Physics I: Mechanics › Rotation › full formula sheet

I = (2/5)MR²

Say it: “the moment of inertia of a solid sphere equals two-fifths M R squared”

Solid sphere

The bottom of the laziness ladder — a ball hides its mass near the axle better than any flat shape can.

Notation on this page: M is the sphere’s total mass, R its radius. The axis is any diameter through the center — a uniform sphere is symmetric in every direction.

Before this lesson: Moment of inertia, Disk / solid cylinder

Where it comes from

The ladder so far: hoop (1) > disk (½) > sphere (࡫). Each step adds a dimension for mass to hide near the axis — and each step shrinks the fraction:

Before reading on: a hoop keeps mass at distance R in a ring; a disk fills the circle; a sphere fills the ball. Rank their r²-averages. Which shape should bottom out the ladder, and why?
hoop
⟨r²⟩ = R²
1D ring: nowhere to hide.
disk
⟨r²⟩ = R²/2
2D: mass can sink toward the center in the plane.
sphere
⟨r²⟩ = 2R²/5
3D: mass hides near the axis in two transverse directions. 2/5 = 0.4 < 0.5 ✓

That’s the whole story: I = M⟨r²⟩, and each dimension gives mass another direction to crowd toward the axis. The sphere, with the most hiding places, gets the smallest fraction — which is exactly why balls beat wheels rolling downhill (more in the rolling lesson).

Derivation

Slice the sphere into thin disks stacked along the z-axis (the rotation axis). At height z the slice is a disk of radius a(z) = √(R²−z²) and thickness dz, with mass dm = ρ π a² dz. Each disk contributes dI = ½ dm a² (the disk formula). Integrate z from −R to R.

dI
=
½ dm a² = ½ ρπ(R²−z²)² dz
Step 1 — one slice. The slice is a disk of radius a = √(R²−z²): dI = ½·(its mass)·a², with dm = ρ·πa² dz.
I
=
½ ρπ ∫−RR (R²−z²)² dz
Step 2 — add the slices. Expand: (R²−z²)² = R⁴ − 2R²z² + z⁴.
∫
=
[R⁴z − 2R²z³/3 + z⁵/5]−RR = 2R⁵(1 − 2/3 + 1/5) = 16R⁵/15
Step 3 — integrate. Odd powers double symmetrically: 2×(R⁵ − 2R⁵/3 + R⁵/5). The bracket is 8/15, so the integral is 16R⁵/15.
I
=
½ ρπ (16R⁵/15) = (2/5)MR²
Step 4 — total mass. M = (4/3)πR³ρ, so ρ = 3M/(4πR³): I = (ρπ/2)(16R⁵/15) = (8/30)·(3M/R³)·R⁵… = (2/5)MR². ∎

Sanity check on the 2/5: it must be less than the disk’s 1/2 (more hiding places in 3D) and more than 0 (mass can’t all sit on the axis). 0.4 threads that needle ✓.

How to use it

The procedure, every time:

  1. Confirm solid sphere: bowling ball, billiard ball, planet (approximately), ball bearing — uniform solid ball.
  2. Any central axis works — pick the convenient one; symmetry guarantees the same I.
  3. Compute I = (2/5)MR².
  4. Use it in Στ = Iα, K = ½Iω², L = Iω, or rolling problems (the sphere’s small I is why it wins downhill races).

Hollow vs solid — don’t cross them

A hollow spherical shell has I = (2/3)MR² — but that formula is not on this sheet (the sheet deliberately omits it). If a problem says “spherical shell” or “hollow sphere,” 2/5 does not apply. Solid only.

Common mistake: writing I = ½MR² for a ball (“it’s round”). ½ is the disk. The ball is 2/5 — smaller, because 3D hides mass better than 2D.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: a bowling ball

  1. Identify. Solid sphere, M = 7 kg, R = 0.11 m.
  2. Compute. I = 0.4 × 7 × 0.11² = 0.4 × 7 × 0.0121 = 0.4 × 0.0847 = 0.0339 kg·m² (3 s.f.).
  3. Sanity check. Small ball → small I; and 0.4 < 0.5 keeps it below the disk ✓.
Common mistake: 2/5 vs 5/2 — writing I = (5/2)MR² = 2.5× too big. The fraction is less than one: two-fifths, 0.4.
Your turn — Solid sphere M = 5 kg, R = 0.1 m. I = ?

Answer: 0.02 kg·m². I = 0.4 × 5 × 0.01 = 0.02 kg·m².

Example 2 — planetary scale: Earth (uniform model)

  1. Identify. Earth as a uniform solid sphere: M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m.
  2. Compute. I = 0.4 × 6.0×10²⁴ × (6.4×10⁶)² = 0.4 × 6.0×10²⁴ × 4.096×10¹³ = 0.4 × 2.4576×10³⁸ = 9.8 × 10³⁷ kg·m².
  3. Read it. Enormous — spinning a planet takes planetary angular momentum (next lessons quantify it). Real Earth’s I is slightly smaller (dense core), but 2/5 is the standard first estimate.
Common mistake: exponent slips — (6.4×10⁶)² = 40.96×10¹² = 4.096×10¹³, not 10¹². Square the mantissa and double the exponent.
Your turn — Moon: M = 7.3 × 10²² kg, R = 1.7 × 10⁶ m. I ≈ ?

Answer: ≈ 8.4 × 10³⁴ kg·m². I = 0.4 × 7.3×10²² × 2.89×10¹² = 0.4 × 2.1097×10³⁹ = 8.44×10³⁴ kg·m².

Before reading on: a 2-kg sphere (R = 0.2 m) and a 2-kg hoop (R = 0.2 m) spin at 15 rad/s. Which holds more rotational KE — and by what factor?

Example 3 — spin energy of the sphere

  1. I. M = 2 kg, R = 0.2 m ⇒ I = 0.4 × 2 × 0.04 = 0.032 kg·m².
  2. K. ω = 15 rad/s ⇒ Krot = ½ × 0.032 × 225 = 3.6 J.
  3. Compare. The hoop (I = 0.08) would hold 9 J — 2.5× more. Same M, R, ω: the sphere is the stingiest energy store on the ladder.
Common mistake: K = ½mv² with v = ωR (treating the ball as a point at the rim). That gives ½ × 2 × 9 = 9 J — the hoop’s answer. Extended bodies need ½Iω².
Your turn — Sphere M = 4 kg, R = 0.15 m, ω = 10 rad/s. Krot = ?

Answer: 1.8 J. I = 0.4 × 4 × 0.0225 = 0.036; K = 0.5 × 0.036 × 100 = 1.8 J.

Example 4 — torque on a sphere

  1. I. M = 10 kg, R = 0.5 m ⇒ I = 0.4 × 10 × 0.25 = 1.0 kg·m².
  2. Torque. Tangential push F = 6 N at the “equator” (r = R): τ = 6 × 0.5 = 3 N·m.
  3. α. α = 3/1.0 = 3 rad/s².
  4. Compare. A hoop of the same M, R (I = 2.5) would manage only 1.2 rad/s² under the same push — the sphere’s small I makes it eager.
Common mistake: using the diameter as the lever arm (6 × 1.0 = 6 N·m). Torque’s r is pivot-to-push-point = one radius.
Your turn — Sphere M = 5 kg, R = 0.4 m; tangential F = 8 N at surface. α = ?

Answer: 10 rad/s². I = 0.4 × 5 × 0.16 = 0.32; τ = 8 × 0.4 = 3.2; α = 3.2/0.32 = 10 rad/s².

Memorization tips

  • Say it aloud: “sphere: two-fifths M R squared.” The 2/5 is the smallest fraction on the sheet — the sphere is the bottom of the ladder.
  • Ladder chant: hoop 1, disk 1/2, sphere 2/5. Descending fractions for increasingly 3D shapes.
  • Any diameter: unlike the rod, there’s no wrong central axis — symmetry means every diameter gives the same I. One less thing to check.
  • Sanity bounds: 0 < 2/5 < 1/2. If your sphere I ever exceeds the disk’s, something’s off.
  • Solid only: “spherical shell” or “hollow” → stop. The 2/3 shell formula isn’t on this sheet — don’t improvise it.

Final challenge

Five mixed questions — basics, judgment calls, and the traps, all in one. Score 5/5 and the solid sphere is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the moment of inertia of a solid sphere?

I = (2/5)·M·R² about any diameter: the smallest of the sheet’s shape formulas, because a sphere packs its mass closest to the axis on average.

Where does the 2/5 come from?

From stacking the sphere as thin disks and integrating: each disk contributes ½·dm·r(z)², and the integral over the sphere gives (2/5)·M·R².

Which axis does I = (2/5)MR² apply to?

Any axis through the center (any diameter) — a uniform sphere looks the same from every direction, so every central axis gives the same I.

Why is the sphere’s I smaller than the disk’s?

In 3D the mass can hide near the axis in two transverse directions, not just one. The r² average drops from R²/2 (disk) to (2/5)R² (sphere).

Does the formula work for planets?

Approximately: treating a planet as a uniform solid sphere gives I = (2/5)MR². Real planets are denser at the core, so their true I is a bit smaller — but 2/5 is the standard first estimate.

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