Moment of Inertia by Integration & the Parallel-Axis Theorem

You’ll be able to

Compute moments of inertia by integrating over a continuous mass distribution
Apply the parallel-axis theorem to shift the axis of rotation
Explain why moment of inertia depends on the axis, not only on the object

Rotational inertia depends on where the mass is

Mass measures resistance to linear acceleration and is a property of the object alone. Moment of inertia measures resistance to angular acceleration and is a property of the object and the chosen axis. Because I = Σm_i r_i², mass far from the axis counts far more heavily — the distance is squared. Two objects of identical mass can have moments of inertia differing by a factor of several, and the same object rotated about a different axis is effectively a different problem.

Moment of inertia, discrete and continuous

I = Σ m_i r_i² I = ∫ r² dm with dm = λ dx, σ dA or ρ dV

The r in the integral is the perpendicular distance from the axis to the mass element, not the distance from the origin. Choosing dm correctly is the whole technique.

Setting up the integral

Three steps, every time. First, choose a mass element dm whose points all lie at the same distance r from the axis. Second, express dm in terms of a coordinate using the density — for a uniform rod of mass M and length L, the linear density is λ = M/L so dm = (M/L)dx. Third, write r in that coordinate and integrate over the object. For a rod rotating about one end, r = x and I = ∫₀^L (M/L)x² dx = (M/L)(L³/3) = ML²/3.

The parallel-axis theorem

Once you know I about an axis through the center of mass, you get it about any parallel axis for free: I = I_cm + Md², where d is the distance between the axes. Two consequences follow immediately. The moment of inertia is always smallest about an axis through the center of mass, since Md² is never negative. And the theorem only works between parallel axes — it says nothing about tilting the axis. Checking it against the rod: I_cm = ML²/12, and shifting to the end gives ML²/12 + M(L/2)² = ML²/12 + ML²/4 = ML²/3, matching the direct integration exactly.

Worked example

Derive the moment of inertia of a uniform rod of mass M and length L about an axis through one end, perpendicular to the rod. Then verify the result using the parallel-axis theorem and I_cm = ML²/12.

1.Linear density is λ = M/L, so a slice of width dx at distance x from the pivot has dm = (M/L)dx.
2.Every point of that slice sits a distance x from the axis, so it contributes x² dm.
3.Integrate: I = ∫₀^L x²(M/L)dx = (M/L)[x³/3]₀^L = (M/L)(L³/3) = ML²/3.
4.Check: the center of mass is at L/2, so I = I_cm + Md² = ML²/12 + M(L/2)² = ML²/12 + 3ML²/12 = ML²/3. The two agree.

Answer: I_end = ML²/3, confirmed independently by direct integration and by the parallel-axis theorem — four times the value about the center, because d = L/2 is squared

Checkpoint

For a given object, the moment of inertia is smallest about an axis that —

Watch out

The parallel-axis theorem requires the reference axis to pass through the center of mass. Applying it between two arbitrary axes — say from the end of a rod to its quarter point — gives a wrong answer. Go from the arbitrary axis back to the center of mass first, then out to the new one.

Checkpoint

A hoop and a solid disk have the same mass and radius. About their central axes, the hoop's moment of inertia is larger because —

Answer the 2 checkpoints as you read.