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 $\rho =ax^{2}+bx+c$ , where $a$ , $b$ and $c$ are constants.

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

Answer 1

Answer: (D)

$\frac{\int_{0}^{ L }\left( ax ^{2}+ bx + c \right) x dx }{\int_{0}^{ L }\left( ax ^{2}+ bx + c \right) dx }$
$\frac{\int_{0}^{ L }\left( ax ^{3}+ bx ^{2}+ cx \right) d x }{\int_{0}^{ L }\left( ax ^{2}+ bx + c \right) dx }$
$=\frac{\frac{ ax ^{4}}{4}+\frac{ b x^{3}}{3}+\left.\frac{ cx ^{2}}{2}\right|_{0} ^{ L }}{\frac{ ax ^{3}}{3}+\frac{ bx ^{2}}{2}+\left. cx \right|_{0} ^{ L }}$
$=\frac{\frac{3 aL ^{4}+4 bL ^{3}+6 cL ^{2}}{12}}{\frac{2 aL ^{2}+3 bL ^{2}+6 cL }{6}}$
$=\frac{3 aL ^{3}+4 bL ^{2}+6 cL }{2\left(2 aL ^{2}+3 bL +6 c \right)}$