A cylinder of mass `m` is kept on the edge of a plank of mass `2m` and length `12 m`, 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 `7 ms^(-1)` but no angular velocity. Find the time after which the cylinder falls off the plank.

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`