Free Centripetal Force Calculator
Calculate the inward (centripetal) force and acceleration needed to keep an object of mass m moving at speed v around a circle of radius r, using F = m·v²/r.
Enter mass, speed and radius to find the inward force and acceleration from F = m·v²/r.
F = m · v² / r and a = v² / r. The force points toward the centre of the circle. Use SI units: kilograms (kg), metres per second (m/s) and metres (m). With angular speed ω (rad/s), use F = m · ω² · r since v = ω · r.
Quick answer
Centripetal force is the net inward force that keeps an object moving on a circular path, given by F = m·v²/r, where m is mass in kilograms, v is speed in metres per second, and r is the radius in metres. The centripetal acceleration is a = v²/r, so F = m·a. For example, a 2 kg object moving at 5 m/s in a circle of radius 4 m needs F = 2 × 5² / 4 = 12.5 N inward and accelerates at a = 5² / 4 = 6.25 m/s² toward the centre.
Formula & method
Centripetal force
F = m · v² / r
- F — Centripetal force directed toward the centre (newtons, N)
- m — Mass of the object (kilograms, kg)
- v — Tangential (linear) speed (metres per second, m/s)
- r — Radius of the circular path (metres, m)
The force points toward the centre of the circle. It does no work because it is always perpendicular to the velocity, so it changes direction but not the magnitude of the speed.
Centripetal acceleration
a = v² / r
- a — Centripetal acceleration toward the centre (m/s²)
- v — Tangential speed (m/s)
- r — Radius (m)
Equivalent forms using angular speed ω (rad/s) are a = ω²·r and F = m·ω²·r, since v = ω·r.
Examples
- Input
- m = 2 kg, v = 5 m/s, r = 4 m
- Result
- F = 12.5 N, a = 6.25 m/s²
- Why
- v² = 5² = 25. Acceleration a = 25 / 4 = 6.25 m/s². Force F = m·a = 2 × 6.25 = 12.5 N directed toward the centre.
- Input
- m = 1000 kg, v = 15 m/s, r = 50 m
- Result
- F = 4500 N, a = 4.5 m/s²
- Why
- v² = 15² = 225. a = 225 / 50 = 4.5 m/s². F = 1000 × 4.5 = 4500 N. This inward force must be supplied by tyre friction; if friction can't reach 4500 N the car skids outward.
- Input
- m = 0.5 kg, v = 4 m/s, r = 2 m
- Result
- F = 4 N, a = 8 m/s²
- Why
- v² = 4² = 16. a = 16 / 2 = 8 m/s². F = 0.5 × 8 = 4 N. The string tension provides this 4 N inward pull; spin it faster or shorten the string and the required tension rises.
- Input
- m = 70 kg, v = 8 m/s, r = 20 m
- Result
- F = 224 N, a = 3.2 m/s²
- Why
- v² = 8² = 64. a = 64 / 20 = 3.2 m/s². F = 70 × 3.2 = 224 N. The runner leans inward so the ground friction supplies this 224 N centripetal force.
When to use this tool
- Finding the tension, friction, or normal force needed to keep an object on a circular path (ball on a string, car on a curve, banked turn).
- Checking whether available friction or tension is enough to prevent skidding or the string breaking at a given speed.
- Solving physics homework on uniform circular motion, satellites, centrifuges, or amusement-park rides where speed and radius are known.
- Estimating the g-forces on a pilot, rider, or sample by comparing a = v²/r against gravitational acceleration g ≈ 9.81 m/s².
Common mistakes
- Forgetting to square the speed: F depends on v², so doubling the speed quadruples the required force, not just doubles it.
- Mixing units — radius in centimetres or speed in km/h. Convert everything to SI (kg, m, m/s) before applying F = mv²/r, or the newton result is wrong.
- Treating centripetal force as a new, separate push. It is the net inward force; the real-world agent is tension, friction, gravity, or the normal force, not an extra force you add.
- Confusing centripetal (real, inward) with 'centrifugal' force. The outward 'centrifugal' force is a fictitious effect felt in the rotating frame and is not used in F = mv²/r.
Frequently asked questions
What is the formula for centripetal force?
Centripetal force is F = m·v²/r, where m is mass (kg), v is the tangential speed (m/s), and r is the radius of the circular path (m). The result is in newtons and points toward the centre of the circle. The associated centripetal acceleration is a = v²/r, so F = m·a.
What is the difference between centripetal force and centripetal acceleration?
Centripetal acceleration a = v²/r is the rate of change of velocity's direction toward the centre, measured in m/s². Centripetal force F = m·v²/r is the net force that produces that acceleration, measured in newtons. They are linked by Newton's second law, F = m·a, so the force is simply the acceleration multiplied by the object's mass.
Is centripetal force the same as centrifugal force?
No. Centripetal force is the real, inward force that keeps an object on a circular path. 'Centrifugal force' is a fictitious outward force that only appears when you analyse motion from the rotating object's own reference frame. In an inertial frame there is no outward force — only the inward centripetal force, supplied by tension, friction, gravity, or the normal force.
What happens to centripetal force if I double the speed?
Because F = m·v²/r depends on the square of the speed, doubling v multiplies the required force by four. For instance, going from 5 m/s to 10 m/s for a 2 kg object at r = 4 m raises the force from 12.5 N to 50 N. This is why cornering at high speed needs far more grip.
How does radius affect the centripetal force?
Force is inversely proportional to radius: F = m·v²/r. At a fixed speed, a tighter circle (smaller r) demands a larger inward force, while a wider, gentler curve (larger r) needs less. Halving the radius doubles the required centripetal force for the same speed and mass.
Can I calculate centripetal force from angular speed or RPM?
Yes. If you know the angular speed ω in rad/s, use F = m·ω²·r and a = ω²·r, since v = ω·r. To convert revolutions per minute to rad/s, multiply RPM by 2π/60. Then either feed v = ω·r into this calculator or apply the angular forms directly.
Sources & references
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