A circular disc of moment of inertia `I_(1)` is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed `omega_(1)`. Another disc of moment of inertia `I_(2)` is placed coaxially on the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed `omega_(2)`. The energy lost by the initially rotating disc to friction is ............

Options

A. `(1)/(2)(I_(2)^(2))/((I_(1)+I_(2)))omega_(1)^(2)`
B. `(I_(2)^(2))/((I_(1)+I_(2)))omega_(1)^(2)`
C. `(I_(2)-I_(1))/((I_(1)+I_(2)))omega_(2)^(2)`
D. `(1)/(2)(I_(2)I_(1))/((I_(1)+I_(2)))omega_(1)^(2)`

Answer

D

Explanation

According to law of conservation of momentum
`therefore I_(1)omega_(1)=(I_(1)+I_(2))omega_(2)`
`therefore omega_(2)=(I_(1))/((I_(1)+I_(2)))omega_(1)....(1)`
Angular velocity of system decreases when another disc is placed on the first disc.
Hence, `therefore omega_(2)ltomega_(1)`
Hence decrease in rotational kinetic energy
= (initial rotational kinetic energy) - (final rotational kinetic energy)
`=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))omega_(2)^(2)`
`=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))((I_(1))/(I_(1)+I_(2)).omega_(1))^(2)`
putting value of equ. (1)
`=(1)/(2)I_(1)omega_(1)^(2)-(1)/(2)(I_(1)+I_(2))(I_(1)^(2))/((I_(1)+I_(2))^(2)).omega_(1)^(2)`
`=(1)/(2)omega_(1)^(2)(I_(1))/((I_(1)+I_(2)))(I_(1)+I_(2)-I_(1))`
`=(1)/(2)omega_(1)^(2)(I_(1)I_(2))/((I_(1)+I_(2)))`