A circular disk of moment of inertia I_t is r | vedclass

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Question

(A) By the principle of conservation of angular momentum,since no external torque acts on the system:$I_t \omega_i = (I_t + I_b) \omega_f$$\omega_f =

Vedclass · Accepted Answer

$\frac{1}{2} \frac{I_b I_t}{(I_t + I_b)} \omega_i^2$

Vedclass · Answer

(A) By the principle of conservation of angular momentum,since no external torque acts on the system:$I_t \omega_i = (I_t + I_b) \omega_f$$\omega_f = \left( \frac{I_t}{I_t + I_b} \right) \omega_i$The initial kinetic energy of the system is $K_i = \frac{1}{2} I_t \omega_i^2$.The final kinetic energy of the system is $K_f = \frac{1}{2} (I_t + I_b) \omega_f^2$.Substituting $\omega_f$:$K_f = \frac{1}{2} (I_t + I_b) \left( \frac{I_t}{I_t + I_b} \omega_i \right)^2 = \frac{1}{2} \frac{I_t^2}{I_t + I_b} \omega_i^2$.The energy lost due to friction is $\Delta K = K_i - K_f$:$\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2}{I_t + I_b} \omega_i^2$$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( 1 - \frac{I_t}{I_t + I_b} \right)$$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( \frac{I_t + I_b - I_t}{I_t + I_b} \right)$$\Delta K = \frac{1}{2} \frac{I_b I_t}{I_t + I_b} \omega_i^2$.

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