Example 1
A car travels at 14 m/s on a curve of radius 40 m. Centripetal acceleration is 4.9 m/s², or 0.4997 g.
a = 4.9 m/s²
Enter the tangential speed v in meters per second and the circle radius r in meters. The calculator finds centripetal acceleration from a = v²/r and reports it in m/s² and in units of standard g, together with angular velocity ω = v/r and the time for one lap. At v = 14 m/s and r = 40 m the result is 4.9 m/s², about 0.50 g. The model does not compute force F = m·v²/r and does not take ω or period T as input fields.
Enter tangential speed and radius. Centripetal acceleration, units of g, angular velocity, and the a(v) chart will appear here.
For the force that goes with this acceleration, use the F = m·v²/r calculator; for straight-line acceleration, a = Δv/Δt; and for linear speed from angular velocity, v = ω·r.
Centripetal acceleration describes how fast the direction of velocity changes in motion along a circle. When the speed v is constant and the path has radius r, the formula is a = v²/r. This is the centripetal acceleration of uniform circular motion. Christiaan Huygens derived a quantity of this magnitude in 1673, and Isaac Newton tied it to a force toward the center in Philosophiae Naturalis Principia Mathematica in 1687. For v = 14 m/s and r = 40 m the numerator is 196 (m/s)², and the result 196/40 = 4.9 m/s².
The calculator works with two numbers: the tangential speed v and the radius r in meters. The calculator squares the speed, so 20 m/s on the same 40 m radius gives 10 m/s², not twice 4.9. A comma and a period are the same number: 14.5 and 14,5. A minus sign on the speed does not change a, because the numerator is a square.
What the result panel shows. Besides a in m/s², the calculator reports the ratio a/g for standard gravity g = 9.80665 m/s² (0.4997 g in the reference example), the numerator v², the angular velocity ω = v/r (here 0.35 rad/s), the time for one lap T = 2πr/v (here 17.95 s), and the acceleration at twice the speed, 19.6 m/s². The panel also repeats the speed and radius you entered. The chart draws the parabola a(v) at the fixed radius, from half the entered speed to four times that speed.
Limits of the model. The formula assumes a constant speed. The radius is the radius of the path, not the width of the road: on a highway curve, r is the local radius of curvature. The calculator does not add weight, does not include banking or friction, and does not report a force. Force F = m·a is covered by the centripetal-force calculator. At the bottom of a loop the track must push harder than m·a, because weight is added. Motion that speeds up along the path has an extra tangential acceleration, which this formula does not include.
The unit switch in the header converts the fields and the panel between m/s and ft/s and between m and ft. The written examples stay in SI as the reference. The radius must be greater than zero, because division by zero does not describe a circle. At v = 0 the result a is 0, and the lap period is not defined.
Examples on this page: 14 m/s and 40 m give 4.9 m/s²; 28 m/s on a 120 m curve gives 6.533 m/s²; a centrifuge sample, 30 m/s on a 0.10 m radius, gives 9,000 m/s²; a rounded low orbit, 7,800 m/s and a radius of 6,771,000 m, gives 8.985 m/s². The button “Fill this example” enters the same numbers and refreshes the result panel and the chart.
a = v2 / r
[a] = m/s²; a = ω²·r with ω = v/r; T = 2πr/v; a/g₀ with g₀ = 9.80665 m/s²
Model: centripetal acceleration of uniform circular motion (Huygens, 1673; Newton’s centripetal force, 1687). The equivalent forms a = ω²·r and a = 4π²r/T² belong to the same model, but this calculator finds a from speed and radius. Order of computation: a = v²/r, then a/g₀, ω = |v|/r, T = 2πr/|v|, and a at speed 2v.
Centripetal acceleration a is the acceleration that changes the direction of velocity when a body moves in a circle. In uniform circular motion the speed v stays constant, and the velocity vector turns with the radius. The magnitude is a = v²/r, directed toward the center of the circle and perpendicular to the tangential velocity. The SI unit is the meter per second squared.
This is not the acceleration that speeds a body up or slows it down along the path. Tangential acceleration changes the speed and equals dv/dt. At constant v the straight-line change in speed is zero, and the direction still changes. That change of direction is what a = v²/r describes. The calculator reports the inward magnitude, not the x and y components.
Christiaan Huygens derived the magnitude v²/r in Horologium oscillatorium in 1673, describing the centrifugal force felt in a frame that rotates with the body. Isaac Newton introduced the name centripetal force, vis centripeta, for the inward force that curves the path in an inertial frame. He published it in Philosophiae Naturalis Principia Mathematica in 1687. The magnitude of the acceleration is the same in both accounts. What changes is the frame in which the reader describes it.
Introductory mechanics (Halliday, Resnick, and Walker; Young and Freedman) uses Newton’s account: acceleration toward the center, and a force F = m·a that keeps the body on the circle. The feeling of being thrown outward in a car on a curve is a description in a non-inertial frame, and it is not a separate result of this calculator.
Step 1. In a short time Δt the body travels an arc of length v·Δt. The radius turns through an angle Δθ = v·Δt/r, in radians. Step 2. The velocity vector has a fixed length v and turns through the same angle Δθ. The change in that vector therefore has length |Δv| = v·Δθ = v²·Δt/r, when the angle is small. Step 3. Acceleration is |Δv|/Δt, so a = v²/r. The direction of Δv, as the velocity vector turns, points toward the center.
Step 4. For v = 14 m/s and r = 40 m: v² = 196 (m/s)² and a = 196/40 = 4.9 m/s². Step 5. The angular velocity of this motion is ω = v/r = 0.35 rad/s, so the same magnitude is a = ω²·r = 0.35²·40 = 4.9 m/s². The time for one lap is T = 2πr/v = 17.95 s, and from that a = 4π²r/T². Twice the speed, 28 m/s, gives a = 19.6 m/s², because the numerator grows by a factor of four.
The formula is exact for uniform motion on a circle, or on an arc of constant radius of curvature, when the speed does not change. If a car on a curve is still speeding up, the full acceleration has two parts: the centripetal part v²/r and the tangential part dv/dt. The calculator computes only the first. The radius has to come from the geometry of the path. Divide a roundabout’s diameter by two before it goes into the radius field.
The model does not include road banking, tire friction, or the slope of the track. It also does not replace a local measurement of gravity. The constant g₀ = 9.80665 m/s², adopted by the 3rd General Conference on Weights and Measures in 1901, is used only to express the result in units of g. It is not the gravity at the reader’s location.
The same pair v and r yields two more kinematic quantities, with no extra measurement. Angular velocity ω = v/r is in radians per second. The lap period T = 2πr/v is the time for one full circuit. At v = 0 it is not defined, because the body never closes a circle. Frequency is f = 1/T, and ω = 2πf. A separate angular-velocity calculator finds ω from the period, and the calculator v = ω·r reverses the step ω = v/r.
Centripetal force has magnitude F = m·a = m·v²/r. This calculator has no mass field and does not report newtons. At the bottom of a vertical loop the track must supply more than m·a, because the weight mg points away from the center when the center is above the body. At the top of the loop, weight helps to curve the path. The calculator adds neither of those forces to a.
On a roundabout of radius 15 m at 8 m/s the acceleration is 4.267 m/s², or 0.4351 g. A car at 28 m/s on a 120 m curve has 6.533 m/s². A carousel seat, 8 m/s and 6 m from the axis, exceeds one standard g: 10.67 m/s². A stone on a 1.2 m string at 5 m/s has 20.83 m/s². The bottom of a loop, 20 m/s and a 10 m radius, gives 40 m/s² of path acceleration alone.
A laboratory centrifuge at 30 m/s and a radius of 0.10 m reaches 9,000 m/s², or 917.7 g. A rounded low orbit, 7,800 m/s on a radius of 6,771,000 m (6,371 km plus 400 km), has a centripetal acceleration of 8.985 m/s², or 0.9163 g. In a circular orbit that value equals the local gravitational acceleration: the craft and the passenger fall at the same a, so the passenger feels no normal force from the floor. Orbital acceleration is not zero.
The definition, the derivation, and the history come from mechanics textbooks and metrology sources: C. Huygens, Horologium oscillatorium, 1673; I. Newton, Philosophiae Naturalis Principia Mathematica, 1687; D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, the chapter on motion in a circle; H. D. Young and R. A. Freedman, University Physics, uniform circular motion; R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Vol. I, Chapter 12; BIPM, The International System of Units (SI), 9th edition; the 1901 declaration of the 3rd CGPM on standard gravity, g₀ = 9.80665 m/s². This page is for educational purposes.
a is centripetal acceleration, v is tangential speed, and r is the path radius. For v = 14 m/s and r = 40 m the result is 4.9 m/s².
A car travels at 14 m/s on a curve of radius 40 m. Centripetal acceleration is 4.9 m/s², or 0.4997 g.
a = 4.9 m/s²
On a curve of radius 120 m, a speed of 28 m/s, or 100.8 km/h, gives an acceleration of 6.533 m/s².
a = 6.533 m/s²
A carousel seat 6 m from the axis moves at 8 m/s. Centripetal acceleration is 10.67 m/s², above one standard g.
a = 10.67 m/s²
A car travels at 8 m/s on a roundabout of radius 15 m. The result is 4.267 m/s².
a = 4.267 m/s²
A centrifuge sample moves at 30 m/s on a circle of radius 0.10 m. Centripetal acceleration is 9,000 m/s², or 917.7 g.
a = 9,000 m/s²
A rounded low orbit: 7,800 m/s on a radius of 6,771,000 m, which is 6,371 km plus 400 km. Centripetal acceleration is 8.985 m/s².
a = 8.985 m/s²
A stone on a 1.2 m string moves at 5 m/s. Centripetal acceleration is 20.83 m/s².
a = 20.83 m/s²
A skater travels at 12 m/s on a curve of radius 25 m. The result is 5.76 m/s².
a = 5.76 m/s²
At the bottom of a loop of radius 10 m the speed is 20 m/s. The path’s centripetal acceleration is 40 m/s². The calculator does not add weight to that value.
a = 40 m/s²
A classroom check: a speed of 10 m/s and a radius of 35 m give an acceleration of 2.857 m/s².
a = 2.857 m/s²
The calculator finds the centripetal acceleration of uniform circular motion from a = v²/r. At 14 m/s and a radius of 40 m the result is 4.9 m/s², or 0.4997 standard g, with an angular velocity of 0.35 rad/s and a lap time of 17.95 s.
Enter speed in meters per second and radius in meters. 50 km/h is about 13.9 m/s, not 50 in the speed field. Divide a path diameter by two. The unit switch converts the fields to ft/s and ft; the written examples stay in SI.
The velocity vector turns faster when the speed itself is higher, and the angle covered in a given time also grows with v. That is why the numerator is v². Twice the speed on the same radius gives four times a: 28 m/s and 40 m give 19.6 m/s².
The formula a = Δv/Δt describes how speed changes with time, usually in a straight line. The formula a = v²/r describes a change in the direction of velocity at constant speed. At constant |v| the first quantity on a straight line is zero, and the second quantity on a curve is positive.
It is the ratio a / g₀, where g₀ = 9.80665 m/s² is the conventional standard gravity from 1901, not local gravity. For 4.9 m/s² the ratio is 0.4997 g. The panel shows that number next to the result in m/s².
Angular velocity is ω = v/r, and the time for one lap is T = 2πr/v. For 14 m/s and 40 m those values are 0.35 rad/s and 17.95 s. The same acceleration equals ω²·r and 4π²r/T². At v = 0 the period is not defined, and the panel shows a dash.
No. There is no mass field. Force has magnitude F = m·a = m·v²/r and is computed on the centripetal-force page. The result here is an acceleration in m/s².
At v = 0 and a positive radius the acceleration is 0, because the direction of velocity is not changing. The lap period is not defined in that case. At r = 0 the calculator shows no result, because division by zero does not describe a circle.
Yes, if the path has a radius of curvature r at that point and the speed is constant. On a road curve, r is the local radius, not the width of the pavement. The formula does not include banking or friction.
a = v²/r is the acceleration of the body toward the center of the circle, not the contact force. At the bottom of a vertical loop the center is above the body, and weight points downward, away from the center. The force from the track must exceed m·a, because the weight also acts downward. The calculator does not add mg to the result. At 20 m/s and r = 10 m, the centripetal acceleration alone is 40 m/s².
It does not change the acceleration, because the numerator is the square of the speed. The field describes the tangential speed. Angular velocity and the period in the panel use the absolute value, so a minus sign does not reverse the reported direction of rotation.
Christiaan Huygens derived the magnitude v²/r in 1673. Isaac Newton described the matching centripetal force in the Principia in 1687. The classroom derivation compares the turning of the velocity vector with the angle swept along the arc: |Δv|/Δt = v²/r.
Divide the square of the tangential speed by the radius. At v = 14 m/s and r = 40 m you get 196/40 = 4.9 m/s², about 0.50 g.
8 m/s², because 400/50 = 8. That is about 0.82 standard g. The panel also shows ω = 0.4 rad/s.
They give the same result for consistent data, because v = ω·r. This calculator takes v and r; ω = v/r appears in the panel, not as a separate input field.
Divide km/h by 3.6 (72 km/h → 20 m/s). For mph, multiply by 0.447. Entering km/h without converting overstates a about thirteen times.
No. The input fields are v and r. Period T = 2πr/v and ω = v/r are companion results. Force F = m·a belongs on the centripetal-force page.
The definition, the unit, the history, and the derivation come from established physics and metrology sources.
Page updated November 3, 2026.