A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.

Options

A. `(1)/(2) (I_(b)^(2))/((I_(t)+I_(b))omega _(i)^(2)`
B. `(1)/(2) (I_(t)^(2))/((I_(t)+I_(b))omega _(i)^(2)`
C. `(I_(b)-I_(t))/((I_(t)+L_(b)))omega _(i)^(2)`
D. `(1)/(2)(I_(b)I_(t))/((I_(t)+I_(b)))omega_(i)^(2)`

Answer

D

Explanation

By Conservation of angular momemtum
`I_(t)omega _(i)(I_(t)+I_(b))omega_(f) implies omega_(f)=((I_(t))/(I_(t)+I_(b)))omega_(i)`
Loss in kinetic energy `=(1)/(2) I_(t)omega_(i)^(2)-(1)/(2)(I_(t)+I_(B))(omega_(f)^(2))`
`=(1)/(2)((I_(b)I_(t))/(I_(b)+I_(t)))omega_(i)^(2)`