Integrate rotational inertia of a nonuniform rod

Define inertia about an actual axis

For a rigid body rotating about a specified fixed axis, I=integral rperp² dm. The distance is perpendicular to that axis, not necessarily distance from the center of mass. The same mass distribution can have different inertias about different axes. Units kg m² follow directly from the integral.

Predict and check a discrete distribution first

For point particles about the same specified axis, I=Σ m_i r_i². Predict whether moving mass outward changes inertia even when total mass stays fixed. Take a 1 kg particle at perpendicular distance 1 m and a 2 kg particle at 2 m: I=1(1²)+2(2²)=9 kg m². Move the 2 kg particle inward to 1 m while keeping the first particle fixed. The total mass remains 3 kg, but I becomes 3 kg m². At the same angular acceleration, the required net torque is now one third as large.

For a changed case, place all 3 kg at 2 m: predict I before checking 3(2²)=12 kg m². Equal mass does not imply equal rotational inertia. A continuous rod is the limit of such a sum over small elements: Σ r_i² Δm_i becomes integral rperp² dm, with each element’s distance measured from the same axis.

Normalize the given line distribution

The line spans 0≤x≤2 m with lambda=2(1+beta x/2) kg/m and length scale 2 m. Integrating gives M=4+2beta kg and first moment Q=4+8beta/3 kg m. The center coordinate is Q/M. The workbench holds the density scale fixed, so increasing beta changes mass as well as shape.

Integrate the end-axis inertia

About a perpendicular axis through x=0, Iend=integral x²lambda dx=16/3+4beta kg m². Squaring the average distance first would not produce the same result; distant material carries disproportionately large weight. A normalized shape factor Iend/(ML²) permits comparison without mistaking extra mass for redistribution alone.

Derive the central parallel-axis relation

Write x=(x−xCM)+xCM in the squared-distance integral. The cross term integrates to zero because integral(x−xCM)dm=0. Therefore Iend=Icm+M xCM². This proof requires parallel axes. At beta=1, M=6, xCM=10/9, Iend=28/3 and Icm=52/27 in SI units.

Apply rotational dynamics conditionally

For constant rigid I about the fixed axis, net torque equals I alpha. At beta=1 and alpha=2, required endpoint-axis torque is 56/3 N m. If mass moves radially or the axis changes, this simple constant-I equation needs re-examination; the integral is not a universal permanent label for the object.

Worked original case

For uniform beta=0, M 4 kg and length 2 m give Iend 16/3 and Icm 4/3 kg m², matching the usual ML²/3 and ML²/12 results derived from the integral.

Investigation

Compare beta=0,1,3. Independently integrate M,Q and Iend, derive Icm and compare normalized shape factors. Then double alpha and predict required torque without reintegrating the unchanged distribution.

Public numerical transfer practice

Try a new practice case

A uniform rod of mass 3 kg and length 2 m rotates about a perpendicular end axis. What is its inertia in kg m²?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.