Question

A thin rod of length $L$ is lying along the $x-$ axis with its ends at $x=0$ and $x=L$. Its linear density (mass/length) varies with $x$ as $\lambda =k{\left(\frac{x}{L}\right)}^n$, when $n$ can be zero or any positive number. If the position $x_{\text{cm}}$ of the centre of mass of the rod is plotted against ${}^{\prime }{n}^{\prime }$, which of the following graphs best approximates the dependence of $x_{\text{cm}}$ on $n$?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

When $n=0,\lambda =k$ where $k$ is a constant. It means that the linear mass density is constant. In this case, the centre of mass will be at the middle of the rod. i.e. at $\frac{L}{2}$
Now, applying the fundamental equation for $\text{COM}$ of system,
$x_{cm}=\frac{{\int }_0^{L}xdm}{{\int }_0^{L}dm}=\frac{{\int }_0^{L}x\lambda dx}{{\int }_0^L\lambda dx}$
$=\frac{{\int }_0^{L}kx{\left(\frac{x}{L}\right)}^{n}dx}{{\int }_0^{L}k{\left(\frac{x}{L}\right)}^{n}dx}=\frac{k.{\left[\frac{x^{n+2}}{(n+2)L^n}\right]}_0^L}{{\left[\frac{k{x}^{n+1}}{(n+1)L^n}\right]}_0^L}$
$\Rightarrow x_{cm}=\frac{L^{(n+2)}}{L^{(n+1)}}\times \frac{(n+1)}{(n+2)}$
$\therefore x_{cm}=\frac{L(n+1)}{(n+2)}$

Putting $n=0,x_{cm}=\frac{L}{2}$
Hence $a$ & $b$ options are incorrect)
For $n=1;x_{cm}=\frac{2L}{3}$
For $n=2,x_{cm}=\frac{3L}{4}$

Finding the value of $x_{cm}$ when $n$ approaches $\infty$
${lim}_{n\to \infty }{x}_{\text{cm}}={lim}_{n\to \infty }\frac{L(1+\frac{1}{n})}{(1+\frac{2}{n})}$
$\Rightarrow {lim}_{n\to \infty }{x}_{\text{cm}}=\frac{L(1+0)}{(1+0)}=L$

Slope of $x_{\text{cm}}\text{vs}n$ is given by:
$\frac{d{x}_{\text{cm}}}{dn}=L\left[\frac{(n+2).1-(n+1).1}{(n+2{)}^2}\right]$
$\frac{d{x}_{\text{cm}}}{dn}=\frac{L}{(n+2{)}^2}$
$\Rightarrow$The slope is always $+ve$, which indicates that the curve of $x_{\text{cm}}\text{vs}n$ follows a continously increasing trend till $n\to \infty$
$\therefore$ option $\left(d\right)$ is correct.