Topic 8 of 8 · Form 6
Circular motion and simple harmonic motion
Angular speed ω; v = rω; acceleration towards the centre = v²/r = rω².
Notes for this topic 1 file, in order
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Horizontal circular motion
Angular speed ω; v = rω; acceleration towards the centre = v²/r = rω².
Worked example: A conical pendulum: a 0.5 kg mass on a string of length 1 m moves in a horizontal circle with the string at 60° to the vertical. Find the tension and speed (g = 9.8).
Vertical: T cos 60° = 0.5g → T = 9.8 N
Horizontal: T sin 60° = mv²/r, with r = sin 60° = 0.866 m → 8.487 = 0.5v²/0.866 → v = 3.83 m s⁻¹
Vehicles on bends
Friction or the horizontal component of the reaction on a banked track provides the centripetal force.
Vertical circles
Use energy conservation for speed, and F = mv²/r at each point. A particle on a string completes the circle if the tension at the top is not negative (v² ≥ gr at the top).
Simple harmonic motion
ẍ = −ω²x; x = a sin(ωt + ε); period T = 2π/ω; v² = ω²(a² − x²); maximum speed aω; maximum acceleration aω².
- Simple pendulum (small angles): T = 2π√(l/g).
- Elastic string or spring: ω² = λ/(ml) about the equilibrium position.
Check yourself
- A particle moves in a circle of radius 2 m at 3 m s⁻¹. Find its acceleration.
- A particle performs SHM with amplitude 0.2 m and period π s. Find its maximum speed.
- Find the period of a simple pendulum of length 2.45 m (g = 9.8).
Show answers
- 9/2 = 4.5 m s⁻²
- ω = 2; vmax = 0.4 m s⁻¹
- 2π√0.25 = π ≈ 3.14 s
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