Two discs of same moment of inertia I are rot | vedclass

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Question

(D) Initial angular momentum $L_i = I\omega_1 + I\omega_2$.Let $\omega$ be the final angular speed of the combined system.Final angular momentum $L_f

Vedclass · Accepted Answer

$\frac{I}{4}{\left( {{\omega _1} - {\omega _2}} \right)^2}$

Vedclass · Answer

(D) Initial angular momentum $L_i = I\omega_1 + I\omega_2$.Let $\omega$ be the final angular speed of the combined system.Final angular momentum $L_f = (I + I)\omega = 2I\omega$.According to the law of conservation of angular momentum,$L_i = L_f$:$I\omega_1 + I\omega_2 = 2I\omega \implies \omega = \frac{\omega_1 + \omega_2}{2}$.Initial rotational kinetic energy $E_i = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 = \frac{1}{2}I(\omega_1^2 + \omega_2^2)$.Final rotational kinetic energy $E_f = \frac{1}{2}(2I)\omega^2 = I \left( \frac{\omega_1 + \omega_2}{2} \right)^2 = \frac{I}{4}(\omega_1^2 + \omega_2^2 + 2\omega_1\omega_2)$.Loss of energy $\Delta E = E_i - E_f = \frac{I}{2}(\omega_1^2 + \omega_2^2) - \frac{I}{4}(\omega_1^2 + \omega_2^2 + 2\omega_1\omega_2)$.$\Delta E = \frac{I}{4} [2\omega_1^2 + 2\omega_2^2 - \omega_1^2 - \omega_2^2 - 2\omega_1\omega_2] = \frac{I}{4}(\omega_1^2 + \omega_2^2 - 2\omega_1\omega_2) = \frac{I}{4}(\omega_1 - \omega_2)^2$.

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