Centrifugal Force of a Rotating Mass

The outward force a spinning mass throws - why rotating gear must be balanced and grinding wheels are speed-rated.

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Example

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Details, formula, and sources

F = (W/g) omega^2 r with omega = 2 pi N/60, plus the acceleration in g and the rim speed v = omega r. A 2 lb part at a 6 in radius at 1,800 rpm throws 1,104 lbf - 552 times its own weight - at 94 ft/s (5,655 ft/min) rim speed; the force climbs with the SQUARE of speed, so 3,600 rpm quadruples it to 4,400 lbf. That is why a small imbalance is violent at speed and a chipped wheel that is safe by hand can burst at rpm. Burst stress, bearing imbalance reaction, and whirl (critical) speed are separate. A design aid; Machinery Handbook and the equipment maker govern.

omega = 2 pi N / 60; F = (W/g) omega^2 r (g = 32.174 ft/s^2, r in ft); a_g = omega^2 r / g; v = omega r.

The centrifugal (centripetal) force F = (W/g) omega^2 r and rim speed v = omega r (standard dynamics; Machinery's Handbook), by name.

The centrifugal-force relation is a standard published dynamics result; the weight, radius, and speed are the user's inputs.

Estimate. AHJ and licensed professional govern.

Field names used by the API: weight_lb, radius_in, speed_rpm, centrifugal_force_lbf, acceleration_g

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