Centripetal Force: Calculator, Formula and Interactive Circular Motion Simulator
Centripetal force is the net force toward the centre that keeps an object moving in a circle. Its size is F = mv²/r. The calculator above solves that equation for any unknown, and the simulator lets you watch it at work in eleven situations, from a ball on a string to a satellite in orbit. Each one has its free-body diagram, live readouts and a graph, in SI or Imperial units.
What is the formula for centripetal force?
An object moving at constant speed v round a circle of radius r is accelerating, because the direction of its velocity keeps changing. That centripetal acceleration is a = v²/r and points to the centre. Newton’s second law turns it into a net force:
| Quantity | Formula | SI unit |
|---|---|---|
| Centripetal force | F = mv²/r = mω²r | N |
| Centripetal acceleration | a = v²/r = ω²r | m/s² |
| Speed from period | v = 2πr/T | m/s |
| Angular speed | ω = v/r = 2π/T = 2πf | rad/s |
| Flat curve, maximum speed | v = √(μsgr) | m/s |
| Banked curve, design speed | v = √(rg tanθ) | m/s |
| Vertical circle, minimum speed at top | v = √(gr) | m/s |
| Loop-the-loop, minimum release height | h = 2.5R | m |
| Conical pendulum period | T = 2π√(L cosθ/g) | s |
| Circular orbit speed | v = √(GM/r) | m/s |
Doubling the speed quadruples the force; doubling the radius at the same speed halves it. In US customary units, divide a mass in pounds by 32.17 to get slugs, and the force comes out in pounds-force.
Worked example: how much force does a car need on a curve?
| Step | Calculation | Result |
|---|---|---|
| Given | m = 1,500 kg, r = 50 m, v = 15 m/s | — |
| Centripetal acceleration | a = 15² / 50 | 4.5 m/s² (0.46 g) |
| Force needed | F = 1,500 × 4.5 | 6,750 N |
| Friction available (dry, μ = 0.70) | 0.70 × 1,500 × 9.81 | 10,300 N: the car grips |
| Fastest safe speed | √(0.70 × 9.81 × 50) | 18.5 m/s (41 mph) |
On ice (μ ≈ 0.1) the same curve is only safe below 7.0 m/s, about 16 mph. Try it in the Flat curve set-up with the surface chips.
Centripetal vs centrifugal force: which one is real?
From the ground, an inertial frame, there is only one horizontal force on a ball whirled on a string: the tension, pulling inward. That inward net force is the centripetal force. Nothing pushes the ball outward. Ride along with the ball, though, and it is at rest; to make the forces balance in that rotating frame you must add an outward force of exactly mv²/r. That is the centrifugal force. It is called fictitious, or inertial, because no object exerts it: it is a consequence of describing the motion from a turning point of view. The Frame switch in the simulator turns the camera with the object so you can watch the dashed centrifugal arrow appear in the free-body diagram, then vanish when you return to the ground frame.
Centripetal and Centrifugal Force Virtual Lab Experiments
This virtual lab runs eleven circular-motion experiments on real apparatus, from an air table to a space station. Each one answers the same question: which real force supplies the centripetal force here? Every experiment shows a live free-body diagram and a graph. Seven also have a Rotating frame switch, where the centrifugal force appears as a dashed, fictitious arrow. The numbers below are each experiment’s starting values, so you can check them on screen.
1. Ball on a string (air table)
A puck on a frictionless air table is held in a circle by a cord tied to a centre post. The tension is the whole centripetal force: a 0.5 kg puck at 4 m/s on a 1 m circle needs T = mv²/r = 8 N. Double the speed and the tension becomes 32 N. Press Cut string and the puck leaves along the tangent at 4 m/s. Its distance from the centre grows, but it never moves along the radius.
2. Car on a flat curve (skidpad)
On a level curve, only static friction from the tyres pushes the car toward the centre. A 1,500 kg car at 15 m/s on a 50 m radius needs 6,750 N, well under the 10,300 N that dry tyres (μ = 0.70) can give. The fastest safe speed is √(μgr) = 18.5 m/s; on ice it drops to 7.0 m/s. Go faster and the car slides wide, leaving tyre marks along a gentler curve.
3. Banked curve (banked oval)
Tilt the road and the normal force leans toward the centre. On a 100 m radius banked at 20°, the design speed √(rg tanθ) = 18.9 m/s needs no friction at all. Add tyre friction to open a band of safe speeds. Too fast and the car slides up the bank; too slow and it slides down.
4. Conical pendulum (retort stand)
A brass bob on a 1.0 m cord swings in a horizontal circle at 30° from vertical. The horizontal part of the tension is the centripetal force, and the vertical part holds up the weight. The period is 2π√(L cosθ/g) = 1.87 s and does not depend on the mass. Cut the cord and the bob flies off horizontally along the tangent, then falls as a projectile.
5. Rotor ride (amusement park)
Riders stand against the wall of a drum 2.5 m in radius spinning at 30 rpm. The wall’s normal force pushes them inward at 2.52 g, and friction holds them up. Drop the floor: they stay pinned if μ ≥ g/(ω²r) = 0.398. Slow the drum and they slide down. In the rotating frame the riders feel the centrifugal push as being thrown against the wall.
6. Vertical circle (ball on a string)
A steel ball whirled in a vertical circle has tension and gravity together as the centripetal force. On a 1 m circle the ball needs at least √(gr) = 3.13 m/s at the top, which means 7.0 m/s at the bottom. Go slower and the string goes slack partway up, so the ball falls inside the circle. Swap the string for a rod and the rod can push. The ball then only has to reach the top at all, which takes 2√(gr) = 6.26 m/s at the bottom.
7. Loop-the-loop (roller coaster)
A frictionless car runs down a drop into an 8 m loop. At the top, the track and gravity both push toward the centre. Energy conservation plus v ≥ √(gR) at the top gives the classic result: release from at least 2.5R = 20 m. From 22 m it makes the loop. From 14 m it leaves the track before the top.
8. Hill and dip (car over a crest)
Over a 30 m crest, the road’s push and gravity together supply the centripetal force, so the seat scale reads less than your weight. At 12 m/s that is 0.51 g at the crest and 1.49 g in the dip. Above √(gr) = 17.2 m/s the normal force reaches zero and the car goes airborne.
9. Ferris wheel (apparent weight)
On a 30 m wheel turning once every 30 s (6.28 m/s), the rider’s seat and gravity provide a centripetal acceleration of 1.32 m/s². The seat scale reads 0.866 g at the top and 1.134 g at the bottom. Speed the wheel up past √(gr) and the rider would lift off the seat at the top.
10. Satellite orbit (Earth, Moon, Mars)
For a satellite, gravity alone is the centripetal force, so v = √(GM/r) and the satellite’s own mass cancels. At the International Space Station’s 400 km that gives 7.67 km/s and one orbit every 92.4 minutes. Raise the altitude to see orbits slow down, and switch to the Moon or Mars to compare planets.
11. Rotating space station (artificial gravity)
A ring 100 m in radius spinning at 3 rpm pushes its floor against the crew with 1.01 g. That floor’s push is the centripetal force, and the crew feel it as weight. Drop a ball from 1.5 m and it lands 0.18 m behind the spot under the hand. No gravity pulls it down: the ball flies straight while the floor turns under it, the Coriolis effect seen from inside.
Banked curves, loops and apparent weight
Centripetal force is a job, not a kind of force, and different real forces do it. On a banked curve the horizontal part of the normal force, N sinθ, does it, giving a design speed of √(rg tanθ) at which no friction is needed. With friction, a band of speeds opens up: too fast and friction must act down the slope, too slow and it acts up the slope. AP Physics 1 calculates the frictionless case; the friction band is standard in calculus-based courses.
In a vertical circle the speed changes, so you need energy conservation as well as F = mv²/r. At the top of a string or a loop, gravity and the tension or normal force both point down, toward the centre. A string cannot push, so the speed there must be at least √(gr). For a frictionless roller coaster released from rest, that leads to the famous result that the start must be at least 2.5 times the loop radius above the bottom, and the riders then feel 6 g at the bottom of the loop. That is why real loops are tear-drop shaped. Over a hill crest the road pushes up less than your weight, N = m(g − v²/r), and above √(gr) the car leaves the road.
Three mistakes students make with circular motion
- Drawing a “centripetal force” arrow. A free-body diagram shows real forces only. The centripetal force is the net of them, so identify which one points inward: tension, friction, gravity or a component of the normal force.
- Thinking a released object flies straight out. With the string cut there is no horizontal force, so the object keeps its velocity and leaves along the tangent. The simulator draws both the tangent path and the radial path it does not take.
- Assuming constant speed means no acceleration. The direction of the velocity changes continuously, so the acceleration v²/r is never zero in circular motion.
Orbits and artificial gravity
For a satellite, gravity is the only force, so GMm/r² = mv²/r and v = √(GM/r), where r is measured from the planet’s centre. The satellite’s own mass cancels. At the International Space Station’s height gravity is still about 88 % of its surface value: astronauts float because they are falling with the station, not because gravity is absent. Squaring the period gives Kepler’s third law, T² = (4π²/GM) r³, which the Data Lab uses to measure a planet’s mass from orbits. A spinning space station works the other way round: its floor pushes the crew toward the hub, and they feel that push as weight. For 1 g, ω = √(g/r).
Who uses this simulator?
High-school physics students working through uniform circular motion, AP Physics 1 students preparing for topic 2.9 (centripetal acceleration from one force, several forces or components of forces, including the vertical loop, the banked curve and the conical pendulum), teachers who want a projector demonstration of the rotating frame, and introductory college and engineering students who need banked curves with friction, rotor rides and orbits. The calculator is also handy for quick checks in vehicle dynamics, centrifuge and machine-design homework.
Frequently asked questions
What is the formula for centripetal force?
Centripetal force is F = mv²/r, where m is the mass, v the speed and r the radius of the circle. It can also be written F = mω²r with the angular speed ω. It is the net force toward the centre that keeps an object moving in a circle, and it is always supplied by a real force such as tension, friction, gravity or a normal force.
What is the difference between centripetal and centrifugal force?
Centripetal force is the real inward net force seen from the ground (an inertial frame). Centrifugal force is an outward force that appears only when you describe the motion from inside the rotating frame, where the object is at rest. It has the same size, mv²/r, but no object exerts it, so it never belongs on a free-body diagram drawn in the ground frame.
What provides the centripetal force on a banked curve?
On a frictionless banked curve the horizontal component of the normal force, N sin θ, is the whole centripetal force while N cos θ supports the car's weight. That gives the design speed v = √(rg tan θ). With friction, static friction acts down the slope above the design speed and up the slope below it, which widens the range of safe speeds.
What is the minimum speed at the top of a vertical circle?
For a ball on a string, or a roller coaster on the inside of a loop, the minimum speed at the top is v = √(gr). At that speed gravity alone supplies the centripetal force and the tension or normal force is zero. By energy conservation the speed at the bottom must then be at least √(5gr), and a frictionless coaster must start from at least 2.5 times the loop radius.
Which way does an object go when the string breaks?
It moves off in a straight line along the tangent to the circle, at the speed it had when the string broke. It does not fly radially outward, because once the string breaks there is no force on it in the horizontal plane, so by Newton's first law it simply keeps its velocity.
Explore Related Simulators
Circular motion builds on Newton’s Laws of Motion and free-body diagrams, and friction on curves is covered in depth in the Friction simulator. For orbits that are not circular, launch a satellite in the Escape Velocity simulator; for the rotational side, try Torque & Rotation, the Projectile Motion simulator for what happens after the string breaks, and the Centrifugal Governor, an engineering machine built on a conical pendulum.
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