I was wondering about something the other day. This is one of those things I was sure I knew but have no idea why, so I'd better confirm it.

Let's say that we have a bullet that is .40 caliber, or 10mm. Then we load this bullet into a gun and fire it in a barrel with a 1 in 7.874 twist. This will make the bullet spin at a rate of 5 revolutions per meter. We fire it at a velocity of 1000 meters per second, and we do it in space where there are no significant effects of air resistance or gravity to deflect it.

If the CG of the bullet is exactly on the rotational axis, it should travel in a straight line for the foreseeable future, right?

What if the CG is 0.5 mm away from the axis of rotation?

My thought is that the barrel will constrain the bullet to rotate on the axis until it clears the barrel, at which point it will begin to rotate around the center of gravity, causing the bullet to wobble. This would lead to the bullet being pulled further and further to one side by the wobble so that it travels in a curved path, or possibly a diagonal away from the original line of sight.

Is this correct?

As I was thinking about it the other day, it occurred to me that it may still travel in a nearly straight path along the line of sight, but that it's exact path would look like a spiral about 0.5 mm around the true LOS.

Thanks.