Circular motion and angular speed
Introduced| English | Meaning |
|---|---|
| radians/ˈreɪdɪənz/ | Units of angle defined by arc length divided by radius; one full turn equals two pi of these units. |
| arc length/ɑːk leŋθ/ | The distance along a curved part of a circle or another curve. |
| angular speed/ˈæŋɡjʊlə spiːd/ | The rate at which angular position changes, measured as angle per unit time. |
| period/ˈpɪərɪəd/ | The time taken for one complete oscillation or cycle. |
| frequency/ˈfriːkwənsi/ | The number of complete cycles per unit time. |
| tangential/tænˈdʒenʃl/ | Directed along the tangent to a curve or circular path at a particular point. |
| linear speed/ˈlɪnɪə spiːd/ | The rate at which distance is travelled along a path. |
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:
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$\theta = \dfrac{s}{r}$, so when the arc $s$ equals the radius $r$, the angle is exactly one radian.
How many radians are there in a complete circle?
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A full circle has arc $s = 2\pi r$, so $\theta = \dfrac{2\pi r}{r} = 2\pi \approx 6.28\ \text{rad}$.
Match each term to the definition the examiner marks.
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Defining the radian as 180/pi degrees, or by the formula theta = s/r, loses the mark. The arc equal in length to the radius is the marked phrase.
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?
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$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.0} = \pi \approx 3.14\ \dfrac{\text{rad}}{\text{s}}$.
Angular speed equals $2\pi$ divided by the ____.
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$\omega = \dfrac{2\pi}{T}$ — 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?
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$v = r\omega = 0.50 \times 4.0 = 2.0\ \dfrac{\text{m}}{\text{s}}$.
On a merry-go-round, a horse further from the centre moves at a higher linear speed.
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Same angular speed $\omega$, but $v = r\omega$, so a larger radius gives a larger linear speed.
A wheel turns once every 4.0 s. What is its angular speed, in rad/s?
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omega = 2 pi / T = 1.57 rad/s. Every formula in this topic assumes radians, so a calculator in degrees quietly ruins v = r omega too.
Describing uniform circular motion
- The exam asks for it "in terms of velocity and acceleration". Say all three things:
- the speed is constant, but the velocity is not — it is along the tangent and its direction changes continuously;
- the acceleration has constant magnitude and is always directed towards the centre, perpendicular to the velocity.
Which statements correctly describe an object in uniform circular motion? Select all that apply.
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The speed is constant but the velocity keeps changing direction, so there is an acceleration — of constant size, always towards the centre.
In uniform circular motion, which are constant? Select all that apply.
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The velocity changes continuously in direction, which is exactly why there is an acceleration at all. The force does no work, so the kinetic energy never changes.
Worked example: standing still, moving fast
Cambridge is at latitude $52.2^\circ$ on an Earth of radius $6.37 \times 10^{6}\ \text{m}$ that turns once a day. Find the radius of the circle Cambridge moves round, its speed, and the resultant force on a $60\ \text{kg}$ student needed for this motion.
- Radius of the circle: the city circles the Earth's axis, not its centre, so $r = R\cos 52.2^\circ = 6.37 \times 10^{6} \times 0.613 = 3.90 \times 10^{6}\ \text{m}$.
- Angular speed: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{86\,400} = 7.27 \times 10^{-5}\ \dfrac{\text{rad}}{\text{s}}$.
- Speed: $v = r\omega = 3.90 \times 10^{6} \times 7.27 \times 10^{-5} = 284\ \dfrac{\text{m}}{\text{s}}$.
- Resultant force: $F = mr\omega^{2} = 60 \times 3.90 \times 10^{6} \times (7.27 \times 10^{-5})^{2} = 1.2\ \text{N}$, towards the axis.
- Check: a tiny force compared with the student's $590\ \text{N}$ weight — which is why we do not notice the spin, even though we are moving faster than a jet aircraft.
$\omega$ must be in radians per second. Revolutions per minute become $\dfrac{\text{rpm} \times 2\pi}{60}$; a full turn is $2\pi$ in every formula here, never $360$. $r$ in $v = r\omega$ is in metres. And the period is the time for one complete revolution, not for a half-turn or a swing.
A drill bit spins at $3000$ revolutions per minute. What is its angular speed, in rad/s?
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$3000$ turns per minute is $50$ per second; each turn is $2\pi$ rad, so $\omega = 50 \times 2\pi = 314\ \dfrac{\text{rad}}{\text{s}}$.
Numbers to get a feel for
- A CD at $500$ revolutions per minute: $\omega = \dfrac{500 \times 2\pi}{60} = 52\ \dfrac{\text{rad}}{\text{s}}$.
- A car wheel of radius $0.30\ \text{m}$ at $20\ \dfrac{\text{m}}{\text{s}}$: $\omega = \dfrac{v}{r} = 67\ \dfrac{\text{rad}}{\text{s}}$, about $11$ turns a second.
- The Earth: one turn in $86\,400\ \text{s}$, so $\omega = 7.3 \times 10^{-5}\ \dfrac{\text{rad}}{\text{s}}$ — slow in angle, fast at the rim.
You've got it
- a radian: $\theta = \dfrac{s}{r}$; a full circle is $2\pi\ \text{rad}$
- angular speed $\omega = \dfrac{\theta}{t} = \dfrac{2\pi}{T} = 2\pi f$ (rad/s — convert rpm)
- linear speed 线速度 $v = r\omega$ (bigger radius → faster); uniform circular motion: constant speed, changing velocity, acceleration towards the centre