Constant Angular Acceleration

The previous topics defined the two rates this one integrates: ω = dθ/dt and α = dω/dt. Now assume α keeps one value from 0 to t. Integrating once, ∫₀ᵗ α dt′ = ω(t) − ω₀, so ω(t) = ω₀ + αt: angular velocity is a straight line in time with slope α. Integrating that line, θ(t) − θ₀ = ∫₀ᵗ (ω₀ + αt′) dt′ = ω₀t + ½αt². On the ω–t graph, ω₀t is the rectangle under the starting value and ½αt² is the triangle sitting on it, which is where the ½ comes from. For a flywheel with ω₀ = 4.00 rad s⁻¹ and α = 1.50 rad s⁻², after 2.00 s: ω = 4.00 + 3.00 = 7.00 rad s⁻¹ and θ − θ₀ = 8.00 + 3.00 = 11.0 rad. This is the constant-acceleration calculation with x, v and a renamed θ, ω and α — the same mathematics, so the same single assumption carries all the weight.

Five quantities — Δθ = θ − θ₀, ω₀, ω, α and t — are tied by two independent equations, so any three fix the other two. Two further forms save algebra. To remove t, use the chain rule: α = dω/dt = (dω/dθ)(dθ/dt) = ω dω/dθ, and with α constant, ∫ω dω = ∫α dθ gives ω² = ω₀² + 2αΔθ. To remove α, note that a quantity linear in time averages to the mean of its end values, so Δθ = ½(ω₀ + ω)t. Each of the four relations leaves out exactly one of the five quantities: name the one you neither know nor want, and use the relation without it. A wheel that slows from 12.0 to 4.00 rad s⁻¹ while turning 20.0 rad has α = (16.0 − 144)/40.0 = −3.20 rad s⁻², and it takes t = 40.0/16.0 = 2.50 s. Neither answer needed the other.

θ₀ and ω₀ are the constants of integration, and the physics does not hand them over: they are the angle and angular velocity at whatever instant you call t = 0. So three choices come before any arithmetic — the positive sense of rotation, the zero of angle, and the zero of the clock. If α switches on at t₁ rather than at 0, replace t by t − t₁ and use ω(t₁), not an earlier value. Motion in stages is stitched the same way: the final θ and ω of one stage are the initial conditions of the next. Across the join θ and ω are continuous, because a finite α cannot change ω in zero time, while α itself may jump. A fan driven from rest at 2.00 rad s⁻² for 5.00 s reaches 10.0 rad s⁻¹ after 25.0 rad; when it then coasts at −0.500 rad s⁻², that stage starts from ω₀ = 10.0 rad s⁻¹ and sweeps another 100 rad in the 20.0 s it takes to stop.

Once the positive sense is fixed, ω and α carry independent signs. The same sign means the rotation is speeding up; opposite signs mean it is slowing, whichever way it turns. When they are opposite, ω = ω₀ + αt passes through zero at tₛ = −ω₀/α, and θ(t) turns around there, after sweeping Δθₛ = −ω₀²/(2α). Take ω₀ = 6.0 rad s⁻¹ and α = −1.5 rad s⁻²: the wheel stops at 4.0 s after 12 rad. Beyond tₛ the equations predict a reversal, and at 6.0 s they give Δθ = 36 − 27 = 9.0 rad, so the wheel has come back 3.0 rad. Whether that happens depends on what supplies α. A string wound on the axle with a hanging mass pulls the same way whichever way the axle turns, so the reversal is real. A friction brake stops acting once the wheel is at rest, so the wheel stays at 12 rad. After a genuine reversal, keep net and total apart: 9.0 rad of displacement came from 15 rad of turning.

Each relation is homogeneous in the angle unit — every term carries the same power of it — so the equations hold in rev, rev s⁻¹ and rev s⁻² just as well as in radians, provided nothing is mixed. The classic slip is ω₀ in rpm alongside α in rad s⁻². Convert once, at the start: 1 rev = 2π rad, so 1 rpm = 2π/60 rad s⁻¹ = 0.1047 rad s⁻¹, and 3000 rpm is 314 rad s⁻¹. A record platter at 45.0 rpm = 0.750 rev s⁻¹ that stops uniformly in 3.00 s has α = −0.750/3.00 = −0.250 rev s⁻², which is −1.57 rad s⁻², and turns ½(0.750 + 0)(3.00) = 1.13 rev, or 7.07 rad, on the way. Radians become compulsory only when an angle is multiplied by a radius to give a length or a speed, the step the next topic takes, which is why radians are the safe default.

Constant α is a claim about the cause, and many real rotations break it. An electric motor's torque falls as it speeds up, so ω creeps towards a ceiling instead of rising in a line; air drag on a fan grows with ω; a pendulum's α depends on its angle. Two tests catch this in data. On an ω–t graph the points must lie on one straight line. With angle readings at equal intervals Δt, the angle swept in successive intervals must step by the same amount, since Δθₙ₊₁ − Δθₙ = α(Δt)² — the rotational form of Galileo's odd-number rule, and a test that needs no derivative of noisy data. Sweeps of 1.10, 1.30, 1.50 and 1.70 rad in successive 0.200 s intervals step by 0.200 rad each time, so α = 0.200/0.0400 = 5.00 rad s⁻². Steps that shrink mean α is falling; the constant-α equations would then overshoot late angles, and the honest options are short stretches with their own α, or integrating the measured α(t).