2026-08-21 15 min

Centripetal Force Calculator

Compute centripetal force using F = mv²/r or F = mω²r for circular motion analysis. Calculate centripetal acceleration, g-force, and angular velocity for vehicle dynamics, rotating machinery, and orbital mechanics.

Calculator

Mass(kg) Help

Mass of the object in circular motion

Tangential Velocity (v)(m/s) Help

m/s

Tangential speed along the circular path

Radius(m) Help

Radius of the circular path

Velocity Mode

Choose input type for velocity

Decimal Precision

Number of decimal places in results

Results update live as you type

Results

Optimal

Centripetal Force

800.00N

F_c = 10.00 × 20.00² / 5.00

Centripetal Acceleration

80.00m/s²

a_c = v²/r = ω²r

G-Force

8.16g

a_c / g = 80.00 / 9.80665

Angular Velocity

4.00rad/s

ω = v/r

Rotational Speed

38.20RPM

ω × 60 / (2π)

Period

1.57s

T = 2π/ω

Results are exact for the given inputs using the standard centripetal force formula. Real-world applications may require additional analysis for the specific force-providing mechanism, dynamic loading, and material limits.

Circular Motion Diagram

F_c

800.00 N

a_c

80.00 m/s\u00B2

g-force

8.16 g

\u03C9

4.00 rad/s

RPM

38.20

Period

1.57 s

Formula

Centripetal force is the net inward force required to keep an object moving in a circular path. It is directed toward the center of rotation and is proportional to mass and the square of velocity, and inversely proportional to the radius of curvature.

F_c = mv²/r = mω²r

Alternative Form

a_c = v²/r = ω²r

Variables

Symbol	Name	Unit	Description
F_c	Centripetal Force	N	The net inward force directed toward the center of the circular path
m	Mass	kg	The mass of the object in circular motion
v	Tangential Velocity	m/s	The instantaneous speed of the object along its circular path
r	Radius	m	The radius of the circular path
ω	Angular Velocity	rad/s	The rate of rotation, related to linear velocity by v = ωr
a_c	Centripetal Acceleration	m/s²	The inward acceleration: a_c = v²/r = ω²r

Explanation

Centripetal acceleration changes the direction of the velocity vector without changing its magnitude. The force required increases quadratically with speed — doubling the speed requires four times the force. This is why high-speed turns are more demanding than low-speed turns.

Derivation

For uniform circular motion, the velocity vector ν is always tangent to the circle. The change in velocity over a small time Δt points radially inward: |Δν| = v × Δθ. Dividing by Δt gives centripetal acceleration: a_c = v × dθ/dt = v × (v/r) = v²/r. Substituting v = ωr gives a_c = ω²r. Newton's second law then gives F_c = ma_c.

Step-by-Step Calculation

Follow each step to understand how the result is calculated from your inputs.

Step 1

Identify the mass m of the object in circular motion.

m = given

Mass (m):10.00 kg

Result: m = 10.00 kg

Step 2

Identify the tangential velocity v in meters per second.

ω = given

v:20.00 m/s

Result: v = 20.00 m/s

Step 3

Identify the radius r of the circular path.

r = given

Radius (r):5.00 m

Result: r = 5.00 m

Step 4

Calculate centripetal acceleration.

a_c = v²/r

v:20.00

r:5.00

Intermediate: a_c = 20.00² / 5.00 = 80.00 m/s²

Result: a_c = 80.00 m/s²

Step 5

Compute centripetal force using Newton's second law.

F_c = m × a_c

m:10.00 kg

a_c:80.00 m/s²

Intermediate: F_c = 10.00 × 80.00 = 800.00 N

Result: F_c = 800.00 N

Step 6

Calculate derived quantities: g-force, angular velocity, RPM, and period.

g-force = a_c/g, ω = v/r, RPM = ω×60/(2π), T = 2π/ω

a_c:80.00 m/s²

g:9.80665 m/s²

Intermediate: g-force = 80.00 / 9.80665 = 8.16 g

Result: 8.16 g, 4.00 rad/s, 38.20 RPM, T = 1.57 s

Final Answer

800.00

Worked Examples

See how the calculator works with different scenarios.

Scenario

A 1500 kg car turns at 13.4 m/s (48 km/h) on a 30 m radius curve. What centripetal force is required, and what friction coefficient is needed?

Inputs

Mass:1500 kg

Velocity:13.4 m/s

Radius:30 m

Calculation

a_c = v²/r = 13.4² / 30 = 5.99 m/s² F_c = m × a_c = 1500 × 5.99 = 8985 N g-force = 5.99 / 9.81 = 0.61 g μ_required = F_c / (mg) = 8985 / (1500 × 9.81) = 0.61

Result

F_c = 8985 N (0.61 g)

Interpretation

The car needs 8985 N of inward force, provided by tire friction. A friction coefficient of 0.61 is required — dry asphalt (μ ≈ 0.7) can handle this, but wet pavement (μ ≈ 0.3) cannot. The driver would skid on a wet road at this speed and radius.

Understanding Your Result

Your Result: Centripetal Force: 800.00 N

Typical range: Car cornering: 1,000–10,000 N. Roller coaster loops: 5,000–50,000 N. Satellite orbit: 10⁳–10⁵ N. Centrifuge: 0.1–1000 N. Fighter jet turn: 10⁵–10⁶ N. The g-force is often more informative than the raw force value.

What This Means

The centripetal force calculator determines the net inward force required to keep an object moving in a circular path of given radius at a given speed. The result represents the real force that must be supplied by some physical mechanism (friction, tension, gravity, etc.) to maintain circular motion.

Why It Matters

Centripetal force is fundamental to vehicle dynamics (cornering limits), rotating machinery (structural loads), amusement park design (ride safety), orbital mechanics (satellite operations), and particle physics (cyclotrons). Understanding it is essential for designing any system involving circular motion.

Professional Interpretation

In professional engineering, centripetal force calculations must account for the actual force-providing mechanism. For cars, this means analyzing tire-road friction and weight transfer. For rotating machinery, it means evaluating bearing loads and structural stress. The g-force metric provides a quick safety assessment: above 4 g for untrained humans, 6–9 g for trained pilots with g-suits.

Force sources vary by application

The centripetal force is not a new type of force — it is the label given to whatever existing force (friction, tension, gravity, normal force) points inward. A car relies on tire friction, a satellite on gravity, a ball-on-string on tension, and a car on a banked road on the horizontal component of the normal force.

Quadratic speed dependence

Force scales with v², so doubling speed quadruples the required force. This is why highway exit ramps have large radii — it keeps the centripetal force (and thus required friction) within safe limits at high speeds.

Centripetal vs. centrifugal interpretation

In an inertial (stationary) reference frame, only centripetal force exists. In a rotating frame (e.g., sitting in a turning car), you feel an apparent outward force called centrifugal force. Both descriptions are valid in their respective frames; engineers use whichever is more convenient for the analysis.

Recommendations

•Always verify the force-providing mechanism can supply the required centripetal force
•Apply a safety factor of 1.5–2.0 for design applications
•Check g-force limits for human-rated systems (typically <4 g for untrained, <9 g for trained pilots)
•Account for dynamic loads: acceleration and deceleration increase instantaneous force requirements
•Consider temperature effects on material strength for high-speed rotating machinery
•For vehicles, analyze both steady-state cornering and transient (entry/exit) conditions

Educational Content

Introduction

Centripetal force is the net inward force required to keep an object moving in a circular path. Without it, Newton's first law dictates the object continues in a straight line. The term comes from Latin centrum (center) and petere (to seek) — literally "center-seeking." It is one of the most important concepts in classical mechanics.

Concept Explanation

Any object moving in a circle must experience a net inward acceleration (centripetal acceleration), which requires a net inward force per Newton's second law. The magnitude depends on speed squared and inversely on radius: a_c = v²/r. The force is always directed toward the center of curvature and is perpendicular to the velocity, changing its direction without changing its magnitude.

Applications

▸Vehicle dynamics: analyzing cornering limits, tire grip requirements, and road design

▸Amusement park engineering: designing safe roller coaster loops, spinning rides, and centrifuge attractions

▸Satellite and orbital mechanics: calculating orbital velocities, transfer orbits, and re-entry conditions

▸Rotating machinery: evaluating bearing loads, shaft stresses, and balancing requirements

▸Centrifuge design: laboratory and industrial centrifuges for separation processes

▸Particle physics: cyclotrons and synchrotrons for accelerating charged particles

▸Washing machine design: spin cycle optimization for water extraction

▸Aircraft and spacecraft: artificial gravity generation through rotation

Benefits

•Fundamental to understanding all circular motion in nature and engineering
•Enables safe design of roads, tracks, rides, and rotating equipment
•Connects linear and rotational dynamics through v = ωr
•Provides the basis for understanding orbital mechanics and satellite operations
•Essential for human safety analysis in high-g environments
•Applicable across scales from subatomic particles to planetary orbits

Limitations

•Assumes uniform circular motion (constant speed); non-uniform motion requires additional tangential force analysis
•Does not account for drag, friction losses, or energy dissipation
•For elastic bodies, centripetal deformation changes the effective radius
•Relativistic effects become significant at very high velocities
•Requires knowledge of the actual force-providing mechanism for practical design

Best Practices

•Always identify the physical force providing the centripetal force in real systems
•Check g-force limits for human-rated systems and apply appropriate safety factors
•Use the quadratic speed relationship to understand why high-speed design is more demanding
•Verify that the available friction, tension, or gravity exceeds the required centripetal force
•Consider both radial and tangential force components for non-uniform circular motion
•Account for dynamic effects: entering and exiting curves creates transient loads

Common Misconceptions

Myth: Centrifugal force is a real outward force pushing objects outward.

Truth: Centrifugal force is a fictitious force that appears only in a rotating reference frame. In an inertial frame, objects move in straight lines (Newton's first law); the centripetal force curves their path inward. What feels like being "pushed outward" is actually your body's inertia trying to continue in a straight line.

Myth: Heavier objects need more centripetal force per unit mass.

Truth: While F_c ∝ m, the required friction coefficient μ = v²/(gr) is independent of mass. A car and a truck need the same friction coefficient for the same speed and radius, even though the car needs less absolute force.

Myth: Centripetal force is a new type of force like gravity or friction.

Truth: Centripetal force is not a fundamental force — it is the label for whatever existing force (friction, tension, gravity, etc.) happens to point inward. There is no "centripetal force" force; there is only the inward component of the actual forces acting.

Myth: An object in uniform circular motion has constant velocity.

Truth: Velocity is a vector with both magnitude and direction. In circular motion, the direction constantly changes, so velocity is never constant. This is why centripetal acceleration exists — it changes the direction of the velocity vector.

Myth: Centripetal force does no work because it is perpendicular to motion.

Truth: This is true for uniform circular motion. Since the force is always perpendicular to the velocity (and thus the displacement), the work done by centripetal force is zero. This is why uniform circular motion requires no energy input after establishing the motion.

Practical Tips

•Use the mode selector: linear velocity (v) for cars and projectiles, angular velocity (ω) for rotating machinery
•Remember that centripetal force scales with v² — doubling speed quadruples the force
•Check the g-force result: above 4 g can cause discomfort, above 9 g can cause unconsciousness
•For cars on curves, verify the required friction coefficient is available: μ = v²/(gr)
•Banked curves reduce the friction requirement by using the normal force component
•For satellite orbits, centripetal force equals gravitational force: F_c = GMm/r²
•Use RPM-to-rad/s conversion: ω (rad/s) = RPM × 2π/60

Common Mistakes

Confusing centripetal (inward) with centrifugal (outward) force

In an inertial frame, only centripetal (inward) force exists. Centrifugal force is apparent only in a rotating reference frame. Always analyze from an inertial frame to avoid errors.

Forgetting that force scales with v², not v

A common error is assuming doubling speed doubles the force. In reality, F_c ∝ v², so doubling speed quadruples the force. Always use v² in calculations.

Using the wrong velocity (linear vs. angular)

Ensure v is in m/s and r in m, or ω is in rad/s and r in m. Mixing RPM with m/s or using diameter instead of radius produces incorrect results.

Ignoring the force-providing mechanism

The formula gives the required force. In practice, you must verify that friction, tension, gravity, or another mechanism can actually supply this force. Insufficient force means the object leaves its circular path.

Using diameter instead of radius

The formula uses radius, not diameter. If given diameter, divide by 2 first. Using diameter doubles the radius, which halves the calculated force — a potentially dangerous error.

Assumptions

This calculator makes the following assumptions to provide estimates:

Uniform Circular Motion: The calculator assumes constant speed around the circle. For non-uniform motion (speeding up or slowing down), additional tangential acceleration must be considered.

Point Mass: The object is treated as a point mass at radius r. For extended objects, the center of mass location and moment of inertia may be needed.

Rigid Path: The circular path is assumed to be fixed. In practice, elastic deformation under load may alter the effective radius.

No Drag or Friction Losses: The calculation gives the required centripetal force without accounting for energy lost to air resistance, rolling resistance, or friction.

Classical Mechanics: Relativistic effects are ignored. At velocities approaching the speed of light, relativistic corrections are needed.

Limitations

Be aware of these limitations when using this calculator:

Uniform Motion Only: Non-uniform circular motion (changing speed) requires additional tangential force analysis. This calculator handles only the radial (centripetal) component.

No Energy Analysis: The calculator does not compute power, kinetic energy, or energy losses. Centripetal force does zero work in uniform circular motion, but real systems have friction and drag.

Material Limits Not Checked: The calculated force is not compared against material strength limits. Structures, strings, and bearings may fail under the computed load.

No Dynamic Effects: Transient conditions (entering/exiting curves, starting/stopping rotation) create additional forces not captured by the steady-state formula.

Frequently Asked Questions

Answers to common questions about this calculator.

What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directed toward the center of rotation. It is given by F_c = mv²/r = mω²r. Despite common misconceptions, centripetal force is a real, inward-directed force required for circular motion — it is not an apparent or fictitious force. Gravity, tension, friction, or normal forces can provide centripetal force.

How do you calculate centripetal force?

Use F_c = mv²/r, where m is mass (kg), v is tangential velocity (m/s), and r is radius (m). For a 10 kg mass moving at 20 m/s in a 5 m radius circle: F_c = 10 × 20² / 5 = 800 N. Alternatively, using angular velocity: F_c = mω²r. This is about 8.16 times the weight of the object (≈ 8 g).

What is the difference between centripetal and centrifugal force?

Centripetal force is a real, inward force acting on the object toward the center of rotation, required for circular motion. Centrifugal force is a fictitious (apparent) force experienced in a rotating reference frame, acting outward. In an inertial frame, only centripetal force exists; centrifugal force appears only when analyzing the situation from a rotating perspective.

What units are used for centripetal force?

The SI unit for centripetal force is the newton (N). The related centripetal acceleration has units m/s². In imperial units: pound-force (lbf). Typical magnitudes: a car rounding a curve experiences ~3000 N, a satellite in orbit ~10⁴ N, and an electron orbiting a nucleus ~10⁻⁸ N.

What is centripetal acceleration?

Centripetal acceleration is the acceleration directed toward the center of a circular path: a_c = v²/r = ω²r. It is sometimes called radial acceleration. For a mass on a 1 m string spinning at 10 m/s: a_c = 10²/1 = 100 m/s² ≈ 10.2 g. Note that centripetal acceleration changes only the direction of velocity, not its magnitude.

How does centripetal force relate to circular motion?

For any object to move in a circle, there must be a net inward force (centripetal force). Without it, the object would continue in a straight line per Newton's first law. The required centripetal force is proportional to the square of speed — doubling speed quadruples the required force. This is why high-speed turns require much larger forces.

What forces can provide centripetal force?

Common sources: 1) Tension — ball on a string, 2) Friction — car tires on road, 3) Gravity — satellite orbiting Earth, 4) Normal force — banked curve wall, 5) Magnetic force — charged particles in magnetic fields, 6) Electromagnetic forces — electrons orbiting nuclei. Often multiple forces combine to provide the net centripetal force.

How do you calculate centripetal force on a car turning a corner?

For a 1500 kg car turning at 15 m/s (54 km/h) on a 50 m radius curve: F_c = 1500 × 15² / 50 = 6750 N. This force is provided by friction between tires and road. The friction coefficient required: μ = F_c / (mg) = 6750 / (1500 × 9.81) = 0.46. On a wet road (μ ≈ 0.3), the car would skid.

What is the centripetal force on a satellite in orbit?

For a satellite in low Earth orbit (r ≈ 6,371,000 + 200,000 = 6,571,000 m), v ≈ 7800 m/s: a_c = v²/r = 7800²/6,571,000 ≈ 9.26 m/s². The centripetal force is provided entirely by gravity: F_c = G × M_earth × m_satellite / r² = m_satellite × a_c.

What happens if centripetal force is insufficient?

If the available centripetal force is less than what is required (mv²/r), the object will move in a larger radius path or fly off tangentially. For a car, this means skidding outward. For a ball on a string, the string breaks and the ball flies off tangent to the circle. For an orbit, the satellite would enter a higher orbit or escape.

How does radius affect centripetal force?

Centripetal force is inversely proportional to radius: F_c ∝ 1/r. Halving the radius doubles the required force for the same speed. This is why tight turns require more friction/grip and why race cars take the widest possible line through corners to minimize required centripetal force.

How does mass affect centripetal force?

Centripetal force is directly proportional to mass: F_c ∝ m. Heavier objects require proportionally more centripetal force for the same speed and radius. However, if the centripetal force is provided by friction (proportional to weight = mg), then the required mass cancels out — all objects need the same friction coefficient for the same speed and radius.

What is the maximum speed for a car on a curve?

The maximum speed without skidding depends on friction: v_max = √(μ × g × r). For a dry road (μ = 0.7) with r = 100 m: v_max = √(0.7 × 9.81 × 100) = √686.7 = 26.2 m/s (94.3 km/h). For a wet road (μ = 0.3): v_max = √(0.3 × 9.81 × 100) = √294.3 = 17.2 m/s (61.8 km/h).

How do banked curves reduce required friction?

Banked curves use the normal force's horizontal component to provide centripetal force. The ideal bank angle satisfies tan(θ) = v²/(rg). For a 100 m radius curve designed for 25 m/s (90 km/h): θ = arctan(25²/(100 × 9.81)) = arctan(0.637) = 32.5°. At this design speed, no friction is needed.

What is the centripetal force on a roller coaster loop?

At the top of a roller coaster loop, the centripetal force is provided by gravity plus the normal force: F_c = mg + N = mv²/r. For a 20 m radius loop, to maintain passenger contact (N ≥ 0): v ≥ √(g × r) = √(9.81 × 20) = 14.0 m/s (50.4 km/h) at the top.

How do you calculate centripetal force in a centrifuge?

In a centrifuge, the centripetal force (providing the radial acceleration) is: F_c = mω²r. A centrifuge with radius 0.15 m spinning at 3000 RPM (ω = 314.2 rad/s) produces a_c = ω²r = 314.2² × 0.15 = 14,800 m/s² ≈ 1510 g. The force on a 0.05 kg sample is F_c = 0.05 × 14,800 = 740 N.

What is the centripetal force on a spinning CD?

For a CD spinning at 500 rad/s (about 4775 RPM) with a 0.06 m outer radius, the centripetal acceleration at the edge is a_c = ω²r = 500² × 0.06 = 15,000 m/s² ≈ 1530 g. The force on a small sector depends on its mass — the CD's structural integrity must withstand these internal forces.

How does centripetal force relate to angular velocity?

Using ω: F_c = mω²r. The force is proportional to ω² — doubling angular velocity quadruples the required centripetal force. This relationship is critical for rotating machinery design because high-speed rotors require robust structural support to withstand the radial forces.

What is the centripetal force on a pendulum bob?

At the bottom of a pendulum swing, the centripetal force is the net of tension upward minus weight downward: T - mg = mv²/r. For a 2 kg pendulum with 1.5 m length swinging through the bottom at 3 m/s: T = m(g + v²/r) = 2(9.81 + 3²/1.5) = 2(9.81 + 6) = 31.62 N.

How do you find the tension in a string for circular motion?

For a mass on a string moving in a horizontal circle, the centripetal force equals the string tension: T = mv²/r. For a vertical circle, tension varies: T_top = m(v²/r - g) and T_bottom = m(v²/r + g). The minimum speed at the top to maintain circular motion: v_min = √(gr).

What is the centripetal force on a proton in a cyclotron?

In a cyclotron, magnetic force provides centripetal force: F_c = qvB = mv²/r. The radius of the path: r = mv/(qB). For a proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) in a 1.5 T magnetic field at 10⁷ m/s: r = (1.67 × 10⁻²⁷ × 10⁷)/(1.6 × 10⁻¹⁹ × 1.5) = 0.07 m.

How does centripetal force affect tire wear?

Centripetal force during cornering increases the lateral load on tires, accelerating wear. The lateral force F_c = mv²/r creates friction at the tire-road contact patch. Cornering repeatedly at high speeds generates heat and abrasion, reducing tire life. Proper tire pressure and alignment minimize uneven wear from lateral forces.

What is the centripetal force required for a 360° loop in a fighter jet?

A fighter jet performing a 360° turn at 900 km/h (250 m/s) with a 1000 m radius: a_c = v²/r = 250²/1000 = 62.5 m/s² ≈ 6.37 g. For a 20,000 kg jet: F_c = 20,000 × 62.5 = 1,250,000 N (1.25 MN). The pilot experiences 6.37 g, near the limit of human tolerance without a g-suit.

How do you calculate centripetal force on a grinding wheel?

For a grinding wheel, centripetal force acts radially outward on the wheel structure. A 0.2 m diameter wheel (r = 0.1 m) at 3000 RPM (ω = 314 rad/s) has a_c = ω²r = 314² × 0.1 = 9860 m/s² ≈ 1005 g. The internal stresses must be below the material's strength to prevent wheel burst.

What is the relationship between centripetal force and period?

Using period T: v = 2πr/T, so F_c = m(2πr/T)²/r = 4π²mr/T². The force is inversely proportional to T² — halving the period quadruples the required force. For a 0.5 kg mass on a 1 m string with 2 s period: F_c = 4π²(0.5)(1)/4 = 4.93 N.

How does centripetal force affect artificial gravity in space?

Rotating a spacecraft creates artificial gravity through centripetal acceleration. For a rotating ring of radius R at angular velocity ω: a_c = ω²R = g. To simulate Earth gravity: ω = √(g/R). For R = 50 m: ω = √(9.81/50) = 0.443 rad/s ≈ 4.23 RPM. The Coriolis effect from the rotating frame causes disorientation at higher rotation rates.

What is the centripetal force on a clothes dryer drum?

In a clothes dryer, the drum spins to create centripetal acceleration that pins clothes against the drum wall. A 0.3 m radius drum at 800 RPM (ω = 83.8 rad/s): a_c = ω²r = 83.8² × 0.3 = 2107 m/s² ≈ 215 g. This high acceleration forces water out of the clothes through centrifugal action.

How does centripetal force affect bearing selection?

Centripetal forces in rotating machinery create radial loads on bearings. For an imbalanced rotor: F_c = m_eccentric × e × ω², where e is the eccentricity. These forces can be significant at high speeds and must be considered in bearing load rating calculations. Dynamic balancing reduces these forces.

What is the centripetal force in a planetary gear system?

In planetary gears, planet gears experience centripetal force from their orbital motion around the sun gear. For a planet gear of mass m at orbit radius r_orb with angular velocity ω_carrier: F_c = m × ω_carrier² × r_orb. This force is carried by the planet bearing and contributes to the overall gear system loading.

How do you calculate centripetal force on a cam follower?

For a cam follower following a curved profile, the centripetal force at any point depends on the radius of curvature of the cam surface: F_c = m × v_follower² / ρ, where ρ is the radius of curvature. This force adds to the spring force keeping the follower in contact and must be considered in cam design to prevent follower jump.

What is the maximum centripetal force a human can withstand?

Human tolerance to centripetal acceleration (g-force): 1-2 g — minor discomfort; 3-4 g — difficulty moving, vision impairment (greyout); 5-6 g — loss of peripheral vision (tunnel vision), possible loss of consciousness without g-suit; 9-10 g — sustained G-LOC (g-induced loss of consciousness) imminent. Trained pilots with g-suits can withstand up to 9 g briefly.

How does centripetal force relate to the banking of aircraft?

When an aircraft banks, the lift vector tilts, providing a horizontal component for turning. For a bank angle φ: tan(φ) = v²/(rg). The centripetal force is L × sin(φ). At 60° bank: the load factor n = 1/cos(60°) = 2 g. The required centripetal force doubles the apparent weight of the aircraft.

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Glossary

Technical terms used in mortgage calculations explained in plain language.

Centripetal Force (F_c): The net inward force required to maintain circular motion, directed toward the center of rotation. Measured in newtons (N).
Centripetal Acceleration (a_c): The inward acceleration experienced by an object in circular motion, given by a_c = v²/r = ω²r. Measured in m/s².
Tangential Velocity (v): The instantaneous velocity of an object moving along a circular path, always tangent to the circle. Measured in m/s.
Angular Velocity (ω): The rate of change of angular position, ω = v/r = 2πf = 2π/T. Measured in rad/s.
Radius of Curvature (r): The radius of the circular path. For non-circular paths, it is the local radius of curvature at each point.
g-force: The ratio of centripetal acceleration to gravitational acceleration: g-force = a_c/g. 1 g = 9.80665 m/s².
Centrifugal Force: A fictitious (inertial) force apparent in a rotating reference frame, directed outward from the center. Not a real force in inertial frames.
RPM (Revolutions Per Minute): A unit of rotational speed. ω (rad/s) = RPM × 2π/60. 1 RPM ≈ 0.1047 rad/s.
Period (T): The time for one complete revolution. T = 2π/ω = 2πr/v. Measured in seconds.
Frequency (f): The number of revolutions per second. f = 1/T = ω/(2π). Measured in Hz.
Normal Force: The perpendicular contact force between surfaces. On banked curves, its horizontal component provides centripetal force.
Static Friction (μ_s): The maximum friction force available before sliding occurs. Determines the maximum speed on unbanked curves: v_max = √(μ_s g r).
Bank Angle (θ): The angle of a road or track surface from horizontal. The ideal bank angle eliminates the need for friction at the design speed.
Load Factor: The ratio of lift to weight in an aircraft turn. Load factor = 1/cos(φ) where φ is the bank angle.
G-LOC: G-induced loss of consciousness, occurring when centripetal acceleration exceeds the brain's blood supply capacity, typically above 6-9 g sustained.

Authoritative References

Trusted sources and standards referenced by this calculator.

ASME — Rotor Dynamics Standards Engineering Toolbox — Centripetal Force MIT OCW — Dynamics NIST — Force Metrology EFunda — Centripetal Force Machine Design — Rotating Machinery HyperPhysics — Circular Motion Khan Academy — Centripetal Force

Applicable Standards

ASME B106.1M — Shaft DesignISO 1940 — Balance Quality RequirementsEN 13814 — Amusement Rides Safety

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Last Updated:

2026-08-21

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2.0.0

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