A uniform rod of length l, mass m, cross-sectional area A and Young's modulus Y is rotated in horizontal plane about a fixed vertical axis passing through one end, with a constant angular velocity $ω$ . Find the total extension in the rod due to the tension produced in the rod.

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$\frac{{mω}^{3} t^{3}}{3 AY}$

$\frac{{mω}^{3} l^{2}}{3 AY}$

$\frac{{mω}^{2} l^{2}}{3 AY}$

$\frac{{mω}^{2} l^{3}}{4 AY}$

answer is A.

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Detailed Solution

Y = $\frac{\frac{T (x)}{A}}{(\frac{∆ e}{∆ x})}; T (x) = \frac{m}{l} (l - x) ω^{2} (\frac{l - x}{2} + x)$

$\Rightarrow T = \frac{{mω}^{2}}{2 l} (l^{2} - x^{2}); de = \frac{{mω}^{2}}{2 AY l} (l^{2} - x^{2}) dx$

$\Rightarrow \frac{{mω}^{2}}{2 AYl} (l^{3} - \frac{l^{3}}{3}) = \frac{{mω}^{2} l^{2}}{3 AY}$

Extension = $\frac{{mω}^{2} l^{2}}{3 AY}$

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