Physics 9702 · Paper 4

Vertical Circles

Turn the circle upright and gravity acts differently at every point. The interesting places are the top and the bottom.
Honest exam note: 2023–2025 Paper 4 asked only horizontal circles and orbits. This lesson covers the syllabus with 9702-style examples, so a vertical circle cannot surprise you.

Top and bottom of the loop

In a vertical circle the centre keeps moving relative to gravity. At the top, weight and tension pull the same way (both toward the centre); at the bottom they pull opposite ways, so the tension must be largest there.

The centre of the circle is below the object at the top and above it at the bottom, so the force equation changes sign around the loop.
At the top, weight and tension (or contact force) both point at the centre: $T + mg = \dfrac{mv^2}{r}$ .
At the bottom, they point opposite ways, and the tension must both hold the weight and supply the resultant: $T - mg = \dfrac{mv^2}{r}$ . The string is most likely to break at the bottom.

Worked example

Smallest case: minimum speed at the top

A truck loops the loop on a track of diameter 8.0 m. At the top of the loop the track pushes on it with zero force. Find its speed there (9702-style, from the coursebook).

Zero contact force at the top means gravity alone provides the resultant: $mg = \dfrac{mv^2}{r}$ .
$v = \sqrt{gr} = \sqrt{9.81 \times 4.0} = 6.3 \text{ m s}^{-1}$ .

Answer

v = 6.3 \text{ m s}^{-1}

Any slower, and gravity is more force than the circle needs: the truck leaves the track and falls.

Worked example

One change: the force at the bottom

A bucket of water (total mass 5.4 kg) swings in a vertical circle of radius 0.90 m at the minimum top speed. Taking the speed as roughly constant, find the force on the hand at the bottom (9702-style).

Minimum top speed: $v^2 = gr = 9.81 \times 0.90 = 8.8 \text{ m}^2\text{s}^{-2}$ .
At the bottom: $T = mg + \dfrac{mv^2}{r} = 5.4\,(9.81 + 9.81) = 106 \text{ N}$ .
With $v^2 = gr$ , the bottom force is exactly $2mg$ : twice the bucket's weight.

Answer

T \approx 106 \text{ N} \; (= 2mg)

Your turn— tap to reveal the worked answer (9702-style)

A ball of mass 50 g on a 1.50 m string is pulled back until it is 0.70 m above its lowest point, then released. Find its speed at the lowest point, and the tension there.

Energy chain (lesson 5.06): $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.70} = 3.7 \text{ m s}^{-1}$ .
Bottom of a circle: $T = mg + \dfrac{mv^2}{r} = 0.050\left(9.81 + \dfrac{3.7^2}{1.50}\right) = 0.95 \text{ N}$ .

Answer: 3.7 m s⁻¹; T = 0.95 N, almost double the 0.49 N weight.

Common mistake

At the bottom, the tension is not equal to the weight, even at constant speed. The object is accelerating upward (toward the centre), so

T = mg + mv^2/r > mg

. Writing T = mg is the standard error here.

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Vertical Circles

Turn the circle upright and gravity acts differently at every point. The interesting places are the top and the bottom.
Honest exam note: 2023–2025 Paper 4 asked only horizontal circles and orbits. This lesson covers the syllabus with 9702-style examples, so a vertical circle cannot surprise you.

Top and bottom of the loop

The centre of the circle is below the object at the top and above it at the bottom, so the force equation changes sign around the loop.
At the top, weight and tension (or contact force) both point at the centre: $T + mg = \dfrac{mv^2}{r}$ .
At the bottom, they point opposite ways, and the tension must both hold the weight and supply the resultant: $T - mg = \dfrac{mv^2}{r}$ . The string is most likely to break at the bottom.

Worked example

Smallest case: minimum speed at the top

A truck loops the loop on a track of diameter 8.0 m. At the top of the loop the track pushes on it with zero force. Find its speed there (9702-style, from the coursebook).

Zero contact force at the top means gravity alone provides the resultant: $mg = \dfrac{mv^2}{r}$ .
$v = \sqrt{gr} = \sqrt{9.81 \times 4.0} = 6.3 \text{ m s}^{-1}$ .

Answer

v = 6.3 \text{ m s}^{-1}

Any slower, and gravity is more force than the circle needs: the truck leaves the track and falls.

Worked example

One change: the force at the bottom

Minimum top speed: $v^2 = gr = 9.81 \times 0.90 = 8.8 \text{ m}^2\text{s}^{-2}$ .
At the bottom: $T = mg + \dfrac{mv^2}{r} = 5.4\,(9.81 + 9.81) = 106 \text{ N}$ .
With $v^2 = gr$ , the bottom force is exactly $2mg$ : twice the bucket's weight.

Answer

T \approx 106 \text{ N} \; (= 2mg)

Your turn— tap to reveal the worked answer (9702-style)

A ball of mass 50 g on a 1.50 m string is pulled back until it is 0.70 m above its lowest point, then released. Find its speed at the lowest point, and the tension there.

Energy chain (lesson 5.06): $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.70} = 3.7 \text{ m s}^{-1}$ .
Bottom of a circle: $T = mg + \dfrac{mv^2}{r} = 0.050\left(9.81 + \dfrac{3.7^2}{1.50}\right) = 0.95 \text{ N}$ .

Answer: 3.7 m s⁻¹; T = 0.95 N, almost double the 0.49 N weight.

Common mistake

At the bottom, the tension is not equal to the weight, even at constant speed. The object is accelerating upward (toward the centre), so

T = mg + mv^2/r > mg

. Writing T = mg is the standard error here.