A thin uniform rod of length L = 2 text{ m}, | vedclass

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(B) The mass of the rod is $M = 	ext{Volume} 	imes 	ext{Density} = (A 	imes L) 	imes d = 2Ad$.The moment of inertia of a uniform rod about an axi

Vedclass · Accepted Answer

$3$

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(B) The mass of the rod is $M = 	ext{Volume} 	imes 	ext{Density} = (A 	imes L) 	imes d = 2Ad$.The moment of inertia of a uniform rod about an axis passing through its center and perpendicular to its length is $I = \frac{1}{12} ML^2$.Substituting $M = 2Ad$ and $L = 2 	ext{ m}$:$I = \frac{1}{12} (2Ad) (2)^2 = \frac{1}{12} (2Ad) (4) = \frac{8Ad}{12} = \frac{2}{3} Ad$.The rotational kinetic energy $E$ is given by $E = \frac{1}{2} I \omega^2$.Substituting $I = \frac{2}{3} Ad$:$E = \frac{1}{2} (\frac{2}{3} Ad) \omega^2 = \frac{1}{3} Ad \omega^2$.Solving for $\omega^2$:$\omega^2 = \frac{3E}{Ad}$.Taking the square root:$\omega = (\frac{3E}{Ad})^{1/2}$.Comparing this with the given expression $\omega = (\frac{\alpha E}{Ad})^{1/2}$, we find $\alpha = 3$.

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