(B) According to the law of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum is equal to the

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(B) According to the law of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum is equal to the final angular momentum.Initial angular momentum $L_i = I_1 \omega$.When the man moves to a distance $r$ from the center,the new moment of inertia of the system becomes $I_2 = I_1 + mr^2$.Let the final angular velocity be $\omega_2$.Final angular momentum $L_f = (I_1 + mr^2) \omega_2$.Equating $L_i = L_f$:$I_1 \omega = (I_1 + mr^2) \omega_2$.Therefore,the final angular velocity is $\omega_2 = \frac{I_1 \omega}{I_1 + mr^2}$.

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

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$A$ rotating table has an angular velocity $\omega$ and a moment of inertia $I_1$. $A$ man of mass $m$ stands at the center of the table. If the man walks a distance $r$ along the radius of the table,what will be his final angular velocity?

A
$\frac{I_1}{I_1\omega + mr^2}$
B
$\frac{I_1\omega}{I_1 + mr^2}$
C
$\frac{I_1\omega}{I_1 + mr}$
D
$\frac{I_1\omega}{I_1 - mr^2}$

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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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$A$ rotating table has an angular velocity $\omega$ and a moment of inertia $I_1$. $A$ man of mass $m$ stands at the center of the table. If the man walks a distance $r$ along the radius of the table,what will be his final angular velocity?

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