Motion in a vertical circle

Speed from energy

For a particle on a string of length a, or inside a smooth surface of radius a, with speed u at the lowest point, after turning through θ from the lowest point it has risen a(1 - cos θ), so

½mv² = ½mu² - mga(1 - cos θ).

At the top it has risen 2a.

Force towards the centre

Resolve along the radius, towards the centre:

• at angle θ from the lowest point: T - mg cos θ = mv²/a
• at the lowest point: T - mg = mv²/a
• at the top: T + mg = mv²/a
• with the string horizontal: T = mv²/a.

Only the radial component of the weight appears; the tangential component changes the speed.

Which forces?

A string can only pull (T ≥ 0). A rod can pull or push. A particle on the inside of a smooth surface feels a normal reaction towards the centre; on the outside, away from the centre. The next lesson uses these to decide whether the particle completes the circle.

Worked example

A particle of mass 0.5 kg on a light inextensible string of length 0.8 m moves in a vertical circle. Its speed at the lowest point is 7 m/s. Find its speed and the tension at the highest point.

1. Energy: v² = 49 - 2 × 9.8 × 1.6 = 17.64, so v = 4.2 m/s.
2. At the top: T + 0.5g = 0.5 × 17.64/0.8 = 11.025.
3. T = 11.025 - 4.9.

Answer: v = 4.2 m/s, T = 6.13 N

Check yourself

1. In the worked example, find the speed at the highest point.
   Show answer

   4.2 m/s. v² = 49 - 31.36 = 17.64, v = 4.2 m/s.

2. In the worked example, find the tension at the highest point, to 3 significant figures.
   Show answer

   6.13 N. T = 11.025 - 4.9 = 6.13 N.

3. In the worked example, find the tension at the lowest point, to 3 significant figures.
   Show answer

   35.5 N. T = mg + mu²/a = 4.9 + 30.625 = 35.5 N.

4. In the worked example, find the speed when the string is horizontal, to 3 significant figures.
   Show answer

   5.77 m/s. v² = 49 - 2(9.8)(0.8) = 33.32, v = 5.77 m/s.

5. In the worked example, find the tension when the string is horizontal, to 3 significant figures.
   Show answer

   20.8 N. T = mv²/a = 0.5 × 33.32/0.8 = 20.8 N.

6. In the worked example, find the tension when the string makes an angle of 60° with the downward vertical, to 3 significant figures.
   Show answer

   28.2 N. v² = 49 - 2(9.8)(0.8)(1 - cos 60°) = 41.16. T = mg cos 60° + mv²/a = 28.2 N.

7. A particle on a string is at the top of a vertical circle. Which equation is correct, with the positive direction towards the centre?
   • T - mg = mv²/a
   • mg - T = mv²/a
   • T = mg
   • T + mg = mv²/a
   Show answer

   T + mg = mv²/a. At the top both the tension and the weight point down, towards the centre.

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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

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