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(B) In this process,the potential energy of the rod is converted into rotational kinetic energy.Initial potential energy of the rod $U = mg \left( \fr

Vedclass · Accepted Answer

$5.4$

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(B) In this process,the potential energy of the rod is converted into rotational kinetic energy.Initial potential energy of the rod $U = mg \left( \frac{l}{2} ight)$,because the center of gravity of the rod is at its midpoint.Rotational kinetic energy $K = \frac{1}{2} I \omega^2$.Here,$I$ is the moment of inertia of the rod about the fixed end $A$,which is $I = \frac{ml^2}{3}$.Let $\omega$ be the angular velocity of the rod when it hits the ground.The velocity of the end $B$ when it hits the ground is $v_B = \omega l$,so $\omega = \frac{v_B}{l}$.According to the law of conservation of energy:$mg \left( \frac{l}{2} ight) = \frac{1}{2} I \omega^2$$mg \left( \frac{l}{2} ight) = \frac{1}{2} \left( \frac{ml^2}{3} ight) \left( \frac{v_B}{l} ight)^2$$mg \frac{l}{2} = \frac{1}{6} m v_B^2$$v_B^2 = 3gl$$v_B = \sqrt{3gl} = \sqrt{3 	imes 9.8 	imes 1} = \sqrt{29.4} \approx 5.42 \ m/s$.Thus,the velocity is approximately $5.4 \ m/s$.

$A$ rod of length $1 \ m$ is held vertically and allowed to fall such that its lower end remains fixed on the ground. The velocity of the other end of the rod when it hits the ground is ......... $m/s$ $(g = 9.8 \ m/s^2)$.

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