Question

If linear density of a rod of length 3 m varies as $\lambda =2+x,$ then the position of the centre of gravity of the rod is:

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

Answer 1

Answer: (B)

$\frac{12}{7}m$

Answer 2

Let rod is placed along $x-$ axis. Mass of element PQ of length
$dx$ situated at $x=x$ is

$dm=\lambda dx=(2+x)dx$
The COM of the element has coordinates $(x,0,0).$ Therefore, $x-$ coordinate of COM of the rod will be
{{x}_{COM}}=\frac{\int_\limits{0}^{3}{xdm}}{\int_\limits{0}^{3}{dm}}
=\frac{\int_\limits{0}^{3}{x(2+x)dx}}{\int_\limits{0}^{3}{(2+x)dx}}
=\frac{\int_\limits{0}^{3}{(2x+{{x}^{2}})dx}}{\int_\limits{0}^{3}{(2+x)dx}}
$=\frac{\left[ \frac{2{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3} \right]_{0}^{3}}{\left[ 2x+\frac{{{x}^{2}}}{2} \right]_{0}^{3}}$
$=\frac{\left[ {{(3)}^{2}}+\frac{{{(3)}^{3}}}{3} \right]}{\left[ 2\times 3+\frac{{{(3)}^{2}}}{2} \right]}$
$=\frac{9+9}{6+9/2}$
$=\frac{18\times 2}{21}$
$=\frac{12}{7}m$