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

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(A) The total initial angular momentum of the system is $L_i = I_1 \omega_1 + I_2 \omega_2$.When the discs are joined,the total moment of inertia beco

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total initial angular momentum of the system is $L_i = I_1 \omega_1 + I_2 \omega_2$.When the discs are joined,the total moment of inertia becomes $I = I_1 + I_2$.Let the common angular speed of the system be $\omega$. By the principle of conservation of angular momentum,$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$.Thus,the final angular speed is $\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$.The initial kinetic energy is $E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$.The final kinetic energy is $E_f = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2 = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.The loss in kinetic energy is $\Delta E = E_i - E_f = \left( \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \right) - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.Simplifying this expression: $\Delta E = \frac{(I_1 + I_2)(I_1 \omega_1^2 + I_2 \omega_2^2) - (I_1^2 \omega_1^2 + I_2^2 \omega_2^2 + 2 I_1 I_2 \omega_1 \omega_2)}{2(I_1 + I_2)}$.$\Delta E = \frac{I_1^2 \omega_1^2 + I_1 I_2 \omega_2^2 + I_1 I_2 \omega_1^2 + I_2^2 \omega_2^2 - I_1^2 \omega_1^2 - I_2^2 \omega_2^2 - 2 I_1 I_2 \omega_1 \omega_2}{2(I_1 + I_2)} = \frac{I_1 I_2 (\omega_1^2 + \omega_2^2 - 2 \omega_1 \omega_2)}{2(I_1 + I_2)}$.Therefore,$\Delta E = \frac{I_1 I_2 (\omega_1 - \omega_2)^2}{2(I_1 + I_2)}$.

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