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A thin rod of mass $m$ and length $l$ is rotated in a horizontal plane about one of its ends with a constant angular speed $ω$ in gravity free space. Then the tension at the midpoint of the rod is:

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$\frac{m l ω^{2}}{4}$

$\frac{3}{8} m l ω^{2}$

$\frac{m l ω^{2}}{8}$

$\frac{m l ω^{2}}{2}$

answer is D.

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Detailed Solution

$T = \int_{r = \frac{l}{2}}^{l} d m r ω^{2}$

$d m = \frac{m}{l} d r$

$T = \int_{r = \frac{l}{2}}^{l} d m r ω^{2} = \int_{r = \frac{l}{2}}^{l} \frac{m ω^{2}}{l} r d r$

$T = \frac{3}{8} m l ω^{2}$

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