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A wire of linear charge density lambda₁ and linear mass density lambda₂ is bent into the shape of a half-ring of radius R and hinged at its two endpoints A and B so that it can rotate freely about the axis AB. The assembly is placed in a uniform electric field E directed along the axis of symmetry (perpendicular to AB). Starting from the unstable equilibrium position, the assembly is given a tiny nudge. What is the maximum angular velocity of the half-ring during the subsequent motion?

1. (A) sqrt(16 * lambda₁ * E / (pi * lambda₂ * R))
2. (B) sqrt(pi * lambda₁ * E / (16 * lambda₂ * R))
3. (C) sqrt(8 * lambda₁ * E / (pi * lambda₂ * R))
4. (D) sqrt(4 * lambda₁ * E / (pi * lambda₂ * R))

Correct answer: (C) sqrt(8 * lambda₁ * E / (pi * lambda₂ * R))

Solution

The total charge on the half-ring is Q = lambda₁ * pi * R. The distance of the charge centroid from AB is x_cm = 2R/pi (standard result for a half-ring). The effective electric dipole moment about AB is p = Q * x_cm = 2 * lambda₁ * R². The PE change from unstable to stable equilibrium is Delta_U = 2 * p * E = 4 * lambda₁ * R² * E. The moment of inertia about AB is I = M * R² = lambda₂ * pi * R³. Energy conservation gives (1/2) * I * omega² = Delta_U, yielding omega = sqrt(8 * lambda₁ * E / (pi * lambda₂ * R)).

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