Spinning up a flywheel.

A cord wound on a flywheel pulled by a force or a hanging mass: τ = Iα, ω builds by ω² = 2αθ, and work done equals ½Iω².

Make the flywheel reach 10 rad/s after unwinding exactly 2 m of cord.

Using rpm in ω equations; assuming hanging mass tension = mg.

• For a fixed axis only the torque components along the axis matter; the bearings supply the constraint forces that cancel the perpendicular components. So only forces in planes perpendicular to the axis, and only the parts of position vectors perpendicular to the axis, need be considered.
• A force F₁ on a particle at distance r₁ from the axis does work dW = F₁r₁ sin α₁ dθ = τ₁ dθ as the body turns through dθ. For all the forces together, dW = τ dθ.
• Power delivered by a torque: P = τω, like P = Fv.
• For a rigid body there is no internal motion, so this work all goes into kinetic energy: τω = d(½Iω²)/dt = Iωα, which gives τ = Iα, Newton's second law for rotation about a fixed axis.
• The angular acceleration is directly proportional to the torque and inversely proportional to the moment of inertia.
• Analogues: x ↔ θ, v ↔ ω, a ↔ α, M ↔ I, F = Ma ↔ τ = Iα, dW = F ds ↔ dW = τ dθ, ½Mv² ↔ ½Iω², Fv ↔ τω, p = Mv ↔ L = Iω.
• A 20 kg flywheel of radius 20 cm (I = MR²/2 = 0.4 kg m²) pulled by a steady 25 N on a cord wound on its rim: τ = 5.0 N m and α = 12.5 s⁻². When 2 m of cord unwinds, the pull does 50 J of work; θ = 2/0.2 = 10 rad, ω² = 2 × 12.5 × 10 = 250 (rad/s)², so K = ½ × 0.4 × 250 = 50 J, equal to the work since the bearings are frictionless.

Teaching this idea? Put the simulation on your own page or LMS with this code.