The moment of inertia of a ring about an axis | vedclass

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(C) Initial state: Moment of inertia of the first ring is $I_1 = I$. Angular velocity is $\omega_1 = \omega$. Initial kinetic energy $K_i = \frac{1}{2

Vedclass · Accepted Answer

$I\omega^2/4$

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(C) Initial state: Moment of inertia of the first ring is $I_1 = I$. Angular velocity is $\omega_1 = \omega$. Initial kinetic energy $K_i = \frac{1}{2} I \omega^2$.Final state: When an identical ring is placed on it, the total moment of inertia becomes $I_f = I + I = 2I$. Since no external torque acts on the system, angular momentum is conserved: $L_i = L_f$.$I \omega = (2I) \omega_f$, which gives $\omega_f = \frac{\omega}{2}$.Final kinetic energy $K_f = \frac{1}{2} (2I) (\frac{\omega}{2})^2 = I \cdot \frac{\omega^2}{4} = \frac{1}{4} I \omega^2$.Loss in kinetic energy $\Delta K = K_i - K_f = \frac{1}{2} I \omega^2 - \frac{1}{4} I \omega^2 = \frac{1}{4} I \omega^2$.

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