A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a

Question

A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_f$. The energy lost by the initially rotating disc to friction is:

Accepted Answer

Let the initial angular momentum of the system be $L_i$ and the final angular momentum be $L_f$. Since there is no external torque on the system, angular momentum is conserved .
By conservation of angular momentum:
$L_i = L_f$
$I_t \omega_i + I_b (0) = (I_t + I_b) \omega_f$
$\omega_f = \frac{I_t \omega_i}{I_t + I_b}$

Initial kinetic energy of the system :
$K_i = \frac{1}{2} I_t \omega_i^2$

Final kinetic energy of the system:
$K_f = \frac{1}{2} (I_t + I_b) \omega_f^2$
$K_f = \frac{1}{2} (I_t + I_b) \left(\frac{I_t \omega_i}{I_t + I_b}\right)^2 = \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}$

Energy lost due to friction is the difference between initial and final kinetic energy:
$\Delta K = K_i - K_f$
$\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}$
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( 1 - \frac{I_t}{I_t + I_b} \right)$
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( \frac{I_t + I_b - I_t}{I_t + I_b} \right)$
$\Delta K = \frac{1}{2} \frac{I_t I_b}{I_t + I_b} \omega_i^2$

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