As there is no slipping at any point, the velocity of points `A and B` on the clinder are equal to the respective velocities of planks, Fig.
i.e., `upsilon_(A) = 10 m//s`
`upsilon_(B) = 4m//s`
Now, `upsilon_(C) = (upsilon_(A) + upsilon_(B))/(2) = (10 + 4)/(2) = 7 m//s`
If `omega` is angular velocity of cylinder, then
`omega = (upsilon_(A) - upsilon_(B))/(2r) = (10 - 4)/(2 xx 0.1) = 30 rad//s`
Kinetic energy of cylinder `= K_(t) + K_(r )`
`K = (1)/(2)m upsilon_(C)^(2) + (1)/(2)I_(C) omega^(2)`
`= (1)/(2)m upsilon_(C)^(2) + (1)/(2)((mr^(2))/(2))omega^(2)`
`= (1)/(2)m upsilon_(C)^(2) + (1)/(4)mr^(2) omega^(2)`
`= (1)/(2) xx 2 xx 7^(2) + (1)/(4) xx 2 xx 0.1^(2) xx 30^(2)`
`K = 49 + 4.5 = 53.5 J`