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 to the wire is

• A. Always radially outwards
• B. Always radially inwards
• C. Radially outwards initially and radially inwards later
• D. Radially inwards initially and radially outwards later

Answer 1

The correct option is D

Radially inwards initially and radially outwards later

Step 1. Given Data,

A wire is bent in the form of a quarter of a circle.

The wire is fixed vertically on the ground as shown in the above figure.

We have to calculate the force, as the bead moves from $A$ to $B$,

Step 2. Free body diagram of the system.

As the bead moves in a circular path

From the free-body diagram of the system, we get

$mg\cos \theta -N=\frac{m{v}^2}{R}\cdots \left(1\right)$

Where,

$\frac{m{v}^2}{R}$is the centripetal force,

$N$is the normal force of the bead,

$m$is the mass of the bead,

Step 3. From the law of conservation of energy,

From the law of conservation of energy, we can say that the loss of potential energy is equal to the gain in kinetic energy

$mgR-mgR\cos \theta =\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{2gR\left(1-\cos \theta \right)}$

Putting this value of $v$ in equation $\left(1\right)$

$$\begin{gathered} mg\cos \theta -N=\frac{m{\left(\sqrt{2gR\left(1-\cos \theta \right)}\right)}^2}{R} \\ \Rightarrow mg\cos \theta -N=\frac{m\left(2gR\left(1-\cos \theta \right)\right)}{R} \\ \Rightarrow mg\cos \theta -N=2mg\left(1-\cos \theta \right) \end{gathered}$$

$$\begin{gathered} \Rightarrow N=2mg\cos \theta +mg\cos \theta -2mg \\ \Rightarrow N=3mg\cos \theta -2mg \\ \Rightarrow N=mg\left(3\cos \theta -2\right) \end{gathered}$$

Force on the bead, $N=mg\left(3\cos \theta -2\right)$

When $\theta =0°$,$N$ is negative. So, it will be in the radially inward direction.

When $\theta$ is large, the value of $N$ is positive. So, it will be in the radially outward direction.

So, the direction of the force on the bead is radially inwards initially and radially outwards later.

Hence, the correct option is$\text{'}D\text{'}$.