Does linear motion have angular momentum? I'm wondering whether an object that is moving in a linear fashion can be considered as having angular momentum. I am thinking that all that would be needed is to pick a reference point that is not on the axis that the object is traveling on.
So for example, if a bullet is flying in a straight line from west to east at 5 feet above the floor. If a pick a point in the middle of the floor, I can say that at the starting point the bullet is at a 170 degree angle, for example, from that reference point. A moment later, it could be directly above--90 degrees. And so on.
Although this makes sense to me, I am working on a problem in which I am trying to use this and it isn't working. I would like to understand why.

The specific problem can be found here, 
and is :


A small 4.50kg brick is released from rest 2.00m above a horizontal seesaw and 1.6m to the right of the fulcrum at the middle of the seesaw.
  A) Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released.
  B) Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.

So I would find the original angle to be about 1.001 radians ($tan^{-1} \frac {2.5}{1.6}$). 
The time of the fall: $$\frac{2d}{g}=t^2=\frac{2*2m}{9.8 \frac{m}{s^2}}$$
$$t=0.6388s$$
$$\omega=\frac{1.001}{0.6388s}=1.566s^{-1}$$
And using $L=I\omega$ and treating the square as a point mass, we get   $$L=4.5kg*1.6^2m^2*1.566s^{-1}=149kgm^2s^{-1}$$
 A: The (orbital) angular momentum of a body is defined by the cross product of the position vector and the linear momentum L = r x p. Thus a body in linear motion has an angular momentum which depends on the reference point of the position vector. The absolute value of the angular momentum is L = r.p. sin(phi), where phi is the angle between the position and the linear momentum vector.
Thus the answers to the two questions are easy:
A) The angular momentum is L = 0 because at t=0 v=0.
B) The velocity of the brick after a free fall of d = 2.0m is (with v = g·t, g = 9.81m/s^2, t = sqrt(2d/g) = 0.6386s) v = g·t = sqrt(2gd) = 6.26m/s, the linear momentum is p = m·v = 28.19 kg·m/s
Because phi = 0, sin(pi/2) = 1, r = 1.60m just before impact, the absolute value of the angular momentum of the brick just before impact on the seesaw is: 
L = 1.60m·28.19kg·m/s = 45.10kg·m^2/s
its direction is perpendicular to the plane of the figure pointing to the back of the figure.
The calculation of the angular momentum with the formula L = I·omega = m·d^2·omega with I = 11.52 kg·m^2 and omega = v/r = 3.915/s gives the same result. You have to take the angular velocity at the impact point. At the starting point the angular velocity and angular momentum is 0. 
