# Why doesn't the number of poles affect the angular velocity of a DC motor?

Angular velocity "$$\ω\$$" of a DC motor is: $$\ω = (2\times π \times n) / 60\$$

Where "$$\n\$$" is RPM.

And not $$\ω = (2 \times π \times n \times p) / (60)\$$ like in synchronous motors, where "$$\p\$$" is the number of pole pairs.

I know that the angular velocity of the DC motor doesn't depend on the number of pole pairs, but I don't understand why.

• @evildemonic Thanks, that was a mistake, I don't need it defined in the question. Feb 19, 2021 at 15:27
• Where did you get that equation for angular velocity of a DC motor? Why would there be a 60 in there if it is DC? Feb 19, 2021 at 15:42
• @evildemonic The 60 is the number of seconds in a minute, because all that equation is is the conversion between rpm and rad/s. Feb 19, 2021 at 16:23
• I should also point out that the symbol for angular velocity is generally $ω$ (the Greek letter omega, ω), not $w$ (the latin letter w, which I can only think to spell "doubleyou"). Feb 19, 2021 at 16:37
• @Hearth Not sure how that didn't occur to me right away. Thanks for the omega tip, edited. I learn something every time you @ me. Feb 19, 2021 at 17:25

Your equation that you list as the angular velocity of a dc motor is nothing more and nothing less than the conversion factor between rad/s and rpm. There's nothing in there that's inherent to the motor.

The reason the number of pole pairs shows up in the equivalent version for the AC motor is because there's a conversion between electrical frequency and mechanical rotational speed happening there too, but there is no such electrical frequency to deal with in a DC motor, as DC is zero frequency by definition.

Basically, the two equations are converting two different things. For the DC motor you're calculating mechanical rad/s from mechanical rpm, but for the AC motor you're calculating mechanical rad/s from electrical rpm (cycles per minute would be a more general word to use).

You're comparing apples to oranges.

Your first equation is a simple conversion, where $$\n\$$ is arbitrary and given: $$\omega = \frac{2\pi\ \mathrm{radians/cycle}}{60\ \mathrm{min/sec}}n$$.

Your second is a calculation where $$\n\$$ is the given line-synchronous speed in RPM, which is a function of line frequency: $$n = f\ \mathrm{\frac{60 RPM}{Hz}}$$.

Then the synchronous frequency calculation takes the above $$\n\$$, which is different from your first $$\n\$$, and plugs it into $$\omega = \frac{2\pi\ \mathrm{radians/cycle}}{60\ \mathrm{min/sec}}\frac{n}{p}$$

First equation $$w=2πn/60$$ is always true (whatever machine it is) because it is equation of an unit conversion which is used for mechanical speed of rotor (in this case) as 'w' is mechanical speed in rad/sec and 'n' is mechanical speed in revolution per minute .

While in second equation -

I think it should be $$w=2nπp/60$$ where 'p' is no. of pole pair and here 'w' is electrical frequency while 'n' is mechanical speed of rotor in rpm , so we can clearly observe that this equation relates electrical frequency to mechanical speed and that is quite obvious why it depends on no. of pole pair