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

Answer

B

Explanation

Initial angular momentum `=I_(t)omega_(1)`
On dropping the disc, final angular momentum `=(I_(t)+I_(b))omega_(f)`
Angular momentum would be conserved,
`L=Iomega=` constant
`:.I_(t)omega_(i)=(I_(t)+I_(b))omega_(f)impliesomega_(f)=(I_(t)omega_(i))/((I_(t)+I_(b)))`
Loss in rotational kinetic energy,
`=(1)/(2)I_(t)omega_(i)^(2)-(1)/(2)(I_(t)+I_(b))omega_(f)^(2)`
`=[(1)/(2)I_(t)omega_(i)^(2)-(1)/(2)(I_(t)^(2)omega_(i)^(2))/((I_(t)+I_(b)))]`
`=(1)/(2)(I_(t)I_(b)omega_(f)^(2))/((I_(t)+I_(b)))`