A uniform rod of mass m and length $l$ is rotating with constant angular velocity $ω$ 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

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$\frac{3 m ω^{2} l}{8 A Y}$

$\frac{m ω^{2} l}{8 A Y}$

$\frac{3 m ω^{2} l}{4 A Y}$

$\frac{m ω^{2} l}{4 A Y}$

answer is B.

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Detailed Solution

$d T = d m (1 - x) ω^{2}$

$d T = \frac{m}{l} d x (l - x) ω^{2}$

$\Rightarrow \int_{0}^{T} d T = \int_{0}^{1 / 2} \frac{m ω^{2}}{l} (l - x) d x$

$= \frac{m ω^{2}}{l} {[1 x - \frac{x^{2}}{2}]}_{0}^{1 / 2} = \frac{m ω^{2}}{l} [\frac{l^{2}}{2} - \frac{l^{2}}{8}]$

$∴$ Tension at mid point is : $T = \frac{3}{8} m l ω^{2}$

$\Rightarrow Stress = \frac{3 m l ω^{2}}{8 A} \Rightarrow strain = \frac{3 m l ω^{2}}{8 A Y}$

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