If linear density of a rod of length 3 m varies as λ =2+ x , then the position of the centre of…

1. $\frac{7}{3}\mathrm{m}$
2. $\frac{12}{7}\mathrm{m}$
3. $\frac{10}{7}\mathrm{m}$
4. $\frac{9}{7}\mathrm{m}$

Solution

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




$dm=\lambda dx=\left(2+x\right) dx$

The CM of the element has coordinates ($x$, 0, 0).

Therefore, $x$-coordinates of CM of the rod will be

$x_{\mathrm{C}\mathrm{M}}=\frac{{\int }_0^{3}xdm}{{\int }_0^{3}dm}$

$=\frac{{\int }_0^{3}x\left(2+x\right)dx}{{\int }_0^3\left(2+x\right)dx}$

$=\frac{{\int }_0^3\left(2x+x^2\right)dx}{{\int }_0^3\left(2+x\right)dx}$

$=\frac{{\left[\frac{2x^2}{2}+\frac{x^3}{3}\right]}_0^3}{{\left[2x+\frac{x^2}{2}\right]}_0^3}$

$=\frac{\left[{\left(3\right)}^2+\frac{{\left(3\right)}^3}{3}\right]}{\left[2\times 3+\frac{{\left(3\right)}^2}{2}\right]}=\frac{9+9}{6+9/2}$

$=\frac{18\times 2}{21}=\frac{12}{7}$m

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