Centripetal force, explained with examples
Anything moving in a circle is accelerating towards the centre, even at constant speed. The net inward force that causes this is called the centripetal force, and identifying where it comes from is the key to every circular motion problem.
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Why circular motion needs a force
Velocity has a direction as well as a size. An object moving in a circle at constant speed keeps changing direction, so its velocity is changing, which means it is accelerating. This acceleration always points towards the centre of the circle and is called centripetal acceleration. By Newton's second law, an acceleration needs a net force in the same direction: the centripetal force.
If the centripetal force suddenly disappeared, the object would not fly outwards. It would move off in a straight line along the tangent, at the speed it had at that moment, exactly as Newton's first law predicts.
The formulas
- ac = v²/r, the centripetal acceleration
- Fc = mv²/r, the net inward force needed
- In terms of angular speed ω: a = ω²r and F = mω²r, with v = ωr
- For one revolution in period T: v = 2πr/T
Doubling the speed needs four times the force, because of the v². Halving the radius at the same speed doubles the force needed.
Centripetal force is not a new kind of force
This is the most important idea in the topic. Centripetal force is not an extra force to draw on a free-body diagram. It is the name for the net force towards the centre, and it is always provided by real forces you already know:
- a car turning on a flat road: static friction between the tyres and the road;
- a ball whirled on a string: tension;
- a satellite or the Moon: gravity;
- an electron in a magnetic field: the magnetic force;
- a car on a banked track: part of the normal force, and possibly friction.
The method for every problem: draw the real forces, choose the direction towards the centre as positive, and write ΣFinward = mv²/r.
What about the outward feeling?
In a turning car, you feel pushed towards the outside of the bend. No outward force is acting on you. Your body tends to keep moving in a straight line, by inertia, while the car turns underneath you. The door or seat belt then pushes you inwards, providing your centripetal force. The “outward push” is how inertia feels from inside the car.
Worked examples
Example 1: a car on a flat curve
A 1200 kg car rounds a flat bend of radius 50 m at 15 m/s. Find the friction force needed.
- F = mv²/r = 1200 × 225/50 = 5400 N of static friction, towards the centre of the bend.
Example 2: maximum safe speed
The coefficient of static friction between the tyres and road is 0.60. Find the maximum speed for this bend. (g = 9.8 m/s²)
- Maximum friction = μmg = mv²/r, so v = √(μgr).
- v = √(0.60 × 9.8 × 50) ≈ 17 m/s. The mass cancels, so the safe speed is the same for any car.
Example 3: a ball on a string
A 0.30 kg ball moves in a horizontal circle of radius 0.80 m, making 2 revolutions per second. Find the tension, ignoring gravity.
- ω = 2π × 2 = 4π rad/s.
- T = mω²r = 0.30 × 16π² × 0.80 ≈ 37.9 N.
Example 4: an orbit
A satellite orbits 6.8 × 106 m from Earth's centre. Find its orbital speed. (GMEarth = 3.99 × 1014 m³/s²)
- Gravity provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r).
- v = √(3.99 × 1014/6.8 × 106) ≈ 7.66 km/s.
Vertical circles
In a vertical circle, gravity acts in a fixed direction, so the forces change around the loop.
- At the top of a loop, gravity and the normal force (or tension) both point down, towards the centre: N + mg = mv²/r.
- At the bottom, the normal force points up and gravity down: N − mg = mv²/r. This is why you feel heavier at the bottom of a dip.
The minimum speed at the top for an object to stay on the track is when N = 0: mg = mv²/r, so vmin = √(gr).
Example 5: the top of a loop
A roller coaster loop has radius 10 m. Find the minimum speed at the top. (g = 9.8 m/s²)
- vmin = √(9.8 × 10) ≈ 9.9 m/s. Slower than this, the track would need to pull the car upwards, which it cannot do.
Example 6: the bottom of a dip
A 60 kg rider passes the bottom of a dip of radius 20 m at 14 m/s. Find the normal force from the seat.
- N = mg + mv²/r = 588 + 60 × 196/20 = 588 + 588.
- N = 1176 N, twice the rider's weight: they feel “2g”.
Banked curves
Race tracks and motorway slip roads are tilted. On a bank at angle θ, the normal force has a horizontal component towards the centre, so the road can provide centripetal force even without friction. With no friction, N sinθ = mv²/r and N cosθ = mg, giving tanθ = v²/(rg). At this design speed, a car can take the curve on ice.
Common mistakes
- Drawing a separate “centripetal force” arrow on a free-body diagram.
- Saying an object flies outwards when released; it moves along the tangent.
- Calling it “centrifugal force” in an inertial frame; there is no outward force on the object.
- Forgetting that the speed is squared.
- Using the same force equation at the top and bottom of a vertical circle.
Practice questions
Try each one before opening the answer.
Q1. A 2 kg object moves in a circle of radius 4 m at 6 m/s. Find the centripetal force.
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Q2. What provides the centripetal force for the Moon orbiting the Earth?
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Q3. The speed of a car on a bend doubles. By what factor does the required friction change?
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Q4. Find the period of a satellite moving at 7.5 km/s in an orbit of radius 7.0 × 106 m.
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Frequently asked questions
Is circular motion on AP Physics 1?
Yes. Uniform circular motion is part of the force and translational dynamics unit in AP Physics 1, and it appears in AP Physics C: Mechanics too.
Is circular motion in A level Physics?
Yes. It is a second-year topic in most specifications, using radians, angular speed and F = mω²r.
Key takeaways
- Circular motion at constant speed is still accelerated motion.
- a = v²/r towards the centre; F = mv²/r.
- Centripetal force is the net inward force from real forces.
- At the top of a loop, vmin = √(gr).
- Without the inward force, the object moves along the tangent.