A bucket of mass `8 kg` is supported by a light rope wound around a solid wooden cylinder of mass `12 kg and radius 20 cm`, free to rotate about its axis. A man holding the free end of the rope with the bucket and the cylinder at rest initially, lets go the bucket freely downward in a well `50 m` deep. Neglecting friction, obtain the speed of bucket and angualr speed of the cylinder just before the bucket enters water. Take `g = 10 m//s^(2)`.

Explanation

Here, mass of bucket, `m_(1) = 8 kg`
mass of cylinder, `m_(2) = 12 kg`
radius of cylinder, `r = 20 cm =0.2m`
When the bucket enters water,
`P.E`. lost by bucket = Translation K.E. of bucket + rotational K.E. of cylinder
`m_(1) gh = (1)/(2)m_(1) v^(2) + (1)/(2)I omega^(2)`
`=(1)/(2)m_(1)v^(2) + (1)/(2) ((1)/(2) m_(2)r^(2))(v^(2))/(r^(2))`
`= (1)/(2)v^(2) (m_(1) + (1)/(2) m_(2))`
`8 xx 10 xx 50 = (1)/(2)v^(2) (8 + (1)/(2) xx12) = 7 v^(2)`
`v = sqrt((8 xx 10 xx 50)/(7)) = sqrt((4000)/(7)) = 23.9 m//s`
Angular speed of the cylinder before the bucket
enters water, `omega = (v)/(r ) = (23.9)/(0.20)= 119.5 rad//s`