Conservation of Angular Momentum · 角动量守恒
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| angular momentum is conserved/ˈæŋɡjʊlə məʊˈmentəm ɪz kənˈsɜːvd/ | 角动量守恒 | jiǎo dòng liàng shǒu héng |
The skater's secret
- A spinning skater pulls in her arms and suddenly whirls faster.
- She pushed off nothing -- no one gave her a shove.
- Yet her spin rate jumps dramatically.
- When nothing twists a spinning system, a hidden quantity stays fixed.
滑冰者的秘密
- 旋转的滑冰者收拢双臂,忽然旋转得更快。
- 她没有蹬任何东西——没人推她一把。
- 可她的旋转速率却猛然跃升。
- 当没有东西去扭转一个旋转系统时,一个隐藏的量保持不变。
When the spin is conserved
- If the net external torque is zero, angular momentum is conserved 角动量守恒:
- The total turning stays the same, however the mass rearranges.
- It is the rotational twin of momentum conservation.
旋转何时守恒
- 若净外力矩为零,角动量守恒:
- 无论质量怎样重新分布,总的旋转量保持不变。
- 它是动量守恒的旋转孪生。
Angular momentum is conserved exactly when... · 角动量恰好守恒当...
Zero net external torque means $dL/dt = 0$, so $L$ stays constant. · 净外力矩为零意味着$dL/dt = 0$，因此$L$保持不变。
Shrink I, spin faster
- Because $I\omega$ is fixed, lowering $I$ forces $\omega$ up.
- The skater pulls her arms in, $I$ drops, and $\omega$ rises.
- She stretches back out, $I$ grows, and the spin slows again.
缩小 I,转得更快
- 因为 $I\omega$ 固定,减小 $I$ 就迫使 $\omega$ 增大。
- 滑冰者收拢双臂,$I$ 减小,$\omega$ 上升。
- 她再次伸展开,$I$ 增大,旋转又慢下来。
A skater with $I_i = 8\ \text{kg}\cdot\text{m}^2$ spins at $2\ \text{rad/s}$, then pulls in to $I_f = 4\ \text{kg}\cdot\text{m}^2$. Her new angular speed (in rad/s)? · 一个转动惯量为$I_i = 8\ \text{kg}\cdot\text{m}^2$的滑冰者以角速度$2\ \text{rad/s}$旋转，然后收拢至$I_f = 4\ \text{kg}\cdot\text{m}^2$。她的新角速度（单位 rad/s）是多少？
$I_i\omega_i = I_f\omega_f \Rightarrow 8\times2 = 4\times\omega_f$, so $\omega_f = 4\ \text{rad/s}$. · $I_i\omega_i = I_f\omega_f \Rightarrow 8\times2 = 4\times\omega_f$，所以 $\omega_f = 4\ \text{rad/s}$。
With angular momentum conserved, decreasing the rotational inertia increases the angular speed. · 在角动量守恒的情况下，减小转动惯量会增加角速度。
$I\omega$ is fixed, so a smaller $I$ forces a larger $\omega$. · $I\omega$是固定的，因此较小的$I$迫使$\omega$变大。
From skaters to stars
- A collapsing star spins up as it shrinks -- the same rule.
- Two spinning discs that lock together share their total $L$.
- Divers and gymnasts tuck to flip faster, then open to slow down.
从滑冰者到恒星
- 坍缩的恒星在收缩时旋转加快——同一条规律。
- 两个旋转的圆盘锁在一起,分享它们的总 $L$。
- 跳水和体操运动员团身翻得更快,再展开减速。
Conservation of angular momentum · 角动量守恒
With no external torque, a spinning skater speeds up by pulling in. · 在没有外力矩的情况下，旋转的滑冰者通过收拢手臂加速。
A star's rotational inertia drops to one-quarter of its old value as it collapses. Its spin rate becomes how many times faster? · 一颗恒星坍缩时其转动惯量降至原来的四分之一。其自转角速度变为原来的多少倍？
$I\omega$ fixed and $I \to I/4$ means $\omega \to 4\omega$ -- four times faster. · $I\omega$固定且$I \to I/4$意味着$\omega \to 4\omega$——快四倍。
A skater has $I_i = 6\ \text{kg}\cdot\text{m}^2$ spinning at $\omega_i = 2\ \text{rad/s}$. She pulls in to $I_f = 2\ \text{kg}\cdot\text{m}^2$.
- $I_i\omega_i = I_f\omega_f \Rightarrow 6 \times 2 = 2 \times \omega_f$.
- So $\omega_f = 6\ \text{rad/s}$ -- three times faster.
一位滑冰者 $I_i = 6\ \text{kg}\cdot\text{m}^2$,以 $\omega_i = 2\ \text{rad/s}$ 旋转。她收拢到 $I_f = 2\ \text{kg}\cdot\text{m}^2$。
- $I_i\omega_i = I_f\omega_f \Rightarrow 6 \times 2 = 2 \times \omega_f$。
- 所以 $\omega_f = 6\ \text{rad/s}$——快了三倍。
The skater's own muscles pulling her arms in are internal forces. Can they change her angular momentum? · 滑冰者自身肌肉拉回手臂是内力。它们能改变她的角动量吗？
Internal forces come in pairs with no net torque; only external torques change $L$. · 内力成对出现且净力矩为零；只有外力矩改变$L$。
When the skater pulls her arms in, her rotational kinetic energy stays exactly the same. · 当滑冰者收拢手臂时，她的转动动能保持不变。
She does work pulling in, so $K_{rot}$ rises even though $L$ is conserved. · 她做功拉回手臂，因此$K_{rot}$上升，尽管$L$守恒。
The spin is conserved only when the net external torque is zero. Internal forces (the skater's muscles) can't change $L$, but friction, a gravity torque, or an outside push can. Note too: kinetic energy is not conserved here -- the skater does work pulling her arms in, so $K_{rot}$ actually rises.
只有在净外力矩为零时旋转才守恒。内力(滑冰者的肌肉)不能改变 $L$,但摩擦、重力力矩或外部的推可以。还要注意:这里动能不守恒——滑冰者收臂时做了功,所以 $K_{rot}$ 其实增大了。
With zero net external torque, angular momentum is conserved: $I_i\omega_i = I_f\omega_f$. Shrinking $I$ forces $\omega$ up -- the skater, the collapsing star, the tucking diver. Only external torques can change $L$; internal forces cannot. And note the spin's kinetic energy need not stay the same.
净外力矩为零时,角动量守恒:$I_i\omega_i = I_f\omega_f$。缩小 $I$ 迫使 $\omega$ 增大——滑冰者、坍缩的恒星、团身的跳水者都是如此。只有外力矩能改变 $L$,内力不能。而且注意旋转的动能未必保持不变。