# Center of mass : Dynamics of system of particles

## What is center of mass?

Let us consider a body consisting of large number of particles whose mass is equal to the total mass of all the particles. When such a body is having transnational motion then displacement is produced in each and every particle.
The center of mass of a body is defined as the point where whole mass of the system can be supposed to be concentrated. The motion of such a system can be defined in terms of the center of mass for the purpose of calculating the effect of external force on the motion of the body. Here in this case the nature of the motion executed by the system remains unaltered, when the force acting on system are applied directly to this point (center of mass).
Consider a stationary frame of reference where we have a body of mass $M$ which is made up of $n$ number of particles. Let $m_{i}$ be the mass and ${{\vec{r}}_{i}}$ be the position vector of $i^{th}$ particle of the body.. Let $C$ be any point in the body whose position vector with respect to the origin $O$ of the frame of reference is ${{\vec{R}}_{c}}$ and position vector of C with respect to $i^{th}$ particle is ${{\vec{r}}_{ci}}$ then,

Figure 1 C is the position of center of mass of body of mass M

\begin{equation*}
\vec{r}_{i} = \vec{R}_{c} + \vec{r}_{ci}
\tag{1}
\end{equation*}

Multiplying both sides of equation 1 from $m_{i}$ we get
$m_{i}\vec{r}_i = m_{i}\vec{R}_{c} + m_{i}\vec{r}_{ci}$
taking summation of above equation for n particles we get

\begin{equation*}
\sum\limits_{i = 1}^n {{m_i}{{\vec r}_i}}  = \sum\limits_{i = 1}^n {{m_i}{{\vec R}_c}}  + {\sum\limits_{i = 1}^n {{m_i}\vec r} _{ci}}
\tag{2}
\end{equation*}

If for a body $\sum\limits_{i=1}^n{m_i}\vec{r}_{ci}=0$ then point $C$ is known as center of mass of the body.

Hence a point in any body w.r.t. which the sum of the product of mass of the particles and their position vector is equal to zero is known as center of mass.

### Position of center of mass

#### (i) Two particle system

Consider a system made up of two particles whose mass are m1 and m2 and their respective position vectors w.r.t. origin O be r1 and r2 and Rcm be the position vector of center of mass of the system as shown below in the figure. So from equation 2

Figure 2 Position of center of mass of two particle system

$\sum\limits_{i = 1}^n {{m_i}{r_{ci}}} = 0$
Hence
$\sum\limits_{i = 1}^2 {{m_i}{r_i} = } \sum\limits_{i = 1}^2 {{m_i}{R_{cm}} = {R_{cm}}\sum\limits_{i = 1}^2 {{m_i}} }$                                     (3)
Or,
${R_{cm}}\sum\limits_{i = 1}^2 {{m_i}} = \sum\limits_{i = 1}^2 {{m_i}{r_i}}$
Or,
${R_{cm}} = \frac{{\sum\limits_{i = 1}^2 {{m_i}{r_i}} }}{{\sum\limits_{i = 1}^2 {{m_i}} }}$
${R_{cm}} = \frac{{{m_1}{r_1} + {m_2}{r_2}}}{{{m_1} + {m_2}}}$                                               (4)
If M=m1+m2=total mass of the system, then
${R_{cm}} = \frac{{{m_1}{r_1} + {m_2}{r_2}}}{M}$                           (5a)
Or
$M{R_{cm}} = {m_1}{r_1} + {m_2}{r_2}$                                               (5b)

#### (ii) Many particle system

Consider a many particle system made up of number of particles as shown below in the figure. Let ${m_1},{m_2},{m_3},{m_4}………….{m_n}$ be the masses of the particles of system and their respective position vectors w.r.t. origin are ${\vec{r}_1},{\vec{r}_2},{\vec{r}_3},{\vec{r}_4}………….{\vec{r}_n}$.

Figure 3 Position of center of mass for many particle systems

Also position vector of center of mass of the system w.r.t. origin of the reference frame be Rcm then from equation 2
$\sum\limits_{i = 1}^n {{m_i}{r_i}} = \sum\limits_{i = 1}^n {{m_i}{R_{cm}}}$
Because of the definition of centre of mass
$\sum\limits_{i = 1}^n {{m_i}{r_{ci}}} = 0$
Or,
$\sum\limits_{i = 1}^n {{m_i}{r_i} = } \sum\limits_{i = 1}^n {{m_i}{R_{cm}} = {R_{cm}}\sum\limits_{i = 1}^n {{m_i}} }$
Or
${R_{cm}}\sum\limits_{i = 1}^n {{m_i}} = \sum\limits_{i = 1}^n {{m_i}{r_i}}$                      (6)
Or
${R_{cm}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{r_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}$
${R_{cm}} = \frac{{{m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} + ……… + {m_n}{r_n}}}{{{m_1} + {m_2} + {m_3} + ………. + {m_n}}}$                              (7)
${R_{cm}} = \frac{{{m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} + ……… + {m_n}{r_n}}}{M}$                    (8)
Where
$M = \sum\limits_{i = 1}^n {{m_i}}$

### Position of center of mass in terms of coordinate components

Let in a system of many particles the co-ordinates of center of mass of the system be $(X_{cmi},Y_{cmj},Z_{cmk})$  then position vector of center of mass would be
$R_{cm}=X_{cmi}+Y_{cmj}+Z_{cmk}$                   (9)
and position vectors of various particles would be
\begin{array}

{{\rm{r}}_{\rm{1}}} = {x_1}{\rm{i}} + {y_1}{\rm{j}} + {z_1}{\rm{k}}\\

{{\rm{r}}_{\rm{2}}} = {x_2}{\rm{i}} + {y_2}{\rm{j}} + {z_3}{\rm{k}}\\

{{\rm{r}}_{\rm{3}}} = {x_3}{\rm{i}} + {y_3}{\rm{j}} + {z_3}{\rm{k}}\\

……..\\

………\\

{{\rm{r}}_{\rm{n}}} = {x_n}{\rm{i}} + {y_n}{\rm{j}} + {z_n}{\rm{k}}

\end{array}                                      (10)
Putting the values from equation 9 and 10 in equation 6, 7 and 8 we get
${X_{cm}} = \frac{{{m_1}{x_{\rm{1}}} + {m_2}{x_{\rm{2}}} + {m_3}{x_{\rm{3}}} + ……… + {m_n}{x_{\rm{n}}}}}{{{m_1} + {m_2} + {m_3} + ………. + {m_n}}} = \frac{1}{m}\sum\limits_{i = 1}^n {{m_i}{x_i}}$                                          (11)
${Y_{cm}} = \frac{{{m_1}{y_{\rm{1}}} + {m_2}{y_{\rm{2}}} + {m_3}{y_{\rm{3}}} + ……… + {m_n}{y_{\rm{n}}}}}{{{m_1} + {m_2} + {m_3} + ………. + {m_n}}} = \frac{1}{m}\sum\limits_{i = 1}^n {{m_i}{y_i}}$                                          (12)
${Z_{cm}} = \frac{{{m_1}{z_{\rm{1}}} + {m_2}{z_{\rm{2}}} + {m_3}{z_{\rm{3}}} + ……… + {m_n}{z_{\rm{n}}}}}{{{m_1} + {m_2} + {m_3} + ………. + {m_n}}} = \frac{1}{m}\sum\limits_{i = 1}^n {{m_i}{z_i}}$                                          (13)
If in any system there are infinite particles of point mass and are distributed continuously also if the distance between them is infinitesimally small then summation in equations 6, 11, 12 and 13 can be replaced by integration. If r is the position vector of very small particle of mass $dm$ of the system then position vector of centre of the system would be
${{\vec R}_{cm}} = \frac{1}{M}\int {\vec rdm}$                 (14)
And the values of its co-ordinates would be,
${X_{cm}} = \frac{1}{M}\int {xdm}$                        (15a)
${Y_{cm}} = \frac{1}{M}\int {ydm}$                        (15b)
${Z_{cm}} = \frac{1}{M}\int {zdm}$                         (15c)
If ρ is the density of the system then dm=ρdV where dV is the very small volume element of the system then,
${{\rm{R}}_{cm}} = \frac{1}{M}\int {\rho {\rm{r}}dV}$                   (16)
${X_{cm}} = \frac{1}{M}\int {\rho xdV}$                               (16a)
${Y_{cm}} = \frac{1}{M}\int {\rho ydV}$                               (16b)
${Z_{cm}} = \frac{1}{M}\int {\rho zdV}$                                (16c)
The centre of mass of a homogeneous body (body having uniform distribution of mass) must coincide with the geometrical centre of the body. In other words we can say that if the homogeneous body has a point, a line or plane of symmetry, then its centre of mass must lie at this point, line or plane of symmetry.
The centre of mass of irregular bodies and shape can be found using equations 14, 15, 16.

### Motion of centre of mass

Differentiating equation 6 we get
$\frac{{d{{\rm{R}}_{cm}}}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{\sum\limits_{i = 1}^n {{m_i}{{\rm{r}}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}} \right) = \frac{{\sum\limits_{i = 1}^n {{m_i}\frac{{d{{\rm{r}}_i}}}{{dt}}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}$
But $\frac{{d{{\rm{R}}_{cm}}}}{{dt}} = {{\rm{V}}_{cm}}$ which is the velocity of the center of mass.
$\frac{{d{{\rm{r}}_i}}}{{dt}} = {{\rm{v}}_i}$ is the velocity of i’th particle of the system
Therefore,
${{\rm{V}}_{cm}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{{\rm{v}}_{\rm{i}}}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} = \frac{1}{M}\sum\limits_{i = 1}^n {{m_i}{{\rm{v}}_{\rm{i}}}}$                      (17)

But, ${m_i}{{\rm{v}}_{\rm{i}}} = {p_i}$ which is the linear momentum of the i’th particle of the system
Therefore
${{\rm{V}}_{cm}} = \frac{1}{M}\sum\limits_{i = 1}^n {{{\rm{p}}_{\rm{i}}}} = \frac{{\rm{p}}}{M}$                              (18)
Or, $M{{\rm{V}}_{cm}} = {\rm{p}}$
Where, ${\rm{p}} = \sum\limits_{i = 1}^n {{{\rm{p}}_{\rm{i}}}}$
p is the vector sum of linear momentum of various particles of the system or it is the total linear momentum of the system. If no external force is acting on the system then its linear momentum remains constant. Hence in absence of external force
$M{{\rm{V}}_{cm}} =$ constant                               (19a)
or,
${{\rm{V}}_{cm}} =$ constant                   (19b)
In the absence of external force velocity of centre of mass of the system remains constant or we can say that centre of mass moves with the constant velocity in absence of external force.
Hence from equation 18 we came to know that the total linear momentum of the system is equal to the product of the total mass of the system and the velocity of the centre of mass of the system which remains constant.
Thus in the absence of the external force it is not necessary that momentum of individual particles of system like p1 , p2 , p3 . . . . . . . . . . . pn etc. remains constant but their vector sum always remains constant.

### Acceleration of center of mass

Differentiating equation 17 we get
$\frac{dV_{cm}}{dt}=\frac{d}{dt}\left[ {\frac{1}{m}\sum\limits_{i = 1}^n {{m_i}{v_i}} } \right] = \frac{1}{m}\sum\limits_{i = 1}^n {{m_i}\frac{{d{v_i}}}{{dt}}}$
but,
that is the $\frac{d\vec{v_i}}{dt}=\vec{a_i}$ acceleration of the i’th particle and
$\frac{dV_{cm}}{dt}=a_{cm}$ is the acceleration of the center of mass of the system.
So,
$\vec{a_cm}=\frac{1}{m}\sum\limits_{i = 1}^n {{m_i}\vec{a_i}}$
If $m_{i}\vec{a_i}=F_i$ which is the force acting on the i’th particle of the system then
$\vec{a_cm}=\frac{1}{m}\sum\limits_{i = 1}^n {F_i}$ (20)
Force acting on the i’th particle,
$\vec{F_i}=\vec{F_{i(ext)}}+\vec{F_{i(int)}}$
Internal force is produced due to the mutual interactions between the particles of teh system.
Therefore,
$\vec{a}_{cm}=\frac{1}{m}\sum\limits_{i = 1}^n\left[ \vec{F}_{i(ext)}+\vec{F}_{i(int)}\right]$
From Newton’s third law of motion,
${\vec F_{ij}} = – {\vec F_{ji}}$
$\therefore \sum\limits_{i \ne j} {{{\vec F}_{ij}}} = 0$
hence,
${{\vec a}_{cm}} = \frac{1}{m}\sum\limits_{i = 1}^n {{{\vec F}_i}_{(ext)}} = \frac{{\vec F}}{m}$
where,
$\vec F = \sum\limits_{i = 1}^n {{{\vec F}_i}} = {{\vec F}_1} + {{\vec F}_2} + ….. + {{\vec F}_n}$
is the total external force acting on the system since internal force on the system , because of the mutual interaction between the system particles, becomes equal to zero. This happens because of action reaction Law of Newton.
Hence form equation 20 it is clear that the center of mass of a system of particles moves as if whole of the mass of the system were concentrated at it and and all the external force were exerted at it. This results holds whether the system is a rigid body with particles in fixed positions or a system of particles with internal motions.