A uniform bar of length 6l and mass 8m lies o | vedclass

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(C) Let the center of the bar be the origin. The moment of inertia of the bar about its center is $I_{bar} = \frac{(8m)(6l)^2}{12} = \frac{8m 	imes 3

Vedclass · Accepted Answer

$\frac{3mv^2}{5}$

Vedclass · Answer

(C) Let the center of the bar be the origin. The moment of inertia of the bar about its center is $I_{bar} = \frac{(8m)(6l)^2}{12} = \frac{8m 	imes 36l^2}{12} = 24ml^2$.After the collision,the masses $2m$ and $m$ stick to the bar at distances $l$ and $2l$ from the center respectively.The moment of inertia of the system about the center $c$ is $I = I_{bar} + I_{2m} + I_{m} = 24ml^2 + (2m)l^2 + (m)(2l)^2 = 24ml^2 + 2ml^2 + 4ml^2 = 30ml^2$.The angular momentum $L$ about the center $c$ is conserved. Initial angular momentum $L_i = (2m)(v)(l) + (m)(2v)(2l) = 2mvl + 4mvl = 6mvl$.Since $L_i = I\omega$,we have $6mvl = (30ml^2)\omega$,which gives $\omega = \frac{6mvl}{30ml^2} = \frac{v}{5l}$.The rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(30ml^2)\left(\frac{v}{5l}ight)^2 = 15ml^2 	imes \frac{v^2}{25l^2} = \frac{15}{25}mv^2 = \frac{3mv^2}{5}$.

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