Question

The density of a linear rod of length $L$ varies as $\rho =A+Bx$ where $x$ is the distance from the left end. Its centre of mass is at a distance (from the left end) of

• A. $\frac{3AL-2B{L}^2}{3(2A-BL)}$
• B. $\frac{3AL+2B{L}^2}{3(2A+BL)}$
• C. $\frac{3AL+2B{L}^2}{2A+BL}$
• D. $\frac{3AL+2B{L}^2}{2A-BL}$

Answer 1

The correct option is B $\frac{3AL+2B{L}^2}{3(2A+BL)}$

In this case, the rod is linear. So, its centre of mass will lie along $x-$axis. For the non-uniform body, the $x-$coordinate of $\text{COM}$ will be,
$x_{\text{com}}=\frac{\int xdm}{\int dm}$
Let us take a very small element of length $dx$ from the left end of the rod as shown in the figure.


$\Rightarrow x_{\text{com}}=\frac{{\int }_0^{L}x\rho dx}{{\int }_0^L\rho dx}$
$[\because \rho =\frac{dm}{dx}\Rightarrow dm=\rho dx]$
$\Rightarrow x_{\text{com}}=\frac{{\int }_0^{L}x(A+Bx)dx}{{\int }_0^L(A+Bx)dx}$
$[\rho =A+Bx\text{given}]$
$\Rightarrow x_{\text{com}}=\frac{{\int }_0^L(Ax+B{x}^2)dx}{{\int }_0^L(A+Bx)dx}$
$\Rightarrow x_{\text{com}}=\frac{{[\frac{A{x}^2}{2}+\frac{B{x}^3}{3}]}_0^L}{{[Ax+\frac{B{x}^2}{2}]}_0^L}$
$\Rightarrow x_{\text{com}}=\frac{[\frac{A{L}^2}{2}+\frac{B{L}^3}{3}-(0+0\left)\right]}{[AL+\frac{B{L}^2}{2}-(0+0\left)\right]}$
$\Rightarrow x_{\text{com}}=\frac{\frac{3A{L}^2+2B{L}^3}{6}}{\frac{2AL+B{L}^2}{2}}$
$\therefore x_{\text{com}}=\frac{3AL+2B{L}^2}{3(2A+BL)}$
Hence centre of mass of rod is at a distance of $\frac{3AL+2B{L}^2}{3(2A+BL)}$ from the left end.