A particle moves in a circle of radius 2\,m at ω = 3\,rad s^-1. Find its speed. [1 mark]

Find the minimum speed at the top of a vertical circle of radius 0.8\,m on a string for it to stay taut. [2 marks]

A car of mass 1000\,kg rounds a bend of radius 50\,m at 20\,m s^-1. Find the centripetal force required. [2 marks]

EnglandFurther MathsSyllabus dot point

How do you analyse motion in a circle at constant and varying speed?

Angular speed, acceleration towards the centre, motion in a horizontal circle, the conical pendulum, and motion in a vertical circle with energy conservation.

A focused answer to the Edexcel A-Level Further Mathematics Further Mechanics content on circular motion, covering angular speed, the acceleration towards the centre, motion in a horizontal circle and the conical pendulum, and motion in a vertical circle using conservation of energy.

Generated by Claude Opus 4.812 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 speed and centripetal acceleration
Horizontal circles and the conical pendulum
Vertical circles
Examples in context
Try this

What this dot point is asking

Edexcel Further Mechanics wants you to use angular speed and the acceleration directed towards the centre, analyse motion in a horizontal circle including the conical pendulum, and analyse motion in a vertical circle by combining Newton's second law along the radius with conservation of energy. The distinction between a string (which can only pull) and a rod (which can also push) is a frequent exam discriminator.

Angular speed and centripetal acceleration

Even when a particle moves in a circle at constant speed, it is accelerating, because the direction of its velocity is constantly changing. That acceleration always points towards the centre and has magnitude $\frac{v^2}{r}$ . By Newton's second law a net inward force of magnitude $\frac{mv^2}{r}$ is needed to produce it; this is not a new kind of force but the resultant of the real forces (tension, gravity, friction, normal reaction).

Horizontal circles and the conical pendulum

In a horizontal circle the height does not change, so the speed is constant and energy conservation is not needed. Resolve the forces into vertical and horizontal directions: the vertical components balance (no vertical acceleration), and the horizontal components provide the centripetal force.

Conical pendulum

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

Step 1: Identify the radius of the circular path

The bob traces a horizontal circle. The string of length $l$ at angle $\theta$ to the vertical provides the hypotenuse of a right-angled triangle, so the horizontal distance from the axis of rotation is:

r = l\sin\theta.

Step 2: Resolve forces vertically

There is no vertical acceleration, so the vertical component of the tension must exactly balance the weight of the bob.

T\cos\theta = mg.

Step 3: Resolve forces horizontally to give the centripetal equation

The horizontal component of tension provides the centripetal force needed to keep the bob moving in a circle. Substituting $r = l\sin\theta$ :

T\sin\theta = mr\omega^2 = m(l\sin\theta)\omega^2.

Dividing both sides by $\sin\theta$ (which is non-zero for a genuine conical pendulum):

T = ml\omega^2.

Step 4: Combine the two equations to find $\omega$

Substitute $T = ml\omega^2$ into the vertical equation $T\cos\theta = mg$ :

ml\omega^2\cos\theta = mg \implies \cos\theta = \frac{g}{l\omega^2}.

Final answer: $\cos\theta = \dfrac{g}{l\omega^2}$ . As $\omega$ increases, $\cos\theta$ decreases, meaning the string rises towards the horizontal. The bob can never reach the horizontal because $\cos\theta > 0$ for all finite $\omega$ .

Vertical circles

In a vertical circle the speed changes with height, so you need two tools together: conservation of energy to find the speed at a given point, and Newton's second law along the radius to find the tension or reaction there.

Speed and tension in a vertical circle

A particle of mass $m$ on a string of length $r$ passes the lowest point at speed $u$ . Find the speed and tension at the highest point, assuming the string stays taut.

Energy from bottom to top (height gained $2r$ ): $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(2r)$ , so $v^2 = u^2 - 4gr$ .

At the top both gravity and tension point downwards (towards the centre), so $T + mg = \frac{mv^2}{r}$ , giving $T = \frac{mv^2}{r} - mg = \frac{m(u^2 - 4gr)}{r} - mg = \frac{mu^2}{r} - 5mg$ .

For the string to be taut at the top, $T \ge 0$ , which requires $v^2 \ge gr$ , i.e. $u^2 \ge 5gr$ . The minimum speed at the bottom for a complete circle on a string is therefore $u = \sqrt{5gr}$ .

Examples in context

Circular motion draws on the energy methods of the work-energy-and-power dot point (conservation of energy is the engine for vertical-circle speeds) and on the force resolution of mechanics generally. The radial equation of motion is a direct application of Newton's second law with the centripetal acceleration substituted in. Banked tracks, satellites in orbit (where gravity supplies the centripetal force), and a bead on a rotating wire are all standard modelling contexts. Simple harmonic motion, studied via differential equations, is the projection of uniform circular motion onto a diameter, linking this dot point to the oscillation solutions of second-order equations.

Try this

Q1. A particle moves in a circle of radius $2\,\text{m}$ at $\omega = 3\,\text{rad s}^{-1}$ . Find its speed. [1 mark]

Cue. $v = r\omega = 2 \times 3 = 6\,\text{m s}^{-1}$ .

Q2. Find the minimum speed at the top of a vertical circle of radius $0.8\,\text{m}$ on a string for it to stay taut. [2 marks]

Cue. $v^2 = gr = 9.8 \times 0.8 = 7.84$ , so $v \approx 2.8\,\text{m s}^{-1}$ .

Q3. A car of mass $1000\,\text{kg}$ rounds a bend of radius $50\,\text{m}$ at $20\,\text{m s}^{-1}$ . Find the centripetal force required. [2 marks]

Cue. $F = \frac{mv^2}{r} = \frac{1000 \times 400}{50} = 8000\,\text{N}$ .

Exam-style practice questions

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

Edexcel 20197 marksA particle of mass

0.5\,\text{kg}

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

0.8\,\text{m}

. The other end is fixed and the particle moves in a horizontal circle as a conical pendulum, with the string at

30^\circ

to the vertical. Find the tension in the string and the angular speed. Take

g = 9.8\,\text{m s}^{-2}

Show worked answer →

Resolve vertically for the tension, then horizontally for the angular speed.

Vertically: $T\cos 30^\circ = mg = 0.5 \times 9.8 = 4.9$ (M1). So $T = \frac{4.9}{\cos 30^\circ} = \frac{4.9}{0.866} \approx 5.66\,\text{N}$ (A1).

The circle radius is $r = l\sin 30^\circ = 0.8 \times 0.5 = 0.4\,\text{m}$ (B1).

Horizontally: $T\sin 30^\circ = mr\omega^2$ (M1). So $5.66 \times 0.5 = 0.5 \times 0.4 \times \omega^2$ , giving $2.83 = 0.2\omega^2$ (A1), so $\omega^2 = 14.15$ and $\omega \approx 3.76\,\text{rad s}^{-1}$ (A1 A1).

Edexcel 20218 marksA particle of mass

m

is attached to the end of a light rod of length

r

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

u

. Show that the speed

v

at the highest point satisfies

v^2 = u^2 - 4gr

, and find the minimum value of

u

for the particle to complete the circle on a rod.

Show worked answer →

Use conservation of energy between lowest and highest points; the rod allows thrust as well as tension.

The height gained from bottom to top is $2r$ (M1). Conservation of energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg(2r)$ (M1 A1).

Dividing by $\frac{1}{2}m$ : $u^2 = v^2 + 4gr$ , so $v^2 = u^2 - 4gr$ as required (A1).

For a rigid rod the particle can complete the circle provided $v^2 \ge 0$ at the top (the rod can push), so the minimum is $v = 0$ (B1), giving $u^2 = 4gr$ (M1), hence $u_{\min} = \sqrt{4gr} = 2\sqrt{gr}$ (A1). (For a string the condition would instead be $v^2 \ge gr$ at the top.)

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)