When the angular acceleration is constant, turning follows the same equations as motion with constant acceleration, with each letter swapped for its rotational twin.
Why it works
The straight-line equations came from two facts: velocity changes at a steady rate, and displacement is the area under the velocity–time graph. Both facts hold for turning. If $\alpha$ is constant, $\omega$ rises along a straight line in time, and the angle turned is the area under the $\omega$–$t$ graph. So the algebra is identical; only the names change.
| Along a line | Turning |
|---|---|
| $v = v_0 + at$ | $\omega = \omega_0 + \alpha t$ |
| $\Delta x = v_0 t + \tfrac{1}{2}at^2$ | $\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$ |
| $v^2 = v_0^2 + 2a\Delta x$ | $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$ |
Key idea
With constant $\alpha$: $\omega = \omega_0 + \alpha t$, $\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$ and $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$. Angles in radians, $\omega$ in rad/s, $\alpha$ in rad/s². Divide radians by $2\pi$ to count revolutions.
Worked examples
Worked example: Speeding up
A wheel turning at $2.0\ \text{rad/s}$ speeds up at $2.0\ \text{rad/s}^2$ for $4.0\ \text{s}$. Find its final $\omega$ and the angle it turns.
- $\omega = 2.0 + (2.0)(4.0) = 10\ \text{rad/s}$.
- $$\Delta\theta = (2.0)(4.0) + \tfrac{1}{2}(2.0)(4.0)^2 = 8.0 + 16 = 24\ \text{rad},$$ which is $24 / 2\pi = 3.8$ revolutions.
Answer: $10\ \text{rad/s}$; $24\ \text{rad}$
Worked example: Coasting to a stop
A grinding wheel at $1800$ rpm is switched off and slows at $5.0\ \text{rad/s}^2$. How many revolutions does it make before stopping?
- $\omega_0 = 1800 \times \dfrac{2\pi}{60} = 188.5\ \text{rad/s}$; slowing means $\alpha = -5.0\ \text{rad/s}^2$.
- No time is given, so use $0 = \omega_0^2 + 2\alpha\Delta\theta$: $\Delta\theta = \dfrac{188.5^2}{2(5.0)} = 3553\ \text{rad}$.
- Revolutions: $3553 / 2\pi = 565$.
Answer: about $565$ revolutions
Worked example: How long it takes
For the same grinding wheel, how long does it take to stop?
- $\omega = \omega_0 + \alpha t$ with $\omega = 0$: $t = \dfrac{0 - 188.5}{-5.0} = 37.7\ \text{s}$.
- Check with the average angular velocity: $\Delta\theta = \tfrac{1}{2}(188.5 + 0)(37.7) = 3553\ \text{rad}$, the same angle as before.
Answer: $37.7\ \text{s}$
Common mistake
Putting a distance where the angle belongs, or mixing revolutions with radians. In $\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$ every term must be in radians, so $\omega$ must be in rad/s, not rpm. A distance (meters) never goes in for $\theta$; if you want the distance a point travels, find $\theta$ first and then use $s = r\theta$. Check the units: rad/s times s gives rad.