The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R > r. The disk rolls without slipping along the inside periphery of the ring. The spring follows the periphery of the ring and obeys Hooke's law. In equilibrium, the disk is at the bottom of the ring. For a small displacement, T = 2π/ω. The correct expression for ω is: Image
Options

Question

The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R > r. The disk rolls without slipping along the inside periphery of the ring. The spring follows the periphery of the ring and obeys Hooke's law. In equilibrium, the disk is at the bottom of the ring. For a small displacement, T = 2π/ω. The correct expression for ω is: Image

Options

G1 Physics · Accepted Answer

For rolling without slipping, v = rωᵣ and the total kinetic energy is K = (1/2)mv² + (1/2)(mr²/2)(v²/r²) = (3/4)mv² = (1/2)(3m/2)v². Hence the effective mass for translational coordinate s is 3m/2. For a small displacement, gravitational potential energy ≈ (1/2)[mg/(R − r)]s², and spring potential energy = (1/2)ks². Therefore ω² = [mg/(R − r) + k]/(3m/2) = (2/3)[g/(R − r) + k/m]. So option A is correct.