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Two uniform circular rough disc of momen...

Two uniform circular rough disc of moment of inertia `I_(1)` and `(I_(1))/(2)` are rotating with angular velocity `omega_(1)` and `(omega_(1))/(2)` respectively in same direction. Now one disc is placed the other disc co-axially. The change in kinetic energy of the system is :

A

`-(I_(1)omega_(1)^(2))/(24)`

B

`-(I_(1)omega_(1)^(2))/12`

C

`3/8 I_(1)omega_(1)^(2)`

D

`1/6 I_(1)omega_(1)^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
From conservation of Angular momentum about the axis, Li = Lf
`rArr I_(1)omega_(1) = I_(1)/2 xx omega_(1)/2 =(I_(1) + I_(1)/2) omega_(f) rArr omega_(f) = (5/4 I_(1)omega_(1))/(3/2 I_(1)) = 5/6 omega_(1)`
`therefore E_(f) - E_(i) =1/2(I_(1) + I_(1)/2) omega_(f)^(2) -[1/2 I_(1)omega_(1)^(2) + 1/2(I_(1)/2)(omega_(1)/2)^(2)]=1/2 xx (3I_(1))/2 (5/6 omega_(1))^(2) -9/16 I_(1)omega_(1)^(2) = -(I_(1)omega_(1)^(2))/24`
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