Quick answer: Moment of inertia is I = ∫r² dm, the sum of each mass element times the square of its distance from the rotation axis. The parallel axis theorem gives I = Icm + Md² for a parallel axis, and torque produces angular acceleration through Στ = Iα.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 5
What is moment of inertia?
Moment of inertia measures how strongly a body resists changes in its rotation about a chosen axis. Each piece of mass contributes its mass times the square of its distance from the axis, so mass far from the axis counts much more.
The moment of inertia of a body about an axis is I = Σ miri², which becomes I = ∫ r² dm for a continuous body. Here r is the perpendicular distance from the axis to the mass element dm. The unit is kg·m².
How do you find I by integration?
Choose an element dm whose points are all at the same distance from the axis, write dm in terms of one variable, then integrate. For a thin rod of mass M and length L, dm = (M/L) dx.
About an axis through one end: I = ∫₀^L x²(M/L) dx = ML²/3. About the centre, integrate from −L/2 to L/2 to get ML²/12.
Ananya in Pune models a rod for a school windmill: M = 0.60 kg and L = 1.0 m. About one end, I = ML²/3 = 0.60 × 1.0²/3 = 0.20 kg·m². About its centre, I = ML²/12 = 0.60 × 1.0²/12 = 0.050 kg·m², four times smaller, because the mass sits closer to the axis.
For a uniform disc, use thin rings of radius r and thickness dr. The ring has area 2πr dr and mass dm = σ(2πr dr) where σ = M/πR². Then I = ∫₀^R r²σ2πr dr = σπR⁴/2 = ½MR².
| Body (uniform) | Axis | I |
|---|---|---|
| Thin hoop | Central, perpendicular to plane | MR² |
| Solid disc or cylinder | Central axis | ½MR² |
| Solid sphere | Through the centre | (2/5)MR² |
| Thin spherical shell | Through the centre | (2/3)MR² |
| Thin rod | Through the centre / one end | ML²/12 / ML²/3 |
What does the parallel axis theorem say?
If you know Icm about an axis through the centre of mass, the moment of inertia about any parallel axis at distance d is larger by Md².
I = Icm + Md². I is the moment of inertia about the new axis (kg·m²), Icm that about the parallel axis through the centre of mass, M the total mass (kg) and d the distance between the axes (m).
Check with the rod: I = ML²/12 + M(L/2)² = ML²/3, which matches the integral. For a disc of M = 2.0 kg and R = 0.20 m, the axis through a point on the rim gives I = ½MR² + MR² = 1.5 × 2.0 × 0.040 = 0.12 kg·m².
How do torque and angular acceleration connect?
The torque of a force is τ = rF sin θ, where θ is the angle between the force and the position vector from the axis (so rF sin θ is force times lever arm). Newton’s second law for rotation reads Στ = Iα, and the angular momentum is L = Iω.
Kofi in Accra winds a rope around a flywheel (solid disc, M = 4.0 kg, R = 0.25 m) and pulls it with a steady 10 N. τ = RF = 0.25 × 10 = 2.5 N·m. I = ½MR² = ½ × 4.0 × 0.0625 = 0.125 kg·m². α = τ/I = 20 rad/s². After 3.0 s from rest, ω = 60 rad/s and K = ½Iω² = 225 J.
Use the perpendicular distance from the axis, not the distance from an origin. The parallel axis theorem must start from an axis through the centre of mass: you cannot add Md² to the moment of inertia about an end.
Learn the rod, disc and sphere results, then build others by adding pieces. A compound body has I equal to the sum of the I values of its parts, each about the same axis.
- I = ∫r² dm; unit kg·m²
- Rod: ML²/12 (centre), ML²/3 (end); disc ½MR²; sphere (2/5)MR²
- Parallel axis theorem: I = Icm + Md²
- Torque τ = rF sin θ; Στ = Iα
- Angular momentum L = Iω; Krot = ½Iω²
Moment of inertia by integration and torque: frequently asked questions
What is the formula for moment of inertia?
I = Σmiri² for separate masses, and I = ∫r² dm for a continuous body. The distance r is measured perpendicular to the rotation axis.
How do you find moment of inertia by integration?
Choose a mass element dm at one distance from the axis, write dm in terms of a single variable such as dx or dr, then integrate r² dm over the whole body.
What does the parallel axis theorem state?
I = Icm + Md², where Icm is the moment of inertia about a parallel axis through the centre of mass and d is the distance between the two axes.
What is the rotational form of Newton’s second law?
The net torque about a fixed axis equals the moment of inertia times angular acceleration, Στ = Iα. It plays the same role as F = ma.
Why does a hoop roll more slowly than a disc?
The hoop has all its mass at the rim, so I is larger (MR² against ½MR²). More of the energy goes into rotation and less into translation, giving a smaller speed.
