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How do you analyse motion in a circle, both at constant speed and in a vertical plane?

Angular speed, the relationship between linear and angular speed, centripetal acceleration and force, horizontal circular motion such as the conical pendulum, and motion in a vertical circle.

A focused answer to the AQA A-Level Further Mathematics circular motion content, covering angular speed, the relationship between linear and angular speed, centripetal acceleration and force, horizontal circular motion such as the conical pendulum, and motion in a vertical circle.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Angular and linear speed
Centripetal force
Motion in a vertical circle

What this dot point is asking

AQA wants you to use angular speed and its link to linear speed, apply the formulae for centripetal acceleration and force, solve horizontal circular motion problems including the conical pendulum and banked tracks, and analyse motion in a vertical circle using energy conservation alongside Newton's second law.

Angular and linear speed

Centripetal force

The centripetal force is never an extra entry on a force diagram. You draw the real forces (tension, weight, normal reaction, friction), then set their resultant towards the centre equal to $mr\omega^2$ or $\frac{mv^2}{r}$ . This is just Newton's second law applied in the radial direction, with the acceleration always pointing inward. Getting this right means resolving forces along and perpendicular to the radius, not along the usual horizontal and vertical, whenever the geometry makes that simpler.

Conical pendulum

A particle of mass $m$ on a string of length $l$ moves in a horizontal circle, the string making angle $\theta$ with the vertical. Find the angular speed $\omega$ .

Resolve vertically

The particle has no vertical acceleration, so the vertical component of tension balances the weight: $T\cos\theta = mg$ .

Resolve horizontally towards the centre

The horizontal component of tension provides the centripetal force, with radius $r = l\sin\theta$ : $T\sin\theta = mr\omega^2 = m(l\sin\theta)\omega^2$ .

Eliminate the tension

Divide the horizontal equation by the vertical one: $\tan\theta = \frac{m l \sin\theta\,\omega^2}{mg}$ . The $\sin\theta$ cancels with $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , leaving $\frac{1}{\cos\theta} = \frac{l\omega^2}{g}$ .

State the result

Rearranging, $\omega^2 = \frac{g}{l\cos\theta}$ , so $\omega = \sqrt{\frac{g}{l\cos\theta}}$ . The angular speed increases as $\theta$ grows, which is why the particle rises as it spins faster.

Motion in a vertical circle

In a vertical circle the speed changes because gravity does work as the height changes, so you cannot treat the motion as uniform. The reliable method has two ingredients used together. First, conservation of energy $\frac{1}{2}mv^2 + mgh = \text{constant}$ relates the speed at any two points by their height difference. Second, Newton's second law in the radial direction at the point of interest relates the speed there to the tension or normal contact force. You almost always apply the radial equation at the top, the bottom, or the level of the centre, because those are where the geometry is simplest.

At the top of a full vertical circle on a string, both the weight and the tension point towards the centre, so the radial equation is $T + mg = \frac{mv^2}{r}$ . The string stays taut only while $T \geq 0$ , so the critical case is $T = 0$ , giving the minimum top speed $v^2 = gr$ . Feeding this into energy conservation over the height $2r$ between bottom and top shows the particle needs a launch speed of at least $\sqrt{5gr}$ at the bottom to get all the way round. For a particle on the inside of a rigid track or rod the condition is different: a rod can push as well as pull, so the body only needs enough energy to reach the top with $v \geq 0$ , giving the smaller requirement $u^2 \geq 4gr$ .

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20207 marksA particle of mass

0.5

kg is attached to one end of a light inextensible string of length

0.8

m. The other end is fixed and the particle moves in a horizontal circle, with the string making a constant angle of

30

degrees with the vertical (a conical pendulum). Taking

g = 9.8

m/s squared, calculate the tension in the string and the angular speed of the particle.

Show worked answer →

Resolve forces vertically and horizontally. The string tension $T$ acts along the string; the only other force is the weight $mg$ downwards.

Vertically the particle does not move, so $T\cos 30 = mg = 0.5 \times 9.8 = 4.9$ . Hence $T = \frac{4.9}{\cos 30} = \frac{4.9}{0.8660} = 5.66$ N.

The radius of the circle is $r = l\sin 30 = 0.8 \times 0.5 = 0.4$ m.

Horizontally the tension's component provides the centripetal force: $T\sin 30 = mr\omega^2$ . So $5.66 \times 0.5 = 0.5 \times 0.4 \times \omega^2$ , giving $2.83 = 0.2\omega^2$ and $\omega^2 = 14.15$ , so $\omega = 3.76$ rad/s.

Markers reward resolving vertically for $T$ , finding the radius, equating the horizontal component to $mr\omega^2$ , and solving for $\omega$ .

AQA 20226 marksA particle of mass

m

is attached to a string of length

r

and swings in a complete vertical circle. At the lowest point its speed is

u

. Using conservation of energy and the radial equation of motion, find the minimum value of

u

for the particle to complete the circle, in terms of

g

and

r

Show worked answer →

At the top of the circle the string is most likely to go slack, so apply the condition there. The minimum case is when the tension just reaches zero at the top.

Let the speed at the top be $v$ . At the top both the weight and the (zero) tension act towards the centre, so the radial equation is $mg = \frac{mv^2}{r}$ , giving $v^2 = gr$ .

By conservation of energy between the bottom and the top (a height gain of $2r$ ): $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(2r)$ .

Substituting $v^2 = gr$ : $\frac{1}{2}u^2 = \frac{1}{2}gr + 2gr = \frac{5}{2}gr$ , so $u^2 = 5gr$ and $u = \sqrt{5gr}$ .

Markers reward the zero-tension condition at the top, the radial equation giving $v^2 = gr$ , the energy equation over height $2r$ , and the result $u = \sqrt{5gr}$ .

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Sources & how we know this

AQA A-level Further Mathematics (7367) specification — AQA (2017)