
Discrete body
\[I=\sum_i m_ir_{\perp i}^2\]
Continuous limit
\[I=\int r_\perp^2\,dm\]
Scale check
\[I\sim MR^2\]
Derivation 2: Choose a mass element before integrating
The integral is only useful after the axis, coordinate, and density model are chosen.
Linear object
\[dm=\lambda\,dx\]
Volume object
\[dm=\rho\,dV\]
Uniform rod about center
\[I=\int_{-L/2}^{L/2}x^2\frac{M}{L}\,dx=\frac1{12}ML^2\]
Rules
These are the compact results from the method above.
Discrete masses
\[I=\sum_i m_ir_{\perp i}^2\]
Continuous body
\[I=\int r_\perp^2\,dm\]
Uniform rod center
\[I_{\mathrm{rod,cm}}=\frac1{12}ML^2\]
Thin hoop center
\[I_{\mathrm{hoop}}=MR^2\]
Solid disk center
\[I_{\mathrm{disk}}=\frac12MR^2\]
Examples
Question
Three point masses
\[2.0\,\mathrm{kg}\]
\[1.0\,\mathrm{kg}\]
and \[3.0\,\mathrm{kg}\]
lie \[0.20\,\mathrm{m}\]
\[0.50\,\mathrm{m}\]
and \[0.40\,\mathrm{m}\]
from an axis. Find \(I\).Answer
\[I=2.0(0.20^2)+1.0(0.50^2)+3.0(0.40^2)=0.81\,\mathrm{kg\,m^2}\]
Checks
- Distances are squared, so far-away mass dominates.
- The axis choice changes \(r_\\perp\) and therefore \(I\).
- Units must be kg m^2.
- A result much larger than \(MR^2\) usually signals a distance error.