Derivation Of Moment Of Inertia Of an Uniform Rigid Rod

Derive the moment of inertia of a uniform rod about an arbitrary perpendicular axis, then test special cases.

University Physics Year 1

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This page gives the UY1 working model/result for Derivation Of Moment Of Inertia Of an Uniform Rigid Rod. You reuse it as part of the mechanics toolkit: choose coordinates, pick a method (Newton/energy/momentum), and check limits and signs.

Module path: Mechanics (UY1)
Full routing: UY1 Assessment Map
Math toolkit: Mathematics for Undergraduate Physics

At a glance

Prerequisites: definition I = ∫ r² dm, uniform linear density.
Outcomes: derive I(h) for an axis at distance h from one end of a rod.
Key result:

I = (1/3)M(L²-3Lh + 3h²)

After shifting the origin to the axis, the rod runs from x = -h to x = L-h (not 0 to L).
In dI = x²dm, x is the perpendicular distance to the axis along the rod (so squaring handles the sign).
Be explicit about what h measures (distance from the left end to the axis).

Setup

Model:

Uniform rod, length L, mass M.
Axis is perpendicular to rod and passes through point O, located h from the left end.

Coordinates and assumptions:

Place origin at the axis, so a rod element at position x contributes dI = x² dm.
Rod spans from x = -h to x = L-h.
Uniform density: λ = M/L, so dm = λ dx = (M/L)dx.

Geometry note:

For a thin rod with an axis perpendicular to the rod, the perpendicular distance from the axis to an element is just r = |x|.
This is why the integral is 1D: I = ∫ x² dm.

Core method

Start with

dI = x² dm = x²(M/L)dx

Integrate over full rod:

I = M/L∫₋ₕ^(L-h)x² dx = (M/L)[x³/3]₋ₕ^(L-h)

I = (M/3L)((L-h)³-(-h)³)

Expand and simplify:

I = (M/3L)(L³-3L²h + 3Lh²) = (1/3)M(L²-3Lh + 3h²)

Non-trivial point: the (-h)³ term becomes + h³ after subtraction, which is where many algebra slips happen.

Worked example

Problem

A uniform rod has M = 2.0 kg and L = 1.5 m. Find its moment of inertia about a perpendicular axis h = 0.40 m from one end, using I = (1/3)M(L²-3Lh + 3h²).

Show full solution

Evaluate the bracket
Method
Substitute L and h.
Reason
Work out each term before multiplying by M/3.
Working
Answer
1.5²-3(1.5)(0.40) + 3(0.40)² = 2.25-1.80 + 0.48 = 0.93 m².
Multiply by M/3
Method
Use M = 2.0 kg.
Reason
The prefactor sets the units, kg m².
Working
Answer
I = (2.0/3)(0.93).
State the answer
Working
Answer
I = 0.62 kg m².

Practice set

Check your understanding 1

Use the general formula to find I when the axis is at one end (h = 0).

Show hint

Substitute h = 0, then simplify.

Show answer

With h = 0 the bracket is L², so I = (1/3)ML².

Check your understanding 2

Find I for a uniform rod with M = 1.2 kg and L = 0.80 m about a perpendicular axis through its centre.

Show hint

The centre means h = L/2.

Show answer

With h = L/2 the bracket is (1/4)L², so I = (1/12)ML² = (1/12)(1.2)(0.80)² = 0.064 kg m².

Check your understanding 3

Why does moving the axis away from the centre of the rod increase I?

Show hint

Think about the average of x² over the rod.

Show answer

More of the mass ends up far from the axis, so the mass-weighted average of x² grows. The centre minimises it, which is also what the parallel-axis theorem says: I = I_cm + Md².

Summary and next steps

For a uniform rod, the arbitrary-axis result comes directly from correct limits and careful expansion. Special cases recover the familiar center and end formulas.

Next derivation: Hollow/solid Cylinder
Applications: Torque & Angular Acceleration, Rigid Body Rotation
Back to overview: Moment Of Inertia
Topic index: Back To Mechanics (UY1)

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Syllabus and review details

No official syllabus alignment is listed for this lesson.