A rod of length $l$ has non-uniform linear mass density given by $\rho \left(x\right)=a+b{\left(\frac{x}{l}\right)}^2,$ where $a$ and $b$ are constants and $0\le x\le l$ The value of $x$ for the centre of mass of the rod is at:

• A

  $\frac{3}{2}\left(\frac{a+b}{2a+b}\right)L$
• Correct answer
  B

  $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$
• C

  $\frac{4}{3}\left(\frac{a+b}{2a+3b}\right)L$
• D

  $\frac{3}{2}\left(\frac{2a+b}{3a+b}\right)L$

Solution

$X_{cm}=\frac{1}{M}\underset{0}{\overset{\iota }{\int }}x.dM$
$dM=\rho .dx=\left(a+b{\left(\frac{x}{ı}\right)}^2\right).dx$
$x_{cm}=\frac{\int xdM}{\int dm}=\frac{\int x\rho dx}{\int \rho dx}=\frac{\underset{0}{\overset{\iota }{\int }}x\left(a+\frac{b{x}^2}{{\iota }^2}\right)dx}{\underset{0}{\overset{\iota }{\int }}\left(a+\frac{b{x}^2}{{\iota }^2}\right)dx}$
$=\frac{a{\left(\frac{x^2}{2}\right)}_0^{\iota }+\frac{b}{{\iota }^2}{\left(\frac{x^4}{4}\right)}_0^{\iota }}{a{\left(x\right)}_0^{\iota }+\frac{b}{{\iota }^2}{\left(\frac{x^3}{3}\right)}_0^{\iota }}$
$=\frac{a\frac{{\iota }^2}{2}+b\frac{{\iota }^2}{4}}{a\iota +\frac{b\iota }{3}}=\frac{\left(2a+b\right)}{\left(3a+b\right)}\frac{\iota }{4}\times 3$
$=\frac{3\iota }{4}\left(\frac{2a+b}{3a+b}\right)$

Hence, the correct answer is option: 'B'.