Vertical Circular Motion

Vertical circular motion is more complex than horizontal because the weight of the object has a component along the direction of motion. This means the speed changes throughout the circle (unless energy is continuously supplied), and the forces vary at different positions. Mastering this topic requires combining Newton's second law with energy conservation.

Forces at Different Points

Consider a ball of mass m on a string of length r being swung in a vertical circle. At any point, there are two forces:

Weight (mg) — always vertically downward
Tension (T) — always along the string, towards the centre

The net centripetal force is the resultant of these two forces directed towards the centre.

At the Top of the Circle

At the top, both weight and tension point downward (towards the centre):

$T_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{r}$

$T_{\text{top}} = \frac{mv_{\text{top}}^2}{r} - mg$

At the Bottom of the Circle

At the bottom, tension points upward (towards the centre) and weight points downward (away from centre):

$T_{\text{bottom}} - mg = \frac{mv_{\text{bottom}}^2}{r}$

$T_{\text{bottom}} = \frac{mv_{\text{bottom}}^2}{r} + mg$

At the Side (3 o'clock / 9 o'clock Position)

At the side, the weight acts perpendicular to the radius — it has no radial component:

$T_{\text{side}} = \frac{mv_{\text{side}}^2}{r}$

Weight only contributes tangentially (changing the speed), not radially.

Key Insight

The tension at the bottom is always greater than at the top (for the same speed) because at the bottom the tension must both provide the centripetal force and support the weight, whereas at the top the weight assists with the centripetal force.

flowchart TD
    A["Vertical Circle:\nForces at Each Position"] --> B["TOP\nT + mg = mv²/r\nBoth forces towards centre\nT is MINIMUM here"]
    A --> C["BOTTOM\nT − mg = mv²/r\nTension towards centre,\nweight away\nT is MAXIMUM here"]
    A --> D["SIDE (3 o'clock)\nT = mv²/r\nWeight perpendicular\nto radius, no radial\ncontribution"]
    B --> E["Critical: If T = 0,\nv_min = √(gr)"]
    C --> F["String most likely\nto break here"]

Minimum Speed at the Top

The critical condition occurs at the top of the circle when the tension just reaches zero. At this point, gravity alone provides the centripetal force:

$mg = \frac{mv_{\text{min}}^2}{r}$

$v_{\text{min}} = \sqrt{gr}$

If the speed at the top drops below √(gr), the object cannot complete the circle — it falls inward.

Worked Example 1

A ball is swung in a vertical circle of radius 0.60 m. What is the minimum speed at the top for the ball to complete the circle?

$v_{\text{min}} = \sqrt{gr} = \sqrt{9.81 \times 0.60} = \sqrt{5.89} = 2.43 \text{ m s}^{-1}$

Finding Speed at the Bottom from the Top Speed

Using conservation of energy (assuming no friction), we can relate the speeds at the top and bottom. The height difference between top and bottom is 2r (the diameter).

$\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)$

$v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr$

Worked Example 2

For the ball above (r = 0.60 m) travelling at exactly the minimum speed at the top (2.43 m s⁻¹), what is its speed at the bottom?

$v_{\text{bottom}}^2 = 2.43^2 + 4 \times 9.81 \times 0.60 = 5.89 + 23.54 = 29.43$ $v_{\text{bottom}} = 5.43 \text{ m s}^{-1}$

And the tension at the bottom (for a 0.20 kg ball): $T = \frac{mv_{\text{bottom}}^2}{r} + mg = \frac{0.20 \times 29.43}{0.60} + 0.20 \times 9.81 = 9.81 + 1.96 = 11.8 \text{ N}$

Compare this to the weight: 0.20 × 9.81 = 1.96 N. The tension at the bottom is about 6 times the weight.

Minimum Speed at the Bottom for a Complete Circle

Combining v_min(top) = √(gr) with v²_bottom = v²_top + 4gr:

$v_{\text{bottom, min}}^2 = gr + 4gr = 5gr$ $v_{\text{bottom, min}} = \sqrt{5gr}$

This is a very useful shortcut for exam problems.

The Bucket of Water

The classic demonstration: a bucket of water swung in a vertical circle. At the top, the water stays in the bucket (does not fall out) provided the centripetal acceleration exceeds g. This occurs when:

$\frac{v^2}{r} \geq g \implies v \geq \sqrt{gr}$

When v = √(gr) at the top, the water is in "free fall" — the bucket accelerates downward at g, the same as the water. The normal contact force between water and bucket is zero, and the water is weightless.

Roller Coasters

At the top of a roller coaster loop, the forces on a rider of mass m are:

Weight mg downward (towards centre)
Normal contact force N from the seat, also downward (towards centre)

$N + mg = \frac{mv^2}{r}$

$N = \frac{mv^2}{r} - mg = m\left(\frac{v^2}{r} - g\right)$

The rider feels the normal force as their "apparent weight." When N = 0, the rider feels weightless. When v²/r > g, the rider feels heavier than normal.

Worked Example 3

A roller coaster loop has radius 15 m. What speed must the car have at the top so that riders experience a normal force equal to their weight?

$N = mg \implies mg = m\left(\frac{v^2}{r} - g\right) \implies g = \frac{v^2}{r} - g \implies \frac{v^2}{r} = 2g$ $v = \sqrt{2gr} = \sqrt{2 \times 9.81 \times 15} = \sqrt{294} = 17.1 \text{ m s}^{-1}$

At this speed, riders feel pushed into their seat with a force equal to their weight (total "apparent weight" = 2mg, often described as "2g").

Car Going Over a Hill

A car travelling over a hill with a circular cross-section is another important vertical circle problem. At the top of the hill:

Weight mg acts downward (towards centre of curvature, which is below the road)
Normal force N acts upward (away from centre)

$mg - N = \frac{mv^2}{r}$

$N = mg - \frac{mv^2}{r} = m\left(g - \frac{v^2}{r}\right)$

The car loses contact when N = 0:

$v_{\text{max}} = \sqrt{gr}$

Worked Example 4

A car travels over a hill with radius of curvature 50 m at the top. At what speed does the car lose contact with the road?

$v = \sqrt{gr} = \sqrt{9.81 \times 50} = \sqrt{490.5} = 22.1 \text{ m s}^{-1}$

This is about 80 km/h. On a humped bridge, exceeding this speed means the car briefly leaves the road surface.

Forces at an Arbitrary Angle

At angle θ measured from the bottom of the circle, the component of weight along the radial direction is mg cos θ.

Position	Angle from bottom	Radial equation	Speed² (energy)
Bottom	0°	T − mg = mv²/r	v²_bottom
Side	90°	T = mv²/r	v²_bottom − 2gr
Top	180°	T + mg = mv²/r	v²_bottom − 4gr
General θ	θ	T − mg cos θ = mv²/r	v²_bottom − 2gr(1 − cos θ)

Common Exam Mistakes

Mistake 1: Using the same formula at top and bottom. At the top, weight ASSISTS the centripetal force (T + mg = mv²/r). At the bottom, weight OPPOSES it (T − mg = mv²/r). Mixing these up is the most common error.

Mistake 2: Forgetting that speed changes around the circle. The ball is fastest at the bottom and slowest at the top. Always use energy conservation to find the speed at each position.

Mistake 3: Confusing the roller coaster (track pushes inward at top) with the string (tension pulls inward at top). The physics is the same but the normal force direction at the top depends on the scenario.

Summary

At the top: T + mg = mv²/r (both forces point to centre)
At the bottom: T − mg = mv²/r (tension towards centre, weight away)
Minimum speed at top for complete circle: v_min = √(gr)
Minimum speed at bottom for complete circle: v_min = √(5gr)
Tension is greatest at the bottom of the circle
Use energy conservation to relate speeds at different heights: v²_bottom = v²_top + 4gr
Roller coaster N = m(v²/r − g) at the top; rider feels "weightless" when N = 0
Car over hill loses contact when v = √(gr)

A-Level Deep Dive: Vertical Circular Motion

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics covers circular motion, including the requirement to apply Newton's second law to objects moving in circular paths and to combine circular-motion analysis with energy conservation when the speed varies (refer to the official specification document for exact wording). Vertical circular motion sits at the intersection of three earlier strands — Topic 3 (Working as a Physicist, where vector decomposition is established), Topic 4 (Mechanics, where weight, normal contact and energy conservation are introduced) and Topic 5 (Materials and waves, only weakly relevant here). Within Topic 6 itself, vertical circular motion is the synoptic capstone: a question can require centripetal-force analysis at one point, energy conservation between two points and a critical-speed condition for the loop to close, all in a single multi-mark item. The Edexcel data and formulae booklet supplies $F = mv^2/r = m\omega^2 r$ and $v = \omega r$ , but it does not give the energy-conservation equation $\tfrac{1}{2}mv^2 + mgh = \text{const}$ — that result is assumed from Topic 4 and must be deployed without prompting.

Worked example with full mark scheme

Question (8 marks):

A small ball of mass $0.20\ \text{kg}$ is attached to a light inextensible string of length $0.80\ \text{m}$ and swung in a vertical circle. Take $g = 9.81\ \text{m s}^{-2}$ .

(a) Show that the minimum speed of the ball at the top of the circle, for the string to remain taut, is approximately $2.80\ \text{m s}^{-1}$ . (3)

(b) Using energy conservation, find the speed of the ball at the bottom of the circle when it is moving at this minimum speed at the top. (3)

Solution with mark scheme:

(a) Step 1 — identify the limiting condition. At the top of the circle, the string can pull but not push. The minimum speed occurs when the string tension is exactly zero, so weight alone supplies the entire centripetal force.

M1 — recognising that $T_\text{top} = 0$ at the limiting condition. Common error: candidates write $T = mg$ at the top, confusing "supports the weight" (vertical equilibrium) with "provides the centripetal force" (radially inward).

Step 2 — apply Newton's second law radially.

$mg = \frac{mv_\text{min}^2}{r} \implies v_\text{min} = \sqrt{gr}$

M1 — equation set up correctly with weight as the centripetal force.

Step 3 — substitute.

$v_\text{min} = \sqrt{9.81 \times 0.80} = \sqrt{7.848} = 2.80\ \text{m s}^{-1}$

A1 — value to 3 sf consistent with the printed answer.

(b) Step 1 — set up energy conservation. Take the bottom of the circle as the zero of gravitational potential energy. The ball rises through a height $2r$ between bottom and top.

$\tfrac{1}{2}mv_\text{bot}^2 = \tfrac{1}{2}mv_\text{top}^2 + mg(2r)$

M1 — energy conservation equation correctly written with $\Delta h = 2r$ , not $r$ . Cancelling $m$ :

$v_\text{bot}^2 = v_\text{top}^2 + 4gr$

M1 — algebraic rearrangement.

Step 2 — substitute.

$v_\text{bot}^2 = (2.80)^2 + 4 \times 9.81 \times 0.80 = 7.84 + 31.39 = 39.23\ \text{m}^2\text{s}^{-2}$

$v_\text{bot} = \sqrt{39.23} = 6.26\ \text{m s}^{-1}$

A1 — value to 3 sf.

(c) Step 1 — Newton's second law at the bottom. At the bottom, tension acts upward (towards the centre) and weight acts downward (away from the centre):

$T_\text{bot} - mg = \frac{mv_\text{bot}^2}{r}$

M1 — equation set up with tension and weight in opposing directions.

Step 2 — substitute.

$T_\text{bot} = m\!\left(\frac{v_\text{bot}^2}{r} + g\right) = 0.20\!\left(\frac{39.23}{0.80} + 9.81\right) = 0.20(49.04 + 9.81) = 11.8\ \text{N}$

A1 — value to 3 sf.

Total: 8 marks (M5 A3). Note that $T_\text{bot} \approx 11.8\ \text{N}$ is six times the weight $mg = 1.96\ \text{N}$ — the difference between top and bottom tensions is exactly $6mg$ when the loop is at the minimum-speed condition, a result candidates can quote as a check.

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (10 marks): A bucket of water of total mass $1.5\ \text{kg}$ is swung in a vertical circle of radius $0.95\ \text{m}$ on a rope. The water remains in the bucket at the top of the circle.

(a) Explain, in terms of forces, why the water remains in the bucket at the top. (3)

(b) Calculate the minimum angular speed $\omega_\text{min}$ for the water to remain in contact with the bucket at the top. (3)

(c) The bucket completes the loop with a speed at the top $30\%$ greater than the minimum. Find the tension in the rope at the bottom of the circle. (4)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1) — at the top, weight acts downward; the water needs a centripetal force directed downward (towards the centre).
B1 (AO1.2) — if the water's weight is sufficient to provide this centripetal force, no extra "support" from the bucket is needed.
B1 (AO2.1) — provided $v^2/r \geq g$ , the bucket is effectively "falling away" from the water at the same rate as gravity pulls the water, so the water stays in contact.

(b)

M1 (AO1.1) — recognise the critical condition is contact-force = 0, so $mg = m\omega^2 r$ .
M1 (AO1.2) — rearrange to $\omega_\text{min} = \sqrt{g/r}$ .
A1 (AO1.2) — $\omega_\text{min} = \sqrt{9.81/0.95} = 3.21\ \text{rad s}^{-1}$ .

(c)

M1 (AO2.1) — find $v_\text{top} = 1.30\sqrt{gr} = 1.30\sqrt{9.81 \times 0.95} = 3.97\ \text{m s}^{-1}$ .
M1 (AO2.1) — apply energy conservation to find $v_\text{bot}^2 = v_\text{top}^2 + 4gr = 15.76 + 37.28 = 53.04\ \text{m}^2\text{s}^{-2}$ .
M1 (AO1.2) — apply Newton's second law at the bottom: $T - mg = mv_\text{bot}^2/r$ .
A1 (AO1.2) — $T = 1.5(53.04/0.95 + 9.81) = 1.5(55.83 + 9.81) = 98.5\ \text{N}$ .

Total: 10 marks split AO1 = 5, AO2 = 5. A typical Topic 6 question pairs an explanation part (AO1-heavy) with two calculation parts that mix AO1 (apply familiar equation) and AO2 (sequence multiple equations).

Synoptic links

Connects to:

Topic 6 — Horizontal circular motion: the centripetal-force equation $F = mv^2/r$ is identical; only the source of the force changes (string tension only in horizontal motion; tension plus a component of weight in vertical motion). The horizontal "conical pendulum" treatment is the warm-up; vertical circular motion is the case where speed is no longer constant.
Topic 4 — Energy conservation: the speed varies around a vertical loop because gravity does work as height changes. Without the principle $\tfrac{1}{2}mv^2 + mgh = \text{const}$ from Year 1, vertical circular questions cannot be solved at all. Examiners often reward the explicit statement "assuming no energy is lost to air resistance".