Centripetal Force Simulator Back
Classical Mechanics Simulator

Centripetal Force Simulator — Uniform Circular Motion

Visualise in real time the centripetal force F_c = m v^2/r that keeps a body moving in uniform circular motion. Adjust the mass, tangential speed, orbit radius and gravitational acceleration to compute the centripetal acceleration a_c = v^2/r, the orbital period T = 2 pi r / v and the apparent g-load. Confirm the v-squared scaling of F_c on the chart and follow the chain from Newton's second law to the dynamics of every rotating body, from cars cornering to satellites orbiting the Earth.

Parameters
Mass m
kg
Tangential speed v
m/s
Radius r
m
Gravity g
m/s²
The body moves along a circle of radius r at tangential speed v. The red inward arrow is the centripetal force F_c, the green tangential arrow is the velocity vector v. The g-load is the centripetal acceleration divided by gravity g and tells you how many times its own weight the body feels, the same metric used to design roller coasters and racing cars.
Results
Centripetal force F_c
Centripetal accel. a_c
Acceleration in g
Period T

Circular motion and centripetal force vector

The blue dashed circle is the orbit of radius r, the orange dot is the body. The red arrow is the inward centripetal force F_c, the green arrow is the tangential velocity v. Press "Spin" to animate the rotation in real time.

F_c vs speed (F_c ∝ v²)

Horizontal axis: tangential speed v (m/s); vertical axis: centripetal force F_c (N). The curve is the parabola F_c = m v² / r for the current mass and radius. The yellow marker shows the current value. Doubling the speed quadruples F_c.

Theory & Key Formulas

Centripetal force (net inward force) and centripetal acceleration:

$$F_c = \frac{m\,v^2}{r} = m\,\omega^2\,r, \qquad a_c = \frac{v^2}{r}$$

Angular velocity and period:

$$\omega = \frac{v}{r}, \qquad T = \frac{2\pi r}{v} = \frac{2\pi}{\omega}$$

Acceleration in g units (apparent gravity multiplier):

$$\frac{a_c}{g} = \frac{v^2}{r\,g}$$

With the tool defaults $m=10\ \text{kg},\ v=20\ \text{m/s},\ r=50\ \text{m},\ g=9.81\ \text{m/s}^2$ you obtain $a_c=8.00\ \text{m/s}^2,\ F_c=80.0\ \text{N},\ a_c/g=0.82\ \text{g},\ T=15.71\ \text{s}$. Doubling $v$ multiplies $F_c$ by four; halving $r$ doubles $F_c$.

What is the centripetal force simulator?

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With the defaults m=10 kg, v=20 m/s, r=50 m I get exactly 80 N of force, 8 m/s² of acceleration and "0.82 g". Is that less than a roller coaster?
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Yes, 0.82 g is 82% of gravity, so it feels like a mild push. The math is F_c = m v²/r = 10 × 400 / 50 = 80 N and a_c = v²/r = 400/50 = 8 m/s², which divided by g = 9.81 m/s² gives 0.815 g. The period is T = 2π × 50/20 ≈ 15.71 s, so circling a 50 m radius at 20 m/s takes just under 16 seconds. Think of running around a 400 m athletics track at about 80 km/h.
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If I double the speed to 40 m/s, F_c jumps to 320 N. That's four times the force, right?
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Exactly — the v² law. F_c scales with the square of v, so doubling the speed quadruples the force. That is the real physical reason a car suddenly loses grip when you push too hard into a corner. Tyre friction has a hard ceiling, and as soon as the required centripetal force exceeds it, you understeer. A 30% speed increase mid-corner needs 1.3² = 1.69 times the lateral force, i.e. 70% more grip — without that margin you're heading for the wall.
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If I shrink r to 20 m and push v to 50 m/s, the tool shows 12.7 g. That's brutal for a person.
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Totally — an untrained human gets gray-out around 5 g (blood drains from the brain) and G-LOC, G-induced loss of consciousness, around 6 to 7 g. Fighter pilots use anti-G suits and the AGSM (Anti-G Straining Maneuver) breathing technique to tolerate 9 g for short periods. Brief impulsive G is different again — crash tests record peaks of 50 g over 0.1 s, near the limit of human survivability. Try v=50, r=20 on the tool and you'll see why F1 corners sit right at the edge of what a trained driver can withstand.
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I always heard that centrifugal force pushes outward with the same magnitude. Why is the red arrow in the tool pointing inward instead?
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Great question. In an inertial frame (a stationary outside observer), the only real force on a body in circular motion is the inward centripetal force; centrifugal force does not exist. The body feels like it wants to fly outward only because Newton's first law tries to keep it moving in a straight line, and the centripetal force is what bends that line into a circle. In a non-inertial co-rotating frame (someone on a merry-go-round) the body looks stationary even though it pushes outward, and to make the math work the observer invents a pseudo-force called centrifugal force. The honest physical picture in an inertial frame is "centripetal force only", which is what the tool draws.
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Satellites are also held in orbit by centripetal force, right? The Moon, the Earth, all of them?
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Yes — in space the centripetal force is almost always Newtonian gravity F = G M m / r². For the ISS at 400 km altitude the orbital speed is 7.66 km/s with a 92.7 minute period, derived from F_g = F_c, i.e. GM_E/r² = v²/r, giving v = √(GM_E/r). The Moon at r = 384,400 km circles at 1.02 km/s with a 27.3 day period, while GPS satellites at r ≈ 26,600 km orbit at 3.87 km/s with a 12 hour period. All of these come out of the same F_c = m v²/r equation as the tool — gravity is just the specific provider of the centripetal force, dynamically equivalent to the tension in a string on a merry-go-round.

Physical model and key equations

Uniform circular motion (UCM) is the simplest curvilinear motion: a body moving at constant speed around a circle. Even though the speed is constant, the direction of the velocity vector changes continuously, so Newton's second law $F = m a$ implies a non-zero acceleration and a force that produces it. That force is called the centripetal force.

If the position of the body in polar coordinates is $\vec{r}(t) = r(\cos\omega t,\ \sin\omega t)$, differentiating twice with respect to time gives the acceleration vector $\vec{a}(t) = -\omega^2 \vec{r}(t)$. Its magnitude is $\omega^2 r = v^2/r$ and its direction is always toward the centre (the negative sign). Hence the centripetal acceleration is

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

and Newton's second law yields the centripetal force

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

The angular velocity is $\omega = v/r$, the period of one revolution is $T = 2\pi r/v = 2\pi/\omega$, and the frequency is $f = 1/T = v/(2\pi r)$.

The g-load $a_c/g$ is a dimensionless number that expresses the apparent gravity multiplier felt by the body, widely used as a design criterion for roller coasters, racing cars and fighter aircraft. With the tool defaults $m=10\ \text{kg},\ v=20\ \text{m/s},\ r=50\ \text{m},\ g=9.81\ \text{m/s}^2$ one finds $a_c=8.00\ \text{m/s}^2,\ F_c=80.0\ \text{N},\ a_c/g=0.82\ \text{g},\ T=15.71\ \text{s},\ \omega=0.40\ \text{rad/s}$.

An important subtlety: "centripetal force" is not the name of a particular physical force, but a label for "whatever net force happens to act radially inward". The actual physical source depends on the situation — string tension for a ball on a string, gravity for an orbiting satellite, friction between tyres and tarmac for a cornering car, the magnetic Lorentz force for charged particles in a cyclotron. The tool computes the magnitude of the required centripetal force from $F_c = m v^2/r$ regardless of its source.

Real-world applications

Cornering design for cars and motorbikes: on a flat curve the centripetal force is supplied by static friction between the tyres and the road. With friction coefficient $\mu$ and weight $W = mg$, the maximum non-slipping speed is $v_{\max} = \sqrt{\mu g r}$. On dry asphalt with $\mu = 0.8$ and $r = 50$ m one finds $v_{\max} = 20$ m/s ≈ 72 km/h, which is the physical basis for posted corner speed limits. F1 cars combine very sticky tyres ($\mu \approx 1.5$) with downforce (which adds aerodynamic weight without adding mass) to take a 30 m radius corner at 200 km/h, demanding more than 10 g of lateral acceleration — exactly why F1 drivers need neck training.

Centrifuges and centrifugal casting: ultracentrifuges spin biological samples at very high rates so that components separate by density difference. A modern Beckman Optima XPN at $r = 10$ cm and 100,000 rpm ($\omega = 10472$ rad/s) generates 1,100,000 g, enough to sediment viruses and ribosomes on a density gradient. In manufacturing, centrifugal casting at $r = 0.5$ m and 1000 rpm gives 56 g and is used to produce cast iron pipes and turbine blades. All of these designs reduce to $a_c = v^2/r = \omega^2 r$.

Space habitats with artificial gravity: long-duration space habitats (such as the proposed Voyager Station or Orbital Reef) consider creating artificial gravity by rotation to prevent bone loss. To reach 1 g via $a_c = \omega^2 r = g$, a 200 m radius requires $\omega = 0.22$ rad/s (about 2 rpm). The Discovery from 2001: A Space Odyssey has $r = 8$ m and would need 1.1 rad/s (10.6 rpm), small enough that gravity differs noticeably between head and feet and Coriolis forces cause motion sickness. Studies show that habitats with $r \gtrsim 100$ m are required to be ergonomically comfortable.

Particle accelerators and cyclotrons: in circular accelerators a charged particle of charge $q$ and mass $m$ is bent by the magnetic Lorentz force $F = qvB$, which plays the role of the centripetal force. Setting $qvB = mv^2/r$ gives the orbit radius $r = mv/(qB)$ and the cyclotron frequency $\omega = qB/m$. At the LHC, 7 TeV protons follow a 4.3 km radius at about $0.999999991\,c$, producing a centripetal acceleration of around $5 \times 10^{12}$ g. The classical equation $F_c = m v^2/r$ is the starting point, dressed with relativistic $\gamma$ factors in production codes.

Common misconceptions and pitfalls

The most common misconception is that "centrifugal force pushes the body outward". In an inertial frame the body would just fly off tangentially by Newton's first law, and the centripetal force (rope tension, friction, gravity, etc.) bends it back inward into a circular path. The outward "centrifugal force" is not a real force, only a pseudo-force introduced in a co-rotating non-inertial frame. When solving problems, work in the inertial frame: write $F_c = m v^2/r$ and identify which physical force (tension, friction, gravity) is actually playing the role of $F_c$. Solving in a rotating frame requires introducing both centrifugal and Coriolis pseudo-forces and is usually much more error-prone.

Next, the belief that "the centripetal force is proportional to v". It is actually proportional to $v^2$, so doubling the speed quadruples the force and tripling the speed multiplies it by nine. This is the physical reason that accelerating mid-corner so easily breaks grip and matches everyday driving experience. Use the chart in this tool to see how steeply the parabola rises. Halving the radius only doubles $F_c$, but quartering it quadruples $F_c$ — hence "slow into tight corners".

Third, the assumption that "centripetal force is its own kind of force". "Centripetal" is a functional label meaning "the net inward force acting on the body", not a unique physical interaction. Different situations supply that role with different physics: gravity for satellites, electrostatic attraction for classical electrons in atoms, friction for cornering cars, tension for whirled balls on a string. Always state the physical source of $F_c$ when solving a problem. By contrast, the centripetal acceleration $a_c = v^2/r$ is a purely kinematic quantity that does not depend on the force source.

Frequently asked questions

The centripetal force is the net inward force required to keep a body moving along a circular path at constant speed. It is given by F_c = m v²/r = m ω² r, where m is the mass, v the tangential speed, r the radius and ω = v/r the angular velocity. With the tool defaults m=10 kg, v=20 m/s, r=50 m you obtain F_c=80 N, centripetal acceleration a_c=8.00 m/s² (0.82 g) and period T=15.71 s.
The centripetal force is a real inward force seen from an inertial frame (rope tension, gravity, friction, magnetic force). It is the net force that satisfies Newton's second law F = m a and produces the circular motion. The centrifugal force is a pseudo-force that only appears in a rotating non-inertial frame, equal in magnitude and pointing outward. An orbiting satellite is acted on only by gravity (centripetal) in an inertial frame; the centrifugal force exists only in a frame co-rotating with the satellite.
Doubling the speed halves the time to circle the orbit, so the rate at which the velocity vector changes direction (the angular velocity) doubles. The magnitude of the velocity vector itself also doubles. The acceleration is the product of these two effects, so it grows by 2 × 2 = 4 and scales as v squared. In the chart, doubling v from 10 to 20 m/s makes F_c jump from 20 N to 80 N. This is why pushing harder on a corner suddenly overwhelms the available grip in a car.
a_c/g is the centripetal acceleration divided by the gravitational acceleration g = 9.81 m/s². It tells you how many times the body's own weight it experiences as apparent gravity. Untrained humans tolerate about 5 g, fighter pilots reach 9 g with a G-suit, F1 drivers feel about 5 g of lateral G in fast corners and roller coasters typically peak at 3 to 4 g. The tool default is 0.82 g (mild), but setting v=50 m/s, r=20 m gives 12.7 g, which would cause G-induced loss of consciousness in an untrained person.