Newton's Second Law Equivalent in rotational dynamics The law that
$$\frac{d\vec{L}}{dt}= \vec{T}$$
where $\vec{T}$ is torque about a frame's origin $o$ and $\vec{L}$ is the angular momentum about that origin $o$. 
Can this law be ultimately (always?) traced backed to Newton's Second Law ?
 A: I'll expand my comment into an answer.
I would take $\mathbf{T}=d\mathbf{L}/dt$ as the definition of torque, but it sounds like the OP takes $\mathbf{T}=\mathbf{r}\times\mathbf{F}$ as the definition. Either way, we need to prove that the two expressions are equivalent for a system of particles.
The total angular momentum is
$$\mathbf{L}_{tot}=\sum \mathbf{r}_i\times \mathbf{p}_i$$
Differentiating with respect to time and applying the product rule, this becomes
$$\sum \frac{d\mathbf{r}}{dt}_i\times \mathbf{p}_i+\sum \mathbf{r}_i\times \frac{d\mathbf{p}_i}{dt}$$
The first sum vanishes term by term. Applying Newton's second law to the second sum, we get
$$\sum \mathbf{r}_i\times \mathbf{F}_i$$
The fact that there was a sum over multiple particles ended up being unimportant, because the manipulations were all term by term.
A: Torque is the rotational equivalent of force. The second law state, the sum of the forces 
$\sum F = ma$.
$\alpha$ is the rotational equivalent for acceleration, so the law would look like $\tau$ (torque) = $m \times \alpha$. The angular momentum would be $m \times \omega$. velocity over time is acceleration. $p/t = F$.
Therefore, angular momentum can be traced to torque, the rotational equivalent of force.
A: No, it is a different law altogether .
Newton's law are true for point masses only
