A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half-angle θ . Find the tension in the chain if it rotates with a constant angular velocity ω about a vertical axis coinciding with the symmetry axis of the cone.

Question

A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half-angle θ . Find the tension in the chain if it rotates with a constant angular velocity ω about a vertical axis coinciding with the symmetry axis of the cone., R ⁡ ω 2 + g cot θ m ⁡ 2 π, R ⁡ ω 2 - g cot θ m ⁡ 2 π, R ⁡ ω 2 + g cot θ m ⁡ π, R ⁡ ω - g cot θ m ⁡ 2 π

MARKS App · Accepted Answer

Every element of chain moves on a circular path. So, every element of the chain experiences centripetal force. We consider a small element Δ m on the chain making angle Δθ at the centre ΔN sin θ = Δmg ...... (i) Net force towards centre is 2 T sin Δ θ 2 - Δ N cos θ = Δ mR ω 2 Δ N ⁡ cos θ = 2 T ⁡ sin Δ θ 2 - Δ mR ω 2 .......... (ii) From eqn.(i) and (ii) we get cos θ = 2 T sin Δ θ 2 - Δ mR ω 2 Δ mg But Δ θ is very small. ∴ sin Δ θ 2 = Δ θ 2 ∵ cot θ = T Δ θ - Δ mR ω 2 Δ mg Here Δm makes an angle Δθ at the centre. The total mass of chain is m. ∴ The mass of chain per unit angle is λ = m ⁡ 2 π ∴ Δ m ⁡ = λ Δ θ = m ⁡ 2 π Δ θ = m ⁡ Δ θ 2 π ∴ cot θ = T Δ θ − m Δ θ 2 π R ω 2 m 2 π Δ θ g ∴ T ⁡ = R ⁡ ω 2 + g cot θ m ⁡ 2 π

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