Tension Is the Only Inward Force: Resolve It Vertically and Horizontally
When a mass moves in a horizontal circle on a string, the tension is the only force with a horizontal component, so it supplies the centripetal force. Resolving the tension gives the angle and the speed.
Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.
Conical pendulum and hanging bobs
Several masses on one string
Three equal masses on a string at distances ℓ, 2ℓ and 3ℓ from the centre, all with the same ω, need centripetal forces m ω² ℓ, 2 m ω² ℓ and 3 m ω² ℓ. The outer section carries only what the outermost mass needs, 3 m ω² ℓ, the middle section adds the middle mass's 2 m ω² ℓ to make 5 m ω² ℓ, and the inner section adds the innermost 1 m ω² ℓ to make 6 m ω² ℓ. From the outermost section inward the tensions are in the ratio 3 : 5 : 6.
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- ✓Conical pendulum: T cos θ = m g and T sin θ = m v²/r, so tan θ = v²/(r g)
- ✓On a rotating platform: tan θ = r ω²/g
- ✓On a horizontal table the tension alone supplies the centripetal force
- ✓Several masses on one string: each section carries the centripetal force of every mass beyond it
Do not give every section of the string the same tension. Each section must supply the inward force for all the masses outside it, so the tension is greatest near the centre and falls toward the end.
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