Question

A rod of length $6\text{m}$ is placed along the $x-$ axis between $x=0$ and $x=6\text{m}$. The linear density (mass/length) $\lambda$ of the rod varies with the distance $x$ from the origin as $\lambda =k(100-x^2)$. Here $k$ is a positive constant. Find the position of the centre of mass of this rod.

• A. $(2.79\text{m},0\text{m},0\text{m})$
• B. $(2\text{m},0\text{m},0\text{m})$
• C. $(3\text{m},0\text{m},0\text{m})$
• D. $(6\text{m},0\text{m},0\text{m})$

Answer 1

The correct option is A $(2.79\text{m},0\text{m},0\text{m})$

Given, length of rod $L=6\text{m}$


Taking element of length $dx$ situated at distance $x$
Mass of this element is $dm=\lambda dx$
$=k(100-x^2)dx$
$COM$ of the element has co-ordinates $(x,0,0)$

Therefore, $x-$ co-ordinate of $COM$ of this rod will be
$x_{COM}=\frac{\underset{0}{\overset{6}{\int }}xdm}{\underset{0}{\overset{6}{\int }}dm}=\frac{\underset{0}{\overset{6}{\int }}x.k(100-x^2)dx}{\underset{0}{\overset{6}{\int }}k(100-x^2)dx}$
$=\frac{k\underset{0}{\overset{6}{\int }}(100x-x^3)dx}{k\underset{0}{\overset{6}{\int }}(100-x^2)dx}=\frac{{[\frac{100x^2}{2}-\frac{x^4}{4}]}_0^6}{{[100x-\frac{x^3}{3}]}_0^6}$
$=\frac{1800-324}{600-72}=\frac{1476}{528}\text{m}$
$\therefore x_{COM}=2.79\text{m}$

$y-$ co-ordinate of COM of the rod = $y_{COM}=\frac{\int ydm}{\int dm}=0$
$z-$ co-ordinate of COM of the rod = $z_{COM}=\frac{\int zdm}{\int dm}=0$
So, center of mass of the rod lies at $(2.79\text{m},0\text{m},0\text{m})$