(A) Since there is no external torque about the pivot $C$,the angular momentum of the system is conserved.Initial angular momentum $L_i = (2m)(v)(L) +

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(A) Since there is no external torque about the pivot $C$,the angular momentum of the system is conserved.Initial angular momentum $L_i = (2m)(v)(L) + (m)(2v)(2L) = 2mvL + 4mvL = 6mvL$.Final moment of inertia $I_f = I_{\text{rod}} + I_{2m} + I_{m} = \frac{(8m)(6L)^2}{12} + (2m)(L)^2 + (m)(2L)^2$.$I_f = \frac{8m \cdot 36L^2}{12} + 2mL^2 + 4mL^2 = 24mL^2 + 2mL^2 + 4mL^2 = 30mL^2$.Using $L_i = I_f \omega$:$6mvL = (30mL^2) \omega$.$\omega = \frac{6mvL}{30mL^2} = \frac{v}{5L}$.

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

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$A$ uniform rod of length $6L$ and mass $8m$ is pivoted at its centre $C$. Two masses $m$ and $2m$ with speeds $2v$ and $v$ respectively,as shown in the figure,strike the rod and stick to it. Initially,the rod is at rest. If it rotates with angular velocity $\omega$ due to the impact,then $\omega$ will be:

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A
$\frac{v}{5L}$
B
$\text{Zero}$
C
$\frac{8v}{6L}$
D
$\frac{11v}{3L}$

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

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$A$ force $\vec F = \alpha \hat i + 3\hat j + 6\hat k$ is acting at a point $\vec r = 2\hat i - 6\hat j - 12\hat k$. The value of $\alpha$ for which angular momentum about the origin is conserved is:

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