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., $$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i}$$


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

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


Try this: take the definition

$$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i},$$

multiply both sides by $M$, and differentiate the equation twice with respect to time.


Clear answers has been given before I have written this comment. Probably the easiest way to derive the equation in question is by definition of center of mass. These answers have been given so I will not do it here. I will however try to arrive to the question from forces acting on a system of particles, as I hope it will give you some insight.

Consider system of n particles. I will represent total force acting on a i'th particle as sum of external force and forces experienced from other particles.

Therefore: $$\text{Force on }i^{th}\text{partilcle:}\qquad\vec F_i=\vec F_i^{ext}+\sum_{j=1,j\neq i}^n \vec F_{ij}$$ $$\text{By Newton's }3^{rd}\,\text {law:}\qquad\vec F_{ij}=-\vec F_{ji},\text{where i, j represent force on }i^{th}\,\text{particle due to}j^{th}\,\text{particle.}$$ Thus total force on n particles is given by: $$\sum_{i=1}^n\vec F_i=\vec F=\sum_{i=1}^n\vec F_i^{ext}+\sum_{i=1}^n\ \Biggl(\sum_{j=1,j\neq i}^n \vec F_{ij}\Biggr)$$ $$\sum_{i=1}^n\ \Biggl(\sum_{j=1,j\neq i}^n \vec F_{ij}\Biggr)=\sum_{j=1}^n\ \Biggl(\sum_{i=1,i\neq j}^n \vec F_{ji}\Biggr)=-\sum_{j=1}^n\ \Biggl(\sum_{i=1,i\neq j}^n \vec F_{ij}\Biggr)=-\sum_{i=1}^n\ \Biggl(\sum_{j=1,j\neq i}^n \vec F_{ij}\Biggr)$$ It implies that $$\sum_{i=1}^n\ \Biggl(\sum_{j=1,j\neq i}^n \vec F_{ij}\Biggr)=0$$

Therefore total force on n particles is sum of all external forces, as internal forces between particle do cancel out.

$$\text{Finally getting back your answer, we know that:}\quad\vec F=M\vec a_c\quad$$ Thus: $$\vec F=\sum_{i=1}^n\vec F^{ext}_i=\sum_{i=1}^nm_i\vec a_i\implies \frac{\sum_{i=1}^nm_i\vec a_i}{M}=\vec a_c\quad \text{where M}=\sum_{i=1}^nm_i$$

The equation tells us that sum of each external force on i'th can be expressed simply as total mass M and acceleration of center of mass. Trivially, if external forces sum out to zero, Newton's First Law applies.


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