Question

If the linear mass density of a rod of length $2\text{m}$ varies as $\lambda =a+bx\text{kg/m}$, where $x$ is the distance (in metres) from its one end, then its centre of mass (in metres) is given by

• A. $x_{\text{com}}=\frac{4(a+b)}{(9a+4b)}$
• B. $x_{\text{com}}=\frac{3a+4b}{3(a+b)}$
• C. $x_{\text{com}}=\frac{2(3a+4b)}{3}$
• D. $x_{\text{com}}=\frac{(4a+12b)}{3}$

Answer 1

The correct option is B $x_{\text{com}}=\frac{3a+4b}{3(a+b)}$


Since the rod is lying along $x-$ axis, as shown in figure, its $\text{COM}$ will have only $x-$coordinate.
$\therefore y_{\text{com}}=0\&z_{\text{com}}=0$
Consider an element of the rod of length $dx$ at a distance $x$ from the origin $\left(\text{O}\right)$
$\Rightarrow dm=\lambda dx$
$\therefore dm=(a+bx)dx$
From the equation:
$x_{\text{com}}=\frac{\int xdm}{\int dm}$
Putting limits of $x=0\to x=2$
$x_{\text{com}}=\frac{{\int }_0^{2}x(a+bx)dx}{{\int }_0^2(a+bx)dx}$
$x_{\text{com}}=\frac{a{\left[\frac{x^2}{2}\right]}_0^2+b{\left[\frac{x^3}{3}\right]}_0^2}{a[x{]}_0^2+b{\left[\frac{x^2}{2}\right]}_0^2}$
$x_{\text{com}}=\frac{2a+\frac{8b}{3}}{2a+2b}$

$\therefore x_{\text{com}}=\frac{3a+4b}{3(a+b)}$