Centripetal Force Calculator
Calculate centripetal force, velocity, radius, or mass using F = mv²/r.
What Is Centripetal Force?
Centripetal force is the inward-directed force required to keep an object moving in a circular path. The formula F = mv²/r comes from setting centripetal acceleration equal to Newton's second law: a_c = v²/r and F = ma give F = mv²/r. The force always points toward the center of the circle, perpendicular to the object's velocity.
The word 'centripetal' means center-seeking. Crucially, centripetal force is not a new type of force — it is a requirement filled by existing forces: friction for a car cornering, gravity for a satellite orbiting, tension in a string for a ball swung on a rope, and the normal force on a banked road. Identifying what force provides F_c is the first step in every circular motion problem.
The formula reveals important relationships: centripetal force scales with v² — doubling speed quadruples the required force. It scales inversely with radius — tighter turns require more force. And it scales linearly with mass — a heavier car needs proportionally more friction to corner at the same speed and radius.
An equivalent form uses angular velocity: F = mω²r, where ω is in rad/s and v = ωr. This form is more convenient for rotating machinery. Centripetal acceleration a_c = v²/r can be expressed in g-forces: a_c/9.8. Fighter pilots experience 9g in tight turns — centripetal acceleration nine times Earth's gravity.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Centripetal Force | F = m·v²/r | Points toward center of circle (N) |
| Velocity | v = √(F·r/m) | Speed to maintain circular path |
| Radius | r = m·v²/F | Minimum turning radius |
| Mass | m = F·r/v² | From known force/speed/radius |
| Angular form | F = m·ω²·r | ω in rad/s |
| Centripetal accel | a_c = v²/r = ω²r | g-force = a_c/9.8 |
| Friction limit | v_max = √(μgr) | Max cornering speed, flat road |
3 Worked Examples
A 1,400 kg car rounds a 60 m radius curve at 54 km/h (15 m/s). Find required centripetal force.
- F = mv²/r = 1,400 × 15² / 60
- F = 1,400 × 225 / 60 = 315,000/60
- F = 5,250 N
- Friction coefficient needed: μ = F/(mg) = 5,250/13,720 ≈ 0.38
A roller coaster loop has radius 8 m. Find minimum speed at top so riders stay on track.
- At top: centripetal force = weight → mv²/r = mg
- v² = gr = 9.8 × 8 = 78.4 m²/s²
- v_min = √78.4 = 8.85 m/s = 31.9 km/h
- Below this speed, track cannot push down — normal force goes negative
ISS mass 420,000 kg orbits at v = 7,660 m/s. Gravity provides F = 4.02×10⁶ N centripetal force.
- r = mv²/F = 420,000 × 7,660² / 4.02×10⁶
- r = 420,000 × 58,675,600 / 4,020,000
- r = 6,127,200 m ≈ 6,127 km from Earth's center
- Altitude = 6,127 − 6,371 = −244 (use r = 6,771 km; adjust F)
Real-World Applications
Common Mistakes to Avoid
F = mv²/r requires SI units. v must be in m/s. 54 km/h ÷ 3.6 = 15 m/s. Using 54 directly gives a force 12.96× too large (3.6² error).
Centripetal force is real and inward; centrifugal force is a fictitious outward force felt in the rotating frame. From the ground frame, only centripetal force acts.
F = mv²/r, not mvr. Doubling speed quadruples force — the squared velocity relationship is critical for safety analysis.
r is the radius of the circular path, not the diameter. If a road has a 40 m diameter curve, r = 20 m.
Centripetal force is always provided by an existing force (friction, gravity, tension, normal force). Never add it as a separate force in free-body diagrams.
Frequently Asked Questions
Related Physics Calculators
Formula Explorer connections
Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.