Conservation of Angular Momentum

The conservation law

If the net external torque about a chosen point is zero, then $d\vec L/dt=0$ and

The three classic settings: (i) a skater pulling her arms in — internal forces only, so $I_1\omega_1=I_2\omega_2$; (ii) two disks pressed together — friction between them is internal, so $I_1\omega_1=(I_1+I_2)\omega_f$; (iii) a bullet embedding in a pivoted rod — the pivot exerts a large force, but that force acts at the pivot, so its torque about the pivot is zero.

Energy is a different question

Since $K=L^2/2I$, conserving $L$ while reducing $I$ increases $K$ — the skater's muscles do the work of pulling her arms inward against the centripetal requirement. In the disk and bullet cases $I$ increases, so $K$ drops: those are perfectly inelastic rotational collisions and the lost energy goes to heat and deformation.

Why linear momentum is usually not conserved here

In case (iii) the pivot delivers a large impulsive external force, so $\vec p$ of the system is not conserved even though $L$ about the pivot is. Choosing the pivot as the reference point is what makes the problem solvable.

Central forces and orbits

Gravity on a satellite always points along $\vec r$ toward the star, so $\vec\tau=\vec r\times\vec F=0$ and $L$ is conserved about the star:

The second equation is Kepler's second law, and it is nothing but conservation of angular momentum written as a rate of swept area.