Derivation Of Moment Of Inertia Of A Thin Spherical Shell

Derive the moment of inertia of a thin spherical shell by integrating circular hoops over polar angle.

University Physics Year 1

On this page

This page gives the UY1 working model/result for Derivation Of Moment Of Inertia Of A Thin Spherical Shell. 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: hoop inertia I = mr², spherical geometry, trig substitution.
Outcomes: derive shell inertia about any central axis.
Key result:

I = (2/3)MR²

The surface strip area is dA = 2π R² sin θ dθ (the sin θ comes from the shrinking hoop radius near the poles).
The hoop radius is r = R sin θ, not R.
This result is for an axis through the center; by symmetry any central axis is equivalent.

Setup

Model:

Thin uniform spherical shell, radius R, mass M.
Rotation axis through center.

Coordinates:

Use polar angle θ from the axis.
A ring (hoop) at angle θ has radius r = R sin θ.

Geometry intuition:

Near the poles (θ ≈ 0 or π), sin θ ≈ 0, so those hoops have tiny radius and contribute little to I.
The equatorial region (θ ≈ π/2) has the largest radius and dominates the inertia.

Surface mass density:

σ = M/(4π R²)

Core method

For each hoop element,

dI = r² dm

Hoop area element (circumference × strip width):

dA = (2π r)(R dθ) = 2π R² sin θ dθ

dm = σ dA = M/2 sin θ dθ

Then

dI = (R sin θ)²(M/2 sin θ dθ) = MR²/2 sin³ θ dθ

Integrate from 0 to π:

I = MR²/2∫₀^π sin³ θ dθ

Using ∫₀^π sin³ θ dθ = 4/3,

I = MR²/2 · 4/3 = (2/3)MR²

Non-trivial point: the extra sin θ in dA is a geometry factor from latitude shrinking near poles.

Worked example

Problem

Find the moment of inertia of a thin spherical shell of mass 3.0 kg and radius 0.25 m about a diameter.

Show full solution

Choose the formula
Method
Use I = (2/3)MR².
Reason
This is the result of summing the shell’s hoops, derived above.
Working
Answer
I = (2/3)MR².
State the answer
Working
Answer
I = (2/3)(3.0)(0.25)² = 0.125 kg m².

Practice set

Check your understanding 1

Find I for a thin spherical shell with M = 1.8 kg and R = 0.10 m.

Show hint

Substitute into I = (2/3)MR².

Show answer

I = (2/3)MR² = (2/3)(1.8)(0.10)² = 0.012 kg m².

Check your understanding 2

The radius of a thin shell triples at fixed mass. How does I change?

Show hint

How does I depend on R?

Show answer

It becomes nine times larger, because I ∝ R².

Check your understanding 3

In the derivation, why do hoops near the poles contribute less area than hoops near the equator?

Show hint

Look at how dA depends on sin θ.

Show answer

A hoop at polar angle θ has radius R sin θ, so its area strip dA = 2π R² sin θ dθ is small near the poles, where sin θ → 0.

Summary and next steps

Hoop decomposition plus the spherical area element yields I = (2/3)MR² in a few lines. The geometry factor sin θ is the critical step.

Previous derivation: Uniform Solid Sphere
Start derivations: Uniform Rigid Rod
Applications: More About Rolling Motion (thin-shell/hoop limit), Rolling Motion
Back to overview: Moment Of Inertia

Spotted an error? Report a correction

Syllabus and review details

No official syllabus alignment is listed for this lesson.