$P = \tau\omega$ — P equals tau omega. $P$ is the power transmitted, in watts; $\tau$ is the torque in the shaft, in newton-metres; $\omega$ is the shaft's angular velocity, in radians per second. It is the rotational cousin of $P = Fv$, and a newton-metre per second is exactly one watt, which is the whole reason the relation is this tidy. Solved the other way, $\tau = \dfrac{P}{\omega}$.

Two disciplines make or break every calculation on this page. First: the power must be in watts, so a nameplate in kilowatts is multiplied by 1000 before it goes near a socket. Second, and the more expensive: $\omega$ is in radians per second, never rpm. The nameplate says rpm; the formula does not speak it. Cross the bridge — $\omega = 2\pi n / 60$ — before the numbers go in, and convert back to rpm only when you report the answer.

Now read the relation for what it tells you about machines: at constant power, torque is inversely proportional to speed. Gear a motor down ten to one and the output shaft carries ten times the torque. That is why the slow end of a gearbox is always the fat end, why low-speed machinery is so heavy for its power — and why a shaft is sized from its torque, because torque is the only thing the steel can actually feel.