Torque and Work · 力矩与功
Cranking a winch — work, but going in circles
- Turn a winch handle round and round and you clearly do work — your arm gets tired, a load rises.
- Yet nothing moves in a straight line; the handle just goes in circles.
- Work done by a torque turning through an angle is the rotational version of $W = Fd$.
- It feeds energy straight into a spinning object's rotational kinetic energy.
摇绞盘——做功,却在打转
- 一圈圈地摇绞盘手柄,你显然在做功——你的手臂累了,重物升起来了。
- 可没有什么沿直线运动;手柄只是在打转。
- 由力矩转过一个角度所做的功,是 $W = Fd$ 的旋转版本。
- 它把能量直接送入旋转物体的转动动能。
Work done by a torque
- A torque $\tau$ turning through an angle $\theta$ does work $W = \tau\,\theta$.
- It is the twin of $W = Fd$, with torque for force and angle (in radians) for distance.
- This work changes the object's rotational kinetic energy: $W = \Delta(\tfrac12 I\omega^2)$.
- Positive torque-work speeds the spin up; negative slows it down.
力矩所做的功
- 力矩 $\tau$ 转过角度 $\theta$ 做功 $W = \tau\,\theta$。
- 它是 $W = Fd$ 的孪生,以力矩代替力、角度(弧度)代替距离。
- 这个功改变物体的转动动能:$W = \Delta(\tfrac12 I\omega^2)$。
- 正的力矩功使旋转加快;负的使它变慢。
A constant torque of $5\ \text{N·m}$ turns a wheel through $4\ \text{rad}$. How much work is done, in joules? · 一个大小为$5\ \text{N·m}$的恒定力矩使轮子转过$4\ \text{rad}$。做了多少功，单位为焦耳？
$W = \tau\theta = 5 \times 4 = 20\ \text{J}$.
Work done by a torque turning through angle $\theta$ is: · 力矩转过角度$\theta$所做的功为：
$W = \tau\theta$ — the rotational twin of $W = Fd$. · $W = \tau\theta$ —— 线性动量$W = Fd$的转动对应量。
The net work done by torques on a rigid body equals the change in its rotational kinetic energy. · 作用在刚体上的合外力矩所做的总功等于其转动动能的变化。
This is the rotational work–energy theorem: $W_{\text{net}} = \Delta(\tfrac12 I\omega^2)$. · 这是转动功能原理：$W_{\text{net}} = \Delta(\tfrac12 I\omega^2)$。
Rotational power
- The rate of doing rotational work is the power $P = \tau\,\omega$.
- It is the twin of $P = Fv$, with torque for force and angular speed for linear speed.
- A car engine's power is often quoted this way — torque times revs.
- More torque or a faster spin both mean more power delivered.
转动功率
- 做转动功的速率是功率 $P = \tau\,\omega$。
- 它是 $P = Fv$ 的孪生,以力矩代替力、角速率代替线速率。
- 汽车引擎的功率常这样表述——力矩乘以转速。
- 更大的力矩或更快的旋转,都意味着输出更大功率。
A torque of $5\ \text{N·m}$ turns a shaft at $\omega = 2\ \tfrac{\text{rad}}{\text{s}}$. What is the power, in watts? · 一个大小为$5\ \text{N·m}$的力矩使轴以$\omega = 2\ \tfrac{\text{rad}}{\text{s}}$旋转。功率是多少，单位为瓦特？
$P = \tau\omega = 5 \times 2 = 10\ \text{W}$.
Rotational power is torque times ____ (the angular quantity). · 转动功率等于力矩乘以____（角量）。
$P = \tau\omega$ — torque times angular velocity. · $P = \tau\omega$ —— 力矩乘以角速度。
Match each rotational energy quantity to its linear twin. · 将每个转动能量量与其线性对应量匹配。
Each rotational energy formula matches a linear one by swapping $\tau\to F$, $\omega\to v$, $I\to m$. · 每个转动能量公式通过替换$\tau\to F$、$\omega\to v$、$I\to m$与一个线性公式相匹配。
The energy picture stays the same
- Rotational work and energy obey the same conservation rules as their linear cousins.
- The net work by all torques equals the change in rotational kinetic energy.
- Energy can flow between translation and rotation (a rolling object), but the total is conserved.
- Every linear energy tool has a rotational partner you already know how to use.
能量图景保持不变
- 转动功和能量遵守与它们直线表亲相同的守恒规则。
- 所有力矩所做的合功等于转动动能的变化。
- 能量可以在平动和转动之间流动(滚动的物体),但总量守恒。
- 每个直线能量工具都有一个你已经会用的转动伙伴。
Work done by a torque · 力矩所做的功
A constant torque turning through an angle does work in proportion to that angle. · 恒定力矩转过一定角度所做的功与该角度成正比。
Rotational work uses the angle in radians, not degrees, and the net torque. $W = \tau\theta$ only gives the right energy when $\theta$ is in radians — the same radian rule as the rest of rotation.
转动功使用以弧度表示的角度,而非度,并使用合力矩。只有当 $\theta$ 以弧度为单位时,$W = \tau\theta$ 才给出正确的能量——与旋转其余部分相同的弧度规则。
A constant torque of $5\ \text{N}\cdot\text{m}$ turns a wheel through $4\ \text{rad}$.
- $W = \tau\theta = 5 \times 4 = 20\ \text{J}$.
If it does this in $2\ \text{s}$ at $\omega = 2\ \tfrac{\text{rad}}{\text{s}}$, the power is $P = \tau\omega = 5 \times 2 = 10\ \text{W}$.
一个 $5\ \text{N}\cdot\text{m}$ 的恒定力矩把轮子转过 $4\ \text{rad}$。
- $W = \tau\theta = 5 \times 4 = 20\ \text{J}$。
如果它在 $2\ \text{s}$ 内、以 $\omega = 2\ \tfrac{\text{rad}}{\text{s}}$ 做到这一点,功率是 $P = \tau\omega = 5 \times 2 = 10\ \text{W}$。
Work done by a torque is $W = \tau\theta$ (angle in radians), the twin of $W = Fd$. It changes the rotational kinetic energy. The rate of doing it is the rotational power $P = \tau\omega$, the twin of $P = Fv$.
力矩所做的功是 $W = \tau\theta$(角度以弧度计),$W = Fd$ 的孪生。它改变转动动能。做功的速率是转动功率 $P = \tau\omega$,$P = Fv$ 的孪生。