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Two wheels A and B are placed coaxially....

Two wheels A and B are placed coaxially. A has moment of inertia `I_(A)` and angular velocity `omega_(A)` while B is stationary. On clubbing them a clutch they move jointly by an angular velcoity `'omega'` then find the M.I of 'B'

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The phenomenon occurs in the absense of enternal torque, so applying law of COAM, we have
`because I_(A)omega_(A)=(I_(A)+_(B))omega rArr I_(B)=(I_(A)omega_(A)-I_(A)omega)/(omega)`
`=I_(A)((omega_(A))/(omega)-1)`
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MOTION-ROTATIONAL MOTION -Exercise - 3 ( Section-B )
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