It is just an expansion of the known Newton's laws of motion to real objects which have size, shape and etc. [where else Newton's laws are defined only for point particles].
When you go from one particle to system of particles, the concept of net external force gets modified, because now you need to define first what is a system and what you meant by external to that system.
Why I am saying this because, we know that the most common force we deal with the movement of macroscopic objects are electrostatic and gravitational, where both of them are central in nature. Now, you are in a situation to distinguish all these central forces into external and internal.
It is not predefined what is external and internal. For example, on playing with marbles, if you say your hands and a marble as a whole system, then there is no external force, but if you just consider the marble alone then the force given by your hands on striking the marble, will be external for the marble.
Thus Newton's second law for a system of particles is written as,
$$ \frac{d^2}{dt^2}\sum_i m_i \vec{r_i} = \sum_i \vec{F_{i(e)}} + \sum_{i,j \,(i\neq j)\,} \vec{F_{ij}} \,\,\,\ldots...eq.(1)$$
Using action - reaction relation, $ \vec{F_{ij}} = -\vec{F_{ji}} $ the second term i.e. internal forces cancel of each other.
It leads to, $$ \sum_i\vec{F_{i(e)}} = \sum_i m_i \left(\frac{d^2}{dt^2}\sum_i \frac{m_i \vec{r_i}}{m_i}\right) $$
The right hand side is defined as the center of mass of the system of particles, $ \sum_i \frac{m_i\vec{r_i}}{m_i} = \vec{R_{cm}} \,\,\,\ldots...eq.(2)$
eq.(1) gives, $$ M\,\frac{d^2}{dt^2} \vec{R_{cm}} = \sum_i \vec{F_{i(e)}} \,\,\,\ldots...eq.(3) $$
From this we understand that, all our Newton's laws are perfectly applicable only for center of mass i.e. Only the point where the center of mass is defined is the only point that moves exactly as it is defined from Newton's laws. All other particles can have motion other than linear translational motion, but not center of mass.
From our definition, it is easily derivable that the total linear momentum is defined as,
$$ \vec{P} = \sum_i m_i \dot{\vec{r}} = M \frac{d\vec{R_{cm}}}{dt} \,\,\,\ldots...eq.(4) $$
where $ \dot{\vec{r}} = \frac{dr}{dt} $
In a similar way, we can describe rotational quantities like angular momentum and torque when the object is in rotational motion.
We can start by finding the angular momentum of a system of particles, $$ \sum_i L_i = \sum_i \vec{r_i}\times \vec{p_i} \,\,\,\ldots...eq.(5)$$
Again making use of the concept of center of mass, position vectors from the origin is written in terms of center of mass coordinate and its relative distance from center of mass. $$ \vec{r} = \vec{R_{cm}} + \vec{r'} \,\,\,\ldots...eq.(5)$$
where $\vec{r'} $ is the position coordinate from center of mass.
Differentiating eq.(5) gives $$ \vec{v} = \vec{v_{cm}} + \vec{v'} \,\,\,\ldots...eq.(6) $$
Angular momentum from eq.(5) becomes,
$$ \sum_i \vec{L_i} = \sum_i \vec{r_i}\times\vec{p_i} = \sum_i m_i (\vec{R_{cm}}+\vec{r'}) \times (\vec{v_{cm}}+\vec{v'}) $$
Expanding with vector multiplication,
$$ \sum_i \vec{r_i}\times\vec{p_i} = \sum_i m_i \vec{R_{cm}}\times\vec{v_{cm}} + \sum_i m_i \vec{r'_i}\times\vec{v'_i} + \sum_i m_i \vec{r'_i}\times\vec{v_{cm}} + \sum_i m_i \vec{R_{cm}}\times\vec{v'_i} $$
From the definition of center of mass, the relative coordinate of center of mass w.r.t. center of mass is zero. $ \sum_i m_i \vec{r'_i} = 0 $ and also $ \sum_i m_i \vec{R_{cm}} \times \vec{v'_i} = \frac{d}{dt} \sum_i m_i \vec{r'_i} = 0 $
We are left with,
$$ \sum_i \vec{L_i} = \sum_i m_i \vec{R_{cm}} \times\vec{v_{cm}} + \sum_i \vec{r'_i}\times\vec{v'_i} $$
It says that, the total angular momentum of system of particles is equal to the angular momentum of the center of mass with respect to origin and the angular momentum of all particles with respect to the center of mass.
In the same way, the kinetic energy of system of particles (with eq.(6)),
$$ \frac{1}{2}\sum_i m_i \vec{v_i}^2 = \frac{1}{2}\sum_i m_i (\vec{v_{cm}}+ \vec{v'_i})^2 $$
$$ \frac{1}{2}\sum_i m_i \vec{v_i}^2 = \frac{1}{2}\left[\sum_i m_i \vec{v_{cm}}^2 + 2 \sum_i m_i \vec{v_{cm}}\cdot\vec{v'_i} + \sum_i m_i \vec{v'_i}^2 \right]$$
By the same token of center of mass, $\,\, 2 \vec{v_{cm}}\cdot \frac{d}{dt} \sum_i m_i \vec{r'_i} = 0 $
Hence, Kinetic energy of the system of particles is equal to the sum of kinetic energy of the center of mass and the kinetic energy about the center of mass.
$$ \frac{1}{2} \sum_im_i\vec{v_i}^2 = \frac{1}{2} \sum_i m_i \vec{v_{cm}}^2 + \frac{1}{2} \sum_i m_i \vec{v'_i}^2 $$
We can use the same procedure to derive the expression for other quantities as well.
When you go from one particle to system of particles, the concept of net external force gets modified, because now you need to define first what is a system and what you meant by external to that system.
Why I am saying this because, we know that the most common force we deal with the movement of macroscopic objects are electrostatic and gravitational, where both of them are central in nature. Now, you are in a situation to distinguish all these central forces into external and internal.
It is not predefined what is external and internal. For example, on playing with marbles, if you say your hands and a marble as a whole system, then there is no external force, but if you just consider the marble alone then the force given by your hands on striking the marble, will be external for the marble.
Thus Newton's second law for a system of particles is written as,
$$ \frac{d^2}{dt^2}\sum_i m_i \vec{r_i} = \sum_i \vec{F_{i(e)}} + \sum_{i,j \,(i\neq j)\,} \vec{F_{ij}} \,\,\,\ldots...eq.(1)$$
Using action - reaction relation, $ \vec{F_{ij}} = -\vec{F_{ji}} $ the second term i.e. internal forces cancel of each other.
It leads to, $$ \sum_i\vec{F_{i(e)}} = \sum_i m_i \left(\frac{d^2}{dt^2}\sum_i \frac{m_i \vec{r_i}}{m_i}\right) $$
The right hand side is defined as the center of mass of the system of particles, $ \sum_i \frac{m_i\vec{r_i}}{m_i} = \vec{R_{cm}} \,\,\,\ldots...eq.(2)$
eq.(1) gives, $$ M\,\frac{d^2}{dt^2} \vec{R_{cm}} = \sum_i \vec{F_{i(e)}} \,\,\,\ldots...eq.(3) $$
From this we understand that, all our Newton's laws are perfectly applicable only for center of mass i.e. Only the point where the center of mass is defined is the only point that moves exactly as it is defined from Newton's laws. All other particles can have motion other than linear translational motion, but not center of mass.
From our definition, it is easily derivable that the total linear momentum is defined as,
$$ \vec{P} = \sum_i m_i \dot{\vec{r}} = M \frac{d\vec{R_{cm}}}{dt} \,\,\,\ldots...eq.(4) $$
where $ \dot{\vec{r}} = \frac{dr}{dt} $
In a similar way, we can describe rotational quantities like angular momentum and torque when the object is in rotational motion.
We can start by finding the angular momentum of a system of particles, $$ \sum_i L_i = \sum_i \vec{r_i}\times \vec{p_i} \,\,\,\ldots...eq.(5)$$
Again making use of the concept of center of mass, position vectors from the origin is written in terms of center of mass coordinate and its relative distance from center of mass. $$ \vec{r} = \vec{R_{cm}} + \vec{r'} \,\,\,\ldots...eq.(5)$$
where $\vec{r'} $ is the position coordinate from center of mass.
Differentiating eq.(5) gives $$ \vec{v} = \vec{v_{cm}} + \vec{v'} \,\,\,\ldots...eq.(6) $$
Angular momentum from eq.(5) becomes,
$$ \sum_i \vec{L_i} = \sum_i \vec{r_i}\times\vec{p_i} = \sum_i m_i (\vec{R_{cm}}+\vec{r'}) \times (\vec{v_{cm}}+\vec{v'}) $$
Expanding with vector multiplication,
$$ \sum_i \vec{r_i}\times\vec{p_i} = \sum_i m_i \vec{R_{cm}}\times\vec{v_{cm}} + \sum_i m_i \vec{r'_i}\times\vec{v'_i} + \sum_i m_i \vec{r'_i}\times\vec{v_{cm}} + \sum_i m_i \vec{R_{cm}}\times\vec{v'_i} $$
From the definition of center of mass, the relative coordinate of center of mass w.r.t. center of mass is zero. $ \sum_i m_i \vec{r'_i} = 0 $ and also $ \sum_i m_i \vec{R_{cm}} \times \vec{v'_i} = \frac{d}{dt} \sum_i m_i \vec{r'_i} = 0 $
We are left with,
$$ \sum_i \vec{L_i} = \sum_i m_i \vec{R_{cm}} \times\vec{v_{cm}} + \sum_i \vec{r'_i}\times\vec{v'_i} $$
It says that, the total angular momentum of system of particles is equal to the angular momentum of the center of mass with respect to origin and the angular momentum of all particles with respect to the center of mass.
In the same way, the kinetic energy of system of particles (with eq.(6)),
$$ \frac{1}{2}\sum_i m_i \vec{v_i}^2 = \frac{1}{2}\sum_i m_i (\vec{v_{cm}}+ \vec{v'_i})^2 $$
$$ \frac{1}{2}\sum_i m_i \vec{v_i}^2 = \frac{1}{2}\left[\sum_i m_i \vec{v_{cm}}^2 + 2 \sum_i m_i \vec{v_{cm}}\cdot\vec{v'_i} + \sum_i m_i \vec{v'_i}^2 \right]$$
By the same token of center of mass, $\,\, 2 \vec{v_{cm}}\cdot \frac{d}{dt} \sum_i m_i \vec{r'_i} = 0 $
Hence, Kinetic energy of the system of particles is equal to the sum of kinetic energy of the center of mass and the kinetic energy about the center of mass.
$$ \frac{1}{2} \sum_im_i\vec{v_i}^2 = \frac{1}{2} \sum_i m_i \vec{v_{cm}}^2 + \frac{1}{2} \sum_i m_i \vec{v'_i}^2 $$
We can use the same procedure to derive the expression for other quantities as well.
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