A bead (m = 5 kg) slides on a fixed vertical circular ring of radius r = 20 cm and is connected by a spring to the topmost point. Natural spring length = r. Initially central angle OCA = 60°. At bottommost point the ring’s normal reaction becomes zero. If spring constant k = 100n N m⁻¹, find n. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A bead (m = 5 kg) slides on a fixed vertical circular ring of radius r = 20 cm and is connected by a spring to the topmost point. Natural spring length = r. Initially central angle OCA = 60°. At bottommost point the ring’s normal reaction becomes zero. If spring constant k = 100n N m⁻¹, find n. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Initial position satisfies ∠OCA = 60°. The initial spring length from topmost point O to A is r, equal to natural length, so initial elastic energy is zero. Step 2 — At lowest point B, spring length is 2r, extension r, so final stored spring energy is (1/2)kr². Gravitational drop is Δh = (3/2)r. Step 3 — Energy: (1/2)mv² + (1/2)kr² = mg(3r/2). At B, the spring tension points upward and gravity downward. Step 4 — Vanishing normal reaction gives mv²/r = kr − mg (radial inward upward). St…