How Fast Could a Cyclone at the North Pole Achieve?
There's a few problems with this question.  The first is, the Coriolis effect isn't a force in and of itself.  It's a velocity dependent effect.   If the wind speed is zero over the North or South Pole, the Coriolis effect would be zero.    If the wind speed is 100 kph over the pole, the Coriolis force would be (if this site is trustworthy) $0.000146 \times V \times \sin(latitude)$, in other words, pretty small, but over long distances, a small force can create a significant change in direction.   But there-in lies the problem to answering your question.   It would depend on the size of the cyclone and the initial wind speed of the cyclone. 

The other problem is, the Coriolis force doesn't form neat little circles.  Around the polls it's more like this, so estimating a presumably circular cycolone entirely due to Coriolis is doubly problematic, because a cyclone wouldn't form.  You'd get some wind direction change but not a cyclone.



The cyclone effect is the most efficient way to move air, as lower mass warm air wants to rise to higher altitudes and higher mass cold air wants to fall to lower altitudes.  The cyclone may be initiated by the Coriolis effect but once begun, it's driven by warm water, evaporation, and variation in air mass.  If you notice, Cyclones lose strength both on land and if they turn north over the ocean.  The colder ocean reduces the wind strength because it evaporates less easily and the air over the ocean cools, so with real cyclones, as the Coriolis effect grows stronger as the cyclone moves north, the cyclone grows weaker.  Cyclones don't represent the Coriolis effect.  It only plays a role in getting them started.  

As you probably know, the Coriolis effect doesn't actually move anything.  Objects move by their momentum and the Earth rotates by it's momentum and it's the earth that moves under the object that gives the impression of a Coriolis effect.   If you stand on the North Pole, the Earth rotates about about 1/2 the speed of the hour hand on your watch.   It's hard to imagine that has much effect on anything and it doesn't unless the distance is significantly large.   

I can't do the precise math on this one, though I find it quite interesting, but the above is a general answer.   If you try the physics board, you might get a more mathematical answer, but I think the question might work better with a bowling ball and a friction-less earth rather than the wind and the earth and a cyclone of unspecified size.   I got curious, so I asked the question with that in mind.