Newton's Second Law in Rotational Form · 牛顿第二定律的转动形式
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| rotational inertia/rəʊˈteɪʃənl ɪˈnɜːʃə/ | 转动惯量 | zhuǎn dòng guàn liàng |
| angular acceleration/ˈæŋɡjʊlə əkˌseləˈreɪʃn/ | 角加速度 | jiǎo jiā sù dù |
| Newton's second law for rotation/ˈnjuːtnz ˈsekənd lɔː fɔː rəʊˈteɪʃn/ | 转动的牛顿第二定律 | zhuǎn dòng de niú dùn dì èr dìng lǜ |
The recipe for spinning something up
- A strong twist spins a wheel up quickly; a gentle one barely stirs it.
- A light wheel responds eagerly; a massive flywheel resists.
- This is exactly like $F = ma$ -- but for turning.
- One equation ties torque, rotational inertia, and angular acceleration together.
把东西转起来的配方
- 大力一扭,轮子迅速转起来;轻轻一拨,它几乎不动。
- 轻的轮子欣然响应;沉重的飞轮则抵抗。
- 这正像 $F = ma$——只不过是关于转动的。
- 一个方程把力矩、转动惯量和角加速度联系在一起。
Newton's second law for rotation
- Newton's second law for rotation 转动的牛顿第二定律 is:
- Net torque equals rotational inertia times angular acceleration 角加速度.
- Give it a torque, and it gains angular acceleration.
转动的牛顿第二定律
- 转动的牛顿第二定律是:
- 净力矩等于转动惯量乘以角加速度。
- 给它一个力矩,它就获得角加速度。
A net torque of $8\ \text{N}\cdot\text{m}$ acts on a wheel with $I = 2\ \text{kg}\cdot\text{m}^2$. Its angular acceleration (in rad/s^2)? · 轮子上作用着$8\ \text{N}\cdot\text{m}$的净力矩，其转动惯量为$I = 2\ \text{kg}\cdot\text{m}^2$。其角加速度（单位 rad/s^2）是多少？
$\alpha = \tau/I = 8/2 = 4\ \tfrac{\text{rad}}{\text{s}^2}$.
The rotational form of Newton's second law is tau = I times ____. · 牛顿第二定律的转动形式为 tau = I × ____。
$\tau_{net} = I\alpha$ -- net torque equals rotational inertia times angular acceleration. · $\tau_{net} = I\alpha$——净力矩等于转动惯量乘以角加速度。
What net torque gives a $5\ \text{kg}\cdot\text{m}^2$ wheel an angular acceleration of $3\ \tfrac{\text{rad}}{\text{s}^2}$ (in N·m)? · 什么净扭矩会给$5\ \text{kg}\cdot\text{m}^2$轮子带来$3\ \tfrac{\text{rad}}{\text{s}^2}$的角加速度（单位为N·m）？
$\tau = I\alpha = 5 \times 3 = 15\ \text{N}\cdot\text{m}$.
The perfect analogy
- Compare with $F = ma$: torque plays the role of force.
- Rotational inertia 转动惯量 $I$ plays the role of mass.
- Angular acceleration $\alpha$ plays the role of acceleration.
完美的对应
- 与 $F = ma$ 对比:力矩扮演力的角色。
- 转动惯量 $I$ 扮演质量的角色。
- 角加速度 $\alpha$ 扮演加速度的角色。
In $\tau = I\alpha$, rotational inertia $I$ plays the role of which quantity in $F = ma$? · 在$\tau = I\alpha$中，转动惯量$I$对应于$F = ma$中的哪个物理量？
Torque ~ force, $I$ ~ mass, $\alpha$ ~ acceleration. So $I$ is the rotational mass. · 力矩 ~ 力，$I$ ~ 质量，$\alpha$ ~ 加速度。因此$I$是转动质量。
Select all · 所有 correct analogies between $F = ma$ and $\tau = I\alpha$. · 选择$F = ma$与$\tau = I\alpha$之间所有正确的类比。
Force ~ torque and mass ~ $I$; acceleration maps to angular acceleration, not torque. · 力 ~ 力矩，质量 ~ $I$；加速度映射到角加速度，而非力矩。
Bigger inertia, gentler response
- For a fixed torque, a larger $I$ gives a smaller $\alpha$.
- A massive, spread-out wheel is hard to spin up or slow down.
- That is why flywheels resist changes in their rotation.
惯量越大,响应越缓
- 对于固定的力矩,更大的 $I$ 给出更小的 $\alpha$。
- 一个沉重、分散的轮子很难转起来或减速。
- 这就是飞轮抵抗其旋转变化的原因。
Newton's second law for rotation · 转动的牛顿第二定律
Torque produces angular acceleration in proportion to the rotational inertia: torque = I times alpha. · 力矩产生与转动惯量成正比的角加速度：力矩 = I × alpha。
For the same net torque, a wheel with a larger rotational inertia gets a smaller angular acceleration. · 对于相同的净力矩，转动惯量较大的轮子获得的角加速度较小。
$\alpha = \tau/I$, so a larger $I$ means a smaller $\alpha$ -- a gentler response. · $\alpha = \tau/I$，因此较大的$I$意味着较小的$\alpha$——响应更平缓。
Translation and rotation together
- A mass on a string over a pulley couples the two worlds.
- Write $F = ma$ for the hanging mass and $\tau = I\alpha$ for the pulley.
- Link them with the rope tension and $a = \alpha r$, then solve together.
平动与转动结合
- 绕过滑轮的绳上挂一个质量,把两个世界耦合起来。
- 对悬挂的质量写 $F = ma$,对滑轮写 $\tau = I\alpha$。
- 用绳的张力和 $a = \alpha r$ 把它们联系起来,再联立求解。
A net torque of $12\ \text{N}\cdot\text{m}$ acts on a wheel with rotational inertia $I = 3\ \text{kg}\cdot\text{m}^2$.
- Angular acceleration: $\alpha = \dfrac{\tau}{I} = \dfrac{12}{3} = 4\ \tfrac{\text{rad}}{\text{s}^2}$.
- Double the rotational inertia and the same torque gives only $2\ \tfrac{\text{rad}}{\text{s}^2}$.
一个 $12\ \text{N}\cdot\text{m}$ 的净力矩作用在转动惯量 $I = 3\ \text{kg}\cdot\text{m}^2$ 的轮子上。
- 角加速度:$\alpha = \dfrac{\tau}{I} = \dfrac{12}{3} = 4\ \tfrac{\text{rad}}{\text{s}^2}$。
- 把转动惯量加倍,同样的力矩只给出 $2\ \tfrac{\text{rad}}{\text{s}^2}$。
You cannot use $F = ma$ for the pulley itself. A pulley with mass has rotational inertia, so the rope tensions on its two sides are different -- that difference provides the net torque $\tau = I\alpha$. Treating it as massless (equal tensions) is a common error.
你不能对滑轮本身用 $F = ma$。有质量的滑轮有转动惯量,所以它两侧绳的张力不同——这个差值提供了净力矩 $\tau = I\alpha$。把它当作无质量(张力相等)是常见的错误。
Newton's second law for rotation is $\tau_{net} = I\alpha$ -- the rotational twin of $F = ma$, with torque for force, rotational inertia for mass, and angular acceleration for acceleration. A bigger $I$ means a gentler response, and coupled mass-pulley systems solve $F = ma$ and $\tau = I\alpha$ together.
转动的牛顿第二定律是 $\tau_{net} = I\alpha$——$F = ma$ 的旋转孪生,用力矩替代力、转动惯量替代质量、角加速度替代加速度。更大的 $I$ 意味着更缓的响应,而耦合的质量-滑轮系统要联立 $F = ma$ 和 $\tau = I\alpha$ 一起求解。