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

A circular disc of moment of inertia `I_(t)` is rotating in a horizont

1.	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.A. `(1)/(2)(I_(b)I_(t))/(I_(t)+I_(b))omega_(1)^(2)`B. `(1)/(2)(I_(b)^(2))/(I_(t)+I_(b))omega_(1)^(2)`C. `(1)/(2)(I_(t)^(2))/(I_(t)+I_(b))omega_(1)^(2)`D. `(I_(b)-I_(t))/((I_(t)+I_(b)))omega_(1)^(2)`
Answer» `I_(t)omega_(1)=(I_(t)+I_(b))omegaimpliesomega=(I_(t)omega_(1))/((I_(t)+I_(b)))` `DeltaK=(1)/(2)I_(t)omega_(1)^(2)-(1)/(2)(I_(t)+I_(b))((I_(t)omega_(1))/(I_(t)+I_(b)))^(2)` `=(1)/(2)(I_(t)I_(b))/((I_(t)+I_(b)))omega_(1)^(2)`

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