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(C) Let $R$ be the radius of the circular path and $	heta$ be the angle the bead makes with the vertical at any instant.By the law of conservation of

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radially outwards initially and radially inwards later

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(C) Let $R$ be the radius of the circular path and $	heta$ be the angle the bead makes with the vertical at any instant.By the law of conservation of energy,the kinetic energy gained by the bead is equal to the loss in potential energy:$\frac{1}{2}mv^2 = mgR(1 - \cos 	heta)$$v^2 = 2gR(1 - \cos 	heta)$The forces acting on the bead are gravity $(mg)$ and the normal force $(N)$ from the wire. The radial component of the net force provides the centripetal acceleration:$mg \cos 	heta - N = \frac{mv^2}{R}$$N = mg \cos 	heta - \frac{mv^2}{R}$Substituting $v^2$:$N = mg \cos 	heta - \frac{m(2gR(1 - \cos 	heta))}{R}$$N = mg \cos 	heta - 2mg + 2mg \cos 	heta$$N = mg(3 \cos 	heta - 2)$When $	heta$ is small (near $A$),$\cos 	heta$ is close to $1$,so $N = mg(3(1) - 2) = mg$,which is positive. $A$ positive normal force means the bead pushes the wire radially outwards,so the wire exerts a force radially inwards on the bead. By Newton's third law,the bead exerts a force radially outwards on the wire.As $	heta$ increases,$N$ decreases. When $\cos 	heta  \arccos(2/3) \approx 48.2^\circ$),$N$ becomes negative. $A$ negative normal force means the bead is pulled inwards by the wire,so the bead exerts a force radially inwards on the wire.Thus,the force the bead applies on the wire is radially outwards initially and radially inwards later.

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