A thin uniform rod of mass M and length L is | vedclass

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(D) Initial moment of inertia of the rod about the axis passing through its centre is $I_1 = \frac{ML^2}{12}$.Initial angular momentum $L_1 = I_1 \ome

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$\frac{1}{3}\omega$

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(D) Initial moment of inertia of the rod about the axis passing through its centre is $I_1 = \frac{ML^2}{12}$.Initial angular momentum $L_1 = I_1 \omega = \frac{ML^2}{12} \omega$.When two objects of mass $\frac{M}{3}$ are attached to the ends,the new moment of inertia $I_2$ is:$I_2 = I_{	ext{rod}} + I_{	ext{objects}} = \frac{ML^2}{12} + \left(\frac{M}{3}ight)\left(\frac{L}{2}ight)^2 + \left(\frac{M}{3}ight)\left(\frac{L}{2}ight)^2$$I_2 = \frac{ML^2}{12} + \frac{ML^2}{12} + \frac{ML^2}{12} = \frac{3ML^2}{12} = \frac{ML^2}{4}$.By the law of conservation of angular momentum,$L_1 = L_2$:$\frac{ML^2}{12} \omega = \frac{ML^2}{4} \omega'$$\omega' = \frac{4}{12} \omega = \frac{1}{3} \omega$.

$A$ thin uniform rod of mass $M$ and length $L$ is rotating about a perpendicular axis passing through its centre with a constant angular velocity $\omega$. Two objects,each of mass $\frac{M}{3}$,are attached gently to the two ends of the rod. The rod will now rotate with an angular velocity of

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