A solid cylinder of mass m and radius R rest on a plank of mass 2m lying on a smooth horizontal surface. String connecting cylinder to the plank is passing over a massless pulley mounted on a movable light block B and the friction between the cylinder and the plank is sufficient to prevent slipping. if the block B is pulled with a constant forceF, the acceleration of the plank is `(2F)/(xm)` then calculate x.

Explanation

For the cylinder, `(F)/(2)-f=ma_(1)` (1)
Taking moment of forces about C

fR=la
`rArrf=(mRalpha)/(2)` (2)
`because 1=mR^(2)//2`
for the plank, `(F)/(2)+f=(2m)a_(2)` (3)
And at the point of contact P, the acceleration of the two bodies must be same

`a_(1)-Ralpha=a_(2)` (4)
From 1 and 3, 2 and 4
`F=(ma_(1)+2ma_(2))` and
`(F)/(2)+(mRalpha)/(2)=2ma_(2)`
Subsitituting value of `alpha` and solving for `a_(1)` and `a_(2)` gives,
Accelecration of plank `a_(2)=(2F)/(7m)`