Derivation Of Moment Of Inertia Of A Hollow/Solid Cylinder

Derive the central-axis moment of inertia for a hollow cylinder and recover hoop and solid-cylinder limits.

University Physics Year 1

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This page gives the UY1 working model/result for Derivation Of Moment Of Inertia Of A Hollow/Solid Cylinder. 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: I = ∫ r² dm, cylindrical geometry, uniform density.
Outcomes: derive I for a hollow cylinder and obtain thin shell/solid cylinder cases.
Key result:

I = (1/2)M(R₁² + R₂²)

The axis here is the central symmetry axis (along the cylinder length). Other axes give different formulas.
Don’t drop the r in dV = 2π rL dr: circumference grows with r.
Keep radii straight: integrate from inner radius R₁ to outer radius R₂.

Setup

Model:

Hollow cylinder of length L, mass M, inner radius R₁, outer radius R₂.
Axis is central symmetry axis (along cylinder length).

Assumptions:

Uniform volume density ρ.
Use thin cylindrical shells of radius r and thickness dr.

Geometry note (what r means):

r is the perpendicular distance from the axis, so dI = r² dm weights outer shells much more strongly.
This is why hollow cylinders (mass farther out) have larger I than solid cylinders for the same M and outer radius.

Core method

For a shell element:

dI = r² dm

Compute dm from shell volume:

dV = (2π rL) dr, dm = ρ dV = 2πρ Lr dr

dI = 2πρ Lr³ dr

Integrate from R₁ to R₂:

I = 2πρ L∫_R₁^R₂r³ dr = (1/2)πρ L(R₂⁴-R₁⁴)

Use

M = ρπ(R₂²-R₁²)L

and factor R₂⁴-R₁⁴ = (R₂²-R₁²)(R₂² + R₁²):

I = (1/2)M(R₂² + R₁²)

Non-trivial point: factoring the quartic difference is what cleanly cancels density and length.

Worked example

Problem

A hollow cylinder has mass M = 4.0 kg, inner radius R₁ = 0.08 m and outer radius R₂ = 0.10 m. Find its moment of inertia about its axis.

Show full solution

Choose the formula
Method
Use I = (1/2)M(R₁² + R₂²).
Reason
This is the result of the derivation above for a uniform cylinder between radii R₁ and R₂.
Working
Answer
I = (1/2)M(R₁² + R₂²).
Substitute
Method
Use SI units.
Reason
Square each radius before adding.
Working
Answer
I = (1/2)(4.0)(0.0064 + 0.0100) kg m².
State the answer
Working
Answer
I = 0.0328 kg m².

Practice set

Check your understanding 1

Show that the general formula gives the solid-cylinder result.

Show hint

Set R₁ = 0.

Show answer

Set R₁ = 0: I = (1/2)M(0 + R²) = (1/2)MR². A solid cylinder is a hollow one with no hole.

Check your understanding 2

Find I for a thin hoop of mass 1.5 kg and radius 0.20 m.

Show hint

Use the hoop limit, R₁ → R₂.

Show answer

In the thin-hoop limit R₁ → R₂ = R, all the mass is at radius R, so I = MR² = (1.5)(0.20)² = 0.060 kg m².

Check your understanding 3

For the same mass and outer radius, which has the larger moment of inertia: a thick hollow cylinder or a solid cylinder?

Show hint

Compare the average of r² for the two mass distributions.

Show answer

The hollow cylinder. Its mass lies farther from the axis on average, and I weights each mass element by r².

Summary and next steps

The shell method gives one compact formula that includes both common limits. Keep geometry and differential volume precise to avoid factor errors.

Next derivation: Uniform Solid Sphere
Previous derivation: Uniform Rigid Rod
Applications: Rolling Motion, More About Rolling Motion
Back to overview: Moment Of Inertia

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Syllabus and review details

No official syllabus alignment is listed for this lesson.