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.

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

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

Dear student,

Please find the attached solution to your problem below.

Here, u = k (coefficient of friction)

Now, during pure rolling

Vs – wr = Vp

v – kgt – 5kgt/2r * r = kgt

or, v = 9/2 kgt

Hence, t = 2v / 9kg

 

Hope it helps.

Thanks and regards,

Kushagra