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.

Dear Student,
 

 

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

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