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.
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.
$5\ \text{N·m}$ の一定トルクが車輪を $4\ \text{rad}$ だけ回転させた。行われた仕事はいくつジュールか？
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$W = \tau\theta = 5 \times 4 = 20\ \text{J}$.
角度 $\theta$ だけ回転するトルクによる仕事は：
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$W = \tau\theta$ — $W = Fd$ の回転における双子である。
剛体に働くトルクによる正味仕事は、その回転運動エネルギーの変化に等しい。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
これは回転の仕事-エネルギー定理である：$W_{\text{net}} = \Delta(\tfrac12 I\omega^2)$ 。
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.
$5\ \text{N·m}$ のトルクが軸を $\omega = 2\ \tfrac{\text{rad}}{\text{s}}$ で回転させている。出力はいくつワットか？
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$P = \tau\omega = 5 \times 2 = 10\ \text{W}$.
回転出力はトルク × ____ （角量）である。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$P = \tau\omega$ — トルク × 角速度。
各回転エネルギー量を直線運動における双子と結びつけよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
各回転エネルギーの式は、 $\tau\to F$ , $\omega\to v$ , $I\to m$ を入れ替えることで直線運動の式に対応する。
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.
トルクによる仕事
一定のトルクが角度だけ回転する場合、その角度に比例して仕事を行う。
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.
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 $P = \tau\omega = 5 \times 2 = 10\ \text{W}$.
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 $P = Fv$.