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debadutta mishra Grade: 11
        

A solid sphere of mass m and radius R is placed on a plank of equal mass, which lies in a smoth horizontal surface.The sphere is given a sharp impulse as a result it  stats slidding with velocity v.Find the time taken by the sphere to start pure roliing on the plank.the coefficient of friction between plank and sphere is k.

6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

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                     

6 years ago
Muskan
12 Points
										Initially velocity of plank is 0 and of disc is v.  friction will act in backward direction ....nowf = kmg                           (f = friction force) ma = kmga = - kg            (a is retardation of center of mass)now , torque = fR = I(alfa)                (alfa = angular accleration)alfa = fR/I alfa = 5gk/2R                        (I sphere = 2MR2/5)On plank friction will act on forward direction hence acceleration is in forward direction. f = makmg = ma `a` of plank is kg Let velocity of plank at time t is V1 v = u + atV1 = kgtSimilarly velocity of disc at time t is V2 = v - kgtAnd W (angular velocity) at time t is W = 2kgt/rSo velocity of contact point is  = (v - kgt) - (2kgt/r)×r = v - 3kgtFor pure rolling kgt = v - 3kgt4kgt = vt = v/4kg
										
9 months ago
Muskan
12 Points
										The above answer become so conjusted so I am sending this again.                                                                f = kmg (f = friction force).                                             ma = kmg.               a = - kg.           (a is retardation of center of mass).   torque = fR = I(alfa).   (alfa = angular acceleration)Alfa = fR/I                         alfa = 5gk/2R              (I sphere = 2MR2/5).                                                                 On plank friction will act on forward direction hence acceleration is in forward direction.                                                f = ma.            kmg = ma.                       a = kg                                    Let velocity of plank at time t is V1.          v = u + at.                       V1 = kgt.                             Similarly velocity of disc at time t is              V2 = v - kgt.                     And W (angular velocity) at time t is.                W = 2kgt/r.                    So velocity of contact point is = (v - kgt) - (2kgt) = v - 3kgt.                For pure rolling                                                                        kgt = v - 3kgt.                             4kgt = v.                                     t = v/4kg
										
9 months ago
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