homework and exercises - Newton's Third law of Motion, who can tell me how to deduct below?

source:http://farside.ph.utexas.edu/teaching/336k/lectures/node11.html#e3.24

Consider a system of N mutually interacting point objects.

Newton's second law of motion applied to the $i$ th object yields:

$$m_i \frac {d^2 \vec {r_i}}{dt^2}=\sum_{j=1,N}^{j\neq i} \vec {f_{ij}}$$ Let us now take the above equation and sum it over all objects. We obtain $$\sum_{i=1,N} m_i \frac {d^2 \vec {r_i}}{dt^2}= \sum_{i,j=1,N}^{j\neq i} \vec {f_{ij}}$$ because of newton's third law of motion , the right side of equation is equal to 0,but the question is that i can't understand how the left side of equation turn to below? $$M\frac {d^2\vec {r_{cm}}}{dt^2}=\vec 0$$ where $M=\sum_{i=1}^{N}m_i$ is the total mass. The quantity $\vec {r_{cm}}$is the vector displacement of the center of mass of the system, which is an imaginary point whose coordinates are the mass weighted averages of the coordinates of the objects that constitute the system: i.e., $$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i}$$

Answer

The position vector of the centre of mass is defined as:

$$\mathbf {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \mathbf {r_i}}{M}$$ i.e., $$M \mathbf {r_{cm}}=m_1 \mathbf r_1+m_2 \mathbf r_2+ m_3 \mathbf r_3+...+m_N \mathbf r_N$$ $$\Rightarrow M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = m_1 \frac {d^2 \mathbf {r_{1}}}{dt^2}+ m_2 \frac {d^2 \mathbf {r_{2}}}{dt^2}+...+m_N \frac {d^2 \mathbf {r_{N}}}{dt^2}$$ (Double Differentiating both sides) Here $M=\sum_{i=1}^{N}m_i$. Since single differentiation of the position vector gives velocity $\mathbf v$ and double differentiation gives acceleration $\mathbf a$. Therefore $$\frac {d^2 \mathbf {r_{cm}}}{dt^2} = \mathbf a$$ This means that $$\sum_{i=1}^{N} m_i \frac {d^2 \mathbf {r_i}}{dt^2} = M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = M \mathbf a = \mathbf {F_{net}}$$

Now if the net force acting on the object is $\mathbf 0$ then

$$M \mathbf a= M\frac {d^2 \mathbf {r_{cm}}}{dt^2}=\mathbf 0$$

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Note that internal forces cannot cause acceleration as they always come in action-reaction pair.