Question

The centre of mass of a non-uniform rod of length $L$ whose mass per unit length $\lambda =\frac{k{x}^2}{L}$ where $k$ is a constant and $x$ is the distance from the one end is:

• A. $\frac{3L}{4}$
• B. $L/8$
• C. $\frac{k}{L}$
• D. $\frac{3k}{L}$

Answer 1

The correct option is A $\frac{3L}{4}$

$\text{Step 1: Element of length dx at a distance x}$ $\text{[Refer Fig.]}$

Consider a small element of length dx at a distance x

Mass of $dx$ element $dm=\lambda dx$

Formula of C.O.M.

$x_{com}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}$

$\Rightarrow$ $x_{com}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}$$=\frac{{\int }_0^{L}K{x}^{3}dx}{{\int }_0^{L}K{x}^{2}dx}$ $\because \lambda =\frac{k{x}^2}{l}$

$\text{Step 2: Apply limit}$

$\Rightarrow$ $x_{com}=\frac{K{\left[\frac{x^4}{4}\right]}_0^L}{K{\left[\frac{x^3}{3}\right]}_0^L}$

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

Hence, Option $A$ is the correct answer.