Actually, it isn't at all reasonable. I'm surprised that someone who is a retired physicist could make this mistake. I was actually surprised that there is so much confusion about this very simple issue even by non-physicisits. Especially after the correct answer was pointed out a number of times.
If you shoot bullets out of a gun, they carry momentum. This causes recoil, because momentum is conserved. That's really all there is too it![/COLOR]**************************************
Hey, let's not get wrapped around the axle on this one. I just thought it was a good brain teaser. I'll give you the last comment and promise not to respond -- we'll agree to disagree. The rifle you talk about is very, very different in two important ways, from the spinning disk:
(1) The rifle is rapidly accelerating the bullet, F = ma. So say the bullet weighs 5 grams and the gun weighs 5000 grams; further the bullet is accelerated to 1000 meter/sec, and the rifle barrel is 1 meter long. By using v = at and s = ½at², by algebra, a = v²/(2S), where a = acceleration, t = time, v = velocity, s = distance.
Anyway a = (1000*1000)/(2*1) = 500,000 m/sec²
F = ma = .005 kg * 500,000 m/sec² = 2500 Newtons which is the recoil.
so the acceleration of the rifle
a = f/m = 2500/5 = 500 m/sec²
However, for the disk, the bullet is sitting on the edge of the disk, spinning. The acceleration occurred earlier, perhaps several minutes earlier, when it was accelerated from the center to the outside of the disk. Now it is just spinning at max velocity -- no linear acceleration (ignoring centripetal force). Then it is released.
(2) For the disk, the force does NOT act through the center of mass, as it does for the rifle. It has a lever arm, equal to the radius of the disk. To maintain equilibrium, the forces and moments must equal zero. With the rifle, the moment is zero, so there is a simple backward recoil. For the disk, there must be an opposite moment -- this occurs by a change in the angular momentum of the disk, i.e. the angular velocity. And this occurs when the torque was initially applied -- as the ball was accelerated from the center of the disk to the outside perimeter.
OK, I'll say no more -- you are probably right. That is just the way I, and some other dudes, see it.