A disc of the moment of inertia `'l_(1)'` is rotating in horizontal plane about an axis passing through a centre and perpendicular to its plane with constant angular speed `'omega_(1)'` . Another disc of moment of inertia `'I_(2)'`. having zero angular speed is placed discs are rotating disc. Now, both the discs are rotating with constant angular speed `'omega_(2)'`. The energy lost by the initial rotating disc is

Options

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

Answer

D

Explanation

When the disc of M.I. `I_(2)` is placed on the rotating disc of M.I., `I_(1)`, the total M.I. of the combined disc `=I_(1)+I_(2)`
By the principle of conservation of angular momentum,
`I_(1)omega_(1)=(I_(1)+I_(2))omega_(2)`
`(omega_(2))/(omega_(1))=(I_(1))/(I_(1)+I_(2))" ...(1)"`
Initial K.E. of the first rotating disc `(E_(1))=(1)/(2)I_(1)omega^(2)`
and the K.E. of the combined disc `(E_(2))=(1)/(2)(I_(1)+I_(2))omega_(2)^(2)`
`therefore` Loss in K.E. `=(1)/(2)I_(1)omega^(2)-(1)/(2)(I_(1)+I_(2))omega_(2)^(2)`
`therefore" DeltaE=(1)/(2)omega_(1)^(2)[I_(1)-(I_(1)+I_(2))(omega_(2)^(2))/(omega_(1)^(2))]`
`=(1)/(2)omega_(1)^(2)[I_(1)-(I_(1)+I_(2))(I_(1)^(2))/((I_(1)+I_(2))^(2))]" ...from (1)"`
`=(1)/(2)omega_(1)^(2)[I_(1)-(I_(1)^(2))/(I_(1)+I_(2))]`
`=(1)/(2)omega_(1)^(2)[(I_(1)^(2)+I_(1)I_(2)-I_(1)^(2))/(I_(1)+I_(2))]`
`=(1)/(2)[(I_(1)I_(2))/(I_(1)+I_(2))]omega_(1)^(2)`