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Question:

A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on the ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is:

always radially outwards.

radially outwards initially and radially inwards later.

always radially inwards.

radially inwards initially and radially outwards later.

Solution:

mgR(1−cosθ)=1/2mv²
So the normal force, N = mgcosθ−mv²/R
So, when θ is small, N is +ve so it is radially inwards.
When the θ gets large, N gets -ve, so it is radially outwards.