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 $$L=4.5kg*1.6^2m^2*1.566s^{-1}=149kgm^2s^{-1}$$