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Question #42492

Difficulty: 8/10 | 🟢 Free

The Question

A smooth circular hoop of mass $m$ and radius $R$ is constrained to rotate freely without friction about its vertical diameter. A small bead, also of mass $m$ , can slide without friction along the hoop. The moment of inertia of the hoop about its vertical diameter is $\frac{1}{2}mR^2$ . Initially, the bead is at an angle $\theta = 60^\circ$ from the lowest point, and the entire system rotates freely with an angular velocity $\omega_0 = \sqrt{\frac{2g}{R}}$ . In this state, the bead is in stable equilibrium and at rest relative to the hoop. The bead is then given a sudden velocity along the hoop towards the lowest point. What is the minimum initial kinetic energy of the bead, measured in the rotating frame of the hoop, required for it to reach the lowest point ( $\theta = 0$ )? (Assume $g$ is the acceleration due to gravity)

Options

A.
$\frac{11}{8} mgR$
B.
$\frac{1}{4} mgR$
C.
$\frac{17}{8} mgR$
D.
$\frac{3}{4} mgR$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Equilibrium Stability|Angular Momentum conservation|Rigid Body Constraints|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze the stability and energetics of a rigid body system with a shifting mass distribution (a variable moment of inertia constraint).

The primary trap is assuming that the angular velocity $\omega$ remains constant during the bead's motion, which would lead to an incorrect application of the centrifugal potential $V_{\text{rot}} = -mgR\cos\theta - \frac{1}{2}m\omega^2 R^2 \sin^2\theta$ (yielding the trap Option B).

Because the system rotates freely, angular momentum $L$ is conserved, not $\omega$ .

Students must construct the correct effective potential energy $V_{\text{eff}}(\theta) = V(\theta) + \frac{L^2}{2I(\theta)}$ and equate the potential well depth to the required kinetic energy to cross the unstable boundary at the lowest point.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Work, Energy, and Power - Class 11 Part 1
Concept	Equilibrium Stability
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free