Question

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

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

${\left(\text{x}\right)}_{{\left(\text{C}\right)}_{\cdot }\text{M}}=\frac{{\int }_0^{\text{L}}\left(\frac{k}{{\left(\text{L}\right)}^n}\cdot {\left(\text{x}\right)}^n\cdot \text{dx}\right)\text{x}}{{\int }_0^{\text{L}}\frac{k}{{\left(\text{L}\right)}^n}\cdot {\left(\text{x}\right)}^n\cdot \text{dx}}$

$\Rightarrow {\left(\text{x}\right)}_{{\left(\text{C}\right)}_{\cdot }\text{M}}=\frac{{\int }_0^{\text{L}}{\left(\text{x}\right)}^{n+1}\text{dx}}{{\int }_0^{\text{L}}{\left(\text{x}\right)}^n\text{dx}}=\frac{{\left(\text{L}\right)}^{n+2}}{n+2}\cdot \frac{\left(n+1\right)}{{\left(\text{L}\right)}^{n+1}}$
$\Rightarrow {\left(\text{x}\right)}_{{\left(\text{C}\right)}_{\cdot }\text{M}}=\frac{\text{L}\left(n+1\right)}{\left(n+2\right)}$

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

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

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