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.

Explanation

Initially the cylinder will slip on the plank, therefore kinetic energy friction will act between the cylinder and the plank.
`a_(c)=(mumg)/(m)=-mug`
`a_(P)=+(mumg)/(2m)=+(mu)/(2)`
`alpha_(C)=((mumg)(R))/((mR^(2)//2))=+(2mumg)/(R)`
For pure rolling
`v_(P)=v_(C)-Romega_(c)`
`therefore(mu)/(2)t=v_(0)-mugt-(R)((2mug)/(R))(t)`
`thereforet=(v_(0))/(3.5mug)=(7)/(3.5xx0.1xx10)=2s`
`therefores_(C)-s_(P)=v_(0)t-(1)/(2)xx(mug)/(t^(2))-(1)/(2)((mug)/(2))(t^(2))`
`=(7xx2)-(1)/(2)(0.1)(10)(4)-(1)/(2)((0.1xx10)/(2))(4)=11m`
Also, `v_(C)-V_(P)=(v_(0)-mugt)-((mug)/(2))(t)`
`=7-0.1xx10xx2-(0.1xx10xx2)/(2)=4m//s`
Hence the remaining distance `(12-11=1m)` is travelled in a time
`t'=(1.0)/(4)=0.25s`
`therefore` total time `=2+0.25=2.25s`