A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on 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

Options

A. (a) always radially outwards.
B. (b) always radially inwards.
C. (c) radially outwards initially and radially inwards later.
D. (d) radially inwards initially and radially outwards later.

Answer

D

Explanation

`mgR(1-costheta)=1/2mv^2`

`:. (mv^2)/(R)=2mg(1-costheta)`
`:. N=mg cos theta-(mv^2)/(R)`
`=mg cos theta-2mg(1-cos theta)`
`=mg cos theta+2mg cos theta-2mg`
`=3mg cos theta-2mg=3mg( cos theta-2/3)`
If `cos thetagt2/3N` is positive
`:.` wire will exert force radially outward. So bead will exert force radially inward initially and radially outward later.