Question

A rod of length $4m$ is placed along positive $x-$axis, with one of its end at origin. If linear density of the rod varies as $\lambda =4+x$, then the position of the centre of gravity of the rod is

• A. $\frac{12}{7}m$
• B. $\frac{20}{9}m$
• C. $\frac{10}{3}m$
• D. $\frac{7}{3}m$

Answer 1

The correct option is B $\frac{20}{9}m$

For homogeneous space ($g=$ constant), centre of gravity and centre of mass are identical. Thus,
$X_{COG}=X_{CM}=\frac{\int xdm}{\int dm}(\because dm=\lambda dx)$
$=\frac{\int \lambda xdx}{\int \lambda dx}(\because \lambda =4+x)$
$=\frac{{\int }_0^4(4+x)xdx}{{\int }_0^4(4+x)dx}(\because l=4m)$
$=\frac{4{\int }_0^{4}xdx+{\int }_0^4{x}^{2}dx}{4{\int }_0^{4}dx+{\int }_0^{4}xdx}$
$=\frac{4\left(\frac{16}{2}\right)+\frac{64}{3}}{4\left(4\right)+\frac{16}{2}}$
$=\frac{160}{24\times 3}$
$\therefore X_{COG}=\frac{20}{9}m.$