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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
A
$\frac{1}{2}\frac{I_bI_t}{(I_t + I_b)}\omega_i^2$
B
$\frac{1}{2}\frac{I_b^2}{(I_t + I_b)}\omega_i^2$
C
$\frac{1}{2}\frac{I_t^2}{(I_t + I_b)}\omega_i^2$
D
$\frac{I_b - I_t}{(I_t + I_b)}\omega_i^2$
Detailed Solution
No external torque acts on the two-disk system, so angular momentum is conserved: $I_t\omega_i = (I_t + I_b)\omega_f$
$\omega_f = \frac{I_t\omega_i}{I_t + I_b}$
Initial kinetic energy: $K_i = \frac{1}{2}I_t\omega_i^2$
Final kinetic energy: $K_f = \frac{1}{2}(I_t + I_b)\omega_f^2 = \frac{1}{2}\frac{I_t^2\omega_i^2}{(I_t + I_b)}$
Energy lost: $\Delta E = K_i - K_f = \frac{1}{2}I_t\omega_i^2\left[1 - \frac{I_t}{I_t + I_b}\right]$
$\Delta E = \frac{1}{2}\frac{I_bI_t}{(I_t + I_b)}\omega_i^2$
$\omega_f = \frac{I_t\omega_i}{I_t + I_b}$
Initial kinetic energy: $K_i = \frac{1}{2}I_t\omega_i^2$
Final kinetic energy: $K_f = \frac{1}{2}(I_t + I_b)\omega_f^2 = \frac{1}{2}\frac{I_t^2\omega_i^2}{(I_t + I_b)}$
Energy lost: $\Delta E = K_i - K_f = \frac{1}{2}I_t\omega_i^2\left[1 - \frac{I_t}{I_t + I_b}\right]$
$\Delta E = \frac{1}{2}\frac{I_bI_t}{(I_t + I_b)}\omega_i^2$