A particle moves in a circle of radius 2\,m at 6\,m s^-1. Find its centripetal acceleration. [2 marks]

An SHM has ω = 5\,rad s^-1 and amplitude 0.2\,m. Find the maximum speed. [1 mark]

CCEA CCEA A2 2021-style practice: A particle of mass 0.5\,kg moves in a horizontal circle of radius 0.8\,m on the end of a light string, completing 2 revolutions per second. Find the angular speed and the tension in the string.

Two revolutions per second means a frequency of 2\,Hz, so the angular speed is ω = 2π f = 2π × 2 = 4π ≈ 12.57\,rad s^-1. For horizontal circular motion the tension provides the centripetal force F = mω^2 r: T = mω^2 r = 0.5 × (4π)^2 × 0.8 = 0.5 × 157.9 × 0.8 = 63.2\,N. Markers reward the angular speed from the frequency, the centripetal force formula mω^2 r, and the tension of about 63\,N.

CCEA CCEA A2 2019-style practice: A particle moves with simple harmonic motion of amplitude 0.15\,m and period π2\,s. Find the maximum speed and the speed when the particle is 0.09\,m from the centre.

The angular frequency is ω = 2πT = 2ππ/2 = 4\,rad s^-1. The maximum speed is at the centre, v_ = ω A = 4 × 0.15 = 0.6\,m s^-1. The speed at displacement x is v = (A^2 - x^2): v = 4sqrt(0.15^2 - 0.09^2) = 4sqrt(0.0225 - 0.0081) = 4sqrt(0.0144) = 4 × 0.12 = 0.48\,m s^-1. Markers reward the angular frequency from the period, the maximum speed ω A, and the speed formula (A^2 - x^2).

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How do you model circular motion and simple harmonic motion in mechanics?

Circular motion with angular speed, centripetal acceleration and force, motion in a horizontal and vertical circle, and simple harmonic motion with its defining equation, period and energy.

A CCEA A2 Further Maths Mechanics answer on circular motion with angular speed, centripetal acceleration and force, motion in horizontal and vertical circles, and simple harmonic motion with its defining equation, period, velocity and energy.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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Quick answer

Circular motion at angular speed $\omega = \frac{v}{r}$ has centripetal acceleration $\frac{v^2}{r} = \omega^2 r$ towards the centre, requiring a centripetal force $\frac{mv^2}{r} = m\omega^2 r$ supplied by a real inward force. For a horizontal circle, resolve so the horizontal component gives $m\omega^2 r$ and the vertical balances weight; in a vertical circle the speed varies and energy conservation links heights. Simple harmonic motion satisfies $\ddot{x} = -\omega^2 x$ , giving $x = A\cos\omega t$ , $v = \omega\sqrt{A^2 - x^2}$ , maximum speed $\omega A$ , period $\frac{2\pi}{\omega}$ independent of amplitude, and constant total energy $\frac{1}{2}m\omega^2 A^2$ .

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What this dot point is asking

CCEA wants you to model circular motion using angular speed, centripetal acceleration and centripetal force, analyse motion in horizontal and vertical circles, and describe simple harmonic motion with its defining equation, period, velocity and energy.

The answer

Circular motion

Horizontal and vertical circles

Simple harmonic motion

A conical pendulum

A particle of mass $0.4\,\text{kg}$ is attached to a string of length $1.0\,\text{m}$ and moves in a horizontal circle so that the string makes $30^\circ$ with the vertical. Take $g = 9.8\,\text{m s}^{-2}$ . Find the tension and the angular speed.

step Resolve vertically

The vertical component of tension balances the weight:

T\cos 30^\circ = mg \Rightarrow T = \frac{0.4 \times 9.8}{\cos 30^\circ} = \frac{3.92}{0.866} = 4.53\,\text{N}.

step Find the radius

r = L\sin 30^\circ = 1.0 \times 0.5 = 0.5\,\text{m}.

step Resolve horizontally for the angular speed

The horizontal component provides the centripetal force:

T\sin 30^\circ = m\omega^2 r \Rightarrow 4.53 \times 0.5 = 0.4\,\omega^2 (0.5).

2.265 = 0.2\,\omega^2 \Rightarrow \omega^2 = 11.3 \Rightarrow \omega = 3.36\,\text{rad s}^{-1}.

Examples in context

Example 1. A car on a banked bend. On a banked track the road's normal reaction has an inward horizontal component that supplies the centripetal force, letting a car corner faster without relying on friction. Resolving the reaction is exactly the horizontal-circle method.

Example 2. A mass bobbing on a spring. A mass on a vertical spring oscillates with SHM about its equilibrium, with period $2\pi\sqrt{\frac{m}{k}}$ independent of how far it is pulled. The energy swap between kinetic and elastic potential is the SHM energy picture.

Try this

Q1. A particle moves in a circle of radius $2\,\text{m}$ at $6\,\text{m s}^{-1}$ . Find its centripetal acceleration. [2 marks]

Cue. $a = \frac{v^2}{r} = \frac{36}{2} = 18\,\text{m s}^{-2}$ .

Q2. State the defining equation of SHM. [1 mark]

Cue. $\ddot{x} = -\omega^2 x$ .

Q3. An SHM has $\omega = 5\,\text{rad s}^{-1}$ and amplitude $0.2\,\text{m}$ . Find the maximum speed. [1 mark]

Cue. $v_{\max} = \omega A = 5 \times 0.2 = 1\,\text{m s}^{-1}$ .

Exam-style practice questions

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

CCEA A2 20216 marksA particle of mass

0.5\,\text{kg}

moves in a horizontal circle of radius

0.8\,\text{m}

on the end of a light string, completing

2

revolutions per second. Find the angular speed and the tension in the string.

Show worked answer →

Two revolutions per second means a frequency of $2\,\text{Hz}$ , so the angular speed is

$\omega = 2\pi f = 2\pi \times 2 = 4\pi \approx 12.57\,\text{rad s}^{-1}.$

For horizontal circular motion the tension provides the centripetal force $F = m\omega^2 r$ :

$T = m\omega^2 r = 0.5 \times (4\pi)^2 \times 0.8 = 0.5 \times 157.9 \times 0.8 = 63.2\,\text{N}.$

Markers reward the angular speed from the frequency, the centripetal force formula $m\omega^2 r$ , and the tension of about $63\,\text{N}$ .

CCEA A2 20197 marksA particle moves with simple harmonic motion of amplitude

0.15\,\text{m}

and period

\dfrac{\pi}{2}\,\text{s}

. Find the maximum speed and the speed when the particle is

0.09\,\text{m}

from the centre.

Show worked answer →

The angular frequency is $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{\pi/2} = 4\,\text{rad s}^{-1}.$

The maximum speed is at the centre, $v_{\max} = \omega A = 4 \times 0.15 = 0.6\,\text{m s}^{-1}.$

The speed at displacement $x$ is $v = \omega\sqrt{A^2 - x^2}$ :

$v = 4\sqrt{0.15^2 - 0.09^2} = 4\sqrt{0.0225 - 0.0081} = 4\sqrt{0.0144} = 4 \times 0.12 = 0.48\,\text{m s}^{-1}.$

Markers reward the angular frequency from the period, the maximum speed $\omega A$ , and the speed formula $\omega\sqrt{A^2 - x^2}$ .

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Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)