AP Physics C: Mechanics · Unit 5: Torque and Rotational Dynamics · Lesson 5.4

Deep Dive: Rotational Inertia

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
5.4.A.1Concept

What Is Rotational Inertia?

Rotational inertia measures a rigid system's resistance to changes in its rotation — the rotational analog of ordinary mass, which measures resistance to changes in linear motion (Unit 2). But rotational inertia has an extra layer of complexity that ordinary mass doesn't: it depends not just on how much mass a system has, but on how that mass is distributed relative to the axis of rotation.

🔑Two objects can have identical total mass and still have wildly different rotational inertia, purely because their mass sits at different distances from the axis. This single idea — that distribution matters, not just total amount — is the whole story of this lesson.
5.4.A.25.4.A.3Math

Point Masses and Discrete Systems

For a single point mass rotating a perpendicular distance r from an axis:

I = mr²
⚠️Notice the square — rotational inertia grows much faster with distance than with mass. Doubling a point mass's radius quadruples its contribution to rotational inertia, while doubling its mass only doubles it.

For a system made of several point masses, total rotational inertia is simply the sum of each individual contribution:

I(tot) = Σ Iᵢ = Σ mᵢrᵢ²

Explore this directly below — the exact three-point-mass system from this unit's header art.

Three point masses on a rotating disk, matching this unit's header art exactly. Adjust each mass and radius — I(total) = Σmᵢrᵢ² updates live.

m₁2.0kg40cm
m₂3.0kg65cm
m₃1.5kg25cm
I₁ = (2.0)(0.40)² = 0.3200 kg·m²I₂ = (3.0)(0.65)² = 1.2675 kg·m²I₃ = (1.5)(0.25)² = 0.0938 kg·m²I(total) = 1.6813 kg·m²

Notice how much more sensitive I is to radius than to mass — doubling a mass doubles its contribution, but doubling its radius quadruples it.

ExampleGuided Example — Total Rotational Inertia of a System

Three point masses sit on a light rod: 2 kg at 0.3 m from the axis, 1 kg at 0.5 m, and 3 kg at 0.15 m. Find the system's total rotational inertia.

Step 1Apply I = mr² to each mass individually
I₁ = (2)(0.3)² = 0.18 kg·m². I₂ = (1)(0.5)² = 0.25 kg·m². I₃ = (3)(0.15)² = 0.0675 kg·m²
5.4.A.4Math

Continuous Objects and Comparing Common Shapes

For a solid object made of continuous, spread-out mass rather than discrete points, the sum becomes an integral over every differential mass element dm:

I = ∫ r² dm

Here, r is each differential mass element's own perpendicular distance from the axis of rotation — exactly the same idea as the discrete sum above, just handled with calculus instead of simple addition.

ExampleWorked Example — Rotational Inertia of a Uniform Rod About Its End

Derive the rotational inertia of a uniform rod of mass M and length L, rotating about an axis through one end, perpendicular to the rod.

This same integration process — done for different shapes — produces the standard rotational inertia formulas you'll see referenced throughout the rest of this unit. Compare several of them directly below, using the same mass and characteristic size for each.

Same mass, same characteristic size — four very different rotational inertias. This is the classic "why does a hoop resist spinning up more than a solid disk" comparison.

M (kg)2.0
R or L (m)0.30
Hoop (about center) (MR²)0.180 kg·m²
Solid disk (about center) (½MR²)0.090 kg·m²
Solid sphere (about diameter) (⅖MR²)0.072 kg·m²
Rod (about center) (1/12 ML²)0.015 kg·m²

All four shapes share the same mass and characteristic radius/length — the hoop wins every time, since every bit of its mass sits at the maximum possible distance r = R from the axis, while the disk and sphere have mass spread closer to the center too.

🔑This directly explains why a hoop has more rotational inertia than a solid disk (or puck) of the identical mass and radius: every bit of the hoop's mass sits at exactly r = R, the maximum possible distance, while the disk spreads its mass across every radius from 0 up to R — most of it much closer to the axis, contributing far less.
← Back to Lesson 5.4Ready for 5.5? Rotational Equilibrium is Newton's first law, spinning — Στ = 0.