Animated Solution for Physics - Work, Energy, and Power: 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

A Bead on a Wire

Imagine you are watching a tiny bead threaded onto a smooth, rigid wire bent into a perfect quarter-circle. The wire stands vertically, like a slice of a Ferris wheel. When you release the bead from the very top, gravity immediately takes over, pulling it down along the curve.

As the bead descends, two things happen simultaneously: it loses height, and it gains speed. But because it is forced to move in a circle, it requires a specific amount of inward force to keep it from flying off in a straight line. This is where the physics gets incredibly interesting. The wire must constantly adjust the force it exerts on the bead to maintain this circular path. Our mission is to figure out exactly how this force behaves.

Gaining Speed

To understand the forces, we first need to know how fast the bead is moving at any given point. Let's define the position of the bead by the angle $\theta$ it makes with the vertical.

As the bead drops through this angle $\theta$, it loses a vertical height $h$. By looking at the geometry of the quarter-circle, we can see that this height is the total radius $R$ minus the vertical projection $R\cos \theta$. Therefore, $h=R(1-\cos \theta )$.

Because the wire is frictionless, mechanical energy is perfectly conserved. The gravitational potential energy lost is entirely converted into kinetic energy. We can write this elegantly as:

Substituting our expression for $h$, we get:

We can rearrange this to isolate a very special term—the centripetal force requirement, $\frac{m{v}^2}{R}$:

Keep this equation safe; it is the key to unlocking the dynamics of the system.

Forces in a Circle

Now, let's freeze time and look at the forces acting on the bead at angle $\theta$. There are only two forces: gravity pulling straight down with magnitude $mg$, and the normal force $N$ from the wire.

Because the bead is moving in a circle, we must analyze the forces in the radial direction (towards the center of the circle). Gravity has a component pointing directly towards the center, which is $mg\cos \theta$.

Let's assume, for the sake of setting up our equation, that the wire pushes the bead radially outwards with a normal force $N$. The net force pointing towards the center must equal the required centripetal force. Therefore, Newton's Second Law gives us:

Solving for Normal Force

We now have two perspectives: the energy perspective giving us the required centripetal force, and the dynamics perspective giving us the actual forces providing it. Let's merge them by substituting our energy result into the force equation:

Now, it's just a matter of simple algebra to solve for the normal force $N$:

This is our master equation. It tells us exactly how hard the wire is pushing on the bead at any angle $\theta$.

When Does the Force Flip?

Look closely at the master equation: $N=mg(3\cos \theta -2)$. Notice that the value inside the parentheses can be positive, zero, or negative depending on the angle $\theta$.

Phase 1: The Initial Descent When the bead is near the top, $\theta$ is small, and $\cos \theta$ is close to 1. Specifically, as long as $\cos \theta >\frac{2}{3}$, the term $(3\cos \theta -2)$ is positive. This means our assumption was correct: $N$ is positive, so the wire is pushing radially outwards on the bead.

Why? Because near the top, the bead is moving slowly, so it doesn't need much centripetal force. However, the inward component of gravity ($mg\cos \theta$) is very strong. Gravity is actually pulling the bead inwards too hard. To prevent the bead from falling into the center, the wire must push it outwards.

Phase 2: The Lower Curve As the bead falls further, it speeds up tremendously, requiring a massive centripetal force. At the same time, the inward component of gravity ($mg\cos \theta$) gets weaker and weaker.

When $\cos \theta$ drops below $\frac{2}{3}$, the term $(3\cos \theta -2)$ becomes negative. This means $N$ is negative. The wire is now pulling radially inwards on the bead. Gravity is no longer strong enough to keep the fast-moving bead in a circle, so the wire has to step in and yank it towards the center.

The Final Trap

We have successfully determined the force on the bead. But wait! Read the question carefully. It asks for the force applied by the bead on the wire.

This is where Newton's Third Law comes in to save the day (or ruin it, if you aren't paying attention). Every action has an equal and opposite reaction.

- Initially, the wire pushes the bead outwards. Therefore, the bead pushes the wire radially inwards. - Later, the wire pulls the bead inwards. Therefore, the bead pulls the wire radially outwards.

So, the force applied on the wire is radially inwards initially, and radially outwards later. This perfectly matches option (D). Physics is beautiful when you follow the logic all the way through!