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 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.
Answer» Here, `I_(i) = I_(t)` Let `omega_(i)` be the initial angular speed `I_(f) = (I_(t) + I_(b))` Let `omega_(f)` be the final angular speed of the system. According to the law of conservation of angular momentum, `I_(f)omega_(f) = I_(i) omega_(i)` `omega_(f) = (I_(t)omega_(i))/(I_(t) + I_(b))` ..(i) Energy lost `= K_(f) - K_(i) = (1)/(2)I_(f) omega_(f)^(2) - (1)/(2)I_(i) omega_(i)^(2)` `= (1)/(2)(I_(t) + I_(b)) omega_(f)^(2) - (1)/(2)I_(t)omega_(t)^(2)` using (i),we get Energy lost `= (1)/(2) (I_(t) + I_(b)) ((I_(t)omega_(i))/(I_(t) + I_(b)))^(2) - (1)/(2)I_(t) omega_(i)^(2)` `= (1)/(2)I_(t) omega_(i)^(2) [(I_(t))/(I_(t) + I_(b)) - 1] = - (1)/(2) (I_(b)I_(t)omega_(i)^(2))/((I_(t) + I_(b)))` Energy lost `=- (1)/(2)I_(b) omega_(i) omega_(f)` [using (i)]

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