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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.

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.

Grade:11

3 Answers

vikas askiitian expert
509 Points
11 years ago

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                     

Muskan
11 Points
6 years ago
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
Kushagra Madhukar
askIITians Faculty 629 Points
2 years ago
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

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