A 2 , kg steel rod of length 0.6 , m is clamp | vedclass

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(A) The rod is initially in a vertical position. When it falls from the upright position to the lowest position,the center of mass of the rod drops by

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$6$

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(A) The rod is initially in a vertical position. When it falls from the upright position to the lowest position,the center of mass of the rod drops by a height $h = \frac{\ell}{2}$,where $\ell = 0.6 \, m$ is the length of the rod.By the principle of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in rotational kinetic energy:$mg \left( \frac{\ell}{2} ight) = \frac{1}{2} I \omega^2$Since the rod rotates about its end,the moment of inertia is $I = \frac{1}{3} m \ell^2$.Substituting $I$ into the energy equation:$mg \left( \frac{\ell}{2} ight) = \frac{1}{2} \left( \frac{1}{3} m \ell^2 ight) \omega^2$$g \ell = \frac{1}{3} \ell^2 \omega^2$$\omega^2 = \frac{3g}{\ell}$$\omega = \sqrt{\frac{3g}{\ell}}$The linear speed $v$ of the free end is given by $v = \omega \ell$:$v = \sqrt{\frac{3g}{\ell}} 	imes \ell = \sqrt{3g \ell}$Substituting the given values $g = 10 \, ms^{-2}$ and $\ell = 0.6 \, m$:$v = \sqrt{3 	imes 10 	imes 0.6} = \sqrt{18} \approx 4.24 \, ms^{-1}$.Wait,re-evaluating the initial potential energy: If the rod starts from the vertical position (upright) and falls to the lowest position,the center of mass drops from $\frac{\ell}{2}$ above the pivot to $\frac{\ell}{2}$ below the pivot. Total change in height $\Delta h = \ell$.$mg \ell = \frac{1}{2} \left( \frac{1}{3} m \ell^2 ight) \omega^2$$2g \ell = \frac{1}{3} \ell^2 \omega^2 \Rightarrow \omega^2 = \frac{6g}{\ell}$$v = \omega \ell = \sqrt{6g \ell} = \sqrt{6 	imes 10 	imes 0.6} = \sqrt{36} = 6 \, ms^{-1}$.

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