A pendulum bob of mass 1 kg carries charge +40 μC and is attached to a thread of length 0.9 m; the support point also carries charge +40 μC. Find the minimum horizontal speed u (m s⁻¹) imparted at lowest point to complete a vertical circle (g = 10 m s⁻²). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A pendulum bob of mass 1 kg carries charge +40 μC and is attached to a thread of length 0.9 m; the support point also carries charge +40 μC. Find the minimum horizontal speed u (m s⁻¹) imparted at lowest point to complete a vertical circle (g = 10 m s⁻²). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Coulomb force between fixed support charge and bob always points radially outward. With string length fixed, this force does no work. Step 2 — Bottom-to-top rise is 2ℓ, so energy needed to just reach the top is mg(2ℓ). Step 3 — Top radial-force equation is T + mg − Fₑ = mv²/ℓ. Here Fₑ = (9 × 10⁹)(40 × 10⁻⁶)²/0.9² ≈ 17.78 N > mg = 10 N, so string tension remains positive even if vₜₒₚ approaches zero. Step 4 — Minimum bottom kinetic energy only needs to match rise: (1/2)mu² = 2mgℓ, or u…