The conical pendulum and horizontal circles

Resolving

With the string of length l at angle θ to the vertical, the particle moves in a circle of radius r = l sin θ.

• Vertically: T cos θ = mg (no vertical acceleration).
• Horizontally (towards the centre): T sin θ = mrω² = m(l sin θ)ω².

Dividing by sin θ gives T = mlω², and then cos θ = g/(lω²).

Variations

• On a table: if the particle moves on a smooth table, the table pushes up with R, so R + T cos θ = mg. It stays on the table while R ≥ 0.
• Elastic string: the tension is also λx/l and the string length is l + x, so the radius depends on x.
• Two strings: resolve with both tensions, and use the geometry to find the angles.

Examiner advice

Pearson's June 2025 report on a horizontal circle question says errors included leaving a term out of the vertical equation and using the length of the string instead of the radius of the circle. Draw the cone and label r.

Worked example

A particle of mass 0.3 kg on a light inextensible string of length 0.8 m moves in a horizontal circle with angular speed 5 rad/s. Find the tension and the angle the string makes with the vertical.

1. T = mlω² = 0.3 × 0.8 × 25 = 6 N.
2. cos θ = mg/T = 2.94/6 = 0.49.

Answer: T = 6 N, θ = 60.7°

Check yourself

1. A particle of mass 0.5 kg on a string of length 1.2 m moves as a conical pendulum with angular speed 4 rad/s. Find the tension.
   Show answer

   9.6 N. T = mlω² = 0.5 × 1.2 × 16 = 9.6 N.

2. In the worked example, find the angle between the string and the vertical, to 1 decimal place.
   Show answer

   60.7 °. cos θ = 0.49, θ = 60.7°.

3. In the worked example, find the radius of the circle, to 3 significant figures.
   Show answer

   0.697 m. r = 0.8 sin θ = 0.697 m.

4. A conical pendulum has a string of length 0.8 m. The particle moves in a horizontal circle with the string inclined to the vertical. Show that ω must exceed a certain value, and find that value.
   Show answer

   3.5 rad/s. cos θ = g/(lω²) must be less than 1, so ω² > g/l = 12.25 and ω > 3.5 rad/s.

5. A particle of mass 0.4 kg is attached by a string of length 1 m to a point 0.6 m above a smooth horizontal table. It moves on the table in a circle of radius 0.8 m with angular speed 3 rad/s. Find the force exerted by the table, to 3 significant figures.
   Show answer

   1.76 N. T = mlω² = 3.6 N (from T sin θ = m(0.8)ω² with sin θ = 0.8). Vertically R + T cos θ = mg, with cos θ = 0.6: R = 3.92 - 2.16 = 1.76 N.

6. For the same particle, find the greatest angular speed for which it stays on the table, to 3 significant figures.
   Show answer

   4.04 rad/s. R = 0 when ml ω² cos θ = mg, so ω² = 9.8/(1 × 0.6) and ω = 4.04 rad/s.

7. A particle of mass 0.2 kg is attached to a fixed point by an elastic string of natural length 0.5 m and modulus 20 N, and moves as a conical pendulum with angular speed 6 rad/s. Find the stretched length of the string, to 3 significant figures.
   Show answer

   0.610 m. T = m(0.5 + x)ω² and T = 20x/0.5. So 7.2(0.5 + x) = 40x, x = 3.6/32.8, length = 0.610 m.

8. In a conical pendulum with string length l at angle θ to the vertical, what is the radius of the circle?
   • l
   • l sin θ
   • l cos θ
   • l tan θ
   Show answer

   l sin θ. The radius is the horizontal distance from the axis: l sin θ.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M3
• Pearson Edexcel IAL Mechanics M3 (WME03) question paper, June 2025: front-cover instruction on g and accuracy
• Pearson Edexcel IAL Mechanics M3 (WME03) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.