IIT JEE , AIEEE Forum - Physics , Mechanics

pavneet

A solid sphere of mass M and radius r is lying on a rough horizontal plane.a constant force F=4Mg acts vertically at point P such that OP makes 60' with horizontal(where O is centre of sphere).find minimum value of coeffecient of friction (u) so that the sphere starts pure rolling

a) 4/7 b)3/7 c)2/7 d)2/5

pavneet

i have asked from many people but no-one gave the right answer

krishna.gopal

I am assuming that P is above O and F acts vertically downwards)

Let f is the limiting frictional force required for pure rolling.
a = f/M

=(fR+FR/2)/I=a/R=f/(MR)

Thus for pure rolling

(fR+4MgR/2)/I=f/(MR)

Solve this to get f

Normal reaction(N) = 5Mg

so minimum coefficient of friction = f/N

solve this to get coefficient of friction

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