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Motion in a Vertical Circle

A ball on a string moving in a vertical circle speeds up at the bottom and slows down at the top, because gravity does work on it. At every point the net force towards the centre must be mv²/r. At the bottom T = mg + mv²/r (largest); at the top T = mv²/r − mg (smallest). The string stays tight at the top only if v_top ≥ √(gr). Using energy conservation, this needs u ≥ √(5gr) at the bottom.

🎬 Step-by-step story

  1. A 0.5 kg ball on a 1 m string swings round in a vertical circle. The centre is O and the radius is r = 1 m. Watch it: fast at the bottom, slow at the top.
  2. Freeze the ball at the bottom. Tension T pulls up, weight mg pulls down. Their difference gives the pull to the centre: T − mg = mv²/r. So T = mg + mv²/r, the largest value.
  3. Freeze it at the top. Now T and mg both point down, towards the centre: T + mg = mv²/r. T can be zero if v² = gr. So the smallest speed at the top is √(gr).
  4. Link top and bottom with energy. Rising 2r turns some K into U: u² = v² + 4gr. With v² = gr at the top, u² = 5gr. So we need u ≥ √(5gr) ≈ 7.07 m/s at the bottom.
  5. Start with only 6 m/s at the bottom. Before reaching the top, tension drops to zero. The string goes slack and the ball leaves the circle, falling like a thrown stone.
  6. Free play: set the bottom speed u and radius r, press ▶ and predict first: full loop or not?

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why is the ball not moving at the same speed all round?

Gravity does work on it: negative when rising, positive when falling. So kinetic energy and speed change along the circle.

Why is tension greatest at the bottom?

At the bottom the string must hold up the weight AND supply the centre-ward pull mv²/r, and v is also the largest there.

Can tension be zero at the top and the ball still go round?

Yes, exactly when v² = gr. Then gravity alone gives the needed mv²/r. Any slower and gravity is too much.

Why √(5gr) and not √(gr) at the bottom?

The ball must climb 2r, which uses up 2mgr of kinetic energy, and it must still have v² = gr left at the top. So u² = gr + 4gr = 5gr.

What happens when the speed is too low?

Somewhere in the upper half T becomes zero, the string goes slack, and the ball falls along a parabola like a thrown stone.

Why speed changes in a vertical circle

In a horizontal circle the speed can stay the same. In a vertical circle, gravity does negative work when the ball goes up and positive work when it comes down. So the ball is fastest at the bottom and slowest at the top. This is non-uniform circular motion.

Take θ as the angle from the lowest point. Energy conservation gives: v² = u² − 2gr(1 − cos θ), where u is the speed at the bottom.

Tension at any point, bottom and top

Towards the centre, the net force must be mv²/r. At angle θ: T − mg cos θ = mv²/r, so T = mv²/r + mg cos θ.

Using u² = v² + 4gr: T_L − T_H = 6mg.

Minimum speeds: √(gr) at the top and √(5gr) at the bottom

  1. A string can pull but not push, so T ≥ 0 everywhere.
  2. T is smallest at the top: mv²/r − mg ≥ 0 → v_top ≥ √(gr).
  3. Energy from bottom to top (height 2r): ½mu² = ½mv² + mg(2r) → u² = v² + 4gr.
  4. Put v² = gr: u_min = √(5gr).
  5. At the side (θ = 90°): v² = u² − 2gr = 3gr, so v = √(3gr).

If a ball is on a rigid rod (which can push), it only needs to reach the top with v ≥ 0, so u ≥ √(4gr) = 2√(gr).

What if the speed is too low?

Try it at home

Tie a small plastic bottle with a little water to a string (outside, safely). Swing it in a vertical circle, first slowly, then fast. Slowly, the string goes loose near the top and water may splash. Fast, the string stays tight and no water falls. Feel the pull in your hand: strongest at the bottom.

Exam corner

CBSE asks: derive the minimum speed at the lowest and highest points (3–5 marks), find T at top and bottom, and "why does water not fall from a bucket" type reasoning. Remember T_L − T_H = 6mg and v at the side = √(3gr).

Key formulas and definitions

Worked examples

1. Find the minimum speed at the top of a vertical circle of radius 0.9 m (g = 10 m/s²).

Step 1: v_top = √(gr). Step 2: = √(10 × 0.9) = √9 = 3 m/s.

2. For r = 2 m, find the minimum speed at the lowest point.

Step 1: u = √(5gr). Step 2: = √(5 × 10 × 2) = √100 = 10 m/s.

3. A 0.2 kg stone on a 1 m string moves at 6 m/s at the bottom. Find the tension there.

Step 1: T = mg + mu²/r. Step 2: = 0.2 × 10 + 0.2 × 36 / 1 = 2 + 7.2. Answer: 9.2 N.

4. A 0.5 kg ball on a 1 m string has 8 m/s at the bottom. Find its speed and tension at the top.

Step 1: v² = u² − 4gr = 64 − 40 = 24, v ≈ 4.9 m/s. Step 2: T = mv²/r − mg = 0.5 × 24 − 5 = 7 N.

5. For the ball above, check T_L − T_H = 6mg.

Step 1: T_L = 0.5 × 64 + 5 = 37 N. Step 2: T_L − T_H = 37 − 7 = 30 N. Step 3: 6mg = 6 × 0.5 × 10 = 30 N. ✔

6. A bucket of water is whirled in a vertical circle of radius 1.6 m. What is the least number of rotations per minute at the top so that no water falls?

Step 1: v = √(gr) = √16 = 4 m/s. Step 2: ω = v/r = 2.5 rad/s. Step 3: rpm = ω × 60 / 2π ≈ 23.9. So at least about 24 rpm (at the top).

7. A string can bear 50 N. A 1 kg ball moves on it in a vertical circle of radius 1 m. Find the largest speed allowed at the bottom.

Step 1: T is largest at the bottom: mg + mu²/r ≤ 50. Step 2: 10 + u² ≤ 50 → u² ≤ 40. Answer: u ≤ 6.32 m/s. (This is less than √(5gr) = 7.07, so this string cannot give a full loop!)

Common mistakes

Practice quiz

1. Tension in the string of a vertical circle is greatest at the:
2. Minimum speed at the top for a string of radius r:
3. Minimum speed at the bottom for a full loop (string):
4. T_bottom − T_top equals:
5. If the string becomes slack in the upper half, the ball then moves along a:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is the minimum velocity at the lowest point of a vertical circle?

√(5gr) for a string. For a rigid rod it is √(4gr).

Is motion in a vertical circle uniform circular motion?

No. The speed changes because gravity does work, so it is non-uniform circular motion.

What is the difference between tensions at the bottom and top?

T_bottom − T_top = 6mg, for any speed that completes the loop.

Where this is taught

CBSE (India)Class 11Work, Energy and Power
England (GCSE, A level)Year 13Optional application 1 Mechanics (part 2)

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