Centripetal force:
formula and worked example

What centripetal force is and is not

An object moving at a constant speed in a circle is still accelerating, because its direction keeps changing. That acceleration points to the centre of the circle, and by Newton's second law there must be a net force pointing the same way. That inward force is the centripetal force.

It is a role, not a separate type of force. For a ball on a string it is the tension; for a car on a bend, friction between tyres and road; for a planet, gravity. The formula tells you how much of that force is required to hold the circular path.

Formula and inputs

• Mass: the mass of the object in kilograms.
• Speed: the speed along the circle in m/s, not the angular rate.
• Radius: distance from the centre of the circle to the object, in metres.

The expression comes from combining acceleration toward the centre, v² ÷ r, with force equals mass times acceleration. If you know angular velocity ω instead of speed, the equivalent form is mass × ω² × radius.

Because speed is squared and radius divides, the three inputs do not behave alike. Mass is linear, speed is quadratic and radius is inverse.

A 2 kg mass at 6 m/s on a 3 m circle

1. 1Square the speed: 6 × 6 = 36.
2. 2Multiply by the mass: 2 × 36 = 72.
3. 3Divide by the radius: 72 ÷ 3 = 24.
4. 4State the result: 24 N directed toward the centre.

Check via acceleration: v² ÷ r = 36 ÷ 3 = 12 m/s², and 2 kg × 12 m/s² gives the same 24 N.

What the value means in a real set-up

The result is the minimum inward force the physical connection must provide. If the string, friction or gravity supplies less than that, the object follows a wider path or leaves the circle along a tangent.

• Rope or string: tension must at least match the calculated value, plus any weight component.
• Road bend: friction must supply it, so the available grip sets a maximum speed.
• Orbit: gravity provides it, which links speed and radius for a given planet.

The apparent outward push you feel on a bend is not an actual force pushing you out. It is your body's tendency to continue in a straight line while the car turns beneath you.

How the three inputs scale

Raising the mass by 10% to 2.2 kg raises the force to 26.4 N, a straight 10% rise. The other inputs are more demanding: a 10% faster speed raises the force by 21%, and doubling speed quadruples it.

Radius works inversely. Halving the radius at the same speed doubles the required force, which is why tight corners feel so much harsher than wide ones taken at the same speed.

Errors seen in circular motion problems

• Forgetting to square the speed, which understates the force at higher speed.
• Using diameter in place of radius, giving an answer half the true value.
• Drawing centripetal force as an extra force in a free-body diagram, on top of tension or friction.
• Treating km/h as m/s, which inflates the result by about 13 times.

Using the number to find a speed limit

Rearranged, a given inward force sets a top speed: v = √(F × r ÷ m). For a bend with a radius of 50 m, the maximum friction force on a 1,000 kg car at a friction coefficient of 0.5 is about 4,905 N, giving a limit of roughly 15.7 m/s, about 56 km/h.

That is why wet or icy bends need slower approach speeds: the friction coefficient falls, and the speed limit falls with its square root.

Vertical circles need extra care, because the string tension changes around the loop as weight adds to or subtracts from the inward requirement. At the bottom, tension equals the centripetal requirement plus the weight; at the top, it equals the requirement minus the weight. The formula gives only the net inward force.