Centripetal Force Calculator

What centripetal force is

Centripetal force is the inward pull that keeps an object moving in a circle. The word comes from two Latin pieces: centrum, meaning center, and petere, meaning to seek. It is not a separate kind of force that appears out of nowhere. It is whatever real force happens to point at the center in a given situation: friction between tires and asphalt, tension in a string, gravity between a planet and a star, the normal force from the wall of a spinning drum.

Because the force is always at right angles to the motion, it changes the direction of travel and not the speed. A car turning at a steady 40 mph is accelerating the whole time, even though the speedometer never moves, because acceleration is a change in velocity and velocity carries a direction. The size of that acceleration is speed squared divided by radius, and multiplying by mass turns it into a force, measured in newtons per the NIST guide to SI units. The NASA Stargazer chapter on rotating frames writes the inward force as m V squared over R and calls it the force that holds a rotating mass to its curved path.

The equal and opposite partner is the one people feel. In a turning car, the door pushes you inward and you feel pushed outward against it. That outward sensation is often called centrifugal force, and in a rotating frame of reference the label is fine. Viewed from the ground, no outward force exists: there is only an inward one that your body resists because of its inertia.

Three forms of the same formula

The version most people learn is F = m x v squared / r. It is the one to use when you know the speed along the path. Two other versions cover the cases where speed is awkward to measure.

If you know the rotation rate, use F = m x omega squared x r, where omega is in radians per second. A car at 20 m/s on a 50 m curve has omega = 20 / 50 = 0.4 rad/s, and 1,500 x 0.4 x 0.4 x 50 = 12,000 N, the same answer in a different currency.

If you know the time for one full lap, use F = 4 x pi squared x m x r / T squared. The same car covers 2 x pi x 50 = 314.16 m per lap, so 314.16 / 20 = 15.71 seconds per lap. Then 4 x 9.8696 x 1,500 x 50 / 246.74 = 12,000 N.

All three assume the speed is constant. In a real corner the driver brakes on the way in, so the tires handle braking force and cornering force at the same time. The available grip is shared between the two, which is why a car that enters a corner too fast loses the front end before it loses the rear.

A car turning a corner, step by step

Take a 1,500 kg sedan moving at 20 m/s, which is 72 km/h or 44.7 mph, through a curve with a 50 m radius. The first step is the acceleration: 400 / 50 = 8 m/s squared. The second step is the force: 1,500 x 8 = 12,000 N.

Divide by 9.80665 to put the acceleration in g: 8 / 9.80665 = 0.816 g. That is the number engineers quote, because tire grip is also quoted in g. A typical all season tire on dry asphalt peaks near 0.8 to 0.9 g, so this corner asks for almost everything the tires have. In the rain the same tire might manage 0.6 g, and the car would slide at this speed.

The friction coefficient needed is the same as the g figure, 0.816, because friction force equals mu times weight and the mass cancels out. That cancellation is worth remembering: on a flat road, the maximum cornering speed depends on the tire and the radius, not on how heavy the car is. A 1,500 kg sedan and a 900 kg hatchback on the same tires corner at the same speed.

At 60 mph, which is 26.82 m/s, the numbers climb fast. The acceleration becomes 719.44 / 50 = 14.389 m/s squared, or 1.467 g, and the force becomes 1,500 x 14.389 = 21,583 N. No street tire reaches 1.47 g on a flat road. A banked curve or a wing is needed, which is why highway ramps are banked and why race cars carry downforce.

Speed and radius do not trade evenly

Force depends on the square of the speed and only the first power of the radius, so speed dominates. Doubling the speed quadruples the force. Halving the radius only doubles it. The table below holds the car and the radius fixed and changes only the speed.

Going from 20 to 30 m/s, a 50 percent speed increase, more than doubles the force, from 12,000 to 27,000 N. Going from 20 to 40 m/s quadruples it. The same asymmetry works in reverse on the radius: at 20 m/s, a 25 m radius needs 24,000 N, while a 100 m radius needs 6,000 N. Four times the radius for a quarter of the force.

This is the arithmetic behind speed limits on ramps and exit loops. Engineers pick a design speed, compute the sideways acceleration, and keep it near 0.2 to 0.3 g for comfort on a highway ramp. Doubling the posted speed on that ramp multiplies the required grip by four, which is why a ramp signed for 25 mph is genuinely dangerous at 50.

Laundry, bicycles and a bucket of water

A front loading washer is a centrifuge. Take a drum with a 0.25 m radius spinning at 1,200 rpm. The rotation rate is 1,200 x 6.2832 / 60 = 125.66 rad/s. The acceleration at the drum wall is 125.6637 x 125.6637 x 0.25 = 3,947.8 m/s squared, which is 3,947.8 / 9.80665 = 402.6 g. A pair of jeans with 1 kg of water left in it presses on the drum wall with 3,948 N, roughly the weight of a small motorcycle. The water escapes through the holes because the drum cannot pull it inward.

A bicycle turn is gentler. A rider and bike together at 80 kg, moving at 5 m/s through a 6 m radius turn, needs 80 x 25 / 6 = 333.3 N. That is 74.9 pounds of sideways force, and the lean angle is set by the same ratio: the tangent of the lean from vertical equals 25 / (6 x 9.80665) = 0.425, which is 23 degrees. Riders lean into the turn without being taught because the alternative is falling outward.

The bucket of water swung in a vertical circle is the classic homework case. At the top of the loop, gravity points inward and helps. The water stays in the bucket as long as the required acceleration is at least 1 g, so the minimum speed at the top of a 1 m loop is the square root of 9.80665, which is 3.13 m/s. Below that, the bucket is falling faster than the water needs to, and the water leaves. At exactly 3.13 m/s the water presses on the bucket with zero force, which is the definition of the threshold.

Orbits use the same equation

Gravity supplies the centripetal force for every orbit, which means the same formula answers questions about satellites and planets. The International Space Station orbits at about 400 km up, so its radius from Earth's center is 6,771 km. Its speed is about 7,660 m/s, so the acceleration is 58,675,600 / 6,771,000 = 8.67 m/s squared, or 0.88 g. NASA's page on side forces makes the same link between a perpendicular force, a speed and a radius of curvature. Astronauts float not because gravity is absent but because they and the station fall at the same rate.

The Earth orbits the sun at 29,780 m/s with a radius of 1.496 x 10^11 m. The acceleration is 886,848,400 divided by 1.496 x 10^11, which is 0.00593 m/s squared. That tiny number, multiplied by Earth's mass of 5.97 x 10^24 kg, comes to about 3.5 x 10^22 N. The Moon needs 0.00272 m/s squared at its 384,400 km distance, moving at 1,022 m/s.

The table shows why orbit tables are written as speeds rather than forces. The acceleration falls with the square of the distance for gravity, and the circular speed falls with the square root, so a satellite twice as far out moves at 0.707 times the speed and takes 2.83 times as long to complete an orbit.

Unit traps that wreck the answer

Most wrong answers on this calculation are unit errors, not formula errors. The formula only works when mass is in kilograms, radius in meters and speed in meters per second. Pounds, feet and miles per hour all need converting first.

Two traps deserve their own line. First, pounds and pounds-force are different units that share a name: a 3,300 lb car has a mass of 1,497 kg, and the 12,000 N of cornering force is 2,698 lbf, not 2,698 lb of mass. Second, diameter is not radius. A 100 ft diameter loop has r = 50 ft = 15.24 m, and using 100 ft as the radius understates the acceleration by half.

Rotation rates need one more step. Revolutions per minute are not radians per second, and 1 rpm equals 0.1047 rad/s. A 33 rpm record is 3.46 rad/s, and a 1,200 rpm washer drum is 125.7 rad/s. Feeding rpm straight into the formula as if it were rad/s understates the force by a factor of 9,549 squared, which is not a rounding error.

Where the number matters, and where it misleads

The design side of this number shows up in places most people never notice. Highway ramps are banked so gravity supplies part of the inward force. Roller coaster loops are clothoid shaped so the radius grows gradually and the rider never meets a sudden spike in g force. Laboratory centrifuges are rated in g because the same arithmetic sets how fast a sample separates. Even a bicycle wheel is built to carry the inward force at the rim, which is why spokes are tensioned rather than compressed.

One limit is worth stating. This page treats the object as a point moving at constant speed on a perfect circle. Real corners tighten, real cars slow down mid turn, and real riders shift their weight. For a first estimate the formula is accurate to a few percent; for a suspension design or a track day, the transient loads matter more than the steady state number.