Question

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

• A. $[\frac{L}{2},0,0]$
• B. $[\frac{3L}{4},0,0]$
• C. $[\frac{L}{4},0,0]$
• D. $[L,0,0]$

Answer 1

The correct option is B $[\frac{3L}{4},0,0]$


Take the element $dx$ of mass$dm$ situated at $x$ from origin.
Then, $dm=\lambda dx=k{x}^{2}dx$

COM of the element has co-ordinates $(x,0,0)$
Therefore, $x-$ co-ordinate of COM of the rod will be
$x_{COM}=\frac{\underset{0}{\overset{L}{\int }}xdm}{\int dm}=\frac{\underset{0}{\overset{L}{\int }}x.k{x}^{2}dx}{\underset{0}{\overset{L}{\int }}k{x}^{2}dx}$
$=\frac{\underset{0}{\overset{L}{\int }}k{x}^{3}dx}{\underset{0}{\overset{L}{\int }}k{x}^{2}dx}=\frac{k\underset{0}{\overset{L}{\int }}{x}^{3}dx}{k\underset{0}{\overset{L}{\int }}{x}^{2}dx}$
$=\frac{{\left[\frac{x^4}{4}\right]}_0^L}{{\left[\frac{x^3}{3}\right]}_0^L}=\frac{3}{4}\frac{L^4}{L^3}$
$\therefore x_{COM}=\frac{3}{4}L$

$\because$Rod is symmetrical about $y$ and $z$ axes, the $y-$ co-ordinate of COM of the rod is $y_{COM}=\frac{\int ydm}{\int dm}=0$
The $z-$ co-ordinate of COM of the rod is $z_{COM}=\frac{\int zdm}{\int dm}=0$
So, the center of mass of the rod lies at $[\frac{3L}{4},0,0]$