Rotational Inertia
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Rotational inertia I measures how hard it is to change an object’s spin. It depends on the mass and how far that mass is from the axis: I = Σmr², or I = ∫r² dm for a continuous object.
1. Calculating I
2. Common shapes
Mass far from the axis counts much more (r²). A hoop has twice the rotational inertia of a disk with the same mass and radius, so the same torque spins it up half as fast.
For any axis, I is smallest through the center of mass. Moving the axis a distance d away adds Md².
3. Example, then try it
A light rod holds a 1.0 kg mass 0.20 m from an axis and a 2.0 kg mass 0.50 m from it. Find I.
I = (1.0)(0.20)² + (2.0)(0.50)² = 0.04 + 0.50 = 0.54 kg·m². The far mass dominates.
Derive the rotational inertia of a uniform rod (mass M, length L) about one end. Check it with the parallel-axis theorem, then evaluate for M = 1.2 kg, L = 0.50 m.
Set up: λ = M/L, so dm = (M/L) dx and I = ∫0L x² (M/L) dx.
Integrate: I = (M/L)(L³/3) = ML²/3.
Parallel axis: Iend = Icm + M(L/2)² = ML²/12 + ML²/4 = ML²/3 ✓.
Numbers: I = (1.2)(0.50)² ÷ 3 = 0.10 kg·m² (about the center: 0.025 kg·m²).
Answer: ML²/3 = 0.10 kg·m².
4. Common mistakes
For a given object, about which axis (out of a set of parallel axes) is the rotational inertia smallest?
I = Icm + Md², and Md² is never negative, so the axis through the center of mass gives the smallest I.
Ready to practice?
5 questions on Rotational Inertia, each with a hint and an explanation for every wrong answer.