Concept of Angular speed

The idea. Imagine a particle going round a circle of radius $r$. Two different "speeds" describe it:

• Linear speed $v$ — how fast it moves along the circular path (metres per second).
• Angular speed $\omega$ (Greek letter omega) — how fast the angle at the centre increases (radians per second).

If the particle sweeps out an angle $\theta$ (in radians) in time $t$ at a steady rate, then:

$\omega = \frac{\theta}{t}$

Why radians, not degrees? All the circular-motion formulae ($v = r\omega$, $a = r\omega^2$, arc length $s = r\theta$) only work when the angle is in radians. A full turn is $2\pi$ radians $= 360°$.

Cambridge tip. Always state $\omega$ in rad s⁻¹. If a question gives a turning rate in any other units (degrees per second, revolutions per minute), convert to rad s⁻¹ as your first line of working.

The most important formula in the whole topic. In time $t$ the particle turns through angle $\theta = \omega t$, and travels an arc of length $s = r\theta$ (arc length $=$ radius $\times$ angle in radians). Its linear speed is

$v = \frac{s}{t} = \frac{r\theta}{t} = \frac{r(\omega t)}{t} = r\omega$

So:

$\boxed{\,v = r\omega\,}$

Reading it both ways:

• Given $\omega$ and $r$, find the speed: $v = r\omega$.
• Given $v$ and $r$, find the angular speed: $\omega = \dfrac{v}{r}$.

A consequence worth noticing. For a fixed $\omega$, points further from the centre move faster: doubling $r$ doubles $v$. This is why the rim of a spinning disc moves faster than a point near its centre, even though both share the same $\omega$.

Units check. $r$ in metres, $\omega$ in rad s⁻¹ gives $v$ in m s⁻¹. (Radians are dimensionless, so they "disappear" — m × rad s⁻¹ reads as m s⁻¹.)

Cambridge tip. When a problem mixes a stated speed with a stated rotation rate, decide which one you are given and use $v = r\omega$ to get the other before substituting into any force formula.

Period $T$ is the time for one complete revolution. In one revolution the particle turns through $2\pi$ radians, so using $\omega = \dfrac{\theta}{t}$ with $\theta = 2\pi$ and $t = T$:

$\omega = \frac{2\pi}{T} \qquad\Longleftrightarrow\qquad T = \frac{2\pi}{\omega}$

Frequency $f$ is the number of revolutions per second (measured in hertz, Hz). It is just the reciprocal of the period:

$f = \frac{1}{T} = \frac{\omega}{2\pi} \qquad\Longleftrightarrow\qquad \omega = 2\pi f$

The three quantities together:

Quantity	Symbol	Meaning	Formula
Angular speed	$\omega$	radians per second	$\omega = \frac{2\pi}{T} = 2\pi f$
Period	$T$	seconds per revolution	$T = \frac{2\pi}{\omega}$
Frequency	$f$	revolutions per second	$f = \frac{1}{T} = \frac{\omega}{2\pi}$

Converting revolutions per minute (rev/min) to rad s⁻¹. This is a very common exam step. Two conversions stack:

1. Each revolution is $2\pi$ radians.
2. "Per minute" → "per second" means dividing by $60$.

$\omega = (\text{rev/min}) \times \frac{2\pi}{60} \ \text{rad s}^{-1}$

For example, $300$ rev/min $= 300 \times \dfrac{2\pi}{60} = 10\pi \approx 31.4$ rad s⁻¹.

Cambridge tip. Write the conversion as a single clear line. Forgetting the $\div 60$ (leaving the answer "per minute") or forgetting the $\times 2\pi$ are the two errors examiners flag most often.