Question

A rod of length $L$ has non-uniform linear mass density given by $\rho \left(x\right)=a+b{\left(\frac{a}{L}\right)}^2$, 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 at:

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

Answer 1

The correct option is B $\frac{3}{4}\left(\frac{2a+b}{3a+b}\right)L$


Linear mass density, $\rho \left(x\right)=a+b{\left(\frac{x}{L}\right)}^2$

We know that, $X_{CM}=\frac{\int xdm}{\int dm}$

Now, $\int dm={\int }_0^L\rho \left(x\right)dx$

$={\int }_0^L[a+b{\left(\frac{x}{L}\right)}^2]dx=aL+\frac{bL}{3}$

So, ${\int }_0^{L}xdm={\int }_0^L(ax+\frac{b{x}^3}{L^2})dx$

$=(\frac{a{L}^2}{2}+\frac{b{L}^2}{4})$

$\therefore X_{CM}=\frac{(\frac{a{L}^2}{2}+\frac{b{L}^2}{4})}{aL+\frac{bL}{3}}$

$\Rightarrow X_{CM}=\frac{3L}{4}\left(\frac{2a+b}{3a+b}\right)$

Hence, $\left(B\right)$ is the correct answer.