Question

The center of mass of a non-uniform rod of length $L=2\text{m}$, whose mass per unit length varies as $\lambda =\frac{K{x}^2}{L}$ where $K$ is a positive constant and $x$ is the distance of any point from the left end of the rod, lies at a distance $\text{in metre}$ of (from the left end) is

Answer 1

Mass per unit length of the rod, $\left(\lambda \right)=\frac{K{x}^2}{L}$

Let us consider an element of length $dx$ located at a distance $x$ away from the left end of the rod and its mass is,

$\Rightarrow dm=\lambda dx$

$dm=\frac{K{x}^2}{L}\times dx$

The $x$-coordinate of center of mass of the rod is,

$x_{com}=\frac{\int xdm}{\int dm}$

Integrating with proper limits we get,

$\Rightarrow x_{com}=\frac{{\int }_0^{L}x\times \frac{K{x}^2}{L}dx}{{\int }_0^L\frac{K{x}^2}{L}dx}$

$x_{com}=\frac{\frac{L^3}{4}}{\frac{L^2}{3}}=\frac{3L}{4}$

$\therefore x_{com}=\frac{3\times 2}{4}=1.5\text{m}$