ohm's law and motor question?

“motor turning at about 340 RPM” Are you sure? Is that after some pulleys or gears that are reducing the speed? An AC motor turning at 340 RPM would be a very unusual motor. If you already have pulleys or gears, that is what you need to change.

It is difficult to change the speed of an AC motor unless it is a universal motor, really a DC motor with a commutator running on AC.

If the motor has a capacitor, it is probably an induction motor that uses the capacitor for starting. Many such motors are designed to switch off the starting circuit when the motor reaches full speed. If you do something to reduce speed, the starting circuit will overheat.

Another answer points out that inserting a resistor will reduce the torque. If the load is a centrifugal pump or a fan, reducing torque probably reduce the speed to a slower stable operating speed. Most other types of loads require about the same torque at lower speeds and may only slow down a little until the resistance in increased to the point that the motor stalls. Reducing the voltage with a light dimmer type device will have the same effect.

Even with a fan or pump load, slowing a motor by reducing the voltage may cause the motor to overheat unless the motor is designed for that duty.

If the load is a fan or pump and if the capacitor is not switched off when the motor reaches full speed, there are electronic frequency changers called variable frequency drives (VFDs) that can be used, but they are rather expensive.

Depending on the type of motor (induction, synchronous, squirrel cage, etc), simply adding a resistor might not change the speed at all, but could drop the torque.

Assuming the motor's speed is proportional to voltage, and that you are able to hold the torque constant, then half the speed would require half the power. These are major assumptions you will need to verify.

You want 440 watts to the motor, but, the only way to reduce the power is with a resistor in series. This means your motor now uses 55 V at 8 A. The other 55 V must drop across your resistor. 55V / 8 A = 6.875 ohms.

The problem, though, is that your resistor will also dissipate 440 W as heat. This is similar to an electric baseboard heater. Finding a 6.875 ohm, 440 W resistor, will be a challenge.

It would be better to switch the motor on and off, really quickly, using a pulse width modulator, so it is only on half of the time.

Another option is to "gear down" the motor. You'll get more torque, less speed, and no resistor is needed.

Another option would be to add an identical motor in series, to turn a flywheel. Adjust the size (or mass) of the flywheel to match the mass of the load on the first motor.

The resistance of the motor can be looked at as R = E/I = 110/8 = 14 ohms. But that is not a constant, the motor draws current depending on the RPM and on the load, so the resistance varies over a wide range. But leaving all that aside...

to drop the power in half, you need an equal value resistor, that is, 14 ohms. Power in the resistor is 55²/14 = 220 watts, so you need a 400 watt resistor, very expensive (you should always double the wattage).

But that value of resistance may not allow the motor to operate at all. First of all, the starting current is 3-5 times more than the operating current, but the resistor limits the current to 8 amps.

Your best bet is to get a light dimmer type control. Get one that is rated for 20 amps and for motor load.

I'd like to know where you think you're going to find an 8 amp resistor to begin with.