Physics · Engineering Physics · A2 Level

Centripetal Force

Whatever is causing an object to accelerate toward the centre of a circular path — a bearing's reaction force holding a shaft in place, the tension in a wire, the internal stress inside a spinning rim — has a name for the role it's playing: centripetal force. It isn't a new fundamental force, just a label for whichever real force is doing this particular job.

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Centripetal force is a role, not a new type of force

By Newton's second law, any acceleration requires a resultant (net) force in the same direction as that acceleration. Since circular motion involves a centripetal acceleration directed toward the centre of the circle, it must also involve a centripetal force — the resultant force, always directed toward the centre, that produces this acceleration.

$$F = \frac{mv^2}{r} = m\omega^2 r$$

This is simply Newton's second law, $F=ma$, with the centripetal acceleration $a = v^2/r = \omega^2 r$ from the previous lesson substituted in. It is crucial to understand that "centripetal force" is not a distinct physical force alongside gravity, tension, friction, and normal contact force — it is a description of the role that some combination of real, physical forces is playing in a given situation. Whatever is actually pulling or pushing an object toward the centre of its circular path — a string's tension, gravity pulling a satellite around a planet, friction between a car's tyres and the road, a bearing's reaction force on a rotating shaft — is the centripetal force in that scenario.

For a mass swung on a string, the string's tension provides the centripetal force.
For a satellite in orbit, gravity provides the centripetal force.
For a car cornering on a flat road, friction between tyres and road surface provides the centripetal force.
For a component fixed to a rotating shaft, the shaft (via its bearings, or the material connecting the component to the axis) provides the centripetal force, as a reaction or structural force.

Worked exampleBearing load on a rotating shaft

A component of mass $2.5$ kg is mounted at radius $r = 0.20$ m on a shaft rotating at $\omega = 150$ rad s$^{-1}$. Find the centripetal force the bearings must supply to hold the component in its circular path.

Known: $m = 2.5$ kg, $r = 0.20$ m, $\omega = 150$ rad s$^{-1}$. Use $F = m\omega^2 r$ since $\omega$ is given directly.

$F = m\omega^2 r = 2.5 \times (150)^2 \times 0.20$

Direct substitution into the formula.

$F = 2.5 \times 22500 \times 0.20 = 11250$ N $\approx 1.1 \times 10^4$ N

This substantial force is what the bearings supporting the shaft must be able to withstand without failing.

By Newton's third law, the component exerts an equal and opposite reaction force on the bearings, directed radially outward.

This outward reaction is what engineers must design the bearing housing to resist — the source of the loads that limit safe rotational speeds in real machinery.

Result

This same idea explains why fast-spinning components like flywheel rims are engineered with such care: every small element of the rim material needs a centripetal force pulling it toward the axis to keep it moving in a circle, and that force is supplied internally by the tensile stress in the material itself, holding the rim together against its own rotation. At high enough rotational speeds, the required centripetal force can exceed what the material's internal strength can supply, and the rim fails structurally — which is exactly why maximum safe operating speed is such an important design parameter for any rotating engineering component.

Always ask: what force is actually supplying it?

When solving any centripetal force problem, the calculation $F = mv^2/r$ (or $m\omega^2 r$) only tells you the size of the required resultant force. The physics is completed by identifying which real force, or combination of forces, is physically supplying it in that specific situation — this is very often exactly what an exam question is testing.

Looking ahead

Across this topic, a complete toolkit has been assembled: the angular kinematic quantities $\theta$, $\omega$, $\alpha$ and the rotational suvat equations connecting them; the dictionary translating between linear and angular descriptions of motion; and now centripetal acceleration and force, describing what's needed to keep something moving in a circle at all. This is the foundation the next topic builds directly on, introducing torque, moment of inertia and rotational kinetic energy — the concepts needed to fully analyse real rotating engineering components like flywheels, where mass isn't concentrated at a single point but distributed across an extended, spinning object.

Key takeaways

Centripetal force is the resultant (net) force directed toward the centre of a circular path, required by Newton's second law to produce centripetal acceleration.
It is not a separate fundamental force — it is the name given to whichever real force (tension, friction, gravity, a bearing's reaction force, internal material stress) is providing the centre-directed resultant.
$F = \dfrac{mv^2}{r} = m\omega^2 r$, obtained directly from $F=ma$ using the centripetal acceleration formula.
In rotating engineering components, bearings or the structure of the component itself must supply this force, which sets practical limits on safe rotational speed.
Solving a centripetal force problem always has two parts: calculating the required magnitude, and identifying the real physical force providing it.
This lesson closes the Rotational Dynamics topic, which together forms the foundation for the next topic on torque, moment of inertia and rotational energy.

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Centripetal Acceleration

Moment of Inertia