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  1. Home
  2. /Physics
  3. /Circular Motion & Rotational Mechanics
  4. /Centripetal Force Calculator

Centripetal Force Calculator

Last updated: March 17, 2026

Calculator

Results

Centripetal Force

250

N

Centripetal Acceleration

50

m/s²

Equivalent Linear Velocity

10

m/s

Equivalent Angular Velocity

5

rad/s

Rotation Frequency

0.7958

Hz

Rotation Period

1.2566

s

Results

Centripetal Force

250

N

Centripetal Acceleration

50

m/s²

Equivalent Linear Velocity

10

m/s

Equivalent Angular Velocity

5

rad/s

Rotation Frequency

0.7958

Hz

Rotation Period

1.2566

s

The Centripetal Force Calculator determines the inward-directed force required to keep an object moving along a circular path. In any form of circular motion, whether a car rounding a curve, a satellite orbiting Earth, or a ball on a string, a net force must act continuously toward the center of the circle. Without this centripetal force, the object would fly off in a straight line due to its inertia, as described by Newton's first law.

This calculator supports two standard formulations. The linear velocity form uses $$F_c = \frac{mv^2}{r}$$, which is ideal when you know the tangential speed. The angular velocity form uses $$F_c = m\omega^2 r$$, which is convenient when working with rotating systems measured in radians per second. Both expressions are mathematically equivalent since $$v = \omega r$$. The tool also computes the centripetal acceleration $$a_c = v^2 / r$$, which is the acceleration directed toward the center regardless of mass.

Understanding centripetal force is essential in mechanical engineering for designing safe roadways and roller coasters, in aerospace for calculating orbital mechanics, and in everyday physics for analyzing anything that spins, turns, or orbits. The concept was formalized by Christiaan Huygens in 1659 and later incorporated into Newton's laws of motion.

Visual Analysis

How It Works

The calculator applies Newton's second law to uniform circular motion. An object of mass m traveling at speed v around a circle of radius r requires a centripetal (center-seeking) force:

$$F_c = \frac{mv^2}{r}$$

Equivalently, if the angular velocity ω is known:

$$F_c = m\omega^2 r$$

The centripetal acceleration is the force per unit mass:

$$a_c = \frac{v^2}{r} = \omega^2 r$$

If you provide an angular velocity value greater than zero, the calculator uses it directly for the angular form. Otherwise, it derives ω from v/r. Note that centripetal force is not a separate force of nature; it is the net inward force provided by tension, gravity, friction, normal force, or any combination thereof.

Understanding Your Results

A larger centripetal force means the object is being pulled more strongly toward the center. If the required force exceeds what the physical constraint can provide (e.g., friction on a road, tensile strength of a string), the object will break away from its circular path. Increasing mass or velocity raises the required force, while increasing the radius reduces it. Engineers use these relationships to set speed limits on curved roads and to design centrifuges.

Worked Examples

Car on a Curved Road

Inputs

mass1200
velocity20
radius50
omega0

Results

fc linear9600
fc angular9600
centripetal accel8

A 1200 kg car traveling at 20 m/s around a 50 m radius curve requires 9600 N of centripetal force, provided by tire friction.

Ball on a String

Inputs

mass0.5
velocity4
radius1
omega0

Results

fc linear8
fc angular8
centripetal accel16

A 0.5 kg ball whirled at 4 m/s on a 1 m string requires 8 N of tension directed inward.

Frequently Asked Questions

Centripetal force is the real inward force that causes circular motion, observed from an inertial (non-rotating) frame. Centrifugal force is a fictitious outward force that appears only in a rotating reference frame. They are equal in magnitude but opposite in direction.

For a car on a curve, friction provides it. For a satellite in orbit, gravity provides it. For a ball on a string, string tension provides it. For a roller coaster loop, the normal force and gravity together provide it.

Because the direction of motion changes more rapidly at higher speeds, requiring a proportionally greater inward acceleration. Doubling the speed quadruples the required force.

No. Centripetal force is always perpendicular to the velocity vector, so it changes the direction of motion but not the speed. Since work equals force times displacement in the direction of force, centripetal force does zero work.

The object continues moving in a straight line tangent to the circle at the point where the force was removed, following Newton's first law of motion.

Dividing centripetal acceleration by 9.81 m/s² gives the g-force. For example, a centripetal acceleration of 49.05 m/s² equals approximately 5g, which is significant in fighter jet maneuvers and roller coasters.

Sources & Methodology

Halliday, Resnick & Walker, Fundamentals of Physics, 12th Edition. Huygens, C. (1659). De vi centrifuga. Serway & Jewett, Physics for Scientists and Engineers, 10th Edition.
R

Roboculator Team

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