| English | 意味 |
|---|---|
| angular momentum/ˈæŋɡjʊlə məʊˈmentəm/ | 角運動量 |
| angular impulse/ˈæŋɡjʊlə ˈɪmpʌls/ | 角インパルス |
The spin that won't quit
- A spinning top stays upright and keeps turning.
- A figure skater glides through a spin for many seconds.
- Something about the motion resists being stopped.
- Spinning objects carry a stubborn "quantity of turning."
Angular momentum
- Angular momentum 角动量 measures how much turning an object carries:
- Rotational inertia times angular speed.
- It is the rotational twin of linear momentum $p = mv$.
A flywheel with $I = 5\ \text{kg}\cdot\text{m}^2$ spins at $\omega = 4\ \text{rad/s}$. Its angular momentum (in kg·m²/s)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$L = I\omega = 5 \times 4 = 20\ \text{kg}\cdot\text{m}^2/\text{s}$.
Angular momentum $L = I\omega$ is the rotational twin of which linear quantity?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$I$ mirrors $m$ and $\omega$ mirrors $v$, so $L = I\omega$ mirrors $p = mv$.
A mass moving in a circle
- A point mass $m$ at radius $r$ moving at speed $v$ has:
- Farther out or faster means more angular momentum.
- This is why pulling in your arms changes a skater's spin.
A $2\ \text{kg}$ ball moves at $3\ \text{m/s}$ in a circle of radius $4\ \text{m}$. Its angular momentum (in kg·m²/s)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$L = mvr = 2 \times 3 \times 4 = 24\ \text{kg}\cdot\text{m}^2/\text{s}$.
Angular impulse changes L
- A torque acting for a time delivers angular impulse 角冲量:
- It is the rotational twin of $F\,\Delta t = \Delta p$.
- Equivalently $\tau = \dfrac{dL}{dt}$: torque is the rate of change of angular momentum.
Angular momentum
Angular momentum is rotational inertia times angular speed.
A torque of $3\ \text{N}\cdot\text{m}$ acts for $5\ \text{s}$. The change in angular momentum (in kg·m²/s)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$\Delta L = \tau\,\Delta t = 3 \times 5 = 15\ \text{kg}\cdot\text{m}^2/\text{s}$.
Torque equals the rate of change of angular momentum, $\tau = dL/dt$.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
This is Newton's second law for rotation in momentum form.
A wheel with $I = 3\ \text{kg}\cdot\text{m}^2$ spins at $\omega = 2\ \text{rad/s}$.
- Angular momentum: $L = I\omega = 3 \times 2 = 6\ \text{kg}\cdot\text{m}^2/\text{s}$.
- A steady torque $\tau = 4\ \text{N}\cdot\text{m}$ for $3\ \text{s}$ adds $\Delta L = 12$, so $L$ rises to $18$.
To state an object's angular momentum meaningfully, you must also give the...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Angular momentum is defined relative to an axis; different axes give different $L$.
Angular momentum depends on the axis you measure it about -- always state the axis. And $L = mvr$ uses the perpendicular distance from the axis to the line of motion; if the velocity is not perpendicular to $r$, keep only the perpendicular component ($L = mv_\perp r$).
Angular momentum $L = I\omega$ (or $L = mvr$ for a point mass) is the rotational twin of $p = mv$. A torque acting over time is an angular impulse that changes it: $\tau\,\Delta t = \Delta L$, or $\tau = dL/dt$. Always name the axis, and use the perpendicular distance.