Centripetal Force Calculator
Everything that turns needs a push toward the center. Find that force, the g-load, and how fast a car can safely take a bend.
By Manzoor Ahmad, pharmacist · Updated How do you calculate centripetal force?
F = m × v² ÷ r. A 1,000 kg car rounding a 50 m radius bend at 20 m/s (72 km/h) needs 1,000 × 20² ÷ 50 = 8,000 N of inward force, which means 8 m/s² (0.82 g) of centripetal acceleration. On a flat road that force comes from tire friction.
Centripetal force and circular motion formulas
| Quantity | Formula | Notes |
|---|---|---|
| Centripetal force | F = mv²/r = mω²r | points to the center |
| Centripetal acceleration | a = v²/r = ω²r | divide by 9.80665 for g |
| Speed from RPM | v = 2πr × RPM ÷ 60 | ω = 2π × RPM ÷ 60 |
| Period | T = 2πr ÷ v | time for one lap |
| Max speed on a flat curve | v = √(μgr) | friction supplies the force |
| Ideal banking angle | θ = arctan(v² ÷ rg) | no friction needed |
Worked example: washing machine spin
A drum of radius 0.25 m spinning at 1,200 RPM has ω = 125.7 rad/s, so a = ω²r = 3,948 m/s², about 403 g. A 1 kg wet towel presses outward with nearly 4 kN, which is why water is flung through the holes. For straight-line problems use the force calculator.
Centripetal force FAQs
F = m × v² ÷ r, where m is mass in kg, v is speed in m/s and r is the radius of the circle in meters. Using angular velocity ω (rad/s), F = m × ω² × r.
Square the speed, multiply by mass and divide by radius. A 1,000 kg car taking a 50 m radius bend at 20 m/s (72 km/h) needs 1,000 × 400 ÷ 50 = 8,000 N of sideways force.
a = v² ÷ r, the acceleration pointing toward the center of the circle. The car above experiences 8 m/s², about 0.82 g.
v = 2πr × RPM ÷ 60. A point 0.3 m from the center of a wheel spinning at 1,000 RPM moves at 2π × 0.3 × 1000 ÷ 60 = 31.4 m/s.
It is not a new kind of force but a role played by real forces: tire friction for a turning car, tension for a ball on a string, gravity for an orbiting satellite, and the normal force for a roller coaster loop.
v_max = √(μgr), where μ is the tire–road friction coefficient. On dry asphalt (μ ≈ 0.7) and a 50 m radius, v_max ≈ 18.5 m/s (67 km/h); on ice (μ ≈ 0.1) it drops to about 7 m/s.
θ = arctan(v² ÷ rg). At this angle a vehicle can take the curve with no friction at all. For 20 m/s on a 50 m radius, θ ≈ 39°.
Centrifugal force is an apparent force felt in a rotating frame of reference. From outside, the only real force is the inward centripetal force; your body's tendency to go straight is what feels like being pushed out.
No, it quadruples it, because force depends on v². Doubling the radius halves the force.
At the bottom of a loop riders feel 1 g plus v²/(rg). A 25 m/s train in a 20 m radius dip pulls about 1 + 3.2 = 4.2 g.
Newtons (N), the same as any force. Centripetal acceleration is in m/s² or multiples of g (9.80665 m/s²).
Lab centrifuges spin samples so fast that their centripetal acceleration reaches thousands of g, expressed as relative centrifugal force (RCF): RCF = 1.118 × 10⁻⁵ × r(cm) × RPM².
Sources
Curve speeds are physics estimates for a flat, unbanked road with steady speed. They are not driving advice: always follow posted limits and slow down well below the theoretical maximum in poor conditions.