Question

A uniform rod of mass m, length L , area of cross-section A and is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity $\omega$ in a horizontal plane. If Y the Youngs modulus of the material of rod, the increase in its length due torotation of rod is :

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

Answer 1

The correct option is C $\frac{m{\omega }^2{L}^2}{3AY}$

$\begin{array}{c}\text{Consider a Small element}\\ \text{of length}dx\text{at a distance}\\ x\text{from the axis of rotation.}\\ \text{Mass of length}L=M\\ dx=\left(\frac{M}{L}\right)dx\end{array}$

$\begin{array}{c}\text{let Tension in the de part be}d\tau \\ dT=\left(dm\right){\omega }^{2}x=\text{Cenlripelal force}\end{array}$

$(\because \text{centripelal force}=m{\omega }^{2}r)$

$\begin{array}{cc}\therefore T& ={\int }_x^L\left(\frac{M}{L}\right)dx{\omega }^{2}x\\ T& ={\int }_L\left(\frac{M}{L}\right){\omega }^{2}xdx\\ T& =\frac{M}{L}\frac{{\omega }^2}{2}(L^2-x^2)\end{array}$

$\begin{array}{c}\text{let}dl\text{be increase in length of the elemant}\\ \text{Then}Y=\frac{T/A}{d1/dx}\Rightarrow dl=\frac{Tdx}{AY}\end{array}$

$\begin{array}{c}dl=\frac{M{\omega }^2}{2LAY}[L^2-x^2]dx\\ \text{So, total elongation of the whole rod is}\\ l={\int }_0^L\frac{M{\omega }^2}{2LAY}[L^2-x^2]dx\end{array}$

$=\frac{1}{3}\frac{m{\omega }^2{L}^2}{AY}$