# Centre of Mass

A body is not made up of a single particle instead it is a collection of large number of particles arranged in a particular way. Here we find a point in a system of a collection of particles that behaves as the centre of the combination of the total mass of all particles which is called centre of mass of the system of particles.

A rigid body can be taken as a collection of particles. Here we find the centre of mass of a rigid body. A net force acts on a rigid body as shown in Figure 1. The body is made up of a large number of particles of masses $m_1$, $m_2$, $m_3$ etc..

The position vector of mass $m_1$ is $\vec r_1$, and that of mass $m_2$ is $\vec r_2$ and so on. The body undergoes a displacement towards positive x-direction under the action of a net force $\vec F$. The total resultant force on the body is the sum of the forces on individual particles. The force on a particle of mass $m_1$ is $\vec F_1$ and on a particle of mass $m_2$ is $\vec F_2$, therefore we can represent the force on any particle of mass $m_i$ as $\vec F_i$ and the total resultant force is the sum of the forces on all particles of the body. Hence we can write,

\begin{align*} {\vec F _{{\rm{res}}}} &= \sum\limits_{i = 1}^n {{{\vec F }_i}} \\ {\rm{or}}\quad M{\vec a _x} &= \sum\limits_{i = 1}^n {{m_i}{{\vec a }_i}} \\ {\rm{or}}\quad M\frac{{{d^2}\vec r }}{{d{t^2}}} &= \sum\limits_{i = 1}^n {{m_i}\frac{{{d^2}{{\vec r }_i}}}{{dt}}} \\ {\rm{or}}\quad \frac{{{d^2}\vec r }}{{d{t^2}}} &= \frac{{{d^2}}}{{d{t^2}}}\left( {\frac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r }_i}} }}{M}} \right)\\ {\rm{or}}\quad \vec r &= \frac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r }_i}} }}{M} \tag{1} \label{1} \end{align*}

The acceleration vector is the double derivative of the position vector. The position vector given by Eq. \eqref{1} is the position vector of the centre of mass of the rigid body. Which can be rewritten as $\vec r_{\text{cm}}$ where the subscript "$\text{cm}$" represents centre of mass. The total mass $M$ of the rigid body is the sum of all masses of individual particles, so $M = \sum\limits_{i = 1}^n {{m_i}}$ and

${{\vec r}_{{\rm{cm}}}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} \tag{2} \label{2}$

The position vector $\vec r_{\text{cm}}$ can be written in component form as ${\vec r _{{\rm{cm}}}} = {x_{{\rm{cm}}}}\hat i + {y_{{\rm{cm}}}}\hat j + {z_{{\rm{cm}}}}\hat k$ where ${x_{{\rm{cm}}}}$, ${y_{{\rm{cm}}}}$ and ${z_{{\rm{cm}}}}$ are the x, y and z-coordinates of the centre of mass respectively. And similarly the position vector of any particle in component form is ${\vec r _{{\rm{i}}}} = {x_{{\rm{i}}}}\hat i + {y_{{\rm{i}}}}\hat j + {z_{{\rm{i}}}}\hat k$. So,

\begin{align*} {{\vec r}_{{\rm{cm}}}} &= \frac{{\sum\limits_{i = 1}^n {{m_i}{{\vec r}_i}} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\\ &= \frac{{\sum\limits_{i = 1}^n {{m_i}({x_{\rm{i}}}\hat i + {y_{\rm{i}}}\hat j + {z_{\rm{i}}}\hat k)} }}{{\sum\limits_{i = 1}^n {{m_i}} }}\\ &= \frac{{\sum\limits_{i = 1}^n {{m_i}{x_{\rm{i}}}\hat i + \sum\limits_{i = 1}^n {{m_i}{y_{\rm{i}}}\hat j} + \sum\limits_{i = 1}^n {{m_i}{z_{\rm{i}}}} \widehat k} }}{{\sum\limits_{i = 1}^n {{m_i}} }} \tag{3} \label{3} \end{align*}

Now after comparing Eq. \eqref{3} with ${\vec r _{{\rm{cm}}}} = {x_{{\rm{cm}}}}\hat i + {y_{{\rm{cm}}}}\hat j + {z_{{\rm{cm}}}}\hat k$, we'll get

${x_{{\rm{cm}}}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{x_{\rm{i}}}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} = \frac{{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + ... + {m_n}{x_n}}}{{{m_1} + {m_2} + {m_3} + ... + {m_n}}}$

${y_{{\rm{cm}}}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{y_{\rm{i}}}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} = \frac{{{m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} + ... + {m_n}{y_n}}}{{{m_1} + {m_2} + {m_3} + ... + {m_n}}}$

${z_{{\rm{cm}}}} = \frac{{\sum\limits_{i = 1}^n {{m_i}{z_{\rm{i}}}} }}{{\sum\limits_{i = 1}^n {{m_i}} }} = \frac{{{m_1}{z_1} + {m_2}{z_2} + {m_3}{z_3} + ... + {m_n}{z_n}}}{{{m_1} + {m_2} + {m_3} + ... + {m_n}}}$

Mechanics