Centrifugal Force Calculator

Centrifugal Force Calculator

Calculate the outward radial force on a rotating mass. Enter mass, radius to the center of gravity, and rotational speed (RPM, rad/s, or linear velocity). Get force in pounds-force or newtons, plus g-force, tangential velocity, and angular velocity. Imperial and metric.

Centrifugal force is the outward radial reaction force a rotating body exerts on its axis or restraint. Its magnitude is F = m × v² / r = m × r × ω², where m is mass, r is the radius to the center of mass, and ω is angular velocity in radians per second. Engineers use it to size flywheel rims, balance rotating shafts, design pulleys and centrifuges, and check bolted joints on rotating machinery.

At standard gravity, weight in lbf equals mass in lbm numerically.
Distance from axis to center of mass.
Revolutions per minute (e.g., motor nameplate).

Centrifugal Force Equations

General Form (any consistent units)

F = m × v² / r = m × r × ω²

where:
F = centrifugal force [force] — e.g. lbf, N, kN
m = mass of revolving body [mass] — e.g. slug, lbm, kg
v = tangential velocity at radius r [length / time] — e.g. ft/s, m/s
r = radius from axis to center of mass [length] — e.g. in, ft, m
ω = angular velocity [angle / time] — e.g. rad/s

Machinery’s Handbook 29th Ed., p. 187

Imperial Form (weight in pounds)

F = W × v² / (g × R)

where:
F = centrifugal force in pounds-force (lbf)
W = weight of revolving body in pounds (lb)
v = tangential velocity in feet per second (ft/s)
g = acceleration due to gravity = 32.174 ft/s²
R = radius in feet (ft)

Machinery’s Handbook 29th Ed., p. 187

RPM Form (small dimensions, radius in inches)

F = 0.000028416 × W × r × n²

where:
F = centrifugal force in pounds-force (lbf)
W = weight of revolving body in pounds (lb)
r = mean radius in inches (in)
n = rotational speed in revolutions per minute (RPM)

Machinery’s Handbook 29th Ed., p. 188

Metric SI Form

F = M × R × ω² = 0.01097 × M × R × n²

where:
F = centrifugal force in newtons (N)
M = mass of revolving body in kilograms (kg)
R = radius in meters (m)
ω = angular velocity in radians per second (rad/s)
n = rotational speed in revolutions per minute (RPM)

Machinery’s Handbook 29th Ed., pp. 187–188

Solving for Speed, Radius, or Mass

The same relationship rearranged for the other variables (per MH29 p. 187):

v = √(F × r / m)      r = F × r / (m × v²)

n (RPM) = √(2933 × F / (W × R))      m = F × r / v²

Machinery’s Handbook 29th Ed., p. 187 (rearranged forms shown explicitly in the source)

G-Force (Relative Centrifugal Force)

g-force = r × ω² / g₀

where:
g₀ = standard gravity = 32.174 ft/s² = 9.80665 m/s²
r, ω in any consistent units

Derived from the centrifugal acceleration a = r × ω² (MH29 p. 187), divided by standard gravity per CGPM 1901 / NIST SP 811.

How to Use This Calculator

Enter the mass (or weight in pounds) of the rotating body, the radius from the axis of rotation to its center of gravity, and the rotational speed. Use RPM if you have a motor nameplate or tachometer reading; switch to angular velocity (rad/s) for analytical work, or to linear velocity if you know the tip speed of a grinding wheel or belt. The calculator returns the centrifugal force in pounds-force (or newtons), along with g-force, tangential velocity, and angular velocity.

For a flywheel rim or pulley rim, use the mean radius — the radius to the centroid of the rim cross-section — not the outer diameter. For a balancing problem, the relevant mass is the unbalanced mass and the radius is the distance from the spin axis to that mass’s center of gravity. To pair this with shaft sizing, use our beam deflection calculator to check whether the resulting radial load causes excessive shaft deflection.

Worked Example

Problem: A cast-iron flywheel has a rim weighing 20 lb (9.07 kg) with a mean rim radius of 9 in (229 mm). It rotates at 800 RPM. What is the centrifugal force on the rim?

Given: W = 20 lb, r = 9 in, n = 800 RPM

Using the RPM form: F = 0.000028416 × 20 × 9 × 800²

F = 0.000028416 × 20 × 9 × 640,000 = 3,274 lbf (14,560 N)

That is roughly 164 times the rim’s static weight — a useful check against the flywheel’s burst speed and against the tensile capacity of the spokes or hub bolts.

Common Engineering Applications

  • Flywheel and pulley rim stress — sizing rim thickness and spoke loads against centrifugal hoop stress.
  • Rotor balancing — computing the corrective mass and angle to offset measured imbalance forces.
  • Grinding wheel safety — checking that operating speed stays well below burst speed for the wheel’s rated peripheral velocity.
  • Lab centrifuges — converting between RPM and Relative Centrifugal Force (RCF) for protocols specified in g-units.
  • Vehicles in curves — sizing tire grip, banking angle, and driver lateral g for cornering analysis.
  • Bolted joints on rotating equipment — calculating the radial load each fastener must clamp against.

Frequently Asked Questions

What is the formula for centrifugal force?

The standard formula is F = m × v² / r, equivalent to F = m × r × ω² when angular velocity is known. In imperial form using weight: F = W × v² / (g × R), where g = 32.174 ft/s². For RPM input, the convenient form is F = 0.000028416 × W × r × n² with W in pounds, r in inches, and n in RPM — force is then in lbf. (Machinery’s Handbook 29th Ed., p. 187.)

How do you calculate centrifugal force from RPM?

Convert RPM to angular velocity first: ω (rad/s) = RPM × 2π / 60, or about RPM × 0.10472. Then F = m × r × ω². For example, a 20 lb flywheel rim section at 9 in radius spinning at 800 RPM produces about 3,274 lbf of outward force per the MH29 worked example.

How do I plug RPM and inches into F = m × r × ω² in imperial units?

Two steps people miss.

1. Convert RPM to angular velocity: ω (rad/s) = n × 2π / 60 ≈ n × 0.10472.

2. Pick a consistent mass-length pair. F = m × r × ω² only gives lbf directly when m is in slugs and r is in feet. If you keep weight in lbm (the lb on a scale) and radius in inches, divide the result by gc = 386.09 in·lbm/(lbf·s²) — that single constant absorbs the lbm-to-slug factor (32.174) and the inch-to-foot factor (12).

Or skip the gymnastics with the MH29 small-dimensions form F = 0.000028416 × W × r × n², which already has all conversions baked in for W in lb, r in inches, and n in RPM. The 386.09 gc is the most common manual-calculation slip-up. (Machinery’s Handbook 29th Ed., pp. 187–188.)

What is the difference between centrifugal and centripetal force?

They are the same magnitude but opposite directions. Centripetal force is the inward force the rotating axis (or rope, or bearing) exerts on the body to keep it on a curved path. Centrifugal force is the equal and opposite reaction that the body exerts outward on the axis or restraint. For sizing bolts, shafts, or housings on rotating equipment, you size for the centrifugal reaction load.

How do I convert centrifugal force to g-force?

Divide the centrifugal acceleration by standard gravity (g = 32.174 ft/s² or 9.80665 m/s²). The acceleration is a = v² / r = r × ω², so g-force = r × ω² / g. For RPM: g-force ≈ 1.118×10⁻⁵ × r(in) × RPM² with r in inches. This is the same value lab centrifuges call Relative Centrifugal Force (RCF).

Why is g in the imperial centrifugal force formula?

Because W in the formula is a weight in pounds-force, not a mass. Newton’s law requires mass: F = m × v² / r. To recover mass from weight you divide by gravity, since W = m × g, so m = W / g (in slugs). The g in the denominator is just that conversion — it strips gravity out of weight to get true mass. Centrifugal force itself has nothing to do with gravity. The SI form has no g because mass is entered directly in kilograms. (Machinery’s Handbook 29th Ed., p. 187.)

What radius should I use for a flywheel or pulley rim?

Use the mean radius of the rim — measured to the centroid (center of gravity) of the rim cross-section, not the outer diameter. Per MH29 p. 187, this is because the rim section behaves as if its mass were concentrated at its centroid, and centrifugal force depends on the radius to the center of mass. For a thin rim, mean radius is close to the outer radius; for a thick rim, it can be noticeably less.

Does centrifugal force depend on direction of rotation?

No. Centrifugal force is purely radial — outward from the axis of rotation — and its magnitude depends only on mass, radius, and the square of speed. Reversing rotation direction does not change the magnitude or the radial direction of the force. This is why balancing problems care about the angular position of unbalanced mass, not the direction of spin.

What is a safe RPM for a flywheel or rotating part?

There is no single safe RPM — it depends on the part’s mass, radius, geometry, and material strength. The hoop stress in a rotating ring is σ = ρ × r² × ω², where ρ is density. For a steel ring, the burst speed corresponds to hoop stress reaching the material’s ultimate strength; design speed should stay well below that with a safety factor of 4–10 depending on the application. Always reference the manufacturer’s rated speed and consult ANSI/ASME standards (e.g., ANSI B7.1 for grinding wheels) before exceeding nameplate values.

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References

  • Oberg, E. et al. Machinery’s Handbook, 29th Edition, Industrial Press, 2012, p. 187: centrifugal force definition and primary formulas F = Wv² / (gR) and rearranged forms.
  • Oberg, E. et al. Machinery’s Handbook, 29th Edition, p. 188: small-dimensions form F = 0.000028416 × Wrn² with worked example (20 lb, 9 in, 800 RPM → 3,273.52 lbf), and metric form F = 0.01097 MRn² with worked example (5 kg, 120 mm, 1,100 RPM → 7,961.50 N).
  • Taylor, B. N., Thompson, A. The International System of Units (SI), NIST Special Publication 811, 2008: standard gravity g₀ = 9.80665 m/s² (32.17404856 ft/s²).

Data last verified: April 2026

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