Equations of Rotational Motion

The equations

For constant angular acceleration $\alpha$, starting with initial angular velocity $\omega_0$:

$\omega = \omega_0 + \alpha t$

$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

$\omega^2 = \omega_0^2 + 2\alpha\theta$

These are exact analogues of the linear equations of motion, with every linear quantity replaced by its rotational counterpart.

Derivation

First equation: By definition, $\alpha = \frac{d\omega}{dt}$. Since $\alpha$ is constant:

$\int_{\omega_0}^\omega d\omega = \int_0^t \alpha \, dt \implies \omega - \omega_0 = \alpha t \implies \boxed{\omega = \omega_0 + \alpha t}$

Second equation: $\omega = \frac{d\theta}{dt} = \omega_0 + \alpha t$. Integrate:

$\int_0^\theta d\theta = \int_0^t (\omega_0 + \alpha t) \, dt \implies \boxed{\theta = \omega_0 t + \frac{1}{2}\alpha t^2}$

Third equation: Eliminate $t$ from equations 1 and 2.

From equation 1: $t = \frac{\omega - \omega_0}{\alpha}$

Substitute into equation 2:

$\theta = \omega_0 \cdot \frac{\omega-\omega_0}{\alpha} + \frac{1}{2}\alpha\left(\frac{\omega-\omega_0}{\alpha}\right)^2 = \frac{\omega_0(\omega-\omega_0)}{\alpha} + \frac{(\omega-\omega_0)^2}{2\alpha}$

$2\alpha\theta = 2\omega_0(\omega-\omega_0) + (\omega-\omega_0)^2 = (2\omega_0 + \omega - \omega_0)(\omega-\omega_0) = (\omega+\omega_0)(\omega-\omega_0) = \omega^2 - \omega_0^2$

$\boxed{\omega^2 = \omega_0^2 + 2\alpha\theta}$

Direct analogy

The substitutions are: $v \to \omega$, $u \to \omega_0$, $a \to \alpha$, $s \to \theta$.

Example

A wheel starts from rest and reaches 600 rpm in 10 seconds. Find angular acceleration and number of revolutions.

$\omega_0 = 0$, $\omega = 600 \text{ rpm} = \frac{600 \times 2\pi}{60} = 20\pi$ rad/s, $t = 10$ s

$\alpha = \frac{\omega - \omega_0}{t} = \frac{20\pi}{10} = 2\pi \text{ rad/s}^2$

$\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(2\pi)(100) = 100\pi \text{ rad} = 50 \text{ revolutions}$