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ARISTOTLE (0)

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a solid sphere of mass m and radius r is free to roll over an inclined surface of a wooden wedge of mass m.wedge lies on a smooth floor .when system is released from rest find frictional force acting between sphere and inclined surface.
    
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someone please solve yaar.........
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koi to solve karo.......
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nammi (53)

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frictional force will be equal to the torque acting o the system=Iw.where w is the angular speed
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plz solveit
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no one there to solve?????????????????
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elessar_iitkgp (2220)

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Let a' be the acceleration of the sphere's cm wrt the wedge. It will be directed down the plane. Let a be the acceleration of the wedge. Let be the angular acceleration  of  the sphere. Now as the sphere is observed in the reference frame of the wedge, there is a pseudo force ma on the sphere directed opposite to the acceleration of the wedge.


If  N be the normal force between the sphere and the wedge,


N=mg cos - ma sin and Nsin = ma


This gives a = gsincos/(1+sin2)


Now, for no slipping a' = R


a cos + g sin - (f/m) = (5f/2m)


Substitute the value of a and find f




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