Discussion Forum

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

Answer:

Explanation:

h= R -Rcos Ω

 Using conservation energy

$mgR(1-\cos\theta)=\frac{1}{2}mv^{2}$

  Radial force equation  is 

  $mg\cos\theta-N=\frac{1}{R}mv^{2}$

 = mg (  3$\cos\theta$-2)

Here, $N=mg \cos\theta-\frac{mv^{2}}{R}$

 N= 0 at   $\cos\theta=\frac{2}{3}$

 $\Rightarrow$    Normal force act radially outward on bead if cos θ  >2/3   and normal  force act radially inward on bead if 

 cos θ < 2/3

$\therefore$  Force ring is opposite to normal force on bead.