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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_(t)I_(b))/((I_(t) + I_(b))) (i)_(i)^(2)`B. `(1)/(2) (I_(t)^(2))/((I_(t) + I_(b))) (i)_(i)^(2)`C. `(I_(b)-I_(t))/((I_(t) + I_(b))) (i)_(i)^(2)`D. `(1)/(2) (I_(b)-I_(t))/((I_(t) + I_(b))) (i)_(i)^(2)` |
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Answer» Correct Answer - A As no external torque is acting, therefore, `I_(t)(i)_(i) = (I_(t) + I_(b)) (i)_(f)` `(i)_(f) = (I_(t)(i)_(i))/(I_(t) + I_(b))` Energy lost `= DeltaK = E_(1) - E_(2)` `DeltaK = (1)/(2)I_(t) (i)_(i)^(2) - (1)/(2)(I_(t) + I_(b))(i)_(f)^(2)` `= (1)/(2)I_(t) (i)_(i)^(2) - (1)/(2) ((I_(t) + I_(b))I_(t)^(2)(i)_(i)^(2))/((I_(t) + I_(b)^(2))` `= (1)/(2)I_(t)(i)_(t)^(2) [1 - (I_(t))/(I_(t) + I_(b))]` `= (1)/(2)I_(t)(i)_(i)^(2) (I_(b))/((I_(t) + I_(b)))` |
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