Physics 9702 · Paper 4

Centripetal Acceleration & Force

One pair of equations turns every circular-motion story into numbers. They are the bridge formulas that appear inside gravity, magnetic-field and electric-field questions all through Paper 4.

The equations

a = \dfrac{v^2}{r} = r\omega^2

Symbols

$a$ = centripetal acceleration (toward the centre) (m s⁻²)
$v$ = speed (m s⁻¹)
$r$ = radius (m)
$\omega$ = angular speed (rad s⁻¹)

Substitute $v = r\omega$ to move between the two forms; combining them differently also gives $a = v\omega$ , a form the 2025 paper asked for directly (9702/41/O/N/25 Q1(b)).
Multiply by the mass (Newton's second law, lesson 3.02) for the resultant force needed:

F = \dfrac{mv^2}{r} = mr\omega^2

Symbols

$F$ = centripetal (resultant) force toward the centre (N)
$m$ = mass (kg)

Worked example

Smallest case: both forms on one wheel

A point on a spinning wheel is at radius 0.85 m and the wheel turns at 140 rad s⁻¹. Find the point's speed and its acceleration. (9702/42/O/N/24 Q1(a))

$v = r\omega = 0.85 \times 140 = 120 \text{ m s}^{-1}$ .
$a = r\omega^2 = 0.85 \times 140^2 = 1.7 \times 10^{4} \text{ m s}^{-2}$ toward the centre.

Answer

v = 120 \text{ m s}^{-1}, \; a = 1.7 \times 10^{4} \text{ m s}^{-2}

Worked example

One change: a spinning disc, radius read carefully

A lump of clay sits on a disc of radius 9.3 cm, with its centre 1.2 cm in from the rim. The clay moves at 0.68 m s⁻¹. Find ω and the acceleration. (9702/42/M/J/24 Q1(c))

The radius of the clay's circle: $r = 9.3 - 1.2 = 8.1 \text{ cm} = 0.081 \text{ m}$ .
$\omega = \dfrac{v}{r} = \dfrac{0.68}{0.081} = 8.4 \text{ rad s}^{-1}$ .
$a = \dfrac{v^2}{r} = \dfrac{0.68^2}{0.081} = 5.7 \text{ m s}^{-2}$ .

Answer

\omega = 8.4 \text{ rad s}^{-1}, \; a = 5.7 \text{ m s}^{-2}

Where a = v²/r comes from

In the Δv triangle (lesson 12.03), the two speeds have the same length v and the angle between them is $\theta$ . For a small angle, $\Delta v = v\theta$ . Divide by the time: $a = \dfrac{\Delta v}{\Delta t} = v\dfrac{\theta}{\Delta t} = v\omega$ . Substituting $\omega = v/r$ gives $a = v^2/r$ ; substituting $v = r\omega$ instead gives $a = r\omega^2$ .

Bridge duty: in Paper 4 these equations mostly appear inside other topics: an electron circling a nucleus (electric force = $mv^2/r$ ) (9702/41/M/J/25 Q2(c)), a satellite in orbit (gravitational force = $mr\omega^2$ ), a charge in a magnetic field. When a question says the path is circular, use these equations.

Pick the form that matches the given data: v and r → $v^2/r$ ; T, f or ω and r → $r\omega^2$ .

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Centripetal Acceleration & Force

One pair of equations turns every circular-motion story into numbers. They are the bridge formulas that appear inside gravity, magnetic-field and electric-field questions all through Paper 4.

The equations

a = \dfrac{v^2}{r} = r\omega^2

Symbols

$a$ = centripetal acceleration (toward the centre) (m s⁻²)
$v$ = speed (m s⁻¹)
$r$ = radius (m)
$\omega$ = angular speed (rad s⁻¹)

Substitute $v = r\omega$ to move between the two forms; combining them differently also gives $a = v\omega$ , a form the 2025 paper asked for directly (9702/41/O/N/25 Q1(b)).
Multiply by the mass (Newton's second law, lesson 3.02) for the resultant force needed:

F = \dfrac{mv^2}{r} = mr\omega^2

Symbols

$F$ = centripetal (resultant) force toward the centre (N)
$m$ = mass (kg)

Worked example

Smallest case: both forms on one wheel

A point on a spinning wheel is at radius 0.85 m and the wheel turns at 140 rad s⁻¹. Find the point's speed and its acceleration. (9702/42/O/N/24 Q1(a))

$v = r\omega = 0.85 \times 140 = 120 \text{ m s}^{-1}$ .
$a = r\omega^2 = 0.85 \times 140^2 = 1.7 \times 10^{4} \text{ m s}^{-2}$ toward the centre.

Answer

v = 120 \text{ m s}^{-1}, \; a = 1.7 \times 10^{4} \text{ m s}^{-2}

Worked example

One change: a spinning disc, radius read carefully

A lump of clay sits on a disc of radius 9.3 cm, with its centre 1.2 cm in from the rim. The clay moves at 0.68 m s⁻¹. Find ω and the acceleration. (9702/42/M/J/24 Q1(c))

The radius of the clay's circle: $r = 9.3 - 1.2 = 8.1 \text{ cm} = 0.081 \text{ m}$ .
$\omega = \dfrac{v}{r} = \dfrac{0.68}{0.081} = 8.4 \text{ rad s}^{-1}$ .
$a = \dfrac{v^2}{r} = \dfrac{0.68^2}{0.081} = 5.7 \text{ m s}^{-2}$ .

Answer

\omega = 8.4 \text{ rad s}^{-1}, \; a = 5.7 \text{ m s}^{-2}

Where a = v²/r comes from

Bridge duty: in Paper 4 these equations mostly appear inside other topics: an electron circling a nucleus (electric force = $mv^2/r$ ) (9702/41/M/J/25 Q2(c)), a satellite in orbit (gravitational force = $mr\omega^2$ ), a charge in a magnetic field. When a question says the path is circular, use these equations.

Pick the form that matches the given data: v and r → $v^2/r$ ; T, f or ω and r → $r\omega^2$ .