We defined angular momentum only for the case of rotating objects. It turns out that angular momentum can be very useful in cases where we do not have simple rotation. The specific case of interest to us will be when there is a combination of rotation and translation. Therefore we need to derive a suitable definition for the angular momentum of a particle moving in a general trajectory.
For simplicity we consider first
a particle moving with velocity v with respect to our reference frame. There are no external forces acting on the particle so according to Newton's
laws its linear momentum cannot change. Nor can by the way its
angular momentum for if there is no external torque acting then
there can be no change in the angular momentum of the particle.
We wish to determine the angular momentum with respect to a specific point of reference. Since no forces act on the particle it
will continue along a straight line and at some point in time will be
at its closest approach, d
to our chosen reference point. This is the time at which
it is possible for us, based on our previous definition of this quantity,
to write an expression for the angular momentum. The reason is that
the velocity of the particle at that instant is indistinguishable
from a particle in rotation about the reference point and so we can
calculate the angular momentum as if the particle was going to
continue in a circular trajectory about the reference point. With this
approach we get
Now this formula should hold any where along the trajectory
since angular momentum is conserved. Specifically when the distance
to the reference point is r and the angle between the vector
from the reference point to the particle is 
then
and the angular momentum is
To give you an example as I walk parallel to your
rows of chairs my angular momentum with respect to all students
in one row is the same. However it is different for each row being larger the larger the smallest distance from the tangential
line of my trajectory to each of you.