I have a rather small motor and a matching energy source.

When the motor starts, it needs a (short) amount of time to 'rev up', to accelerate at its max speed.

Is there any cause for this other than mechanical inertia and friction? Would, say, a pullcord attached wound in the rotation's direction, pulled as the motor start, serve to bypass this delay?

Please explain it like I'm five (but good at googling terminology).

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    \$\begingroup\$ "Is there any cause for this other than mechanical inertia and friction?" Measure the voltage on your supply and see if it drops. If so it indicates that you've reached the current limit of the supply. "Would, say, a pullcord attached wound in the rotation's direction, pulled as the motor start, serve to bypass this delay?" Probably. \$\endgroup\$ – Transistor Jul 4 '19 at 21:49
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    \$\begingroup\$ Just make sure the cord connects to the shaft through a one way bearing or something like that or you'll get a cord that likes to whip around. Think...lawn mower starter. \$\endgroup\$ – DKNguyen Jul 4 '19 at 22:12

This delay is due to inertia, not to friction.

A spinning motor has rotational kinetic energy, which is equal to \$0.5\omega^2 I\$, where \$I\$ is the moment of inertia of the motor, and \$\omega\$ is its angular speed.

To make a motor spin, you have to supply this energy.

The rate at which you can supply energy is limited. All motors and power sources have some limit to the power, that is the rate of change of energy, that they can produce. This means that for any given motor and power source, there's a lower limit to the time it takes to spin up to a given speed.

If you can increase the rate at which you supply energy to the motor, by using a pull-cord wrapped round the shaft for instance, then you can reduce the time it takes for the motor to get up to speed.

There may be other more convenient ways to increase the power. If the limitation is the power supply, then using an energy store like a capacitor to temporarily deliver more power can help. If the limitation is the continuous thermal rating of the motor, then it's usually OK to overload the motor briefly at startup and rely on its thermal mass to limit the temperature rise. If the limitation is the maximum rated current (or the stall current at the rated voltage) of the motor, then it's a bad idea to supply more current, as it could risk demagnetising the field magnets.

  • \$\begingroup\$ Lovely info, particularly the capacitor idea! \$\endgroup\$ – Thanos Maravel Jul 5 '19 at 12:34

How small is 'rather small'? Obviously any motor requires 'some amount of time' to reach operating speed. It cannot go from stationary to full speed instantly. If you want to know why, you will need to post on a physics site.

But if we are talking about your typical 2cm hobby motor (which we would expect to reach speed in a time that isn't perceptible to a human), and the stationary load doesn't exceed its torque rating to move then it's most likely you aren't providing enough current. When it first starts, it will draw more current than its standard operating amount. If you cannot supply this current, the voltage will drop causing the motor to spin slower. As it spins up, the current required will lower and the voltage will increase until it reaches a point that your supply can handle.

Spinning it up with external force is equivalent to lowering the amount of torque it needs to deliver, which means it would need less current. So in theory that should work.

You could also use a supply with a higher current rating or charge up a capacitor to deliver the required start up current.

  • \$\begingroup\$ I know of the physics involved. The only issue was that I had read something rather offhand and badly worded about stall current, leaving me with the impression that the wiring itself provided some form of electrical resistance, thus the question. It is now obvious that this isn't the case. \$\endgroup\$ – Thanos Maravel Jul 5 '19 at 12:37
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    \$\begingroup\$ @ThanosMaravel Stall current is just how much current the motor will draw when the load applied to it is greater than it's ability to move. \$\endgroup\$ – hekete Jul 5 '19 at 12:46

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