A smooth wire of length 2πr is bent into a circle and kept in a vertical plane. A bead can slide smo

Q: Why does the normal reaction act towards the center of the circular wire?

For a smooth circular wire, the normal force is always perpendicular to the tangent at any point. Geometrically, the perpendicular to a circle's tangent always passes through its center.

Q: Why is the radius of the circular path $r/2$ and not $r$?

The bead rotates around the vertical axis $AB$, not the center of the ring $O$. The radius of its horizontal circular path is its perpendicular distance from the axis of rotation, which is given as $r/2$.

Q: Why doesn't the bead slide down due to gravity?

The vertical component of the normal reaction ($N \cos\theta$) perfectly balances the downward gravitational force ($mg$), keeping the bead in vertical equilibrium.

Q: What provides the centripetal force for the bead's circular motion?

The horizontal component of the normal reaction ($N \sin\theta$) acts towards the axis of rotation, providing the exact centripetal force required to keep the bead moving in a circle.

Q: What would happen if the angular velocity $\omega$ was increased?

If $\omega$ increases, the required centripetal force increases. To provide this, the bead would slide further up the wire, increasing the angle $\theta$ and the horizontal component of the normal force.

Question

A smooth wire of length $2\pi r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter $AB$, as shown in figure, the bead is at rest with respect to the circular ring at position $P$ as shown. Then, the value of $\omega^2$ is equal to

Options:

(A) $\frac{\sqrt{3}g}{2r}$

(B) $2g / (r\sqrt{3})$

(C) $(g\sqrt{3}) / r$

(D) $2g / r$

Sigma Percentile · Accepted Answer

The problem presents a fascinating scenario: a bead resting on a smooth circular wire that is rotating about its vertical axis. At first glance, it might seem like the bead should slide down due to gravity. However, the rotation introduces a dynamic equilibrium. Let's break down the physics step-by-step.

### Analyzing the Setup

When the circular wire rotates, the bead moves along with it in a horizontal circle. Because the bead is at rest relative to the wire, it is in a state of equilibrium in the vertical direction, while simultaneously undergoing uniform circular motion in the horizontal plane.

To understand this, we must first identify the real forces acting on the bead. There are only two:

1. The **gravitational force** ($mg$) acting vertically downwards.
2. The **normal reaction** ($N$) exerted by the smooth wire. Since the wire is circular, this normal force acts radially inward, directly towards the center of the circle $O$.

### The Geometry of the Problem

Before diving into the force equations, let's establish the geometry. The bead is located at a horizontal distance of $r/2$ from the vertical axis of rotation.

If we draw a line from the center $O$ to the bead $P$, it makes an angle $	heta$ with the vertical axis. We can form a right-angled triangle where the hypotenuse is the radius $r$ and the opposite side is $r/2$.

Using basic trigonometry:

$$ \sin	heta = \frac{	ext{Opposite}}{	ext{Hypotenuse}} = \frac{r/2}{r} = \frac{1}{2} $$

This tells us that the angle $	heta$ is exactly $30^\circ$.

### The Master Equations

Now, we resolve the normal reaction $N$ into its vertical and horizontal components. Because the normal force points towards the center, the angle it makes with the vertical is also $	heta$.

The vertical component is $N \cos	heta$ (acting upwards), and the horizontal component is $N \sin	heta$ (acting towards the axis of rotation).

Since the bead does not slide up or down the wire, the vertical forces must perfectly balance:

$$ N \cos	heta = mg $$

In the horizontal direction, the bead is moving in a circle of radius $r/2$. The horizontal component of the normal force provides the necessary **centripetal force**:

$$ N \sin	heta = m \left(\frac{r}{2}ight) \omega^2 $$

### Final Calculation

We now have a system of two equations. To eliminate the unknown normal reaction $N$, we divide the horizontal equation by the vertical equation:

$$ \frac{N \sin	heta}{N \cos	heta} = \frac{m (r/2) \omega^2}{mg} $$

This simplifies beautifully to:

$$ 	an	heta = \frac{r \omega^2}{2g} $$

We already determined that $	heta = 30^\circ$, and we know that $	an 30^\circ = \frac{1}{\sqrt{3}}$. Substituting this value into our equation:

$$ \frac{1}{\sqrt{3}} = \frac{r \omega^2}{2g} $$

Finally, rearranging the terms to solve for $\omega^2$, we get our final answer:

$$ \omega^2 = \frac{2g}{r\sqrt{3}} $$

This elegant result shows how the rotational speed is intrinsically linked to the geometry of the bead's position. If the ring were to rotate faster, the bead would naturally slide higher up the wire to find a new equilibrium!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Identifying Forces$

$Geometry of the Position$

$Resolving the Normal Reaction$

$Vertical Equilibrium$

$Horizontal Centripetal Force$

$Eliminating Normal Reaction$

$Substituting the Angle$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Dynamics of Circular Motion

Analyzing the Setup

The Geometry of the Problem

The Master Equations

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resolving the Normal Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vertical Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Centripetal Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eliminating Normal Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting the Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward