Angular Displacement (θ = ω₀t + ½αt²)

Under constant angular acceleration the total angle turned splits cleanly into two parts: the angle you would have covered coasting at the initial rate, $\omega_0 t$, plus the extra contributed by speeding up, $\tfrac{1}{2}\alpha t^2$. The second term grows with the square of time, for the same reason the linear version does — the extra angular velocity builds linearly from zero, so its average over the interval is half its final value.

A washing machine drum accelerating from rest at 15 rad/s² turns through $\theta = 0 + \tfrac{1}{2} \times 15 \times 2^2 = 30$ rad in its first two seconds. To read that as revolutions, divide by $2\pi$: about 4.8 turns. A centrifuge spinning up from 1000 rpm (105 rad/s) at 50 rad/s² for 8 s covers $105 \times 8 + \tfrac{1}{2} \times 50 \times 64 = 840 + 1600 = 2440$ rad, roughly 388 revolutions.

That total-revolutions figure is the practical reason the equation exists. It is how you work out how many turns a spool takes up during a start ramp, how far a workpiece rotates while a spindle comes to speed, and how much cable a winch pays out before it reaches its running rate. It is the exact twin of $d = v_0 t + \tfrac{1}{2}at^2$, and the two can be checked against each other on any rolling wheel: $d = r\theta$ should hold at every instant.

Watch the unit on the answer, because radians are unforgiving here. The output is in radians, and 30 rad is not 30 revolutions and not 30 degrees — it is 4.8 revolutions, or about 1719 degrees. Dividing by 360 instead of $2\pi$ is a common slip and gives an answer 57 times too small. Note also that this page solves for $\theta$, $\omega_0$ and $\alpha$ but deliberately not for $t$: rearranged for time it becomes a quadratic, $\tfrac{1}{2}\alpha t^2 + \omega_0 t - \theta = 0$, which can have two positive roots, and rather than silently choosing one for you the calculator leaves that case to the angular acceleration page, where time comes out unambiguously. Last, the formula assumes $\alpha$ is genuinely constant throughout. A motor whose torque falls off as it approaches synchronous speed does not accelerate uniformly, and applying this over a whole start ramp will overstate the angle turned.