Angular acceleration


Angular acceleration is any change in the angular velocity vector. That means an increase in angular velocity, a decrease in angular velocity, or a change of direction of the angular velocity vector.

Recall that the angular velocity vector lies along the axis of rotation (at a right angle to the rotational plane) and, by convention, follows the right hand rule.

Angular acceleration is usually denoted by the Greek letter alpha, α. It has SI units of radians/s2 (rad·s-2) or, less frequently, degrees·s-2.

In this section, we'll consider both angular acceleration and centripetal acceleration.

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The Greek alphabet

alphaΑα
betaΒβ
gammaΓγ
deltaΔδ
epsilonΕε
zetaΖζ
etaΗη
thetaΘθ
iotaΙι
kappaΚκ
lambdaΛλ
muΜμ
nuΝν
xiΞξ
omicronΟο
piΠπ
rhoΡρ
sigmaΣσ
tauΤτ
upsilonΥυ
phiΦφ
chiΧχ
psiΨψ
omegaΩω

The math of angular acceleration


Angular acceleration is the change in angular velocity over time.

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

As defined, the angular acceleration is the average angular acceleration, which we denote by placing a bar over the alpha:

$$\bar{\alpha} = \frac{\omega_f - \omega_i}{\Delta t}$$

Of course, the average accleration can also be determined by simply taking the numerical average of two accelerations, assuming a smooth transition from one to the next:

$$\bar{\alpha} = \frac{a_1 + a_2}{2}$$

The instantaneous angular acceleration is the limit of Δω/Δt as Δt approaches zero:

$$\alpha = \lim_{\Delta t \to 0} \frac{\Delta \omega} {\Delta t}$$

If you know a little differential calculus, this is just the first derivative of the angular velocity, ω with respect to time.

$$\alpha = \frac{d\omega}{dt}$$

And finally, because angular velocity is the first derivative of the angle with respect to time, angular acceleration is the second derivative.

$$\alpha = \frac{d^2 \theta}{dt^2}$$

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SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units

lengthmeter(m)
massKilogram(Kg)
timesecond(s)
forceNewtonN
energyJouleJ

Relationship to linear acceleration


To find the linear acceleration of an object traveling in a circle (also called tangential acceleration because the vector is always tangent to the circle), we can step back to an earlier development of linear velocity from angular velocity.

The arc S can be determined simply if we work in radians. We simply set up the proportion: "The angle θ is to a full circle () as the arc S is to the full circumference (2πr):

$$\frac{\theta}{2\pi} = \frac{S}{2\pi r}$$

The 's cancel, giving the angle in terms of the arc length and radius.

$$\theta = \frac{S}{r}$$

Now the average angular velocity is:

\bar{\omega} = \frac{\Delta \theta}{\Delta t} = \frac{\Delta S}{r \Delta t}&&

In the last step we substituted S/r for θ. Now we showed that

$$\frac{\Delta S}{\Delta t} = \bar{v}$$

Therefore the angular velocity is related to the velocity (average or instantaneous) by:

$$\omega = \frac{v}{r}$$

Now angular acceleration is the change in angular velocity over time,

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

So we can combine that with the previous result to get:

$$\omega = \frac{v}{r} = \frac{1}{r} \frac{\Delta v}{\Delta t}$$

Note that r is constant, so we can exempt it from the delta (it doesn't change). Substituting linear acceleration for Δv/Δt, we get the relationship between angular and linear or tangential acceleration:

$$\alpha = \frac{a}{r}$$


Centripetal acceleration


Let's insert here a few comments about centripetal acceleration, the only acceleration needed to keep a turning body turning.

You are familiar with the feeling of going around a corner as a passenger in a car, and feeling like if the door wasn't there, you'd fly out of the car (you would). That's often confused with an outward force (away from the axis of rotation), but no such force exists. Such a "centrifugal" force is a pseudo-force. It's not a real force, only the impression of one that really stems from the centripetal or "center-seeking" force.

What you're feeling when you're going around that corner is the car pushing you into the turn. Without the car there to push on you, you'd continue in the path of your inertia, which is a straight line tangent to the curve. The figure here illustrates that.

A body, like the green circle in the figure, has centripetal acceleration just because it is moving in a circle, because its velocity vector is always changing direction.

If its velocity vector is also changing in length, then it is accelerating tangentially, too. Otherwise it has no linear or tangential acceleration.

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