Question

A uniform thin rod $AB$ of length $L$ has linear mass density $\mu \left(x\right)=a+\frac{bx}{L},$ where $x$ is measured from $A$. If the $CM$ of the rod lies at a distance of $\left(\frac{7}{12}\right)L$ from $A$, then $a$ and $b$ are related as :

• A. a = b
• B. a = 2b
• C. 2a = b
• D. 3a = 2b

Answer 1

Answer: (C)

$x_{CM}=\frac{1}{M}\int xdm;M=\int dm$
$={\int }_0^L\left(a+\frac{bx}{L}\right)dx$
$=\left(aL+\frac{bL}{2}\right)$
$\Rightarrow \int xdm={\int }_0^L\left(ax+\frac{b{x}^2}{L}\right)dx$
$=\left(\frac{a{L}^2}{2}+\frac{b{L}^2}{3}\right)$
Given, $x_{CM}=\frac{7L}{12}$
Therefore,
$\frac{7/L}{12}=\frac{\frac{a/L^2}{2}+\frac{b/L^2}{3}}{a/L+\frac{/bL}{2}}=\frac{3a+2b}{6\left(a+\frac{b}{2}\right)}$
$\Rightarrow 6a+4b=7a+\frac{7b}{2}$
$\Rightarrow a=\frac{b}{2}$
$\Rightarrow b=2a$