Physics 9702 · Paper 4

What Provides the Centripetal Force

“Centripetal force” is a job title, not a new kind of force. Some real force must do the job in every situation.

A job, not a force

Mark schemes never credit a force labelled just “centripetal force”. They want the real force (or component) whose resultant points at the centre (9702/42/O/N/25 Q1(b)).

Situation	The real force doing the job
car cornering on a flat road	friction from the road on the tyres
planet or satellite in orbit	the gravitational pull
bung whirled on a string	the tension in the string
electron circling a nucleus	the electric attraction
rider against a fairground drum wall	the normal contact force of the wall

The release test: cut the string, and the bung flies off along the tangent at steady speed. Without a resultant force, Newton's first law takes over.
The same logic explains a car on oil: with no friction there is no centripetal force, so the car continues in a straight line and leaves the bend (9702-style explain).
The satellite wording that scored in 2024: the gravitational force acts perpendicular to the motion, and it provides the centripetal acceleration (9702/41/O/N/24 Q1(b)(i)).

The 2025 scenario: riding the Earth's spin

A person at latitude λ rides a circle of radius R cos λ around the spin axis. The centripetal resultant points horizontally at the axis — not at the Earth's centre.

Worked example

Exam version: standing at latitude 52.2°

A person of mass 58.6 kg stands in Cambridge, at latitude 52.2°. The Earth (radius 6.37 × 10⁶ m) spins once per day, carrying them around a circle. Find the radius of that circle, their speed, and the resultant force needed. (9702/42/O/N/25 Q1)

The circle sits at the latitude, around the spin axis: $r = R\cos\lambda = 6.37 \times 10^{6} \times \cos 52.2^\circ = 3.90 \times 10^{6} \text{ m}$ .
Period = 24 hours = 86 400 s: $v = \dfrac{2\pi r}{T} = \dfrac{2\pi \times 3.90 \times 10^{6}}{86\,400} = 280 \text{ m s}^{-1}$ .
$F = \dfrac{mv^2}{r} = \dfrac{58.6 \times 280^2}{3.90 \times 10^{6}} = 1.2 \text{ N}$ .

Answer

r = 3.90 \times 10^{6} \text{ m}, \; v = 280 \text{ m s}^{-1}, \; F = 1.2 \text{ N}

The resultant points horizontally at the spin axis, not at the Earth's centre. It is the small difference between the weight and the contact force from the ground.

First name the real forces, then write: their resultant (or component) toward the centre = $mv^2/r$ . That sentence structure is the mark scheme's own.

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