Question

The density of a non-uniform rod of length $1m$ is given by $\rho (x)=a(1+b{x}^2)$ where a and b are constants and $0\le x\le 1$. The centre of mass of the rod will be at

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

Answer 1

Answer: (A)

Mass of a small element of length $dx$ of the rod at a distance $x$ from the one end of the rod is
$dm=\rho dx=a(1+b{x}^2)dx$
The centre of mass of the rod is
$X_{CM}=\frac{\underset{0}{\overset{1}{\int }}xdm}{\underset{0}{\overset{1}{\int }}dm}=\frac{\underset{0}{\overset{1}{\int }}xa\left(1+b{x}^2\right)dx}{\underset{0}{\overset{1}{\int }}a\left(1+b{x}^2\right)dx}$
$=\frac{\underset{0}{\overset{1}{\int }}\left(x+b{x}^{3}dx\right)}{\underset{0}{\overset{1}{\int }}\left(1+b{x}^2\right)dx}=\frac{{\left[\frac{x^2}{2}+\frac{b{x}^4}{4}\right]}_0^1}{{\left[x+\frac{b{x}^3}{3}\right]}_0^1}$
$=\frac{\left[\frac{1}{2}+\frac{b}{4}\right]}{\left[1+\frac{b}{3}\right]}=\frac{3\left(2+b\right)}{4\left(3+b\right)}$