Question

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 $0\le \text{x}\le \text{L}.$ The value of $x$ for the centre of mass of the rod is at

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

Answer 1

The correct option is D $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$

Taking an elemental mass at a distance $x$ from left end of the rod

$x_{cm}=\frac{\int xdm}{\int dm}$
$=\frac{\int \left(\rho dx\right)x}{\int dm}$
Position of center of mass:
$=\frac{{\int }_0^L(a+\frac{{bx}^2}{L^2})xdx}{{\int }_0^L(a+\frac{{bx}^2}{L^2})dx}$
$=\frac{\frac{{aL}^2}{2}+\frac{b}{L^2}.\frac{L^4}{4}}{aL+\frac{b}{L^2}.\frac{L^3}{3}}$
$=\frac{\left(\frac{4a+2b}{8}\right)L}{\left(\frac{3a+b}{3}\right)}$
$\frac{3\left(2a+b\right)L}{4\left(3a+b\right)}$
Final Answer: $\left(c\right)$