Definition respective derivation of angular momentum formula I am reading An Introduction to Mechanics by Kleppner and Kolenkow (2014). On page 241 is the definition of the angular momentum: 

Here is the formal definition of the angular momentum $\vec{L}$ of a particle that has momentum $\vec{p}$ and is at position $\vec{r}$ with respect to a given coordinate system: $$\vec{L}=\vec{r} \times \vec{p}$$

In the book there is no explanation why this formula should be true. From this equation the formulas for torque and moment of inertia are derived. 
My question is: Why is the formula above correct? Why isn't the formula for angular momentum something completely different, like $\vec{L}=\sqrt{(\vec{r} \times \vec{p})2\pi M}$? Is the formula for angular momentum just a mere definition and if not, how to derive it? How did people come across that particular formula?
 A: You know Newton's second law:
$$\vec F = m \vec a~.$$
Now you multiply both sides by "$\vec r \times~$", then you get
$$\vec r \times \vec F = \vec r \times m \vec a = \frac{d}{dt} (\vec r \times \vec p)~.\tag{1}$$
(If you carry out the time derivative, the first term from the product rule is $\vec v\times\vec v$, which of course is $0$.)
This is very similar to the other form of Newton's second law:
$$\vec F = \frac{d}{dt} \vec p~,$$ 
which can be read as: a force gives us the rate of change of momentum.
Likewise, equation (1) reads: a force applied about an axis produces a rate of change of the momentum about the same axis. This means we identify the left hand side of (1) with the torque $\vec \tau$ and the the quantity in the brackets of the right hand side with the angular momentum $\vec L=\vec r \times m\vec v$, so eqn. (1) can be written as
$$\vec \tau=\frac{d}{dt}\vec L~.$$
So in a sense it's in comparison with Newton's second law that you define this quantity as angular momentum.   
