A circular disk of moment of inertia I t is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ωi . 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 ωf . The energy lost by the initially rotating disc to friction is :
Answer & explanation
Correct answer: option 1
Loss of energy : \(\Delta E = \frac{1}{2} I_t \omega^2 - \frac{I_t^2 \omega^2}{2(I_t+I_b)} \)
\(\Delta E = \frac{1}{2} \frac{I_bI_t}{I_t+I_b} \omega^2\)