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. always radially outwards
B. always radially inwards
C. raidally outwards initially and inwards later
D. radially inwards initially and radially outwards later

Answer

D

Explanation

As the bead is moving in the circular path
`:.` `mg cos theta-N=(mv^2)/(R)`
`:.` `N=mg cos theta - (mv^2)/(R) ` …(1)
By energy conservation, `1/2mv^2=mg[R-Rcostheta]`
`:.` (v^2)/(R)=2g(1-costheta)` ...(2)
From (1) and (2)
`N=mg cos theta-m[2g-2gcos theta]`
`N=mg cos theta-2mg+2mgcostheta`
`N=3mg cos theta-2mg`
`implies N=mg(3costheta-2)`
Clearly N is positive (acts radially outwards) when `costhetagt2/3`
Similarly, N acts radially inwards if `cos thetalt2/3`