Question

A uniform rod of mass m and length $l$ is rotating with constant angular velocity $\omega$ about an axis which passes through its one end and perpendicular to the length of rod. The area of cross section of the rod is A and its young's modulus is Y. Neglect gravity. The strain at the mid point of the rod is :

• A. $\frac{m{\omega }^{2}l}{8AY}$
• B. $\frac{3m{\omega }^{2}l}{8AY}$
• C. $\frac{3m{\omega }^{2}l}{4AY}$
• D. $\frac{m{\omega }^{2}l}{4AY}$

Answer 1

The correct option is B $\frac{3m{\omega }^{2}l}{8AY}$

$\begin{array}{c}\text{Stress}=y\left(\text{strain}\right)\\ \text{Strain}=\frac{T_{\text{mid-point}}}{AY}\end{array}$

$\begin{array}{c}\text{Let us assume a small element of}\\ \text{length}dx\text{and mass dm.}\\ \text{Mass per unit length}=\frac{m}{L}\\ \begin{array}{cc}d& m=\left(\frac{m}{L}\right)\left(dx\right)\\ dT& =\left(dm\right)(l-x){\omega }^2\\ & =\left(\frac{m}{L}\right)(1-x){\omega }^{2}dx\end{array}\end{array}$

$\begin{array}{c}\text{Now, centripetal force at mid point is}\\ \begin{array}{cc}{\int }_0^{l/2}dT& ={\int }_0^{l/2}\frac{m}{L}(l-x)w^{2}dx\\ & =\frac{m{w}^2}{L}{\left[lx\frac{x^2}{2}\right]}_0^{1/2}\end{array}\end{array}$

$\begin{array}{c}{T}_{\text{mid-point}}=\frac{M{\omega }^2}{L}(\frac{L^2}{2}-\frac{L^2}{8})=\frac{3M{\omega }^{2}L}{8}\\ \text{Strain}=\frac{3m{\omega }^{2}L}{8AY}\end{array}$