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. `(1)/(2) (I_b - I_t)/((I_t + I_b))omega_i^2`
D. `(1)/(2) (I_b I_t)/((I_t + I_b))omega_i^2`

Answer

D

Explanation

(d) Loss of energy,
`DeltaE = K_(initial) - K_(final) = (1)/(2) I_1 omega_1^2 - (1)/(2) (I_t^2 omega_i^2)/((I_t + I_b))`
=`(1)/(2)' (I_b I_t omega_i^2)/((I_t + I_b))`.