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 circular disk of moment of inertia I_(t) is rotating in a horizontal

1.	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
Answer» `(1)/(2) (I_(B)^(2))/((I_(t) + I_(b))) omega_(i)^(2)` `(1)/(2) (I_(t)^(2))/((I_(i) + I_(b)))"" omega_(i)^(2)` `(I_(b)- I_(t))/((I_(t) + I_(b))) "" omega_(i)^(2)` `(1)/(2) (I_(b) I_(t))/((I_(t) + I_(b))) "" omega_(i)^(2)` ANSWER :D