Rotational dynamics

Spinning objects obey rules that mirror straight-line motion. Torque plays the part of force, moment of inertia plays the part of mass, and angular velocity plays the part of velocity. These ideas explain why skaters spin faster with their arms pulled in and how flywheels store energy.

Key facts

  • Torque τ = rF (force × perpendicular distance from the axis).
  • Moment of inertia I = Σmr²: point mass mr², solid disc ½MR², hoop MR².
  • τ = Iα (the rotational version of F = ma).
  • Rotational kinetic energy = ½Iω²; angular momentum L = Iω.
  • With no external torque, angular momentum is conserved: I₁ω₁ = I₂ω₂.
  • Rotational kinematics: ω = ω₀ + αt, θ = ω₀t + ½αt²; work W = τθ, power P = τω.

Worked examples

Example 1

A torque of 12 N m acts on a wheel with I = 0.40 kg m². Find its angular acceleration and its speed after 5 s from rest.

  1. Angular acceleration = torque ÷ moment of inertia. α = 12 ÷ 0.40 = 30 rad/s²
  2. It gains 30 rad/s every second. ω = 0 + 30 × 5 = 150 rad/s

Answer: 30 rad/s²; 150 rad/s

Example 2

A skater spins at 2.0 rad/s with I = 4.0 kg m². She pulls her arms in so I = 1.6 kg m². Find her new angular speed and the change in kinetic energy.

  1. No outside torque, so Iω stays the same. 4.0 × 2.0 = 1.6 × ω, ω = 5.0 rad/s
  2. Kinetic energy before and after: ½ × 4.0 × 2.0² = 8 J → ½ × 1.6 × 5.0² = 20 J
  3. The extra 12 J comes from the work she does pulling her arms in.

Answer: 5.0 rad/s; KE rises from 8 J to 20 J

Example 3

A solid cylinder rolls from rest down a slope, dropping 1.2 m. Find its speed at the bottom. (g = 9.8 m/s²)

  1. GPE becomes kinetic energy of moving along AND of spinning. mgh = ½mv² + ½Iω²
  2. For a solid cylinder I = ½mr² and ω = v/r, so the spinning part is ¼mv². mgh = ¾mv²
  3. The mass cancels. v = √(4 × 9.8 × 1.2 ÷ 3) = 3.96 m/s

Answer: 3.96 m/s