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A circular disc of moment of inertia I(t...

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)`

Text Solution

Verified by Experts

The correct Answer is:
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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