Question

Find the $x$ coordinate of the centre of mass of the non-uniform rod of length $L$ given below. The origin is taken at the left end of the rod. The density of the rod as a function of its $x$ -coordinate is $\rho =a{x}^2+bx+c$ , where $a$ , $b$ and $c$ are constants.

• A. $\frac{2{\left(aL\right)}^2+3{\left(bL\right)}^2+6cL}{2(3{\left(aL\right)}^2+4bL+8c)}$
• B. $\frac{4{\left(aL\right)}^3+3{\left(bL\right)}^2+2cL}{2(3{\left(aL\right)}^2+2bL+c)}$
• C. $\frac{3a{L}^2+4b{L}^2+6cL}{4a{L}^2+6bL+8c}$
• D. $\frac{3{\left(aL\right)}^3+4{\left(bL\right)}^2+6cL}{2(2{\left(aL\right)}^2+3bL+6c)}$

Answer 1

Answer: (D)

$\frac{{\int }_0^L\left(a{x}^2+bx+c\right)xdx}{{\int }_0^L\left(a{x}^2+bx+c\right)dx}$
$\frac{{\int }_0^L\left(a{x}^3+b{x}^2+cx\right)dx}{{\int }_0^L\left(a{x}^2+bx+c\right)dx}$
$=\frac{\frac{a{x}^4}{4}+\frac{b{x}^3}{3}+{\frac{c{x}^2}{2}|}_0^L}{\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+{cx|}_0^L}$
$=\frac{\frac{3a{L}^4+4b{L}^3+6c{L}^2}{12}}{\frac{2a{L}^2+3b{L}^2+6cL}{6}}$
$=\frac{3a{L}^3+4b{L}^2+6cL}{2\left(2a{L}^2+3bL+6c\right)}$