You already know that something moving in a circle is always accelerating, because its direction keeps changing. And by Newton’s second law, wherever there’s an acceleration there must be a resultant force causing it. That force is the centripetal force — the inward pull that constantly bends an object’s path into a circle instead of letting it fly off in a straight line. Here’s the crucial thing to get straight from the start: it always points to the centre.
📘 What you need to know
An object in uniform circular motion accelerates towards the centre, so a resultant force must act towards the centre too
This inward resultant force is the centripetal force — it keeps the object moving in a circle
It always acts perpendicular to the velocity, pointing to the centre of the circle
Its magnitude is F = mv2/r = mrω2
Centripetal force is not a new kind of force — it’s the role played by whatever real force acts inward (tension, friction, gravity, and so on)
Because the force is perpendicular to the motion, it does no work — the object’s speed and kinetic energy stay constant
Why there must be an inward force
Velocity and acceleration are both vectors. An object going round a circle at steady speed still has a changing velocity, because its direction is never the same for two instants running. A changing velocity is an acceleration — and this particular acceleration points straight to the middle of the circle. It’s called the centripetal acceleration.
Newton’s second law then does the rest: if the object accelerates towards the centre, a resultant force must push it towards the centre. That resultant is the centripetal force.
The velocity (red) is always tangent to the circle; the centripetal force (blue) always points inward to the centre. The two are at right angles to each other at every point.
The name gives it away: centripetal comes from Latin for “centre-seeking”. So whenever a question mentions circular motion, your first instinct should be to ask: “what is providing the inward pull towards the centre?” That single question unlocks most circular-motion problems.
The centripetal force equation
The size of the centripetal force needed to keep a mass m moving in a circle of radius r comes in two equivalent forms — one using linear speed, one using angular speed:
Centripetal forceF = mv2 ÷ r = mrω2
where F is the centripetal force (N), m is the mass (kg), v is the linear speed (m s−1), ω is the angular speed (rad s−1) and r is the radius (m). The two forms are linked by v = rω from the previous topic — use whichever matches the information you’re given.
Reading the equation tells you a lot. A faster object needs much more inward force (it depends on speed squared). A tighter circle (smaller r) also needs more force. That’s why a car has to slow right down for a sharp bend, and why whirling a conker faster makes the string pull harder.
Centripetal force is a role, not a new force
This is the idea examiners test most. The centripetal force is never a separate, extra force you add to a diagram. It’s simply the name for whichever real force happens to be pointing towards the centre and doing the job of keeping the object on its circular path.
Car on a bend → friction
Ball on a string → tension
Planet orbiting → gravity
Depending on the situation, the centripetal force might be provided by tension, friction, gravity, an electrostatic pull, a magnetic force, or a normal reaction. When you solve a problem, you set that real force equal tomv2/r.
A word of warning about “centrifugal force” — the feeling of being flung outward on a fairground ride. There is no real outward force acting on you. What you feel is your own inertia: your body wants to carry straight on (Newton’s first law), while the ride wall pushes you inward. The only real force is the inward, centripetal one. Don’t put an outward arrow on your free-body diagram.
No work is done
Because the centripetal force is always perpendicular to the velocity, it never has a component along the direction of motion. Work is force along the displacement, so a force at 90° to the motion does no work at all. That’s why an object in uniform circular motion keeps a constant speed and constant kinetic energy, even though a force is acting on it the whole time.
🛠️ Solving circular-motion force problems
Identify the real force pointing towards the centre — tension, friction, gravity, and so on.
Set it equal to the centripetal force: real force = mv2/r (or mrω2).
Pick the right form — use mv2/r if you have a speed, mrω2 if you have an angular speed.
Rearrange for the unknown, keeping units consistent (mass in kg, radius in m).
WE 1
A car of mass 1200 kg drives around a roundabout of radius 25 m at a steady 12 m s−1. Calculate the centripetal force needed to keep it on this path.
Step 1 — choose the form with speed
F = mv² ÷ r
Step 2 — substitute the valuesF = (1200 × 12²) ÷ 25F = 172800 ÷ 25F = 6900 N (6.9 kN)This inward force is supplied by friction between the tyres and the road.
WE 2
A washing-machine drum of radius 0.25 m spins at 8.0 rad s−1. Find the centripetal force on a wet shirt of mass 0.40 kg pressed against the drum wall.
Step 1 — we have an angular speed, so use mrω²
F = mrω²
Step 2 — substitute the valuesF = 0.40 × 0.25 × 8.0²F = 0.40 × 0.25 × 64F = 6.4 NThe drum wall pushes inward on the shirt — a normal reaction force doing the centripetal job.
Horizontal circular motion: cars on curves
A very common exam setting is a car driving round a flat, curved road. Two real forces act: its weight pulling down, and the friction between the tyres and the road. On a flat road, it’s the friction that points towards the centre of the curve — so friction provides the centripetal force.
Looking down on a car rounding a bend: it moves along the road (red, tangent) while friction (blue) points inward towards the centre of the curve. That friction is the centripetal force.
Setting friction equal to the centripetal force gives a neat result. The most friction the road can supply is μmg, so:
Friction as the centripetal forcemv2 ÷ r = μmg
The mass cancels, and rearranging for speed gives the maximum speed the car can take the bend without skidding:
Maximum cornering speedvmax = √(μgr)
Go faster than this and the friction can no longer supply enough centripetal force — the car slides off the curve. Notice the mass dropped out entirely: a heavy lorry and a light car skid at the same speed on the same bend (for the same tyres).
WE 3
A car takes a flat curve of radius 40 m. The coefficient of friction between the tyres and the road is 0.70. Find the maximum speed at which the car can corner without skidding. (Take g = 9.81 m s−2.)
Step 1 — friction provides the centripetal force
mv² ÷ r = μmg (mass cancels)
Step 2 — rearrange for maximum speed
vmax = √(μgr)
Step 3 — substitute the valuesvmax = √(0.70 × 9.81 × 40)vmax = √274.7vmax = 16.6 m s⁻¹That’s about 60 km/h — and the answer doesn’t depend on the car’s mass.
A mass on a string in a horizontal circle
Whirl a ball on a string in a flat, horizontal circle and something interesting happens: the string is never quite horizontal. Two forces act on the ball — its weight straight down, and the tension along the string. The weight has to be balanced by the vertical part of the tension, so the string tilts up to an angle. It’s the horizontal part of the tension that points to the centre and provides the centripetal force.
The tension (teal) has a vertical part that balances the weight (purple) and a horizontal part that points to the centre (blue). The string can never be perfectly horizontal, because something must hold the ball’s weight up.
Splitting the tension into components, the vertical part balances the weight while the horizontal part is the centripetal force:
Vertical balance & horizontal (centripetal)T cos θ = mgT sin θ = mv2/r
💡 Top tips
Ask what points inward. Always identify the real force (tension, friction, gravity) that provides the centripetal force before reaching for the equation.
No outward arrows. There is no centrifugal force in your free-body diagram — the only force is inward.
Match the form to the data. Use mv2/r for a speed, mrω2 for an angular speed.
The force does no work. Speed and kinetic energy stay constant in uniform circular motion — a favourite exam catch.
Quick recap: Circular motion needs an inward resultant force — the centripetal force — of size F = mv2/r = mrω2. It’s not a new force but the role played by tension, friction or gravity, always aimed at the centre, always perpendicular to the velocity, and doing no work.
⚠ Common mistakes
Adding a centrifugal (outward) force to a free-body diagram — only the inward force is real
Treating centripetal force as an extra force instead of the role of tension, friction or gravity
Forgetting the speed is squared in mv2/r — doubling the speed quadruples the force
Thinking the force does work and changes the speed — it’s perpendicular to the motion, so it does none
Leaving a radius in cm or a mass in g when substituting into the equation
We’ve kept saying the object accelerates towards the centre, but we haven’t proved how big that acceleration is. That’s the very next step: Centripetal Acceleration, where we show that a = v2/r = ω2r and see exactly where the centripetal force equation comes from.
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