Question

A rod of length $L$ is placed along the $\text{x-axis}$ between $x=0$ and $x=L$. The linear mass density $\left(\text{mass/length}\right)$ $\rho$ of the rod varies with the distance $x$ from the origin as $\rho =a+bx$. Here, $a$ and $b$ are constants. Find the position of centre of mass of this rod.

• A. $[\frac{3aL+2b{L}^2}{6a+3bL},0,0]$
• B. $[\frac{3aL+b{L}^2}{6bL},0,0]$
• C. $[\frac{aL}{bL},0,0]$
• D. $[\frac{3aL+5b{L}^2}{5a+2bL},0,0]$

Answer 1

The correct option is A $[\frac{3aL+2b{L}^2}{6a+3bL},0,0]$

Let us consider an element of length $PQ$ at distance $x$ from the left end.


Mass of the considered element of the rod will be

$dm=\rho dx=(a+bx)dx$

The $COM$ of the element will have co-ordinates $(x,0,0)$.

Therefore, $x-$co-ordinate of $COM$ of the rod is given by

$x_{COM}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}=\frac{{\int }_0^L\left(x\right)(a+bx)dx}{{\int }_0^L(a+bx)dx}$

$x_{COM}=\frac{{[\frac{a{x}^2}{2}+\frac{b{x}^3}{3}]}_0^L}{{[ax+\frac{b{x}^2}{2}]}_0^L}$

$x_{COM}=\frac{3a{L}^2+2b{L}^3}{6aL+3b{L}^2}=\frac{3aL+2b{L}^2}{6a+3bL}$

Similarly, $y-$co-ordinate of $COM$ of the rod is

$y_{COM}=\frac{\int ydm}{\int dm}=0[\because y=0]$

And for $z-$ co-ordinate,

$z_{COM}=0(\because z=0)$

Hence, the centre of mass of the rod lies at
$[\frac{3aL+2b{L}^2}{6a+3bL},0,0]$

So, option $\left(a\right)$ is the correct answer.