Question

The linear mass density of a rod of length $5\text{m}$ varies with distance from its end fixed at the origin as represented in the graph shown below:

Then find the position of centre of mass of the rod w.r.t origin.

• A. $\frac{25}{16}\text{m}$
• B. $\frac{5}{9}\text{m}$
• C. $\frac{25}{9}\text{m}$
• D. $\frac{5}{16}\text{m}$

Answer 1

The correct option is C $\frac{25}{9}\text{m}$

From the graph of linear mass density of rod, writing it in $y=mx+c$ format:

$\lambda =(5+x)\text{kg/m}$

Let us consider a small element of length $dx$ and mass $dm$, which is at a distance $x$ from the origin.


Let $x_{\text{CM}}$ be the position of $COM$ of rod, so by applying basic formula and integrating with proper limits:
$x_{\text{CM}}=\frac{\int xdm}{\int dm}...\left(i\right)$

$\lambda =\frac{dm}{dx}=5+x$
$\Rightarrow dm=(5+x)dx....\left(ii\right)$

Substituting $dm$ in Eq $\left(i\right)$ and setting the values of limits as $x=0$ to $x=5$:

$x_{\text{CM}}=\frac{{\int }_0^{5}x(5+x)dx}{{\int }_0^5(5+x)dx}=\frac{{\int }_0^5(5x+x^2)dx}{{\int }_0^5(5+x)dx}$
$=\frac{{[\frac{5x^2}{2}+\frac{x^3}{3}]}_0^5}{{[5x+\frac{x^2}{2}]}_0^5}$
$=\frac{[\frac{5(5{)}^2}{2}+\frac{(5{)}^3}{3}]}{\left[5\right(5)+\frac{(5{)}^2}{2}]}$
$\Rightarrow x_{\text{CM}}=\frac{25}{9}\text{m}$