Centripetal acceleration

A stone whirled on a string moves at constant speed, yet something is unmistakably changing every instant: the direction of its velocity. Because velocity is a vector, a change in direction alone is an acceleration — even though the speedometer reading never moves. That acceleration always points from the stone toward your hand, at the centre of the circle, and it has a name: centripetal acceleration. This single idea, born purely from geometry, is what lets Newton's second law explain why planets orbit, why a car grips a curve, and why a coin sits still on a spinning turntable until it doesn't. In this article we derive $a_c = v^2/r = \omega^2 r$ from first principles, connect it to the force that must supply it, and apply both to the two ISC-syllabus staples: a vehicle on a level road and a vehicle on a frictionless banked road.

The core insight

A particle moving at constant speed $v$ on a circle of radius $r$ is still accelerating, because its velocity keeps changing direction, and that acceleration has magnitude $a_c = v^2/r = \omega^2 r$ and always points toward the centre.

The core idea

Why constant speed does not mean zero acceleration. Velocity has both magnitude and direction. In uniform circular motion the magnitude $v$ never changes, but the direction rotates continuously — the velocity vector is always tangent to the circle, so as the particle sweeps around, the tangent direction sweeps around with it. A changing direction, at fixed magnitude, is exactly what acceleration measures.

Angular quantities. Let the particle sweep through a small angle $\theta$ in time $t$ . Angular velocity is the rate of that sweep,

\omega = \frac{d\theta}{dt}

and angular acceleration is $\alpha = d\omega/dt$ (zero for uniform circular motion, since $\omega$ itself is constant). Linear speed and angular speed are tied together by the arc-length relation

v = \omega r

where: $v$ is linear speed (m/s), $\omega$ is angular speed (rad/s), and $r$ is the radius (m). Doubling $\omega$ at fixed $r$ doubles $v$ ; doubling $r$ at fixed $\omega$ also doubles $v$ — the outer horse on a merry-go-round moves faster than the inner one even though both share the same $\omega$ .

The direction of the acceleration. Over a short interval $\Delta t$ , the velocity vector rotates by the same small angle $\Delta\theta$ that the particle itself sweeps, while its length $v$ stays fixed. Two vectors of equal length separated by a small angle differ by a third vector that is, to a very good approximation, perpendicular to both — and for a particle going around a circle, "perpendicular to the tangent" points straight at the centre. So the change in velocity, however small, is centre-directed; dividing by $\Delta t$ and shrinking it to zero keeps that direction. This is the geometric heart of why $a_c$ points inward even though nothing about the motion looks like it is "falling" anywhere.

Figure

Velocity vectors tangent to a circle at two nearby points, with the change between them drawn as an arrow pointing from the circle toward the centre

Common traps

"Constant speed means no acceleration." Wrong for circular motion — speed is constant but direction is not, so $a_c = v^2/r \ne 0$ throughout.
Drawing "centripetal force" as an extra arrow on a free-body diagram. It is not a new force; it is the label for whatever real inward force (friction, tension, normal reaction, gravity) is already present. Adding it separately double-counts.
Mixing up centripetal with centrifugal. The outward "centrifugal force" only appears in a rotating (non-inertial) frame; in the ground (inertial) frame used throughout this syllabus, the only real force is centripetal, pointing in.
Applying the banking formula $v = \sqrt{rg\tan\theta}$ when friction is present. That formula is strictly for the frictionless, ideal-speed case; a banked road with friction is outside this syllabus.

Derivation ADeriving centripetal acceleration and force

Show, starting only from the definition of acceleration as the rate of change of velocity and simple vector geometry of two equal-length vectors separated by a small angle, that a particle moving at constant speed $v$ on a circle of radius $r$ has acceleration of magnitude $a_c = v^2/r = \omega^2 r$ directed toward the centre, and that Newton's second law then fixes the centre-directed force needed to sustain that motion.

Step 1. Set up the geometry. Let the particle be at point P at time $t$ with velocity $\vec v$ tangent to the circle, and at nearby point Q at time $t+\Delta t$ with velocity $\vec v\,'$ , also tangent, with $|\vec v| = |\vec v\,'| = v$ constant. Because the radius to P and the radius to Q have rotated by the small angle $\Delta\theta$ , and each velocity is perpendicular to its own radius, the velocity vector itself has also rotated by $\Delta\theta$ .

Step 2. Find the magnitude of $\Delta \vec v = \vec v\,' - \vec v$ . Two vectors of equal length $v$ separated by a small angle $\Delta\theta$ form an isosceles triangle whose short side, for small $\Delta\theta$ , has length $|\Delta \vec v| \approx v\,\Delta\theta$ (the arc of a circle of radius $v$ subtending angle $\Delta\theta$ ). This is the same small-angle approximation used for arc length $s = r\theta$ , applied here to the tip of the velocity vector instead of to the particle's own position.

Step 3. Find the direction of $\Delta \vec v$ . As $\Delta\theta \to 0$ , the chord joining the tips of $\vec v$ and $\vec v\,'$ becomes perpendicular to $\vec v$ itself — and since $\vec v$ is tangent to the circle, "perpendicular to the tangent at P" is exactly the direction from P toward the centre. So $\Delta \vec v$ points centre-ward.

Step 4. Take the limit to get acceleration. By definition, $a_c = \lim_{\Delta t \to 0} |\Delta \vec v|/\Delta t = v\,\dfrac{d\theta}{dt} = v\omega$ . Using $v = \omega r$ from the core idea, this can be written two equivalent ways:

a_c = \frac{v^2}{r} = \omega^2 r

directed toward the centre at every instant — hence "centripetal."

Step 5. Apply Newton's second law. A net force must produce this acceleration, so

F_c = ma_c = \frac{mv^2}{r} = m\omega^2 r

This $F_c$ is not an independent force of nature; it is simply the name given to the resultant of whatever real forces (tension, friction, normal reaction, gravity) happen to sum to a centre-directed push of this magnitude. On a level road that resultant is friction; on a frictionless banked road it is the horizontal component of the normal reaction — worked out below.

Worked example

A car of mass $1200\ \text{kg}$ rounds a level circular curve of radius $50\ \text{m}$ at $15\ \text{m/s}$ , with coefficient of static friction $\mu_s = 0.4$ between tyres and road (take $g = 9.8\ \text{m/s}^2$ ). Find (a) the centripetal acceleration and force needed, (b) the maximum safe speed, and (c) whether $15\ \text{m/s}$ is safe.

Step 1. Centripetal acceleration at $15\ \text{m/s}$ : $a_c = v^2/r = 15^2/50 = 225/50 = 4.5\ \text{m/s}^2$ , directed toward the centre of the curve. The force this requires is $F_c = ma_c = 1200 \times 4.5 = 5400\ \text{N}$ .

Step 2. Maximum friction available. On a level road, friction alone supplies the centripetal force, and its ceiling is $f_{s,\max} = \mu_s N = \mu_s mg = 0.4 \times 1200 \times 9.8 = 4704\ \text{N}$ .

Step 3. Maximum safe speed. Setting $mv_{max}^2/r = \mu_s mg$ and cancelling $m$ gives $v_{max} = \sqrt{\mu_s g r} = \sqrt{0.4 \times 9.8 \times 50} = \sqrt{196} = 14\ \text{m/s}$ .

Step 4. Compare. The car needs $5400\ \text{N}$ of centripetal force at $15\ \text{m/s}$ , but friction can supply at most $4704\ \text{N}$ (equivalently, $15\ \text{m/s}$ exceeds $v_{max} = 14\ \text{m/s}$ ). The car would skid at 15 m/s on this level road; it must slow to 14 m/s or below, or the curve must be banked.

Competitive stretch

Suppose instead the same curve (radius $60\ \text{m}$ ) is banked at $30°$ for vehicles moving at the design speed, with no reliance on friction. Then $v = \sqrt{rg\tan\theta} = \sqrt{60 \times 9.8 \times \tan 30°} = \sqrt{60 \times 9.8 \times 0.577} = \sqrt{339.3} \approx 18.4\ \text{m/s}$ (about $66\ \text{km/h}$ ), and the corresponding angular velocity is $\omega = v/r = 18.4/60 \approx 0.307\ \text{rad/s}$ . At exactly this speed the horizontal component of the normal reaction alone provides the required centripetal force — no friction needed at all.

Ask the tutor: Want a walkthrough of why $\Delta \vec v$ points exactly toward the centre, or a fresh numeric problem mixing friction and banking to test yourself?

If the string breaks while I am whirling a stone, which way does the stone fly?

It flies off in a straight line tangent to the circle at the point where the string broke — not radially outward. This is Newton's first law: once the centripetal force (the string's tension) disappears, there is no force left to bend the path, so the stone continues in whatever direction it was moving at that instant, which is always tangent to the circle. People often imagine it flying straight out from the centre, but that only looks plausible because early in the flight the tangent path does carry the stone away from the circle.

Why does the maximum speed formula not have the car's mass in it?

Because mass cancels. The requirement is $mv_{max}^2/r = \mu_s mg$ , and $m$ appears on both sides, so it drops out to give $v_{max} = \sqrt{\mu_s g r}$ . Physically, a heavier car needs proportionally more centripetal force, but it also has proportionally more available friction (since $f_{s,\max} = \mu_s mg$ also scales with mass), so the two effects cancel and the safe speed depends only on the road's friction, its radius, and $g$ — not on how loaded the car is.

Is the banking angle formula still valid if there is some friction on the banked road?

No, not as written. $v = \sqrt{rg\tan\theta}$ is derived assuming the normal reaction is the only force supplying the horizontal centripetal component — i.e., zero friction. If friction is present, a range of speeds (not just one) becomes safe, bounded below and above by friction acting down-slope or up-slope respectively; deriving those bounds needs an extra force term in both equations. That fuller analysis, and the related conical pendulum, are explicitly outside the ISC Class 11 syllabus, which restricts banking to the frictionless, ideal-speed case shown here.

Centripetal acceleration

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads