Question

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

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

$\left(\textit{x}\right)_{\left(\text{C}\right)_{\cdot } \text{M}} = \frac{\displaystyle \int _{0}^{\textit{L}} \left(\frac{k}{\left(\textit{L}\right)^{n}} \cdot \left(\textit{x}\right)^{n} \cdot \textit{dx}\right) \textit{x}}{\displaystyle \int _{0}^{\textit{L}} \frac{k}{\left(\textit{L}\right)^{n}} \cdot \left(\textit{x}\right)^{n} \cdot \textit{dx}}$

$\Rightarrow \left(\textit{x}\right)_{\left(\text{C}\right)_{\cdot } \text{M}} = \frac{\displaystyle \int _{0}^{\textit{L}} \left(\textit{x}\right)^{n + 1} \textit{dx}}{\displaystyle \int _{0}^{\textit{L}} \left(\textit{x}\right)^{n} \textit{dx}} = \frac{\left(\textit{L}\right)^{n + 2}}{n + 2} \cdot \frac{\left(n + 1\right)}{\left(\textit{L}\right)^{n + 1}}$
$\Rightarrow \left(\textit{x}\right)_{\left(\text{C}\right)_{\cdot } \text{M}} = \frac{\textit{L} \left(n + 1\right)}{\left(n + 2\right)}$

The variation of the mass with $\textit{x}$ is given by

$\frac{\textit{dx}}{\textit{dn}} = \textit{L} \left\{\right. \frac{\left(n + 2\right) 1 - \left(n + 1\right)}{\left(n + 2\right)^{2}} \left.\right\} = \frac{\textit{L}}{\left(n + 2\right)^{2}}$

If the rod has the same density as at $\textit{x}$ = 0 i.e., $\textit{n}$ = 0, therefore uniform, the cenre of mass would have been at $\textit{L} / 2$ . As the density increases with length, the centre of mass shifts towards the right. Therefore it can only be (b).