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Calculating moments of inertia

Key idea: Find moment of inertia by integrating r² over mass about a stated axis. The parallel-axis theorem shifts a centre-of-mass value by Md²; composite parts must use the same axis.

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  • GCE A-Level H3 Physics 2027
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For a body made of point masses, I = ∑ mᵢrᵢ². A continuous body is made of infinitely many small mass elements, so the sum becomes an integral:

I = ∫ r² dm

where r is the perpendicular distance of the element dm from the axis. The work is in choosing dm so that every part of it is at the same distance from the axis.

Choose a mass element

Check your understanding 1 Predict first

A uniform disc is cut into thin rings of equal radial width. How does the mass of a ring near the rim compare with the mass of a ring near the centre?

Ring near the rim
Show answer

More mass. A thin ring of radius r and width dr has area 2π r dr, so at twice the radius it holds twice the mass. That is why a disc cannot be treated as a rod along its radius.

Choosing a mass element

For a uniform thin rod of mass M and length L, the linear density is λ = M/L. A short element has dm = λ dx.

For a uniform thin disc of mass M and radius R, the surface density is σ = M/(π R²). A thin annulus of radius r and radial width dr has area dA = 2π r dr, so dm = 2π rσ dr. The annulus gets heavier as r increases because its circumference increases. Equal radial widths do not contain equal masses.

Choose mass elements that match the geometryA uniform rod has a short element of length dx at coordinate x from its midpoint axis, giving dm equals lambda dx and distance magnitude x. Parallel axes through the midpoint and left end point out of the page, separated by half the rod length. A uniform disc has a thin annulus at radius r with radial width dr; its area is circumference 2 pi r times dr, so dm equals sigma times 2 pi r dr.
Choose mass elements that match the geometryA uniform rod has a short element of length dx at coordinate x from its midpoint axis, giving dm equals lambda dx and distance magnitude x. Parallel axes through the midpoint and left end point out of the page, separated by half the rod length. A uniform disc has a thin annulus at radius r with radial width dr; its area is circumference 2 pi r times dr, so dm equals sigma times 2 pi r dr.Uniform rod: dm = λ dx⊙ denotes an axis out of the pageEnd axisCOM axisElement dmdx−L/20xL/2Axis separation d = L/2λ = M/L; distance = |x|Uniform disc: dm = σ (2πr dr)rRdrσ = M/(πR²)Thin annulus area: dA = 2πr drGreater r means greater circumferenceEqual dr does not mean equal dm
Schematic differential elements. Element widths are exaggerated; these are not finite measured samples. Both rod axes and the disc symmetry axis point out of the page.
Illustrative numerical equivalents

The diagrams use schematic widths. These values describe uniform model bodies, not experimental measurements.

Body and axesDensityMoment of inertia
Rod: M = 3 kg, L = 2 m; perpendicular axesλ = 1.5 kg m⁻¹COM: 1.0 kg m²; end: 4.0 kg m²
Disc: M = 2 kg, R = 1 m; symmetry axisσ = 0.637 kg m⁻²1.0 kg m²

For each integral, identify the perpendicular distance, express dm using the correct density, and integrate over the whole body. The limits and the density must describe the same object.

Uniform thin rod about its centre

Choose the axis through the centre and perpendicular to the rod. Put x = 0 at the centre, so -L/2 ≤ x ≤ L/2 and the perpendicular distance is |x|:

I_cm = ∫_(-L/2)^(L/2)x²λ dx; = (M/L)[x³/3]_(-L/2)^(L/2); = (1/12)ML².

Negative coordinates still give positive contributions because the distance is squared.

Worked example 1

Integrating a uniform rod

Problem

A uniform thin rod has mass 0.60 kg and length 1.2 m. Find its inertia about an axis through its centre and perpendicular to the rod.
Show full solution

  1. Choose the element and limits

    Method

    dm = 0.50 dx in SI units, with x from -0.60 to + 0.60 m.

    Reason

    The mass is uniform along the rod and the axis passes through its midpoint.

    Working

    Answer

    λ = M/L = 0.60/1.2 = 0.50 kg m⁻¹

  2. Integrate

    Method

    I = 0.072 kg m².

    Reason

    Each element contributes x²dm.

    Working

    Answer

    I = 0.50∫_(-0.60)^(0.60)x² dx = 0.072 kg m²

Check your understanding 2 Derive the end-axis result

On paper, integrate I = ∫ x²λ dx for a uniform rod of mass M and length L about a perpendicular axis through one end, with x from 0 to L. Enter the coefficient of ML² as a decimal; keep the derivation on paper.

Show answer
I_end = ∫₀^L x²M/L dx = (M/L)[x³/3]₀^L = (1/3)ML²

Compare your limits and density: the element runs over the whole rod from the axis, so the distance is x itself. The result is four times I_cm, because more of the mass is far from the axis.

Uniform disc about its central axis

Every point of an annulus is the same distance r from the disc’s central axis, so the annulus is the right element for a disc. Treating the disc as if its mass were spread evenly along a radius gives the wrong answer.

Spot the mistake 2

A disc is not a rod in the radial direction

About 7 min

Learner claim

A learner models a uniform disc with dm = (M/R)dr and obtains I = MR²/3. Explain why that mass element is wrong and derive the correct central-axis inertia on paper. Compare your reasoning and integral with the solution.

Try this before viewing the solution

Mass element for a uniform disc
Show solution step by step

  1. Use annular area

    Method

    dm = 2π rσ dr, with σ = M/(π R²).

    Reason

    A disc has uniform mass per area, rather than per radial distance.

    Working

    Answer

    dA = 2π r dr

  2. Integrate the inertia

    Method

    I = (1/2)MR².

    Reason

    The distance of each annulus from the symmetry axis is r.

    Working

    Answer

    I = 2πσ∫₀^R r³ dr = (1/2)MR²

The parallel-axis theorem

The parallel-axis theorem is I = I_cm + Md². Here I_cm refers to a parallel axis through the centre of mass of the whole object, and d is the perpendicular separation of the axes. It does not let you add Md² to an arbitrary non-centroidal inertia.

For the rod’s end axis, d = L/2: I_end = (1/12)ML² + M(L/2)² = (1/3)ML². Within a set of parallel axes, the COM axis gives the smallest inertia because Md² ≥ 0. This comparison keeps the axis direction fixed.

Guided practice 3

A disc about a parallel axis through its rim

About 5 min

Problem

Find a uniform thin disc’s inertia about an axis through a point on its rim and perpendicular to its plane.

Try this before viewing the solution

Hints

Hint 1

identify the centre axis result

Which central-axis result do you need, and how far apart are the two parallel axes?

Show solution step by step

  1. Set the reference

    Method

    I_cm = (1/2)MR² and d = R.

    Reason

    The rim axis is parallel to and one radius from the symmetry axis.

    Working

    Answer

    I = I_cm + Md²

  2. Combine

    Method

    I = (3/2)MR².

    Reason

    Add the parallel-axis shift MR².

    Working

    Answer

    I = (1/2)MR² + MR² = (3/2)MR²

Exam-style question 1

A rod rotating about its end

3 marks

Examination question

A uniform thin rod has mass 3.0 kg and length 2.0 m. Calculate its inertia about an axis through one end and perpendicular to the rod, using the parallel-axis theorem. Show your method on paper. [3 marks]

Try this before viewing the solution

Unit: kg m²
Unit: kg m²
Show the steps Optional working
Unit: kg m²
Unit: kg m²
Unit: kg m²
Show solution step by step

  1. Find the COM result

    1 mark

    Method

    I_cm = 1.0 kg m².

    Reason

    The reference axis passes through the rod’s midpoint.

    Working

    Answer

    (3.0)(2.0)²/12 = 1.0

  2. Set the separation

    1 mark

    Method

    d = 1.0 m.

    Reason

    The end axis is half a rod length from the parallel COM axis.

    Working

    Answer

    d = L/2

  3. Apply the theorem

    1 mark

    Method

    I_end = 4.0 kg m².

    Reason

    Use the full rod mass in the shift term.

    Working

    Answer

    I = 1.0 + (3.0)(1.0)² = 4.0 kg m²

Check your understanding 3 Can you shift from any parallel axis?

A uniform rod has mass 1.2 kg, length 1.0 m and end-axis inertia 0.40 kg m². A parallel axis lies 0.25 m from its COM. A learner calculates its inertia as 0.40 + 1.2(0.25)². Find the centre-of-mass inertia and the correct inertia, and identify the error. All axes are perpendicular to the rod.

Show the steps Optional working
Unit: kg m²
Unit: kg m²
Show the steps Optional working
Unit: kg m²
Unit: kg m²
Unit: kg m²
The learner’s error
Show answer

First recover the COM-axis inertia: I_cm = 0.40-1.2(0.50)² = 0.10 kg m². Then I = 0.10 + 1.2(0.25)² = 0.175 kg m², or 0.18 kg m² to two significant figures. The learner used the end-axis inertia as though it were the COM-axis reference, counting the old shift as well as the new one.

Composite bodies

To find the moment of inertia of a body made of several parts, find each part’s moment of inertia about the same axis and add them. If the parts’ inertias are easier to find about the centre of mass of the whole body, add them about that axis first and then shift the total, using the whole body’s mass.

Always state the axis in full. A disc’s central axis is its symmetry axis, perpendicular to its faces; the rod axes in this lesson are perpendicular to the rod.

Try it yourself 4

Shift the axis of a whole composite body

Minimal support

Problem

A uniform thin rod of mass 1.2 kg and length 1.0 m has a 0.25 kg point mass attached at each end. Find the whole body’s inertia about an axis through one end and perpendicular to the rod. Show your method on paper, starting from its COM.

Try this before viewing the solution

Unit: kg
Show the steps Optional working
Unit: kg m²
Unit: kg m²
Unit: kg m²
Show the steps Optional working
Unit: kg m²
Unit: kg
Unit: kg m²

Hints

Hint 1

locate the whole-body COM

Equal end masses preserve symmetry about the midpoint.
Show solution step by step

  1. Add inertias about the COM

    Method

    I_cm = 0.225 kg m² before rounding.

    Reason

    The rod and both masses use the same midpoint axis.

    Working

    Answer

    I_cm = 1.2(1.0)²/12 + 2(0.25)(0.50)² = 0.225

  2. Shift the whole object

    Method

    I_end = 0.65 kg m².

    Reason

    The total mass is 1.70 kg and the axis separation is 0.50 m.

    Working

    Answer

    I = 0.225 + (1.70)(0.50)² = 0.650 kg m²

You will not need the perpendicular-axis theorem or the derivation of a sphere’s moment of inertia. When a question involves a sphere, it gives the moment of inertia and its axis.

Common mistakes

  • Using dm = (M/R) dr for a disc. A disc has uniform mass per area, so use the annulus dm = 2π rσ dr.
  • Integrating over only half of a rod, or using limits that do not match the density.
  • Adding Md² to a moment of inertia that is not about the centre-of-mass axis.
  • Using one part’s mass instead of the whole body’s mass in the shift term.
  • Adding inertias that are about different axes.

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