Circular motion and angular speed · 圆周运动与角速度
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| radians/ˈreɪdɪənz/ | 弧度 | hú dù |
| arc length/ɑːk leŋθ/ | 弧长 | hú zhǎng |
| angular speed/ˈæŋɡjʊlə spiːd/ | 角速度 | jiǎo sù dù |
| period/ˈpɪərɪəd/ | 周期 | zhōu qī |
| frequency/ˈfriːkwənsi/ | 频率 | pín lǜ |
| tangential/tænˈdʒenʃl/ | 切向 | qiè xiàng |
| linear speed/ˈlɪnɪə spiːd/ | 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.
旋转木马
- 旋转木马上的两匹马一起转完一圈。
- 然而外圈的马显然比内圈的马跑得 更快。
- 我们在这里建立的正是"转圈"与"速率"之间的联系。
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
用弧度表示的角
- 弧度(radian) 是弧长等于半径时所对的角:$\theta = \dfrac{s}{r}$。
- 一整圈是 $2\pi\ \text{rad}$。(这个主题要把计算器设成 弧度 模式。)
摩天轮:每个座舱以稳定的角速度(angular speed)做圆周运动
Angular speed · 角速度
s = rθ
Angular speed turns angle · 角度 per time; arc length s = rθ. · 角速度是单位时间转过的 角;弧长 s = rθ。
One radian is the angle for which the arc length equals the: · 一弧度是弧长等于以下哪个量的那个角:
$\theta = \dfrac{s}{r}$, so when the arc $s$ equals the radius $r$, the angle is exactly one radian. · $\theta = \dfrac{s}{r}$,所以当弧 $s$ 等于半径 $r$ 时,角恰好是一弧度。
How many radians are there in a complete circle? · 一整圈有多少弧度?
A full circle has arc $s = 2\pi r$, so $\theta = \dfrac{2\pi r}{r} = 2\pi \approx 6.28\ \text{rad}$. · 一整圈的弧 $s = 2\pi r$,所以 $\theta = \dfrac{2\pi r}{r} = 2\pi \approx 6.28\ \text{rad}$。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
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. · 把弧度定义成 180/pi 度,或者只写公式 theta = s/r,都得不到分。得分的说法是弧长等于半径。
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
角速度
- 角速度 $\omega$ 是角度变化的快慢:$\omega = \dfrac{\theta}{t}$。
- 单位:$\dfrac{\text{rad}}{\text{s}}$。
一弧度是弧长等于半径时所对的角
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
周期与频率
- 转一圈用时为 周期(period) $T$,所以 $\omega = \dfrac{2\pi}{T} = 2\pi f$。
- $f = \dfrac{1}{T}$ 是每秒转的圈数(Hz)。
半径转过一个角度时,物体以速率 v 走过一段弧
An object goes once round every $2.0\ \text{s}$. What is its angular speed? · 一个物体每 $2.0\ \text{s}$ 转一圈。它的角速度是多少?
$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.0} = \pi \approx 3.14\ \dfrac{\text{rad}}{\text{s}}$. · $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2.0} = \pi \approx 3.14\ \dfrac{\text{rad}}{\text{s}}$。
Angular speed equals $2\pi$ divided by the ____. · 角速度等于 $2\pi$ 除以 ____。
$\omega = \dfrac{2\pi}{T}$ — the whole turn ($2\pi$) divided by the time for one turn. · $\omega = \dfrac{2\pi}{T}$——整圈($2\pi$)除以转一圈所用的时间。
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.
线速度与角速度
- 实际的(切向)速率是 $v = r\omega$。
- $\omega$ 相同、$r$ 更大 → $v$ 更大——这就是外圈的马更快的原因。
A point at radius $0.50\ \text{m}$ turns at $\omega = 4.0\ \dfrac{\text{rad}}{\text{s}}$. What is its linear speed? · 一个点在半径 $0.50\ \text{m}$ 处以 $\omega = 4.0\ \dfrac{\text{rad}}{\text{s}}$ 转动。它的线速度是多少?
$v = r\omega = 0.50 \times 4.0 = 2.0\ \dfrac{\text{m}}{\text{s}}$. · $v = r\omega = 0.50 \times 4.0 = 2.0\ \dfrac{\text{m}}{\text{s}}$。
On a merry-go-round, a horse further from the centre moves at a higher linear speed. · 在旋转木马上,离中心更远的马有更大的线速度。
Same angular speed $\omega$, but $v = r\omega$, so a larger radius gives a larger linear speed. · 角速度 $\omega$ 相同,但 $v = r\omega$,所以更大的半径给出更大的线速度。
A wheel turns once every 4.0 s. What is its angular speed, in rad/s? · 一个轮子每 4.0 s 转一圈。它的角速度是多少 rad/s?
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. · omega = 2 pi / T = 1.57 rad/s。这个主题的每个公式都假定弧度,所以计算器停在角度模式也会悄悄毁掉 v = r omega。
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.
描述匀速圆周运动
- 考试要求"用速度和加速度"来描述。三点都要说:
- 速率 恒定,但 速度 不恒定——它沿切线,方向不断改变;
- 加速度 大小恒定,始终指向 圆心,与速度垂直。
Which statements correctly describe an object in uniform circular motion? Select all that apply. · 哪些说法正确描述了做匀速圆周运动的物体?选出所有正确的。
The speed is constant but the velocity keeps changing direction, so there is an acceleration — of constant size, always towards the centre. · 速率恒定,但速度的方向不断改变,所以存在加速度——大小恒定,始终指向圆心。
In uniform circular motion, which are constant? Select all · 所有 that apply. · 在匀速圆周运动中,哪些量是恒定的?选出所有适用的。
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. · 速度的方向在不断改变,这正是有加速度的原因。力不做功,所以动能从不改变。
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.
例题:站着不动,却跑得很快
剑桥位于纬度 $52.2^\circ$,地球半径 $6.37 \times 10^{6}\ \text{m}$,每天自转一圈。求剑桥绕行的圆的半径、它的速率,以及一个 $60\ \text{kg}$ 的学生做这种运动所需的合力。
- 圆的半径: 城市绕的是地球的 自转轴,不是地心,所以 $r = R\cos 52.2^\circ = 6.37 \times 10^{6} \times 0.613 = 3.90 \times 10^{6}\ \text{m}$。
- 角速度: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{86\,400} = 7.27 \times 10^{-5}\ \dfrac{\text{rad}}{\text{s}}$。
- 速率: $v = r\omega = 3.90 \times 10^{6} \times 7.27 \times 10^{-5} = 284\ \dfrac{\text{m}}{\text{s}}$。
- 合力: $F = mr\omega^{2} = 60 \times 3.90 \times 10^{6} \times (7.27 \times 10^{-5})^{2} = 1.2\ \text{N}$,指向自转轴。
- 检查: 与学生 $590\ \text{N}$ 的重力相比这是个很小的力——所以我们察觉不到自转,尽管我们比喷气式飞机跑得还快。
$\omega$ 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.
$\omega$ 必须用 弧度每秒。每分钟转数换算为 $\dfrac{\text{rpm} \times 2\pi}{60}$;在这里的每个公式中一整圈都是 $2\pi$,绝不是 $360$。$v = r\omega$ 中的 $r$ 用 米。另外,周期是 完整转一圈 的时间,不是半圈或一次摆动的时间。
A drill bit spins at $3000$ revolutions per minute. What is its angular speed, in rad/s? · 一个钻头以每分钟 $3000$ 转的速度旋转。它的角速度是多少(单位 rad/s)?
$3000$ turns per minute is $50$ per second; each turn is $2\pi$ rad, so $\omega = 50 \times 2\pi = 314\ \dfrac{\text{rad}}{\text{s}}$. · 每分钟 $3000$ 圈就是每秒 $50$ 圈;每圈是 $2\pi$ rad,所以 $\omega = 50 \times 2\pi = 314\ \dfrac{\text{rad}}{\text{s}}$。
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.
找找数量感
- 每分钟 $500$ 转的 CD:$\omega = \dfrac{500 \times 2\pi}{60} = 52\ \dfrac{\text{rad}}{\text{s}}$。
- 半径 $0.30\ \text{m}$、速率 $20\ \dfrac{\text{m}}{\text{s}}$ 的车轮:$\omega = \dfrac{v}{r} = 67\ \dfrac{\text{rad}}{\text{s}}$,约每秒 $11$ 圈。
- 地球:$86\,400\ \text{s}$ 转一圈,所以 $\omega = 7.3 \times 10^{-5}\ \dfrac{\text{rad}}{\text{s}}$——角度上慢,边缘上快。
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
你掌握了
- 弧度:$\theta = \dfrac{s}{r}$;一整圈是 $2\pi\ \text{rad}$
- 角速度 $\omega = \dfrac{\theta}{t} = \dfrac{2\pi}{T} = 2\pi f$(rad/s——记得换算 rpm)
- 线速度(linear speed)$v = r\omega$(半径越大越快);匀速圆周运动:速率恒定、速度变化、加速度指向圆心