Rotational Work & Power

Take a rigid body on a fixed axis and push on it at radius r with a force F. During a small turn dθ the point you push moves along an arc of length r dθ, tangent to its circle, so the work is F ⋅ ds = Fₜₐₙ r dθ: only the tangential component counts, and the radial component, however large, moves nothing along its own line. But Fₜₐₙ r is the torque about the axis, so dW = τ dθ, and over a finite motion W = ∫τ_z dθ. Three things are packed into that line. The angle must be in radians, because r dθ is an arc length only then. The sign is the sign of τ_z dθ: a torque in the sense of the rotation does positive work, one against it negative. And although torque and work are both printed in N m, a torque is not an energy: it is the energy delivered per radian turned. Push a playground roundabout tangentially with 100 N at 1.5 m, a constant 150 N m, through a quarter turn and W = τΔθ = 150 × π/2 = 236 J; the rim agrees, since your hands travel an arc of 1.5 × π/2 = 2.36 m and 100 × 2.36 = 236 J.

When the torque changes as the body turns, W = ∫τ dθ is a genuine integral: the area between the τ(θ) curve and the θ axis, counted negative wherever the torque opposes the rotation. Two cases recur. A torsion spring obeys τ = −κθ, the rotational Hooke's law, so twisting it slowly from its relaxed angle to θ takes ½κθ² of work from you and stores it: with κ = 0.80 N m rad⁻¹ wound through 2.5 rad that is ½ × 0.80 × 2.5² = 2.5 J, the area of a triangle of base 2.5 rad and height 2.0 N m. Gravity on a uniform rod pivoted at one end gives τ = (mgL/2) cos θ, with θ measured below the horizontal — largest when the rod is level, zero when it hangs down. Integrating from θ₁ to θ₂ gives (mgL/2)(sin θ₂ − sin θ₁), which is mg times the drop of the centre of mass. That is no coincidence: gravity is conservative, so its torque integral must equal −ΔU of the rod–Earth system, and the two routes are one calculation in different clothes. The graphical rule also handles torque data with no formula behind it — tabulate τ against θ and the trapezium rule gives W.

Differentiate dW = τ dθ with respect to time and P = τω, the rotational form of P = F ⋅ v. It is an instantaneous statement: at every moment the power flowing into the body equals the torque on it times its angular velocity, both about the same axis and with ω in rad s⁻¹. That last condition is where most errors live, because engines are quoted in rev min⁻¹. A motor delivering 180 N m at 3000 rpm has ω = 3000 × 2π/60 = 314 rad s⁻¹ and P = 180 × 314 = 5.65 × 10⁴ W, a 56.5 kW motor; feeding in 3000 instead of 314 overstates the power by a factor of 9.55. The product also explains gearing: a gearbox that halves the speed doubles the torque and leaves τω, and so the power, untouched apart from losses. Average power over a process is a different quantity — total work divided by the time taken — and it matches the instantaneous value throughout only if the product τω never changes, most simply when τ and ω are both steady. In a spin-up from rest under a constant net torque, ω climbs linearly with time, so the power at the end is twice the average over the run.

Combine the two rotational laws you already have. Newton's second law for rotation says τₙₑₜ = Iα, and the chain rule rewrites α = dω/dt as (dω/dθ)(dθ/dt) = ω dω/dθ. So τₙₑₜ dθ = I ω dω, and integrating both sides between the start and the end of the motion gives Wₙₑₜ = ∫τₙₑₜ dθ = ½Iω₂² − ½Iω₁². This is the work–energy theorem for a fixed axis — the same statement as ΣF ⋅ ds = Δ(½mv²) summed over every particle of the body — and the previous lesson's ½Iω² is exactly the quantity the torque integral changes. Use it as you use its linear twin: it delivers angular speeds at given angles without α or the time. A flywheel with I = 0.60 kg m² taken from rest to 100 rad s⁻¹ needs W = ½ × 0.60 × 100² = 3000 J of net work, whatever torque supplies it; if that torque is a constant 12 N m the angle is 3000/12 = 250 rad, or 39.8 revolutions. Kinematics agrees — α = 12/0.60 = 20 rad s⁻², and ω² = 2αθ gives θ = 10000/40 = 250 rad — but the energy route never needed α.

Real spin-ups have more than one torque: a motor drives, bearing friction resists. Each contributes its own ∫τ dθ and the theorem sums them with their signs, so Wₘₒₜₒᵣ + W_friction = ΔK with W_friction negative. That is often the cleanest way to measure a friction torque: spin the wheel up, remove the drive, count the revolutions to rest, convert them to radians, and τ_f = ½Iω²/Δθ is the friction torque averaged over that angle. A conservative torque can be moved to the other side of the ledger as potential energy: a uniform rod of length 1.2 m released from horizontal about a frictionless end pivot has I = ⅓mL², and mg(L/2) = ½(⅓mL²)ω² gives ω = √(3g/L) = √(3 × 9.81/1.2) = 4.95 rad s⁻¹ at the bottom, the mass cancelling. Know the boundaries of the method. It assumes a rigid body on a fixed axis, so one θ, one ω and one I describe everything. It gives states, not durations — for the time taken you must return to τ = Iα. And a body whose axis itself translates, a rolling wheel or a swung bat, carries translational kinetic energy too; that accounting is the next lesson.