3.6.1.1
Angular speed is the rate of change of angular displacement
Circular Motion — AQA A-Level Physics
Key Definition
Angular speed — The rate of change of angular displacement with respect to time. Symbol: omega. Unit: rad \(s^{-1}\).
$$\begin{aligned}
\omega &= \frac{\Delta \theta}{\Delta t} \\
&= \frac{v}{r} \\
&= 2\pi f \\
&= \frac{2\pi}{T}
\end{aligned}$$
- $\omega$: angular speed (rad \(s^{-1}\))
- $v$: linear speed (m \(s^{-1}\))
- $r$: radius of orbit (m)
- $f$: frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). (Hz)
- $T$: time periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (s)
- The linear speed is related to angular speed by $v = r \omega$.
- The greater the angular displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m). in a given time, the greater the angular speed.
- An object further from the centre (larger r) at the same angular speed has a greater linear speed.
Worked Example
A bird flies in a horizontal circle of radius 650 m with an angular speed of 5.25 rad \(s^{-1}\). Calculate (a) the linear speed and (b) the frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz)..
Show Solution
1
Use v = r omega for linear speed
$$v = r\omega = 650 \times 5.25 = 3410 \text{ m s}^{-1} \text{ (3 s.f.)}$$
2
Use omega = 2 pi f for frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz).
$$f = \frac{\omega}{2\pi} = \frac{5.25}{2\pi} = 0.836 \text{ Hz (3 s.f.)}$$
Answer
$v = 3410$ m s$^{-1}$, $f = 0.836$ Hz