Centripetal force and acceleration from mass, speed and radius; rpm to rad/s and period; moment of inertia of nine shapes; torque from force and lever arm.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
Keeping 1,000 kg on a 50 m circle at 20 m/s takes 8,000 N directed at the centre: an acceleration of 0.816 g.
Centripetal force is the inward pull that keeps a mass on a circular path: F = mv²/r, with the centripetal acceleration a = v²/r = ω²r pointing at the centre. The other modes convert a rotation rate between rpm, rad/s, hertz and period (ω = 2πf), give the moment of inertia I of nine standard shapes, and work out torque τ = rF sin θ and angular acceleration α = τ/I.
The default is a 1,000 kg car taking a 50 m radius curve at 20 m/s (72 km/h). The tyres must supply 8,000 N of sideways grip, an acceleration of 8 m/s² or 0.816 g. In torque mode, a 50 N push at 90° on a 0.3 m wrench gives 15 N·m.
Results assume uniform circular motion and rigid bodies of uniform density. Mass farther from the axis raises I, so a thin hoop has twice the moment of inertia of a solid disc with the same mass and radius.
Centripetal force is F = mv²/r: mass times speed squared, divided by the radius of the circle. A 1,000 kg car taking a 50 m curve at 20 m/s (72 km/h) needs 1,000 × 20² ÷ 50 = 8,000 N of inward force, an acceleration of 8 m/s² or 0.816 g.
The acceleration points at the center of the circle: a = v²/r = ω²r. The speed stays constant, but the direction of the velocity changes every instant, and that change needs a net inward force.
Centripetal force is a job other forces do. Tire friction supplies it for a cornering car, string tension for a whirled ball and gravity for a satellite.
Speed matters most because it is squared. At 40 m/s the same car needs 32,000 N, four times as much. On a 100 m curve at 20 m/s it needs 4,000 N, half as much.
Solved for speed, v = √(Fr/m). A 500 N inward force on a 10 kg mass moving on a 2 m circle allows 10 m/s.
One revolution is 2π radians, so ω = 2π × rpm ÷ 60, or rpm × 0.10472. 3,000 rpm is 314.159 rad/s, 50 revolutions per second, and one turn takes 0.02 s. To go back, multiply rad/s by 9.5493.
A point at radius r moves at v = ωr. At 0.1 m from the axis of a 3,000 rpm rotor, that is 31.42 m/s, and the centripetal acceleration ω²r is 9,870 m/s², about 1,006 g.
Moment of inertia, I, plays the part of mass in rotation: the farther the mass sits from the axis, the larger I is. For a uniform body of mass M (Serway and Jewett, Table 10.2):
For 5 kg at R = 0.2 m, a hoop gives 0.2 kg·m², a solid disc 0.1 kg·m² and a solid sphere 0.08 kg·m². A 2 kg, 3 m rod has 1.5 kg·m² about its center but 6 kg·m² about one end (OpenStax).
A spinning body stores KE = ½Iω² and carries angular momentum L = Iω. The 2 kg rod turning about its end at 60 rpm (ω = 2π rad/s) stores 118.4 J and has L = 37.70 kg·m²/s.
Torque is τ = rF sin θ, where r is the distance from the axis to where the force acts and θ is the angle between the arm and the force. A 50 N push at 90° on a 0.3 m wrench gives 15 N·m, or 11.06 lbf·ft.
At 30° the same push gives 7.5 N·m, and along the handle it gives none. Dividing a net torque by the moment of inertia gives the angular acceleration, α = τ/I: 15 N·m on 0.5 kg·m² is 30 rad/s².
Default is the 1,000 kg car at 20 m/s on a 50 m curve. Angular rpm is 3,000 rpm, inertia sphere is a 5 kg solid sphere of radius 0.2 m, and torque is 50 N at 90° on a 0.3 m arm.
| Scenario | Selected result |
|---|---|
| Default | 8,000 N |
| Angular rpm | 314.159 rad/s |
| Inertia sphere | 0.08 kg·m² |
| Torque | 15 N·m |
For the forces behind a curve, use the force and friction calculator. To turn torque and rpm into engine power, use the horsepower calculator, and for orbits, the gravity and orbit calculator.
Checked against: a = 400/50 = 8, F = 8000, ω = v/r = 0.4
Checked against: v = √(Fr/m) = √100
Checked against: ω = 2π·3000/60, v = ωr
Checked against: Serway Table 10.2: I = 2/5 MR² = 0.4 × 5 × 0.04
F = mv²/r, with mass m in kilograms, speed v in metres per second and radius r in metres, giving newtons. Equivalent forms are F = mω²r and F = 4π²mr/T². Doubling the speed quadruples the force and doubling the radius halves it: a 1,000 kg car at 20 m/s on a 50 m curve needs 8,000 N, and at 40 m/s it would need 32,000 N.
Multiply rpm by 2π/60, about 0.10472. 3,000 rpm is 314.159 rad/s, or 50 revolutions per second (50 Hz), so one revolution takes 0.02 s. To go back, multiply rad/s by 60/(2π), about 9.5493. The radian is the coherent SI unit of plane angle (BIPM SI Brochure), so rad/s is the SI unit of angular velocity.
Not in an inertial, non-rotating frame of reference: there, the only horizontal force on a cornering car is the inward centripetal force from the tyres. Centrifugal force appears only when motion is described from inside the rotating frame, where it has the same size, mv²/r, pointing outward. The outward push a passenger feels is their body's inertia carrying it in a straight line while the car turns.
I = ½MR² about its central axis, so a 5 kg disc of radius 0.2 m has I = 0.5 × 5 × 0.2² = 0.1 kg·m². Other standard results are MR² for a thin hoop, ⅖MR² for a solid sphere, ⅔MR² for a thin spherical shell, and ML²/12 or ML²/3 for a thin rod about its centre or one end (Serway and Jewett, Table 10.2).
Torque is τ = rF sin θ, where r is the distance from the axis to the point where the force acts and θ is the angle between the arm and the force. A 50 N push at 90° on a 0.3 m wrench gives 15 N·m; the same push along the handle (θ = 0°) gives none. One newton-metre is 0.7376 pound-force feet.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.