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(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.For a thin uniform rod of length $\ell$ rotating about an axis through its

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$3$

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(D) The rotational kinetic energy is given by $E = \frac{1}{2} I \omega^2$.For a thin uniform rod of length $\ell$ rotating about an axis through its center,the moment of inertia is $I = \frac{m \ell^2}{12}$.The mass $m$ of the rod is given by $m = 	ext{density} 	imes 	ext{volume} = d 	imes (A \ell) = d A \ell$.Substituting $m$ into the expression for $I$,we get $I = \frac{(d A \ell) \ell^2}{12} = \frac{d A \ell^3}{12}$.Now,substitute $I$ into the kinetic energy formula: $E = \frac{1}{2} \left( \frac{d A \ell^3}{12} ight) \omega^2 = \frac{d A \ell^3}{24} \omega^2$.Given $\ell = 2 \ m$,we have $E = \frac{d A (2)^3}{24} \omega^2 = \frac{8 d A}{24} \omega^2 = \frac{d A}{3} \omega^2$.Rearranging for $\omega$,we get $\omega^2 = \frac{3 E}{d A}$,which implies $\omega = \sqrt{\frac{3 E}{d A}}$.Comparing this with the given expression $\omega = \sqrt{\frac{\alpha E}{Ad}}$,we find $\alpha = 3$.

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