Circular motion and angular speed
| English | Chinese | Pinyin |
|---|---|---|
| radian | 弧度 | hú dù |
| arc length | 弧长 | hú zhǎng |
| angular speed | 角速度 | jiǎo sù dù |
| period | 周期 | zhōu qī |
| frequency | 频率 | pín lǜ |
| tangential | 切向 | qiè xiàng |
| linear speed | 线速度 | xiàn sù dù |
The merry-go-round
- Two horses on a merry-go-round both go round once together.
- Yet the outer horse clearly moves faster than the inner one.
- The link between "going round" and "speed" is what we set up here.
Angles in radians 弧度
- A radian is the angle whose arc length 弧长 equals the radius: $\theta = \dfrac{s}{r}$.
- A full circle is $2\pi\ \text{rad}$. (Set your calculator to radians for this topic.)
A Ferris wheel: each car moves in a circle at a steady angular speed 角速度
Angular speed
s = rθ
Angular speed turns angle per time; arc length s = rθ.
One radian is the angle for which the arc length equals the:
$\theta = \dfrac{s}{r}$, so when the arc $s$ equals the radius $r$, the angle is exactly one radian.
How many radians are there in a complete circle?
A full circle has arc $s = 2\pi r$, so $\theta = \dfrac{2\pi r}{r} = 2\pi \approx 6.28\ \text{rad}$.
Angular speed
- Angular speed $\omega$ is how fast the angle changes: $\omega = \dfrac{\theta}{t}$.
- Unit: $\dfrac{\text{rad}}{\text{s}}$.
One radian is the angle whose arc length equals the radius
Period 周期 and frequency 频率
- One turn takes the period $T$, so $\omega = \dfrac{2\pi}{T} = 2\pi f$.
- $f = \dfrac{1}{T}$ is the number of turns per second (Hz).
As the radius turns through the angle the object moves an arc at speed v
An object goes once round every $2.0\ \text{s}$. What is its angular speed?
$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.0} = \pi \approx 3.14\ \dfrac{\text{rad}}{\text{s}}$.
Angular speed equals $2\pi$ divided by the ____.
$\omega = \dfrac{2\pi}{T}$ — the whole turn ($2\pi$) divided by the time for one turn.
Linear and angular speed
- The actual (tangential 切向) speed is $v = r\omega$.
- Same $\omega$, bigger $r$ → bigger $v$ — which is why the outer horse is faster.
A point at radius $0.50\ \text{m}$ turns at $\omega = 4.0\ \dfrac{\text{rad}}{\text{s}}$. What is its linear speed?
$v = r\omega = 0.50 \times 4.0 = 2.0\ \dfrac{\text{m}}{\text{s}}$.
On a merry-go-round, a horse further from the centre moves at a higher linear speed.
Same angular speed $\omega$, but $v = r\omega$, so a larger radius gives a larger linear speed.
You've got it
- a radian: $\theta = \dfrac{s}{r}$; a full circle is $2\pi\ \text{rad}$
- angular speed $\omega = \dfrac{\theta}{t} = \dfrac{2\pi}{T} = 2\pi f$
- linear speed 线速度 $v = r\omega$ (bigger radius → faster)