Angular momentum about a moving point I was used to see the Angular Momentum Principle as:
$\dot{\underline{H}_O} = \underline{M}_O^{ext}$
that is, the change in angular momentum equates the external torque w.r.t. a point $O$.
Now, I've found a book where the same principle is expressed as:
$\dot{\underline{H}_O} + \underline{v}_O \times \underline{P} = \underline{M}_O^{ext}$
where $\underline{v}_O$ is the velocity of the point $O$, and $P$ the total linear momentum.
Is the first equation still valid if the point $O$ is moving? What is the difference between the two otherwise? 
Thanks!
 A: Taking the answer in advance, the second formula is the general one. Thus the first is not always valid.
But let's derive shortly the angular momentum principle (AMP) for a point mass. To do so, let's look at this general situation:

The definition of angular momentum $\underline H_B$ w.r.t. a generic point B is:
$$\underline H_B(t):=\underline\varrho_B(t) \times \underline P(t)$$
and the definition of the resultant torque around w.r.t a point $B$ is:
$$\underline M_B:=\underline \varrho_B\times\underline F$$
where F denotes the summ of all forces on the point mass.
to get the AMP, we need to differentiate $\underline H_B$ with respect to time. Using the product rule we get:
$$\underline{\dot H} = \frac{\mathrm d}{\mathrm d t}\left(\underline\varrho_B \times \underline P\right) = \underline{\dot\varrho_B} \times \underline P + \underline\varrho_B \times \underline{\dot P} = \underline{\dot\varrho_B} \times \underline P + \underline\varrho_B \times \underline F$$
now, let's use $\underline \varrho = \underline r - \underline r_B$ and thus $ \underline {\dot\varrho} = \underline {\dot r} - \underline {\dot r_B} = \underline v - \underline  v_B$ to get:
$$\underline{\dot H} = (\underline v - \underline  v_B)\times \underline P + \underline M_B$$
but since the cross product of two parallel vectors vanishes, $\underline v \times \underline P = \underline v \times (m\underline v) = \underline 0$, we can simplify it to the following:
$$\underline {\dot H_B} + \underline v_B \times \underline P = \underline M_B$$
For the other formula mentioned to be valide, it must hold $\underline v_B \times \underline P = 0$. We only get this if either $\underline v_B = 0$ (B is fixed) or $\underline v_B || \underline P$.
