Real application- Spinning flywheel and the kinetic energy of a moving rod So I think I've decided on a weapon for my robot, but now I'm unsure of how to go about the last bit of physics. I've modeled a spinning flywheel which is to be made out of a 4" iron caster which is clad in urethane. The flywheel is spun up, and a 16" steel rod, 5/8" in diameter, is thrown between the wheel and an idler. Rotational kinetic energy is converted to translational kinetic energy. But ignoring losses due to slipping, what is the final kinetic energy of the rod after a distance s, just before the rod meets a stopper and the whole wheel/rod system comes to a stop?

At first, I thought all of the kinetic energy would naturally be converted. But now I'm not so sure of myself. After all, the rod has to be accelerated over the distance s. As the wheel slows down, the rod picks up speed. It should be possible to calculate the final velocity of the rod just prior to position s, and thereby the final angular velocity of the wheel at the same time. 
But on the other hand, since the system will inevitably come to a full stop, can we not say that all of the energy of the flywheel (barring frictional losses) is delivered to the spike? By this reasoning, the last bit of rotational energy is converted at the point where the rod stops, whether that be at point s or at the point where the rod mercilessly perforates the enemy. Does this sound like good reasoning? 
For S&Gs, I'll go ahead and include my figures:
Flywheel at max angular velocity:


*

*$M = 1.13$kg

*$r = 0.0508$m

*$I = 0.000815$ (approximated as a simple disk)

*$ω = 556.1$ rads/s (maximum)

*$E_k = 126.02$J (rotational kinetic energy)


Rod, starting from zero initial velocity:


*

*M = .632kg

*s = 0.3048m (12" of travel distance)

 A: In the simplest approximation the two separate bodies become one at the instant of contact, which isn't realistic but gives an upper bound on the final speed of the rod. 
The kinetic energy of the flywheel is then shared by the combined mass of the flywheel and rod, the angular velocity of which can be approximated by conserving the momentum in the system.
A more accurate velocity estimate would require knowledge of the friction between the disc and flywheel, which gives a resultant force, acceleration and final velocity. This would also require far more maths than I can manage at this time of night.
All of this assumes infinitely stiff shafts i.e. the shaft doesn't act as a spring smoothing the transfer of energy from disc to rod.
No doubt there is someone else here with considerably better maths than I have who can develop this further.
A: As with inelastic collisions where there is linear motion - eg when two objects stick together after collision - KE will not be conserved but, in the absence of external forces which might do work on the system, momentum is conserved. You should be calculating on this basis, rather than conservation of kinetic energy. You can get the KE of the rod from its linear momentum $p=mv$ using $E=\frac{p^2}{2m}$.
There are 2 collisions here : (1) when the rod is gripped by the flywheel and idler wheel, and (2) when the rod hits the stopper.
(1) When the rod "collides" with the flywheel and idler wheel, the relative speed between the three bodies is initially high, but after the rod is gripped the relative speed becomes zero. This is equivalent to a totally inelastic linear collision in which the objects stick together and the speed of separation is zero. In such cases KE is not conserved. However, momentum is still conserved - in this case angular momentum. The initial AM of the rotating flywheel is shared with the rod and the rotating idler, in such a way that there is no relative motion (no slipping) at the points of contact $(v_2=r\omega_2)$ :
$I\omega_1+mv_1r=2I\omega_2+mv_2r$  
Probably the initial velocity of the rod $v_1$ is small and can be ignored.
The length of the rod makes no difference to the calculation, provided that it is long enough for the relative speed between contacting surfaces to become zero (the no-slipping condition). If, as here, the friction is large, there will not be much slipping so the rod is accelerated over a very short distance.
(2) When the rod hits the stopper (or target), and the rod, flywheel and idler wheel eventually come to rest, there might again be some slipping between the rod and wheels. In that case not all of the KE of the wheels will be transferred to the target through the rod. It depends on the amount of friction there is between wheels and rod and how quickly the rod is slowed down by the target. 
This is not an easy calculation to make, I think, because the target provides an unknown external force. However, if the target is relatively soft - ie deformable - but the rod is not, and friction between wheels and rod is high, then there will be no slipping so the whole KE of the wheels as well as the rod will be deposited in the target.
