Skip to content

Moment of Inertia

Source lecture(s): SC133 Lec 12

Intuition

Spin a pencil about its long axis: trivial. Spin the same pencil end-over-end: harder. Same mass — different distribution of mass around the axis. The moment of inertia \(I\) is rotation's version of mass, but with a twist: distance from the axis counts squared. Mass far from the axis resists spinning ferociously; mass on the axis doesn't resist at all.

Definition

\[\boxed{\,I = \sum_i m_i r_i^2 \;\;\longrightarrow\;\; I = \int r^2\, dm\,}\]

where \(r\) is each element's perpendicular distance from the rotation axis — \(I\) is meaningless until you name the axis.

The standard catalogue (mass \(M\), about the center unless noted)

Body Axis \(I\)
Hoop / thin ring center, ⊥ plane \(MR^2\)
Solid disk / cylinder central axis \(\tfrac12 MR^2\)
Solid sphere diameter \(\tfrac25 MR^2\)
Hollow sphere diameter \(\tfrac23 MR^2\)
Thin rod, length \(L\) center, ⊥ rod \(\tfrac1{12} ML^2\)
Thin rod, length \(L\) end, ⊥ rod \(\tfrac13 ML^2\)

The pattern: the more mass lives near the rim, the closer the prefactor is to 1. Ranking objects by \(I/MR^2\) predicts the classic race — rolling downhill, the solid sphere beats the disk beats the hoop, regardless of mass or radius.

Parallel-axis theorem

Know \(I_\text{cm}\) about the center of mass? Any parallel axis a distance \(d\) away:

\[I = I_\text{cm} + Md^2\]

Check with the rod: \(\tfrac1{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2\) ✓. The theorem also proves the CM axis is always the easiest axis of a given direction to spin about.

Sample derivation: solid disk

Slice into rings of radius \(r\), width \(dr\): \(dm = \dfrac{M}{\pi R^2}\,2\pi r\,dr\).

\[I = \int_0^R r^2\,dm = \frac{2M}{R^2}\int_0^R r^3\,dr = \frac{2M}{R^2}\cdot\frac{R^4}{4} = \tfrac12 MR^2\]

Every entry in the catalogue is this integral with different geometry — and this same \(\int r^2 dm\) resurfaces as the area moment \(I_{xx}\) in hydrostatic forces on surfaces (PC316).

Common mistakes

  • Quoting \(I\) without an axis. The rod's \(I\) changes by a factor of 4 between center and end.
  • Treating \(I\) like mass in every formula without checking the axis — the parallel-axis theorem exists precisely because axes matter.
  • Adding \(Md^2\) to a non-CM value. The theorem only jumps from the CM axis.
  • Assuming heavier means harder to spin. A heavy solid sphere can out-accelerate a light hoop of equal radius under equal torque.

Knowledge graph position

Prerequisites: Rotation, Center of mass. Leads to: Rolling, torque & angular momentum, pendulum (physical pendulum).

Quiz

Q1 (computational). Four 1 kg point masses sit at the corners of a square of side 2 m. Find \(I\) about the axis through the center, ⊥ to the plane.

Answer

Each mass is \(r = \sqrt2\) m from the center: \(I = 4\times1\times2 = 8\,\text{kg·m}^2\).

Q2 (conceptual). A figure skater pulls in her arms and spins faster. What changed, and what stayed constant?

Answer

Pulling mass toward the axis reduces \(I\); with no external torque, angular momentum \(L = I\omega\) is constant, so \(\omega\) rises. (Her kinetic energy \(\tfrac12 I\omega^2 = L^2/2I\) increases — her muscles supply it.)

Q3 (multiple choice). Which reaches the bottom of a ramp first (rolling without slipping): (a) hoop (b) solid disk (c) solid sphere

Answer

(c). Acceleration \(a = g\sin\theta/(1 + I/MR^2)\): smallest \(I/MR^2\) (\(2/5\)) wins. Mass and radius cancel — only the shape races.