The faster you spin the motor, the more volts you'll get - somewhat!... it depends on the circuit you attach it to! A low resistance will increase the current and decrease the volts.
Think more about the power you are putting into rotate the rotor of this motor. What happens to that power when you attach a load?
You'll end up with [pure theory, discounting losses, &c.] Power-in=Volts x Amps, and Load resistance = V/I. Two variables (V and I), two knowns (Power and load). [This is very idealised.]
So you can see that volts and amps are dependent on power and load.
One thing to remember about using a DC generator to, e.g., charge a battery is that you have to include some means to prevent the battery from discharging back into the generator when it is no longer 'generating'. The output volts have to stay above the volts of the battery (and not too far above either, else the battery will overcharge).
Preventing the discharge can be done with just a simple diode, but if you are, for example, charging a 6V battery with a diode with a 0.6V forward voltage drop, you are losing 10% of your input power to the diode. Similarly, if you end up generating 10V because your generator is running fast, then if you are running a simple linear circuit you'd need to dump 40% of the power.
Clever switching circuits are needed to efficiently manage the power coming out of, particularly, wind generators for reasons of their (obvious) speed variability...and the emphasis is, indeed, usually on 'efficiency', because 'energy optimisation' is the name-of-the-game if you're already bothered to build a wind generator!
I would imagine that what you'd want to do for an efficient DC wind generator is to find one that generates high volts, thus low current and the diode wouldn't suck up such a big percentage of losses, then feed capacitors from which you use a switching scheme to draw the power off them. As the generator drops below the level where it is feeding volts onto the capacitors, it'd 'unload' a little and the output volts would go up again. This should therefore reach some balance where it runs pretty much at a rate that generates the volts enough to cross the diode to the caps, and current to supply the power demand on the other side of the switching circuit. This is just a guess, I am sure there are heaps and heaps of pages on DIY wind power management on the www.
