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

Step 1: Set up the equation of motion for the bead.

Let $R$ be the radius of the quarter-circle wire, and $\theta$ be the angle made by the radius vector of the bead with the vertical.

Using conservation of mechanical energy, the speed $v$ of the bead after descending through an angle $\theta$ is given by:

$$v^2 = 2gR(1 - \cos\theta)$$

Step 2: Apply Newton's second law in the radial direction.

Let $N$ be the normal force exerted by the wire on the bead, directed radially outwards (away from the center of the circle). The component of gravity directed towards the center is $mg\cos\theta$.

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

Step 3: Express the normal force $N$ as a function of $\theta$.

Substitute the expression for $v^2$ into the radial equation of motion.

$$N = mg\cos\theta - \frac{m}{R}\left[2gR(1 - \cos\theta)\right]$$ $$N = mg(3\cos\theta - 2)$$

Step 4: Analyze the direction of the normal force.

The normal force $N$ (exerted by the wire on the bead) changes sign when $N = 0$.

$$3\cos\theta - 2 = 0$$ $$\cos\theta = \frac{2}{3}$$

• For $\theta < \arccos\left(\frac{2}{3}\right)$, we have $\cos\theta > \frac{2}{3}$, which means $N > 0$ (the wire exerts a radially outward force on the bead).
• For $\theta > \arccos\left(\frac{2}{3}\right)$, we have $\cos\theta < \frac{2}{3}$, which means $N < 0$ (the wire exerts a radially inward force on the bead).

Step 5: Determine the force applied by the bead on the wire.

By Newton's third law, the force applied by the bead on the wire is equal and opposite to the force applied by the wire on the bead.

$$F_{\text{bead on wire}} = -N$$

• Initially (when $\theta$ is small), $F_{\text{bead on wire}} < 0$, which corresponds to a radially inward force.
• Later (when $\theta$ is large), $F_{\text{bead on wire}} > 0$, which corresponds to a radially outward force.

Answer: The force applied by the bead on the wire is radially inward first, and radially outward later.