Q5.40:

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{g / R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega = \sqrt{2g / R}$ ? Neglect friction.

Step 1: Establishing the Coordinate System and Forces

Consider a circular loop of radius $R$ rotating about its vertical diameter (z-axis) with angular frequency $\omega$. Let a bead of mass $m$ be located at an angular position $\theta$ measured from the vertical downward direction. The forces acting on the bead are:

• The gravitational force $mg$ acting vertically downward.
• The normal reaction $N$ from the wire, directed radially inward toward the axis of rotation or outward depending on the geometry.

The bead undergoes circular motion with a horizontal radius $r = R \sin\theta$. The centripetal acceleration is $a_c = \omega^2 r = \omega^2 R \sin\theta$.

Step 2: Equation of Motion

Resolving forces along the radial and vertical directions:

Vertical (y-axis): $N \sin\theta - mg = 0 \implies N \sin\theta = mg$ [Per Newton's Second Law]

Horizontal (radial): $N \cos\theta = m \omega^2 (R \sin\theta)$ [Centripetal Force Requirement]

Step 3: Deriving the Condition for Equilibrium at $\theta = 0$

Dividing the radial equation by the vertical equation:

$\frac{N \cos\theta}{N \sin\theta} = \frac{m \omega^2 R \sin\theta}{mg}$

$\cot\theta = \frac{\omega^2 R \sin\theta}{g}$

$\frac{\cos\theta}{\sin\theta} = \frac{\omega^2 R \sin\theta}{g} \implies \cos\theta = \frac{\omega^2 R \sin^2\theta}{g}$

Using $\sin^2\theta = 1 - \cos^2\theta$:

$g \cos\theta = \omega^2 R (1 - \cos^2\theta)$

$\omega^2 R \cos^2\theta + g \cos\theta - \omega^2 R = 0$

At the lowermost point ($\theta = 0$), $\cos\theta = 1$. Substituting $\cos\theta = 1$ into the force balance reveals the threshold: the bead stays at the bottom if the outward force generated by rotation is insufficient to lift the bead against gravity. The stability condition $\omega \leq \sqrt{g/R}$ emerges from the derivative of the effective potential energy $U(\theta) = -mgR\cos\theta - \frac{1}{2}m\omega^2R^2\sin^2\theta$.

Step 4: Solving for $\omega = \sqrt{2g/R}$

Given $\omega^2 = \frac{2g}{R}$, substitute this into the equation derived in Step 3:

$g \cos\theta = (\frac{2g}{R}) R (1 - \cos^2\theta)$

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

$2 \cos^2\theta + \cos\theta - 2 = 0$

Applying the quadratic formula $\cos\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$\cos\theta = \frac{-1 \pm \sqrt{1^2 - 4(2)(-2)}}{2(2)} = \frac{-1 \pm \sqrt{17}}{4}$

Since $\theta$ is between $0$ and $\pi/2$, $\cos\theta$ must be positive:

$\cos\theta = \frac{\sqrt{17} - 1}{4} \approx \frac{4.123 - 1}{4} \approx 0.78075$

$\theta = \arccos(0.78075) \approx 38.66^\circ$

Final Solution: The bead remains at the lowest point for $\omega \leq \sqrt{g/R}$. For $\omega = \sqrt{2g/R}$, the angle is $\theta = \arccos\left(\frac{\sqrt{17}-1}{4}\right) \approx 38.7^\circ$.