Simulation · Physics · Class 11
Spinning up a flywheel
From the lesson Dynamics of rotation in System of Particles and Rotational Motion. Change the values and watch what happens.
Spinning up a flywheelPhysics · Class 11
The idea behind it
NCERT §6.11
- 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.