Question

A uniform thin rod of length $10\text{m}$ is placed along $x$- axis such that one of its ends is at the origin. The linear mass density of the rod varies with respect to distance from the end as $\lambda =l_{0}x$, where $l_0$ is a constant. Then the position of the centre of mass of the rod w.r.t the origin will be:

• A. $\frac{10}{3}\text{m}$
• B. $\frac{20}{3}\text{m}$
• C. $10\text{m}$
• D. $3\text{m}$

Answer 1

The correct option is B $\frac{20}{3}\text{m}$


Let $dm$ be the mass of a small element of the rod of length $dx$ at a distance $x$ from it's end.
Linear mass density $\lambda =\frac{dm}{dx}$
$\Rightarrow dm=\lambda dx=l_{0}xdx.....\left(i\right)$

Let $x_{\text{CM}}$ be the position of $COM$ of the rod.
Applying the basic formula and taking limits from $x=0$ to $x=L=10\text{m}$

$x_{\text{CM}}=\frac{\int xdm}{\int dm}=\frac{{\int }_0^{10}x\left(l_{0}x\right)dx}{{\int }_0^{10}\left(l_{0}x\right)dx}=\frac{l_0{\int }_0^{10}{x}^{2}dx}{l_0{\int }_0^{10}xdx}$
$\Rightarrow x_{\text{CM}}=\frac{{\left[\frac{x^3}{3}\right]}_0^{10}}{{\left[\frac{x^2}{2}\right]}_0^{10}}=\frac{20}{3}\text{m}$