A rod of length 1 m is held vertically. What | vedclass

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(B) Let the mass of the rod be $M$ and its length be $\ell = 1 \ m$. The rod is hinged at the bottom end. When it falls,the potential energy lost by t

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$\sqrt{29.4} \ m/s$

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(B) Let the mass of the rod be $M$ and its length be $\ell = 1 \ m$. The rod is hinged at the bottom end. When it falls,the potential energy lost by the center of mass is converted into rotational kinetic energy.The loss in potential energy is $\Delta U = Mg \Delta h = Mg \left( \frac{\ell}{2} ight)$.The rotational kinetic energy is $K = \frac{1}{2} I \omega^2$,where $I = \frac{M \ell^2}{3}$ is the moment of inertia about the hinge.Equating the two: $\frac{1}{2} \left( \frac{M \ell^2}{3} ight) \omega^2 = Mg \frac{\ell}{2}$.Simplifying,we get $\frac{1}{6} M \ell^2 \omega^2 = \frac{1}{2} Mg \ell$,which leads to $\omega^2 = \frac{3g}{\ell}$.The linear speed $V$ of the free end is $V = \omega \ell$. Therefore,$V^2 = \omega^2 \ell^2 = \left( \frac{3g}{\ell} ight) \ell^2 = 3g \ell$.Substituting $g = 9.8 \ m/s^2$ and $\ell = 1 \ m$:$V = \sqrt{3 	imes 9.8 	imes 1} = \sqrt{29.4} \ m/s$.

$A$ rod of length $1 \ m$ is held vertically. What will be the speed of the other end when it touches the ground without slipping?

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