# What is the centripetal force?

Centripetal force calculator: F = m v² ÷ r from mass, speed and radius, or with rpm (F = m r ω²), or the speed, radius or mass for a force, with the acceleration and bank angle.

- Page: https://www.acalculator.org/physics/centripetal-force-calculator
- JSON spec: https://www.acalculator.org/physics/centripetal-force-calculator.json
- Version: 962c6f020ccc

## Default answer

Example with the default inputs (Find Force, Motion as Speed (v), Mass (m) 1,984.1603596639 lb, Radius (r) 1,640.41994750656 ft, Speed (v) 55.9234073013601 mph): F = 1,125 N (252.91 lbf), with a centripetal acceleration of 1.25 m/s².

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Which value to work out from F = m v² ÷ r. |
| by | Motion as | Give the motion as the speed along the circle or as a rotation rate such as rpm. |
| m | Mass (m) | The mass moving in the circle. |
| r | Radius (r) | The radius of the circle (of the curve, for a car). |
| v | Speed (v) | The speed along the circle. |
| n | Rotation rate (ω) | How fast it goes around, in the unit chosen below. |
| nu | Rate unit | The unit of the rotation rate. |
| F | Centripetal force (F) | The net force toward the centre of the circle. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The value worked out, to 6 significant figures, in SI and US units. |
| force | Centripetal force (N) | F = m v² ÷ r, in newtons. |
| lbf | Centripetal force (lbf) | F in pounds-force. |
| accel | Centripetal acceleration (m/s²) | a = v² ÷ r = rω², toward the centre. |
| gs | Acceleration (g) | a ÷ 9.80665 m/s²; on a flat road, also the friction coefficient needed. |
| speed | Speed (m/s) | v along the circle, in metres per second. |
| mph | Speed (mph) | v in miles per hour. |
| omega | Angular velocity (rad/s) | ω = v ÷ r. |
| rpm | Rotation rate (rpm) | Turns per minute: ω × 60 ÷ 2π. |
| period | Time for one lap (s) | T = 2πr ÷ v. |
| radius | Radius (m) | r in metres. |
| mass | Mass (kg) | m in kilograms. |
| bank | Bank angle with no friction (°) | θ = arctan(v² ÷ (rg)): the road tilt that needs no friction. |

## Method

F = m v² ÷ r = m r ω²; a = v² ÷ r = rω²; v = √(F r ÷ m); ω = √(F ÷ (m r)); r = m v² ÷ F; m = F r ÷ v²; bank angle θ = arctan(v² ÷ (rg)).

## Assumptions

- Uniform circular motion: a steady speed on a circle of fixed radius; F is the net force toward the centre.
- The bank angle is for a frictionless road, with g = 9.80665 m/s²; on a flat road the friction coefficient needed is a ÷ g.

## Worked examples

1. find = force, by = speed, m = 900, r = 500, v = 25 gives F = 1,125, accel = 1.25, gs = 0.127465, lbf = 252.910061, bank = 7.264008, result = F = 1,125 N (252.91 lbf). Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19).
2. find = motion, by = speed, F = 1,125, m = 900, r = 500 gives v = 25, result = v = 25 m/s (55.9234 mph). Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19).
3. find = force, by = rate, m = 1, r = 1, n = 60, nu = rpm gives F = 39.478418, omega = 6.283185, period = 1. Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19) (F = m r ω²).
4. find = radius, by = speed, F = 1,125, m = 900, v = 25 gives r = 500. Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19).
5. find = mass, by = speed, F = 1,125, r = 500, v = 25 gives m = 900. Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19).
6. find = force, by = speed, m = 1,000, r = 100, v = 24.3 gives bank = 31.053501. Source: OpenStax, University Physics Volume 1, §6.3 Centripetal Force (a_c = v² ÷ r = rω², F_c = m v² ÷ r = m r ω²; Example 6.15: 900.0 kg at 25.00 m/s on a 500.0 m curve needs 1,125 N and μ_s = 0.13; Example 6.16: tan θ = v² ÷ (rg)), https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force (retrieved 2026-10-19), Example 6.16 (θ = 31.0° gives 24.3 m/s with g = 9.80 m/s²).

## FAQ

### How do you calculate centripetal force?

Multiply the mass by the square of the speed and divide by the radius: F = m v² ÷ r. A 900 kg car at 25 m/s on a curve of radius 500 m needs F = 900 × 25² ÷ 500 = 1,125 N toward the centre.

### How do you calculate centripetal force from rpm?

Turn the rpm into radians per second (ω = rpm × 2π ÷ 60), then F = m r ω². A 1 kg mass on a 1 m string at 60 rpm has ω = 2π rad/s and F = 1 × 1 × (2π)² = 39.5 N.

### What is centripetal acceleration?

The acceleration toward the centre that keeps a body on a circle: a = v² ÷ r = rω². The car above has a = 25² ÷ 500 = 1.25 m/s², or 0.127 g.

### What friction does a car need on a flat curve?

Static friction supplies the centripetal force, so μ must be at least v² ÷ (rg), the acceleration in g. For 25 m/s on a 500 m curve that is 1.25 ÷ 9.80665 = 0.127; OpenStax gets 0.13 with g = 9.80 m/s².

### What is the ideal bank angle of a curve?

On a curve banked so that no friction is needed, tan θ = v² ÷ (rg). For 24.3 m/s on a 100 m curve, θ = arctan(24.3² ÷ (100 × 9.80665)) = 31.05°, the 31° curve of OpenStax Example 6.16.

### Is centripetal force a new kind of force?

No. It is the name for whatever net force points to the centre: friction on a car tire, tension in a string, gravity on a satellite, or the normal force on a banked road.

## Sources

- OpenStax, University Physics Volume 1, §6.3 Centripetal Force: a_c = v² ÷ r = rω²; F_c = m v² ÷ r = m r ω²; Example 6.15 (900.0 kg at 25.00 m/s on a 500.0 m curve: 1,125 N, μ_s = 0.13); Example 6.16 (ideal banking, tan θ = v² ÷ (rg); 31.0° and 100.0 m give 24.3 m/s), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/6-3-centripetal-force
- OpenStax, University Physics Volume 1, §10.1 Rotational Variables: v = rω; 250 rpm = 250 × 2π ÷ 60 rad/s, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/10-1-rotational-variables
- NIST, CODATA 2022: standard acceleration of gravity g_n = 9.80665 m/s² (exact), retrieved 2026-10-19. https://physics.nist.gov/cgi-bin/cuu/Value?gn
- NIST Special Publication 811 (2008), Appendix B.8: 1 lb = 0.45359237 kg, 1 lbf = 4.4482216152605 N, 1 kgf = 9.80665 N, 1 ft = 0.3048 m, 1 mi = 1,609.344 m, 1 mph = 0.44704 m/s, retrieved 2026-10-19. https://www.nist.gov/pml/special-publication-811