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Centripetal Force Calculator

Find the centripetal force.

How it is calculated

The tool first computes the acceleration toward the centre, a = v² ÷ r, then multiplies by mass: F = m × a = m·v² ÷ r. Here m is the mass in kilograms, v the speed along the circle in metres per second and r the radius in metres. The engine does no unit conversion, so values must already be in these SI units to get newtons (N) and m/s². A radius of zero or less, or any missing field, gives no result. Because speed is squared, doubling the speed quadruples the force, while doubling the radius halves it. The result is the net inward force needed, not a separate extra force.

Worked example

A 1200 kg car takes a curve of radius 50 m at 20 m/s. First square the speed: 20 × 20 = 400. Divide by the radius: 400 ÷ 50 = 8.00 m/s² of acceleration. Multiply by the mass: 1200 × 8 = 9,600. The inward force required is 9,600.00 N, which the tyre friction on the road must supply.

Try this example →

Common mistakes

  • Entering speed in km/h. 72 km/h is 20 m/s; typing 72 gives a force almost thirteen times too large.
  • Entering the diameter instead of the radius. This halves the true radius and doubles the force.
  • Forgetting that mass must be in kilograms, not grams or pounds.
  • Treating centripetal force as an additional force. It is the sum of real forces (friction, tension, gravity, normal force) pointing toward the centre.
  • Expecting the force to grow linearly with speed. It grows with the square of speed.

About this calculator

This calculator finds centripetal force using F = mv²/r (or F = mω²r), the force that keeps an object moving in a circle instead of flying off in a straight line. It also reports the centripetal acceleration, a = v²/r.

Physics students use it for textbook problems on circular motion, while engineers apply the same formula when designing roller coaster loops, banked highway curves, centrifuges, and satellite orbits — anywhere something has to be constantly pulled toward a center to keep turning.

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Comments (1)

  • Editorial team

    On a flat road, friction must supply the whole inward force, so dividing the acceleration shown by 9.81 gives the friction coefficient the tyres need. In the 20 m/s example, 8.00 divided by 9.81 is about 0.82, which is already demanding for ordinary tyres. If your figure exceeds what the surface offers, the car slides outward: slow down, widen the curve or bank it. Which circular motion are you analysing, a road, a ride or a rotating machine?

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