Anything that moves along a circle is constantly changing direction, and changing direction is a form of acceleration. The force that causes that inward acceleration is the centripetal force. This calculator solves the relationship for whichever quantity you are missing: the force, the mass, the speed or the radius of the circle. You can describe the motion either as a linear speed (m/s, km/h, mph) or as a rotation rate (rpm, rad/s), which is handy for centrifuges, washing machines and spinning wheels.
How to use the centripetal force calculator
- Under Solve for, pick Force, Mass, Speed or Radius.
- If you are not solving for speed, choose how to Give the motion: as a linear speed (v) or as a rotation rate (ω).
- Enter the known values with their units: centripetal force in N, kN, lbf or kgf; mass in kg, g, t, lb or US tons; speed in m/s, km/h, mph, ft/s or km/s; radius in m, cm, mm, km, in, ft or mi.
- Choose the unit of the answer under Show the result in.
- Read the answer, the centripetal acceleration (also in g), the rotation rate, the time for one lap and, for road-scale turns, the tire grip needed and the ideal banking angle.
Centripetal force formula
Here m is the mass, v the speed along the circle, r the radius and ω the angular speed in radians per second. The two forms are linked by v = ωr, and one revolution per minute equals 2π/60 ≈ 0.10472 rad/s. Dividing by the mass gives the centripetal acceleration:
The force always points toward the center of the circle, at right angles to the velocity. Because it never pushes along the direction of motion, it changes the direction of the velocity but not its size.
Worked example
A car in a highway curve
A 1,500 kg car takes a curve of radius 80 m at 25 m/s (90 km/h, about 56 mph).
F = 1,500 × 25² ÷ 80 = 11,718.75 N, and a = 625 ÷ 80 = 7.8125 m/s², which is 0.797 g.
On a flat road, friction must supply all of that, so the tires need a friction coefficient of at least 0.797, which is close to the limit of dry pavement. Banking the road at about 38.5° would let the car round the curve with no sideways grip at all.
A ball on a string. Swing a 0.15 kg ball in a horizontal circle of radius 0.8 m at 120 rpm. In rad/s that is 4π ≈ 12.566, so F = 0.15 × 12.566² × 0.8 = 18.95 N of string tension, and the ball experiences an acceleration of 12.9 g.
Where the centripetal force comes from
Centripetal force is not a separate kind of force; it is the job title of whatever real force points toward the center:
| Situation | Force that does the job |
|---|---|
| Car on a flat curve | Static friction between tires and road |
| Ball on a string | Tension in the string |
| Satellite or Moon | Gravity (see the orbital period calculator) |
| Roller-coaster loop | Normal force from the track plus gravity |
| Electron in a classical atom model | Electric attraction |
| Clothes in a spin cycle | The drum wall; water escapes through the holes |
When the available force is smaller than mv²/r, the object cannot follow the circle and drifts outward along a straighter path: a car skids wide, a spinning bucket spills, a string snaps.
Speed matters more than anything else
Because speed is squared, small changes in speed have a big effect. Raising speed by 10% raises the required force by 21%. Halving the radius doubles it. That is why warning signs for sharp curves give speed limits rather than suggesting lighter vehicles: mass scales the force, but it also scales the friction available, so mass cancels out of the grip requirement μ ≥ v²/(rg).
Limits of the model
The calculator assumes uniform circular motion: constant speed on a circle of fixed radius. For a curve that tightens, use the local radius at the point of interest. At speeds above about 10% of the speed of light, relativistic effects make the classical formula inaccurate, and the calculator flags such inputs. For straight-line forces, the force calculator applies F = ma directly, and the friction calculator estimates how much grip a surface provides.
Frequently asked questions
What is centripetal force in simple terms?
It is the net inward push or pull that keeps an object moving on a curved path. Without it, the object would continue in a straight line. Tension in a string, friction from tires, gravity on a satellite and the normal force from a loop-the-loop track can all supply it.
Why does doubling the speed quadruple the force?
Speed appears squared in F = mv²/r. A car at 25 m/s on an 80 m curve needs 11,719 N per 1,500 kg; at 50 m/s it would need 46,875 N, four times as much, which is far more grip than tires can provide.
How do I use rpm instead of speed?
Choose “Rotation rate (ω)” and enter rpm, rad/s, rev/s or degrees per second. The calculator converts to rad/s and uses F = mω²r. A 0.15 kg ball on a 0.8 m string at 120 rpm needs about 18.95 N of tension.
Is centrifugal force the same thing?
Not quite. Centrifugal force is the outward force you seem to feel when you describe the motion from inside the rotating object. Seen from outside, only the inward centripetal force acts; your body simply tends to keep going straight.
What do the friction and banking results mean?
On a flat road, tire friction must supply the whole centripetal force, so the friction coefficient must be at least v²/(rg). The ideal banking angle, arctan(v²/(rg)), is the tilt at which the road alone holds the car in the turn with no sideways friction.