Physics · Engineering Physics · A2 Level

Angular Acceleration

A turbine spinning up from rest, a flywheel braking to a stop, a drill bit reaching full speed in a fraction of a second — none of these rotate at constant angular velocity throughout. Angular acceleration describes how fast the spin rate itself is changing, and reading its behaviour off a graph is a skill worth mastering thoroughly.

7 min read

Defining angular acceleration

Angular acceleration $\alpha$ (alpha) is the rate of change of angular velocity:

$$\alpha = \frac{\Delta \omega}{\Delta t}$$

It is measured in radians per second squared (rad s$^{-2}$), and is the direct rotational analogue of linear acceleration $a = \Delta v/\Delta t$. A positive $\alpha$ means $\omega$ is increasing (the rotation is speeding up, assuming the standard sign convention), while a negative $\alpha$ means $\omega$ is decreasing — which could mean slowing down, or speeding up in the opposite sense of rotation, exactly as with linear acceleration and deceleration.

Engineering situations that call for angular acceleration are common: a motor's rotor accelerating up to its running speed after being switched on, a flywheel decelerating under a braking torque, a turbine's blades speeding up as steam pressure builds. In every case, $\alpha$ captures how quickly the angular velocity itself is changing.

Graphical treatment: reading rotation from graphs

Just as linear kinematics uses displacement-time and velocity-time graphs, rotational kinematics uses angular displacement-time ($\theta$–$t$) and angular velocity-time ($\omega$–$t$) graphs, and the same gradient and area rules apply directly.

  • The gradient of a $\theta$–$t$ graph at any instant gives the angular velocity $\omega$ at that instant, since $\omega = d\theta/dt$.
  • The gradient of an $\omega$–$t$ graph at any instant gives the angular acceleration $\alpha$ at that instant, since $\alpha = d\omega/dt$.
  • The area under an $\omega$–$t$ graph, between two times, gives the angular displacement $\theta$ that occurred in that interval — the rotational equivalent of area under a velocity-time graph giving linear displacement.

A straight-line $\omega$–$t$ graph has a constant gradient, meaning constant (uniform) angular acceleration — this is the case the four rotational suvat equations, covered in the next lesson, apply to directly. A curved $\omega$–$t$ graph, by contrast, has a gradient that changes from point to point, meaning $\alpha$ itself is changing: this is non-uniform angular acceleration, and reading a value of $\alpha$ off such a graph means finding the gradient of a tangent to the curve at the instant of interest, rather than simply dividing rise by run over the whole graph.

Uniform vs non-uniform — the distinction that gets tested

Uniform angular acceleration means $\alpha$ is constant: the $\omega$–$t$ graph is a straight line, and the $\theta$–$t$ graph is a smooth curve (a parabola, for constant non-zero $\alpha$) with a steadily changing gradient. Non-uniform angular acceleration means $\alpha$ itself varies with time: the $\omega$–$t$ graph is curved, not straight. A common error is reading a changing gradient on an $\omega$–$t$ graph as if it represented uniform acceleration — always check whether the graph is straight or curved before deciding which suvat-style equations, if any, can be applied.

Worked exampleReading a uniform-acceleration ω–t graph
1
A motor's rotor has an $\omega$–$t$ graph that is a straight line, rising from $\omega = 0$ at $t=0$ to $\omega = 84$ rad s$^{-1}$ at $t = 6.0$ s. Find the angular acceleration.
A straight-line $\omega$–$t$ graph means constant $\alpha$, equal to the gradient.
2
$\alpha = \dfrac{\Delta \omega}{\Delta t} = \dfrac{84 - 0}{6.0 - 0} = 14$ rad s$^{-2}$
Gradient of the straight line = rise over run.
3
Find the angular displacement over this 6.0 s interval.
Area under an $\omega$–$t$ graph gives $\theta$; here the graph is a triangle (starting from rest).
4
$\theta = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 6.0 \times 84 = 252$ rad
Standard triangle-area formula, since the line starts at the origin.
Result

Key takeaways

  • Angular acceleration $\alpha = \Delta\omega/\Delta t$ is the rate of change of angular velocity, measured in rad s$^{-2}$.
  • It is the rotational analogue of linear acceleration, and the sign follows the same convention as $\theta$ and $\omega$.
  • The gradient of a $\theta$–$t$ graph gives $\omega$; the gradient of an $\omega$–$t$ graph gives $\alpha$; the area under an $\omega$–$t$ graph gives $\theta$.
  • A straight-line $\omega$–$t$ graph means uniform (constant) angular acceleration; a curved $\omega$–$t$ graph means non-uniform angular acceleration.
  • For non-uniform acceleration, $\alpha$ at an instant is found from the gradient of the tangent to the curve at that point, not the overall rise-over-run.
  • This completes the three core angular quantities — $\theta$, $\omega$, $\alpha$ — that feed directly into the rotational equations of motion covered next.