Calculation
F = m · v² / r
Force grows with speed squared and falls with larger radius.
F = m·v²/r
Centripetal Force Calculator
Force grows with speed squared and falls with larger radius.
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Result
Select a target and calculate.
Centripetal force F versus path speed v
F = m·v²/r rises with the square of the speed: at the same radius, twice the speed needs four times the force, for example as tyre grip in a bend or as rope tension.
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Understand the inputs
- Centripetal force F — The force directed inward toward the centre of the circle that forces the circular motion (e.g. rope tension, cornering friction, or rail thrust) – it is not an extra outward-acting "centrifugal force" but its counterpart.
- Mass m — The moving mass. Source: weigh it, or take it from a drawing/parts list.
- Tangential speed v — The path speed (magnitude of velocity along the circular path), not the angular speed.
- Circle radius r — The radius of the circular path, measured from the centre of rotation to the body.
Example
100 kg, 10 m/s and 10 m radius give 1,000 N.
Assumptions and limits
Uniform circular motion; bearing and friction forces are excluded.
Understand Centripetal Force Calculator
This calculator finds the missing quantity among centripetal force F, mass m, tangential speed v and circle radius r from the other two. It addresses one of the most persistent misconceptions in physics: the 'centrifugal force' you seem to feel pushing outward in a turn is not a real force acting on the body.
What does this quantity describe?
Centripetal force is the resultant force directed toward the centre of a circle that keeps a body on its circular path: F = m·v²/r. Without this inward force, the body would continue in a straight line, per the principle of inertia.
In a turn, static friction between tyres and road surface supplies the centripetal force that keeps the vehicle on its circular path — if friction isn't enough, the vehicle slides tangentially out of the curve rather than being 'flung' outward. The sensation of being pushed outward in a turn arises only because your own body, viewed in a co-rotating, non-inertial frame, lags behind the vehicle's actual circular motion; no real, outward-acting force is actually present.
Formula and variables
F = m · v² / r
F = m · v² / rv = √(F·r/m)r = m·v²/F
| Symbol / input | Meaning |
|---|---|
| Centripetal force F | The force directed inward toward the centre of the circle that forces the circular motion (e.g. rope tension, cornering friction, or rail thrust) – it is not an extra outward-acting "centrifugal force" but its counterpart. |
| Mass m | The moving mass. Source: weigh it, or take it from a drawing/parts list. |
| Tangential speed v | The path speed (magnitude of velocity along the circular path), not the angular speed. |
| Circle radius r | The radius of the circular path, measured from the centre of rotation to the body. |
Choose the inputs correctly
Enter two of the three quantities. Centripetal force requires mass, speed and radius (two of these together with F to find the third). F is always the resultant radial force actually acting toward the centre — in cornering, usually tyre friction.
How to use the calculator
Select the target quantity and enter the two known values with units. Compare the result against the maximum available force (e.g. maximum static friction) afterward, to check whether circular motion at this speed and radius can be sustained at all.
Worked example
Given m = 100 kg, v = 10 m/s and r = 10 m, substitution gives F = 100 kg · 10² m²/s² / 10 m = 1,000 N.
Understand the result and units
A centripetal force of 1,000 N is the force needed to hold the 100 kg mass on a 10 m radius circular path at 10 m/s. If the actually available force (e.g. friction or rope tension) falls short, the body leaves the circular path tangentially, in the direction of its instantaneous velocity.
Force in newtons (N), mass in kilograms (kg), speed in m/s, radius in metres. The formula requires consistent SI units, since v enters squared.
Why isn't centrifugal force a real force?
Viewed from a stationary frame outside the rotating body, only the real centripetal force acts on it, directed inward toward the centre — without it, the body would continue in a straight line, as predicted by Newton's first law. 'Centrifugal force' appears only when viewing the motion from a co-rotating, accelerating reference frame: in this non-inertial frame, an additional fictitious force must be introduced for Newton's equations of motion to still hold. This fictitious force corresponds to no actual physical interaction between two bodies and does not appear in an inertial frame.
Typical applications
Centripetal force calculations are used in road and rail curve design, sizing centrifuges and carousels, and safety assessment of rotating machine parts.
Assumptions, limits and common mistakes
The formula applies to uniform circular motion with constant tangential speed and constant radius. For a changing radius or non-uniform speed, radial and tangential acceleration must be considered separately.
Common mistake: Do not treat 'centrifugal force' as a real force acting on the body — it is a fictitious force that appears only in a co-rotating reference frame. In a stationary (inertial) frame, only the real centripetal force acts, directed inward. Also, do not confuse F with the tangential force that would be responsible for changing the tangential speed itself.
Frequently asked questions
Is centrifugal force a real force?
No. From a stationary observer's viewpoint, only the real centripetal force acts, directed inward. 'Centrifugal force' is a fictitious force that appears only in a co-rotating, accelerating reference frame.
What supplies the centripetal force in a car turning a corner?
Usually static friction between tyres and road. If that isn't enough, the vehicle slides tangentially out of the curve.
What happens if available force is less than the calculated centripetal force?
The body can no longer hold the circular path and instead continues approximately in a straight line along its instantaneous direction of travel.