A bob of mass m on a light string of length ℓ is given horizontal speed v₀ at the bottom of a vertical circle. At point P, where OP makes angle θ above the horizontal, the string just becomes slack. Find vₚ/v₀. Image
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Question

A bob of mass m on a light string of length ℓ is given horizontal speed v₀ at the bottom of a vertical circle. At point P, where OP makes angle θ above the horizontal, the string just becomes slack. Find vₚ/v₀. Image

Options

G1 Physics · Accepted Answer

Step 1: At the instant the string becomes slack, tension is zero. The inward component of gravity provides centripetal force: mg sinθ = mvₚ²/ℓ. Step 2: Therefore vₚ² = gℓ sinθ. Step 3: From bottom to P, the bob rises by h = ℓ(1 + sinθ). Step 4: Conservation of energy: (1/2)mv₀² = (1/2)mvₚ² + mgℓ(1 + sinθ). Step 5: Substitute vₚ²: v₀² = gℓ sinθ + 2gℓ(1 + sinθ) = gℓ(2 + 3sinθ). Step 6: Divide and take square root: vₚ/v₀ = √[sinθ/(2 + 3sinθ)]. Final answer: Option D.