Chuck,
it looks like I am forced to break out the physics.
It's all about torque and the drag force.
1) Let me represent a bullet with a vector for simplicity. It goes from (0,0) to (0,10) (so that means its vertical pointing down). It has a center of gravity at (0,2).
2) This bullet is travelling to the left (X approaches negative infinity) at some velocity which produces a drag force which is uniform over the entire length of the bullet (assume that the bullet is simply a uniform cone). We cannot calculate the drag force, but since it is constant, we can give it an arbitrary value of 2.
3) We can calculate the torque about both sides of the radius which is the center of gravity.
4) Torque = RxF (r [i]cross[/i] f, this is a cross product, not a multiplication). This is equivalent to |r||f|sin(theta), where theta is the angle between the direction of the radius and the direction of force being applied. Since the radius is going vertically and the force is being applied from the left, the theta is 90 degrees. Sin90 = 1, so our equation can be further simplified: T = |r||F|
5) Torque on the lower half would then be the integral of |r||f|dr from r = 0 to r = 2. f we already said was a constant of 2. This constant can be moved out of the integral. So we have T = 2 integral(rdr)from r=0 to r =2. This very simple integration is obviously 4 (2^2-0^2).
6) Torque on the upper half is 2 integral(rdr)from r = 2 to 8, so that comes out to 60 (8^2-2^2).
7) Thus, since the torque on the upper half is greater, the bullet will rotate clockwise with the upper half coming down until it is parallel. Since it rotated about the center of gravity, the bullet is now in the position (-2,2) to (8,2)
8) At this point, the torque about the upper and lower half are both 0, so it will not rotate any more. This is the stable position.
9) At first, it might seem that this same logic would mean that it would rotate due to torque of gravity. This is false, because remember that force due to gravity = m*g, and since torque is about the CM (center of mass) point, that means that force due to gravity will be zero on both sides of the CM.