Uniform circular motion

Swing a stone tied to a string in a horizontal circle at a steady speed and something odd is true: the stone's speed never changes, yet it is accelerating the entire time. That sounds like a contradiction until you remember that velocity has a direction as well as a magnitude — and in a circle, the direction is turning every instant, even while the speed stays fixed. Wherever velocity changes, Newton's second law demands a force, and here that force points sideways to the motion, constantly toward the centre of the circle. This one idea — uniform circular motion — explains the tension pulling on a whirled stone, why a car needs friction to take a curve without skidding, and why a satellite can orbit forever without its engines firing. The topic sits inside the ICSE Class 10 unit on Force, Work, Power and Energy and rests on a single formula, $F = mv^2/r$ . This article builds that formula, applies it to two everyday situations, and clears up the most common exam mix-up — the false idea of an outward "centrifugal force" acting on the object itself.

The core insight

An object moving at constant speed along a circle is still accelerating because its velocity's direction keeps changing, so a real net force of magnitude $F = mv^2/r$ must act toward the centre at every instant — the outward-feeling centrifugal force is only a pseudo-force noticed by an observer riding along inside the turning frame.

The core idea

Constant speed does not mean constant velocity. Velocity is a vector — it has size (speed) and direction. In uniform circular motion the size never changes, but the direction sweeps around continuously, so the velocity vector itself is changing every instant. A changing velocity is, by definition, an acceleration, even when the speedometer reading is frozen.

Newton's second law says a net force must exist in the same direction as that acceleration — toward the centre, never forward along the path and never outward. This force is called the centripetal force:

F = \frac{mv^2}{r}

where: $F$ is the centripetal force (newton, N), $m$ is the mass of the moving object (kg), $v$ is its constant speed along the circle (m/s), and $r$ is the radius of the circular path (m). Centripetal is a description of direction and role, not a new type of force — it is always supplied by a familiar force already on your syllabus: tension for a whirled stone, friction between tyres and road for a car rounding a bend, gravity for an orbiting satellite, or the normal reaction of the rails for a train on a curved track.

Inertia (Newton's first law) says a moving object left alone travels in a straight line. A circle is a straight-line path continuously bent inward, so bending it requires a continuous sideways pull — the centripetal force. Remove that force at any instant and the object stops curving, flying off along the tangent to the circle at that point, not straight out from the centre.

Centrifugal force is where most students slip. The outward push felt when a car swings around a bend, or mud flying off a spinning bicycle wheel, feels exactly like an outward force. This centrifugal force is a pseudo-force: it involves no real physical push or pull on the object, and is felt only by an observer inside the rotating frame. Viewed from outside the car, nothing pushes the passenger outward — their own inertia carries them in a straight line while the seat and seatbelt supply the real, inward centripetal force. The "outward force" is an illusion created by watching from a frame that is itself accelerating.

Figure

Diagram of a stone whirled on a string in a circle showing the centripetal force arrow pointing inward toward the centre at four positions, and a dashed tangent line showing the straight path the stone would take if the string broke

Derivation AGetting the centripetal force formula onto the exam page

The goal here is to show precisely why the sideways force needed to hold a mass $m$ on a circular path of radius $r$ at constant speed $v$ must equal $mv^2/r$ , using nothing beyond Newton's second law and the fact that velocity is a vector, so you can reproduce the reasoning rather than only recall the finished formula.

Step 1. Start from Newton's second law. For any object, the net force equals mass times acceleration: $F = ma$ . To find the centripetal force, we only need to find the acceleration $a$ of an object moving at constant speed $v$ around a circle of radius $r$ — the direction of that acceleration will turn out to be the direction of $F$ too.

Step 2. Track how the velocity vector changes. As the object moves around the circle, its speed $v$ stays the same, but its direction rotates. The velocity vector is always tangent to the circle, perpendicular to the radius joining the object to the centre, so as the object sweeps around, that perpendicular direction sweeps around too — the tip of the velocity vector traces its own small circle of "radius" $v$ , just as the object's position traces a circle of radius $r$ .

Step 3. Compare the two circles. The position vector (length $r$ ) and the velocity vector (length $v$ ) rotate together, through the same angle in the same time. So the rate of change of velocity (the acceleration) compares to $v$ in the same ratio that $v$ compares to $r$ . This ratio works out to $a = v^2/r$ , directed toward the centre — the standard centripetal acceleration result, which Class 10 ICSE expects you to know and apply rather than derive geometrically in an exam.

Step 4. Substitute into Newton's second law. Since $a = v^2/r$ points toward the centre, the net force required is

F = ma = \frac{mv^2}{r}

also directed toward the centre. This is the centripetal force formula. No assumption was made about what supplies this force — that is decided separately by the physical setup (a string, friction, gravity, a rail) — the formula only fixes how large that supplying force must be for the given $m$ , $v$ , and $r$ .

Worked example

A stone of mass $0.2$ kg is tied to a string and whirled in a horizontal circle of radius $0.8$ m at a constant speed of $4$ m/s. A car of mass $1000$ kg later goes around a curve of radius $50$ m at a constant speed of $10$ m/s. Find the centripetal force in each case, and name the real force supplying it.

Step 1. List the knowns for the stone. $m = 0.2$ kg, $v = 4$ m/s, $r = 0.8$ m.

Step 2. Apply the formula for the stone.

F = \frac{mv^2}{r} = \frac{0.2 \times 4^2}{0.8} = \frac{0.2 \times 16}{0.8} = \frac{3.2}{0.8} = 4\ \text{N}

This $4$ N is supplied by the tension in the string, directed from the stone toward the centre at every instant. If the string snapped, the stone would fly off along the tangent to the circle at that instant, not straight outward.

Step 3. List the knowns for the car and apply the same formula. $m = 1000$ kg, $v = 10$ m/s, $r = 50$ m.

F = \frac{mv^2}{r} = \frac{1000 \times 10^2}{50} = \frac{100000}{50} = 2000\ \text{N}

Step 4. Identify the supplying force and interpret. This $2000$ N is supplied by the friction between the tyres and the road, acting toward the centre of the curve. On a wet or icy road, friction may fall short of the $2000$ N needed, and the car skids outward along a straighter path than the curve — the effect a passenger describes as being "thrown outward" by the centrifugal pseudo-force.

The final answers: $F_{\text{stone}} = 4\ \text{N}$ (tension) and $F_{\text{car}} = 2000\ \text{N}$ (friction).

Competitive stretch

If the same car's speed were doubled to $20$ m/s on the same curve, the required centripetal force would not double — it would quadruple, since $F \propto v^2$ : $F = 1000 \times 20^2 / 50 = 8000$ N. Because the road's friction cannot supply an unlimited force, this is exactly why speed limits on curved roads are so much stricter than on straight ones — doubling your speed demands four times the grip.

Common traps

Saying an object moving at constant speed around a circle has no acceleration — speed is constant, but direction is not, so velocity (and hence acceleration) is always changing.
Treating centrifugal force as a real force acting on the object itself — at this level it is only a pseudo-force felt by an observer inside the rotating frame; the only real force present is the centripetal one.
Believing a released or unstrung object flies straight outward, away from the centre — inertia sends it off along the tangent to the circle at the release point, not radially.
Forgetting that $F = mv^2/r$ only gives the size of the needed force — you must still separately identify which real force (tension, friction, gravity, normal reaction) is actually supplying it in a given situation.

Ask the tutor: want a walkthrough of why the escape path is tangential and not radial, or more practice plugging different masses, speeds, and radii into $F = mv^2/r$ ? Ask the AI tutor to work through it with you.

If centrifugal force is not real, why do I feel pushed outward on a fast turn

You feel it because your own body has inertia and wants to keep moving in a straight line while the car curves beneath and around you. The seat, seatbelt, or door pushes you inward (toward the centre) to keep you on the curved path, and it is that inward push you actually feel as pressure against your side. From inside the car it feels like an outward force, but no such force is actually acting on you — it is your inertia resisting the inward pull, described from a frame that is itself accelerating.

Does the centripetal force do any work on the object

No. The centripetal force always points toward the centre, which is exactly perpendicular to the object's velocity (tangent to the circle) at every instant. Since work depends on the component of force along the direction of motion, a force perpendicular to velocity does zero work. That is precisely why the object's speed stays constant in uniform circular motion — the force changes direction, not speed.

What happens if the required centripetal force is bigger than what is available, like weak string or low friction

The available force sets a limit on how fast or how tightly an object can turn. If the true requirement $mv^2/r$ exceeds what the string, friction, or other agent can supply, that agent breaks or slips, the object stops being held onto the circular path, and it moves off along the tangent instead — a snapped string, a skidding car, or a satellite drifting away from too weak a gravitational pull are all versions of the same idea.

Uniform circular motion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads