Power in Rotation

Derivation

Instantaneous power = rate of doing work:

$P = \frac{dW}{dt} = \frac{\tau \, d\theta}{dt} = \tau\omega$

$\boxed{P = \tau\omega}$

The analogy

$P = Fv \quad \text{(linear)} \leftrightarrow \quad P = \tau\omega \quad \text{(rotational)}$

Applications

Engine power and torque:

$P = \tau\omega \implies \tau = \frac{P}{\omega}$

At low speed (small $\omega$), an engine with fixed power $P$ delivers high torque — this is why vehicles have more pulling force at low speeds. At high speed, torque decreases.

Example: A motor delivers 5 kW at 500 rpm. Torque:

$\omega = \frac{500 \times 2\pi}{60} \approx 52.4 \text{ rad/s}$

$\tau = \frac{P}{\omega} = \frac{5000}{52.4} \approx 95.4 \text{ N·m}$

Maximum power and maximum torque are not at the same speed

In real engines:

• Maximum torque occurs at low-to-mid RPM (when combustion efficiency is highest)
• Maximum power occurs at higher RPM (power = torque × speed, and even if torque decreases, speed increase wins for a while)

The "power band" of an engine is the RPM range where $P = \tau\omega$ is maximised.

Summary of rotational analogies