Feed a three-phase stator and the windings build a magnetic field that rotates. Its speed is not a design choice you make later — it is set the moment you choose a supply frequency and a pole count, and it is called the synchronous speed.

$N_s = \dfrac{120 f}{p}$, read aloud N-sub-s equals 120 f over p. $N_s$ is the synchronous speed in revolutions per minute; $f$ is the supply frequency in hertz; and $p$ is the number of poles the stator is wound for — a plain count, always even, and poles rather than pole pairs. The 120 is not magic: it is 2 (a pole pair takes two poles) times 60 (seconds in a minute), and that is the entire derivation. At 60 Hz the family is 3600, 1800, 1200, 900 and 720 rpm for 2, 4, 6, 8 and 10 poles. Learn those five and half this lesson is already done.

Now the catch. An induction rotor makes its own current by being cut by that moving field, and a rotor turning at exactly synchronous speed is cut by nothing — no current, no torque, no motor. So it always runs a little slower, and that shortfall is slip: $s = \dfrac{100(N_s - N_r)}{N_s}$, s equals a hundred times N-s minus N-r, over N-s. Here $N_r$ is the measured rotor speed in rpm — subscript s is synchronous, subscript r is rotor — and $s$ is a percentage, so it carries no unit of its own.

Two nuggets worth keeping. The base is always the SYNCHRONOUS speed, never the rotor speed; dividing by the wrong one gives an answer that is close, always high, and wrong. And a nameplate reading 1750 rpm is telling you two things at once: the machine is 4-pole on 60 Hz, so synchronous is 1800, and it slips about 2.8% at rated load. Nobody prints that on the plate. You read it off.