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

A circular disk of moment of inertia `I_(t)` is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed `omega_(i)` Another disk of moment of inertia `I_(b)` is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed `omega_(f)` The energy lost by the initially rotating disc to friction is

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)`

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(t) omega_(i)= (I_(t) + I_(b)) omega_(f), omega_(f)= (I_9t) omega_(i))/(I_(t) + I_(b))`
`K_(i)= (1)/(2) I_(t) omega_(i)^(2)`
`K_(f)= (1)/(2) (I_(t) + I_(b)) omega_(f)^(2)`
loss of energy `= K_(i)- K_(f)`
`=(1)/(2) I_(t) omega_(i)^(2) - (1)/(2) ((I_(t) omega_(i))/(I_(t) + I_(b)))^(2) (I_(t) + I_(b))`
`= (1)/(2) I_(t) omega_(i)^(2) [1- (I_(t))/(I_(t) + I_(b))] = (1)/(2) ((I_(b) I_(t))/(I_(b) + I_(t))) omega_(i)^(2)`
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