Does a rotating disk develop a potential difference between the centre and rim? This stems from thinking about the question If a perfect conductor were to move, what happens to the electrons?.
Suppose we have a rotating disk with no external magnetic field, so this is not a homopolar generator/Faraday disk experiment. The rotation creates a difference in potential energy between the centre and the rim, so does this mean that if we connect a wire (with suitable brushes) between the centre and the rim electrons will flow from the centre through the disk to the rim then back through the wire? That is, does the rotation create an electrical potential difference between the centre and the rim?
It seems obvious to me that the answer is yes, however I have never seen the calculation done. Attempts to Google it fail because the results are swamped by articles on Faraday disks and/or homopolar generators.
 A: Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is
$$ m_e\omega^2 r = -e{d\phi\over dr} $$
After integrating, we get a potential difference between the center and a point R out
$$ \Delta\phi = -{m_e\omega^2 R^2\over 2e} $$
A conducting disk spinning at a rate of six million radians per second should generate about one volt of potential ten centimeters out from the center. I hope this was helpful. ;)
A: This answer is valid under the assumption that the wire also rotates with the disk. 
@cag's answer reveals two things :
(1) that the electric field is independent of the distribution of charge in the disk. We know that this distribution will vary for materials with different conductivities. 
(2) that the field is independent of the shape of the conducter under rotation. 
Since the wire is under rotation, a centripetal force exists in the wire as well. And thus, from @cag's formula we get to know that the potential difference in the wire is exactly same as that of the disk and in the same direction. 
Now, since you have specified the conductivity of the wire, being zero, we know that the distribution of charge is markedly different from that of the disk. 
For a wire having same conductivity as that of the disk, we know that no current would flow as soon as you connect the wire, since the net emf of the two cancel each other.
For a wire of different conductivity than that of the disk, the distribution difference between the two causes a kind of osmosis current which flows until it reaches a state where the distribution of charge along the radius of the disk is same for both.
Forgive me for my simplistic approach at solution, I'm not particularly adept at mechanisms on the microscopic scale.
