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 $c$. The moment of inertia of the bar about $c$ is $I_{bar} = \frac{(8m)(6l)^2}{12} = 24ml^2$.After the co

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

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(C) Let the center of the bar be the origin $c$. The moment of inertia of the bar about $c$ is $I_{bar} = \frac{(8m)(6l)^2}{12} = 24ml^2$.After the collision,the masses $m$ and $2m$ stick to the bar at distances $2l$ and $l$ from the center respectively.The total moment of inertia of the system about $c$ is $I = I_{bar} + m(2l)^2 + (2m)(l)^2 = 24ml^2 + 4ml^2 + 2ml^2 = 30ml^2$.The angular momentum of the system about $c$ before the collision is $L = (2m)(v)(l) + (m)(2v)(2l) = 2mvl + 4mvl = 6mvl$.Since there are no external torques,angular momentum is conserved: $L = I\omega$.$6mvl = (30ml^2)\omega \Rightarrow \omega = \frac{6mvl}{30ml^2} = \frac{v}{5l}$.The total energy about the center of mass is the rotational kinetic energy: $E = \frac{1}{2}I\omega^2$.$E = \frac{1}{2}(30ml^2)\left(\frac{v}{5l}ight)^2 = 15ml^2 	imes \frac{v^2}{25l^2} = \frac{15}{25}mv^2 = \frac{3mv^2}{5}$.

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