I recently came across a question asking the following:

If a motor is switched on, it quickly reaches a top speed. Why does it not just go faster and faster and faster?

I thought it might be related to the fact that the electrical power is constant, so I tried approaching with $P=\tau \cdot \omega$. However I still do not have any clue to reaching the conclusion.

Is it somehow related to the energy loss (due to friction)? Like the terminal speed of a parachute or something? But I was thinking that friction should be constant ($f=\mu \cdot N$) while air drag is related to the speed, so this may not be the answer.

Thank you.


2 Answers 2


Surpringingly the top speed is not necessarily anything to do with friction, though friction will of course have some effect.

A motor acts as a generator, i.e. if you turn a motor it will generate a potential difference just like a generator, and this potential difference (usually called the back EMF) is proportional to the motor speed. So if you connect a X volt battery to an unloaded motor the armature will accelerate until the back EMF rises to the same value as the applied voltage. At this point the motor will settle to a steady angular velocity.

The equations describing this are given in the Wikipedia article on the DC motor, though this is far from the best written article I've seen. Basically the back EMF, $E_b$, is given by:

$$ E_b = \omega \Phi k_b $$

where $\phi$ and $k_b$ are dependant on the motor design and for the sake of this discussion can be treated as constants. So ignoring friction, the internal resistance of the motor and with no load on the motor the top speed will be when $E_b$ is the same as the applied voltage $V$ so:

$$ \omega = \frac{V}{\Phi k_b} $$

So it predicts angular velocity is proportional to applied voltage.

The effect of internal resistance, $R_i$, is easily included since it just creates a potential drop of $IR_i$ and the equation changes to:

$$ \omega = \frac{V - IR_i}{\Phi k_b} $$

  • $\begingroup$ Can you please give me a little hint on why "if you connect a X volt battery to an unloaded motor the armature will accelerate until the back EMF rises to the same value as the applied voltage" is true? (Is it just because all the emf from the cell has to be used up by the motor?) Thank you. $\endgroup$
    – user12205
    Commented Feb 18, 2014 at 18:33
  • $\begingroup$ (Just to clarify: after reading your answer I now understand why there is a top speed. I just want to know why it reaches that top speed but not some lower speed) $\endgroup$
    – user12205
    Commented Feb 18, 2014 at 18:42
  • $\begingroup$ I would love to see this worked out in greater detail. Can you post a good reference if you find one? $\endgroup$
    – webb
    Commented Feb 18, 2014 at 18:54

John Rennie's answer is great for DC motors but I thought I'd mention why AC induction motors reach a top speed. For AC induction motors, top speed is limited by the speed of the rotating magnetic field set up by the stator (aka, the synchronous speed of the motor). The rotor in an AC induction motor can only rotate as fast as rotating magnetic field and the rotating magnetic field is purely a function of the frequency of the input power and the number of poles the motor is wound as.


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