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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.

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.

10 years ago

Answers : (3)

vikas askiitian expert
509 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                     

10 years ago
Muskan
11 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
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
4 years ago
Kushagra Madhukar
askIITians Faculty
618 Points
							
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
6 months ago
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