A uniform circular disc has radius `R` and mass `m`. A particle also of mass `m` is fixed at a point `A` on the wedge of the disc as in fig. The disc can rotate freely about a fixed horizontal chord `PQ` that is at a distance `R//4` from the centre `C` of the disc. The line `AC` is perpendicular to `PQ`. Initially the disc is held vertical with the point `A` at its highest position. It is then allowed to fall so that it tarts rotating about `PQ`. Find the linear speed of the particle at it reaches its lowest position.
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Answer

`v = sqrt(5 gR)`

Explanation

Moment of inertia of system about `PQ`

`I = I_(disck) + I_(mass)`
`I = [(1)/(4) mR^(2) + m((R)/(4))^(2) ] +m ((5R)/(4))^(2)`
`I = (15 mR^(2))/(8) `....(1)
From Work Energy Theorem
`W_(g) = Delta KE`
`mg(R)/(4) xx 2 +mg[R + (R)/(4)] xx 2=(1)/(2) I omega^(2) rArr omega =sqrt((16g)/(5R))`
velocity of particle at lowest position
`V = ((5R)/(4))omega rArr V = (5R)/(4) sqrt((16 g)/(5 R)) rArr V =sqrt(5gR)`.