Two discs of moments of inertia I_1 and I_2 a | vedclass

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Question

(A) Let the moment of inertia of disc $I$ be $I_1$ and its angular speed be $\omega_1$. Let the moment of inertia of disc $II$ be $I_2$ and its angula

Vedclass · Accepted Answer

$\frac{I_1 I_2 (\omega_1 - \omega_2)^2}{2(I_1 + I_2)}$

Vedclass · Answer

(A) Let the moment of inertia of disc $I$ be $I_1$ and its angular speed be $\omega_1$. Let the moment of inertia of disc $II$ be $I_2$ and its angular speed be $\omega_2$. The initial angular momentum of the system is $L_i = I_1 \omega_1 + I_2 \omega_2$. When the discs are brought into contact,the final moment of inertia of the system is $I = I_1 + I_2$. Let $\omega$ be the final angular speed of the system. By the law of conservation of angular momentum: $I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$ $\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$ Initial kinetic energy $K_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$. Final kinetic energy $K_f = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2 = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$. Loss in kinetic energy $\Delta K = K_i - K_f = \left( \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \right) - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$. Simplifying this expression: $\Delta K = \frac{I_1 I_2 (\omega_1 - \omega_2)^2}{2(I_1 + I_2)}$.

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