Two discs have moments of inertia I_{1} and I | vedclass

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Question

(C) From the law of conservation of angular momentum,the total angular momentum remains constant:$I_{1} \omega_{1} + I_{2} \omega_{2} = (I_{1} + I_{2}

Vedclass · Accepted Answer

$\frac{I_{1} I_{2}}{2(I_{1}+I_{2})}(\omega_{1}-\omega_{2})^{2}$

Vedclass · Answer

(C) From the law of conservation of angular momentum,the total angular momentum remains constant:$I_{1} \omega_{1} + I_{2} \omega_{2} = (I_{1} + I_{2}) \omega$where $\omega$ is the common angular speed after contact.Thus,$\omega = \frac{I_{1} \omega_{1} + I_{2} \omega_{2}}{I_{1} + I_{2}}$.The initial kinetic energy is $K_{i} = \frac{1}{2} I_{1} \omega_{1}^{2} + \frac{1}{2} I_{2} \omega_{2}^{2}$.The final kinetic energy is $K_{f} = \frac{1}{2} (I_{1} + I_{2}) \omega^{2}$.The loss in kinetic energy is $\Delta K = K_{i} - K_{f} = \frac{1}{2} I_{1} \omega_{1}^{2} + \frac{1}{2} I_{2} \omega_{2}^{2} - \frac{1}{2} (I_{1} + I_{2}) \left( \frac{I_{1} \omega_{1} + I_{2} \omega_{2}}{I_{1} + I_{2}} \right)^{2}$.Simplifying this expression:$\Delta K = \frac{1}{2} \left[ I_{1} \omega_{1}^{2} + I_{2} \omega_{2}^{2} - \frac{(I_{1} \omega_{1} + I_{2} \omega_{2})^{2}}{I_{1} + I_{2}} \right]$$\Delta K = \frac{1}{2(I_{1} + I_{2})} [ (I_{1} \omega_{1}^{2} + I_{2} \omega_{2}^{2})(I_{1} + I_{2}) - (I_{1}^{2} \omega_{1}^{2} + I_{2}^{2} \omega_{2}^{2} + 2 I_{1} I_{2} \omega_{1} \omega_{2}) ]$$\Delta K = \frac{1}{2(I_{1} + I_{2})} [ I_{1}^{2} \omega_{1}^{2} + I_{1} I_{2} \omega_{1}^{2} + I_{1} I_{2} \omega_{2}^{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} ]$$\Delta K = \frac{I_{1} I_{2}}{2(I_{1} + I_{2})} [ \omega_{1}^{2} + \omega_{2}^{2} - 2 \omega_{1} \omega_{2} ]$$\Delta K = \frac{I_{1} I_{2}}{2(I_{1} + I_{2})} (\omega_{1} - \omega_{2})^{2}$.

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