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UY1: Work, Energy & Power Of Rotating Object

Rotational work, energy, and power relations with clear guidance on when to integrate torque over angle.

  • University Physics Year 1
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Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the archived enrichment page for rotational work-energy identities in the UY1 mechanics pathway.

At a glance

dW = τ dθ, Δ Kᵣₒₜ = ∫τ dθ, P = τω
  • Common trap: replacing W with τθ when torque is not constant.

Setup

Model:

  • Rigid body rotates about a fixed axis.
  • Consider net external torque about that axis.

Coordinate + sign convention (do this before the algebra):

  • Choose a rotation axis and define positive rotation using the right-hand rule.
  • Measure angular displacement θ in radians.
  • Work is positive when the torque component along the axis has the same sign as dθ.
Pitfall: degrees vs radians

In W = ∫ τ dθ, the angle must be in radians (because it comes from ds = r dθ). If you use degrees, you will be off by a factor of π/180.

Method choice: when this is faster than dynamics

Use rotational work-energy when you need speeds after an angular displacement and you don’t care about the detailed time dependence.

  • If torque is given as a function of angle τ(θ), energy is usually the cleanest route: Δ Kᵣₒₜ = ∫ τ(θ) dθ.
  • If you need ω(t) or θ(t) explicitly, use τ = Iα with α = dω/dt (dynamics).

Core method

Tangential force does rotational work:

dW = vector F · d vector s = Fₜ ds

Since ds = r dθ and τ = rFₜ,

dW = τ dθ

Integrate:

W = ∫_θ₀^θτ dθ

Using τ = Iα with constant I and α = dω/dt:

τ = Idω/dt = Iωdω/dθ

so

dW = Iω dω ⇒ W = (1/2)Iω²-(1/2)Iω₀²

Hence rotational kinetic energy is

Kᵣₒₜ = (1/2)Iω²

Power:

P = dW/dt = τdθ/dt = τω

Non-trivial point: P = τω is instantaneous power; use average values only when conditions are steady.

Worked example

A motor applies constant net torque τ = 6.0 N m to a flywheel with I = 0.50 kg m², starting from rest. Find angular speed after rotating through θ = 10 rad.

Work done:

W = τθ = (6.0)(10) = 60 J

Set W = Δ Kᵣₒₜ:

60 = (1/2)(0.50)ω² ⇒ ω² = 240 ⇒ ω = 15.5 rad s⁻¹

If instantaneous speed is 15.5 rad s⁻¹, then

P = τω = (6.0)(15.5) = 93 W

Checks:

  • Units check: τθ → N m = J, and τω → N m s⁻¹ = W.
  • Sanity check: if θ = 0, then W = 0 and rotational kinetic energy does not change.

Mini-example: variable torque as an area under τ–θ

Suppose the net torque varies with angle as τ(θ) = τ₀ + kθ, applied from θ = 0 to θ = Θ.

Then the work done is the τ–θ area: W = ∫₀^Θ(τ₀ + kθ) dθ = τ₀Θ + (1/2)kΘ².

Checks: if k = 0 you recover W = τ₀Θ (constant torque); units are (N m)(rad) = J since radians are dimensionless.

Practice set

  1. A constant torque of 4.0 N m acts through 12 rad. Find work. Hint: constant torque case. Answer: 48 J.

  2. A rotor has I = 0.20 kg m² and speeds up from 10 to 20 rad s⁻¹. Find Δ Kᵣₒₜ. Hint: use (1/2)I(ω_f²-ωᵢ²). Answer: 30 J.

  3. At an instant, τ = -3.0 N m and ω = 40 rad s⁻¹. Interpret P. Hint: sign of P. Answer: P = -120 W, so torque removes rotational energy (braking).

Summary and next steps

Rotational work and power follow directly from torque: integrate τ over angle for work, multiply by ω for instantaneous power. Always check whether torque is constant before simplifying integrals.