A cylinder of mass `m` is kept on the edge of a plank of mass 2m and length 12m, which in turn is kept on smooth ground. Coefficient of friction between the plank and the cylinder is 0.1. The cylinder is given an impulse, which imparts it a velocity `7m//s` but no angular velocity. find the time after which the cylinder falls off the plank. `(g=10m//s^(2))`

Answer

`2.25s`

Explanation

Initially the cylinder wil slip on the plank therefore kinetic friction will act between the cylinder and the plank.
`FBD` of cylinder

`a_(c)=(mumg)/m=-mug, a_(p)=(mumg)/(2m)=(mug)/2`
`alpha_(c)=(2mumgR)/(mR^(2))=(2mug)/R,v_(c)=v_(0)-mu"gt"`

`v_(p)=(mug)/2t, omega_(c)=(2mu"gt")/2`
Also `s_(c)=v_(0)t-(1/2)mu"gt"^(2),`
`s_(p)=(1/2)(mug)/2t^(2)`
`v_(0)-mu"gt"-2mugt=m"gt"/2implies =(2v_(0))/(7mug)=2s`
Putting the value of `t` we get `=(12v_(0)^(2))/(49mug),s_(p)=(v_(0)^(2))/(49mug)`
`s_c-s_p=(11v_0^2)/(49mug)=11m`
Hence, remaining distance `=12m-11m=1m`
Also, `v_(c)=5/7v_(0),v_(p)=(v_(0))/7impliesv_(c)-v_(p)=(4v_(0))/7=4m//s`
After the start of pure rolling, the velocities become constant because fricton vanishes. the time after the starts of pure rolling when the cylinder falls`=(1m)/(4m//s)=0.25s`
`:.` Total time `=2+0.25=2.25s`