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 $0\le x\le L.$ The value of $x$ for the centre of mass of the rod is a t :

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

Answer 1

Answer: (C)

$x_{cm}=\frac{\int xdm}{\int dm}=\frac{\int \left(\lambda dx\right)x}{\int dm}$
$=\frac{{\int }_0^L\left(a+\frac{b{x}^2}{L^2}\right)xdx}{{\int }_0^L\left(a+\frac{b{x}^2}{L^2}\right)dx}$
$=\frac{\frac{a{L}^2}{2}+\frac{b}{L^2}.\frac{L^4}{4}}{aL+\frac{b}{L^2}.\frac{L^3}{3}}$
$=\frac{\left(\frac{4a+2b}{8}\right)L}{\frac{\left(3a+b\right)}{3}}=\frac{3}{4}\frac{\left(2a+b\right)L}{\left(3a+b\right)}$