Newtons law on a rigid body, the center of mass I've read that if the sum of the forces on a rigid body is zero, then the center of mass will not accelerate(the body may rotate around the center of mass even though the sum of the forces is zero), and that Newtons second law $F=ma$ holds for the center of mass. But is this proven from Newtons second law used on all the small masses making up the rigid body?, or is this a law that has only been shown to be correct experimentally?
If this version of Newtons second law can be proven by the simpler form on particle masses, how do we show it?
 A: First we need to define what we mean by sum of forces on the rigid body equals zero. One way to do this is to say that the sum of the forces on each particle that makes up the rigid body is zero. That is to say:
$\sum{\mathbf{F_{i}}}=0$,
where $\mathbf{F_{i}}$ is the force on the ith particle in the rigid body. Now the centre of mass is calculated according to the formula:
$\mathbf{R}=\frac{1}{M}\sum{m_i\mathbf{r_i}}$
Now to find the acceleration of the centre of mass $\mathbf{R}$ take the second time derivative:
$\ddot{\mathbf{R}}=\frac{1}{M}\sum{m_i\ddot{\mathbf{r_i}}}$
But we can identify the right hand side with being the sum of the forces due to Newton's second law:
$\mathbf{F}=m\mathbf{a}=m\ddot{\mathbf{r}}$
And so
$\ddot{\mathbf{R}}=\frac{1}{M}\sum{\mathbf{F_{i}}}=0$
Where we've equated the net force with zero.
Furthermore from this we can see that the centre of mass obeys an equation of motion which is exactly identical to that of a point particle with the total mass of the body at the location of the centre of mass with a net force equal to the net force on the extended body, since:
$\mathbf{F_{net}}=\sum{\mathbf{F_{i}}}=M\ddot{\mathbf{R}}$
A: Yes Newton's 1nd law can be proven by the summation of all the particles in a rigid body. The talks about the motion of the center of mass only.
The story goes as to define the total momentum of a rigid by as the sum of the each particles momentum and noting that by tracking the velocity of the center of mass we can arrive at
$$ \boldsymbol{p} = \left( \sum_i m_i \right) \boldsymbol{v}_{\rm cm} $$
Now to apply Newton's second law
$$ \sum_i \boldsymbol{F}_i = \frac{\rm d}{{\rm d}t} \boldsymbol{p} $$
and note that the derivative of momentum only contains the acceleration of the center of mass since momentum only depends on the motion of the center of mass.
