Lesson 3 · 14 min
Rotation of rigid bodies
Mass is not enough. Where the mass sits decides how hard a shaft is to spin.
Moment of inertia
The rotational analogue of F = ma is τ = Iα. I = ∫ r² dm. Mass far from the axis contributes more. A flywheel is a deliberate I. A robot arm is an accidental one, and it changes as the arm extends — which is why joint motors are sized at worst-case inertia, not average.
τ = Iα KE_rot = ½ Iω² I_parallel = I_cm + Md²
Solid cylinder about its axis: ½MR². Thin rod about center, perpendicular: (1/12)ML². Hoop about center: MR².
Worked example
Grinding wheel
- Given: I = 0.25 kg·m²
- Given: spins up from rest to 1800 rpm in 8.0 s
Find: Required torque (constant)
- ω = 1800 × 2π / 60 = 188.5 rad/s
- α = ω/t = 188.5 / 8.0 = 23.6 rad/s²
- τ = Iα = 0.25 × 23.6
5.9 N·m
Check
Two wheels have equal mass and radius. One is a hoop, one is a disk. Same torque, from rest. After 2 s,