Q.2. A plank of length 20 m and mass 1 kg is kept on a horizontal smooth surface. A cylinder of mass 1 kg is kept near one end of the plank. The coefficient of friction between the two surfaces is 0.5. Plank is suddenly given a velocity 20 m/s towards left. Find the time which plank and cylinder separate.
velocity of center of mass is v & angular velocity of sphere is 0 initially...
friction will act in backward direction ....now
f = kmg (f = friction force)
ma = kmg
a=-kmg (a is retardation of center of mass)
now , torque = fR = I(alfa) (alfa = angular accleration)
alfa = fR/I
= 5kg/2R (Isphere = 2MR2/5)
now after time t let velocity of ceneter of mass is V then
V = U + at (initial linear velocity is v)
V = v - kgt ............1
let at this time angular velocity os W then
W = Wo + (alfa)t (initial angular velocity is 0)
W = 5kgt/2R ............2
now if pure rolling has started then V = WR
so , v - kgt = 5kgt/2
t = 2v/7kg
t=40/7 sec
Regards