Running a 6V DC Motor on 0-30V variable supply?

If the motor draws 30mA at 6V then it has a load of 200 ohms. To reduce 30v to 6v you'd need a 4:1 ratio, so you'd have to put 800 ohms in series.with the supply.

 

Shall we assume that the motor is a permanent magnet motor with commutator brushes rather than something exotic like a brushless DC motor? If that is the case then one thing to realize is that the motor is also a generator. Attach a voltmeter to the motor terminals and manually turn the shaft in the same direction that the power supply would make it turn, and notice that the generated voltage polarity is in opposition to the power supply. See attachment for a discussion of this generated voltage - the Back EMF (BEMF).

The current through the motor is proportional to the shaft torque, and the torque is equal to the load plus internal friction. Since you are running the motor with no load, torque equals friction. I would expect that the internal friction is independent of motor speed so that the 30 mA no-load current is also the minimum current needed to start shaft rotation.

So if you put an 800 ohm resistor in series with the 30 volt variable power supply, the voltage would need to be raised to 800 x 0.030 = 24 volts before the motor would start turning. But even a very small increase in the load torque would be enough to stall the motor at 24 volts. These are just generalizations that ignore the effect of air resistance on the turning rotor, so the starting friction might be less than at 2000 RPM. Without more detailed performance measurements on the motor it would be difficult to predict what might actually happen when adding an 800 ohm resistance, but it is generally considered bad practice to add resistance to a motor circuit.