2.1 Rod

  Centre of Mass and Conservation of Linear Momentum

Suppose a rod of mass $M$ and length $L$ is lying along the $x-$ axis with its one end at point $A$ i.e. $(0,0)$ and another end at point $B$ i.e. $(L,0)$.

Mass per unit length $\left( {{\lambda _L}} \right) = \frac{M}{L}$

Let $A$ be our reference point.

Consider an infinitesimal small mass $dm$ at a distance $x$ from point $A$. The co-ordinate of the point $P$ is given by $(x,0)$.

So, the position vector of the centre of mass $(COM)$ of point $P$ is ${\overrightarrow r _P} = x\widehat i + 0\widehat j$

Note: The small section at point $P$ is infinitesimally small, so, the centre of mass of that section is assumed to be at point $P$ only.

So, the centre of mass $(COM)$ of the rod is given by, $${\overrightarrow r _{COM}} = \frac{{\int {{{\overrightarrow r }_P}dm} }}{{\int {dm} }} = \frac{{\int {\left( {x\widehat i + 0\widehat j} \right)dm} }}{{\int {dm} }} = \frac{{\int {xdm} }}{{\int {dm} }}\widehat i\quad ...(i)$$

Length of the small mass $dm$ is $dl$. Therefore, $$dm = {\lambda _L}dl = \frac{M}{L}dx\quad ...(ii)$$ From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\frac{M}{L}\int {xdx} }}{{\frac{M}{L}\int {dx} }}\widehat i \\ {\overrightarrow r _{COM}} = \frac{{\int {xdx} }}{{\int {dx} }}\widehat i \\\end{aligned} \end{equation} $$

Since, we want to find the centre of mass $(COM)$ of the rod, so we will integrate from 0 to $L$.

$$\begin{equation} \begin{aligned} {\overrightarrow r _{COM}} = \frac{{\int\limits_0^L {xdx} }}{{\int\limits_0^L {dx} }}\widehat i \\ {\overrightarrow r _{COM}} = \frac{L}{2}\widehat i \\\end{aligned} \end{equation} $$

So, the position centre of mass of the uniform rod of length $L$ is at $\left( {\frac{L}{2},0} \right)$, i.e. at its midpoint.