Question

A uniform rod having mass density $\sigma$, length L, cross-sectional area A is rotating with a uniform angular velocity $\omega$ about a fixed vertical axis through its end. Neglecting the effect of gravity, The normal stress developed in the rod at its midpoint is found to be $\frac{x}{8}\sigma L^2{\omega }^2$. Find the value of x.

• A. 3
• B. 4
• C. 5
• D. 1

Answer 1

Answer: (D)

1

$\begin{array}{c}\text{Let us consider a small element of differential length}\\ \text{dx at a distance}x\text{from free end of the rod.}\end{array}$

$\begin{array}{c}\text{mass of the rod of length}dx=\sigma dx\\ \begin{array}{cc}\text{stress =}& \text{s}\\ ds& =\frac{\left(\sigma dx\right){\omega }^{2}x}{A}\end{array}\end{array}$

$s=\int ds={\int }_0^{L/2}\frac{\sigma {\omega }^{2}xdx}{A}=\frac{\sigma L^2{\omega }^2}{8}$

$\text{so}x=1$