Surely if I have a rod about a fixed axis and a moving particle hits the end it will cause the rod to spin and therefore create angular momentum?
First off, there is no reason to expect that any of the conservation laws apply to the rod. A moving particle collides with the rod, and the rod has constraints that act on it to keep one end fixed. The collision and those constraint forces are external forces, some of which result in external torques. The conservation laws don't apply to the rod. They apply to the rod+particle+Earth system.
In general,
- A system conserves energy if there is no transfer of energy between the system and the surrounding environment.
- A system conserves linear momentum if no external forces act on the system and if all forces internal to the system obey the weak form of Newton's third law.
- A system conserves angular momentum if no external torques act on the system and if all forces internal to the system obey the weak form of Newton's third law.
Secondly, you are ignoring that even point masses can have non-zero angular momentum. Angular momentum is always measured with respect to a point, not an axis. The angular momentum of a point mass is easily computed: It is $\vec L = m \vec r \times \vec v$, where $m$ is the mass of the point mass, $\vec r$ is the displacement vector from the central point to the point mass, and $\vec v$ is the velocity of the point mass. When viewed from the right perspective, the rod+particle system does conserve angular momentum. This "right perspective" is one in which the constraint forces on the rod exert zero torque.
The "trick" only works for elastic collisions, for which the relative velocities of approach and separation are equal and opposite, when measured along the common normal at the point of contact :
$v_2-v_1 = u_2-u_1$.
In general we can write :
$v_2-v_1=e(u_2-u_1)$
where $e$ is the Coefficient of Restitution. For elastic collisions $e=1$; for inealstic collisions $0 \le e \lt 1$.
Since relative velocities are being compared, the "trick" also works in the frame of reference of the rod. It can be applied in other reference frames also, but in such frames it is more difficult to distinguish between normal and tangential components of motion/forces.
The "trick" can be deduced from both conservation of kinetic energy and linear momentum (see Wikipedia article on Elastic Collisions). Therefore using it is equivalent to using one of the two conservation equations. You can combine it with either, whichever is the more convenient - usually (IMO) conservation of momentum is the easier to apply.
I think the key which answers your doubt over why the above works for extended bodies when it is derived for linear collisions between point particles, is your own observation that the collision occurs between rigid bodies. Because of this there is a clear point or plane of contact, and the collision is instantaneous compared with the time required for any other motion, such as rotation. Under these conditions, the collision can be modelled as a linear collision between a point particle and a plane - the spike being the point particle and the rod the plane.
If there were significant deformation of either body, contact would be along a non-planar surface, and would take a finite time, during which one or both bodies rotate, with the result that the collision would be considerably more difficult to model.
Best Answer
Assuming that there are no external forces/torques acting then angular momentum is conserved about any point.
Here is an analysis of your example where the incoming mass sticks to a rod.
If there had been a hinge involved this would mean that there was an external force/torque applied on the rod and incoming mass system.
In such a case making the point about which to find the angular momentum the hinge would mean that angular momentum would be conserved because the force exerted by the hinge would exert no torque about the hinge.