Uniform circular motion

Spin a stone on a string and something strange is true: its speed never changes, yet a physicist insists it is accelerating the whole time. That is not a trick of words. Velocity is a vector, and on a circle its direction turns continuously even while its magnitude (the speed) stays fixed — and any change in a vector, including a change of direction alone, counts as acceleration. This is uniform circular motion, and untangling "constant speed but non-zero acceleration" is one of the first real conceptual leaps in kinematics. This article stays inside the Class 11 syllabus scope: uniform circular motion is introduced here purely as a worked example of accelerated motion, using only speed, period, radius and the centripetal acceleration formula $a_c = v^2/r$ . The deeper dynamics — angular velocity $\omega$ , angular acceleration, and the centripetal force that causes this acceleration — belong to the Laws of Motion unit that follows, and are only named here as a forward pointer. We build the geometric picture, derive $a_c = v^2/r$ from first principles, and work two numerical examples of very different scale.

The core insight

In uniform circular motion, constant speed hides a continuously changing velocity direction, and this direction-only change produces a centre-pointing centripetal acceleration of magnitude $a_c = v^2/r$ , always perpendicular to the velocity.

The core idea

Speed is constant, velocity is not. A body in uniform circular motion covers equal arc lengths in equal times, so its speed $v$ (the magnitude of velocity) never changes. But velocity is a vector with both magnitude and direction, and on a circular path the direction is always tangent to the circle at the body's current position — so as the body moves around, this tangent direction rotates continuously. A vector that keeps rotating is a vector that keeps changing, even if its length is fixed. Since acceleration is defined as the rate of change of velocity, this body is accelerating despite never speeding up or slowing down.

The acceleration points inward. This acceleration is called centripetal acceleration ("centre-seeking"): at every instant it points from the body straight toward the centre of the circle. Its magnitude is

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

where: $a_c$ is the centripetal acceleration (m/s $^2$ ), $v$ is the constant speed (m/s), and $r$ is the radius of the circular path (m). A larger speed or a tighter (smaller-radius) circle both demand a larger inward acceleration — intuitively, faster turning or sharper turning both fight inertia harder.

Perpendicularity is the whole story. Because the centripetal acceleration always points toward the centre while the velocity is always tangent to the circle, the two vectors are always at right angles. This is the geometric reason speed stays constant: a component of acceleration parallel to velocity would change the speed, but an acceleration that is purely perpendicular to velocity can only rotate it, never stretch or shrink it. This is the opposite situation from straight-line accelerated motion, where acceleration is parallel (or antiparallel) to velocity and changes only the speed.

Period, frequency, and speed. The period $T$ is the time for one complete revolution, and the frequency $f = 1/T$ counts revolutions per unit time. Since the body covers the full circumference $2\pi r$ in one period at constant speed,

v = \frac{2\pi r}{T} = 2\pi r f.

where: $T$ is the period (s) and $f$ is the frequency (revolutions per second, or Hz). This relation is what lets a problem state "one revolution every 0.4 s" instead of quoting speed directly — the two descriptions carry the same information.

Figure

A circle with a body at four positions showing velocity arrows tangent to the circle and acceleration arrows pointing inward toward the centre at each position

Derivation ACentripetal acceleration from the change in velocity direction

The goal, in one careful sentence, is this: starting only from the definition of speed as constant and the geometric fact that the velocity vector's direction rotates by the same small angle as the position vector, show that the magnitude of the resulting acceleration is exactly $a_c = v^2/r$ , with the direction landing perpendicular to velocity and therefore toward the centre of the circle.

Step 1. Set up two nearby instants. Let the body be at point $P_1$ at time $t$ , moving with velocity $\vec{v}_1$ , and at a nearby point $P_2$ a short time $\Delta t$ later, moving with velocity $\vec{v}_2$ . Since the motion is uniform, $|\vec{v}_1| = |\vec{v}_2| = v$ ; only the direction has changed. Let $\Delta\theta$ be the small angle between the two velocity vectors — this is also the small angle swept by the radius as the body moves from $P_1$ to $P_2$ , because velocity is always perpendicular to the radius.

Step 2. Find the length of the velocity-change vector. Draw $\vec{v}_1$ and $\vec{v}_2$ tail to tail; since both have the same length $v$ and the angle between them is $\Delta\theta$ , the vector $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ is the base of an isosceles triangle with two equal sides of length $v$ . For small $\Delta\theta$ (measured in radians), the base of this thin triangle has length $|\Delta\vec{v}| \approx v\,\Delta\theta$ — this is the same small-angle approximation used for arc length, since a thin isosceles triangle with apex angle $\Delta\theta$ and equal sides $v$ has a base very close to the arc $v\,\Delta\theta$ of a circle of radius $v$ .

Step 3. Bring in the geometry of the path itself. Over the same short time $\Delta t$ , the body itself sweeps the same small angle $\Delta\theta$ along the actual circle of radius $r$ , covering an arc length $\Delta s = r\,\Delta\theta$ . Because the body moves at constant speed, this arc length also equals $v\,\Delta t$ , so $r\,\Delta\theta = v\,\Delta t$ , which rearranges to $\Delta\theta = \dfrac{v\,\Delta t}{r}$ .

Step 4. Combine the two results and take the limit. Substituting Step 3 into Step 2 gives $|\Delta\vec{v}| \approx v\left(\dfrac{v\,\Delta t}{r}\right) = \dfrac{v^2}{r}\,\Delta t$ . The magnitude of acceleration is the rate of change of velocity, $a_c = \lim_{\Delta t \to 0} \dfrac{|\Delta\vec{v}|}{\Delta t}$ , and dividing through by $\Delta t$ gives exactly $a_c = \dfrac{v^2}{r}$ , with the approximation becoming exact as $\Delta t \to 0$ .

Step 5. Fix the direction. As $\Delta t \to 0$ , the two points $P_1$ and $P_2$ merge, and the isosceles triangle from Step 2 becomes vanishingly thin with its base direction rotating to become perpendicular to $\vec{v}_1$ itself. Since $\vec{v}_1$ is tangent to the circle, a direction perpendicular to it and pointing "inward" (toward smaller radius, consistent with the triangle's orientation as the body turns) is exactly the direction from the body toward the centre. So $\Delta\vec{v}$ , and hence the acceleration, points toward the centre of the circle — confirming both the magnitude $a_c = v^2/r$ and the centripetal direction stated at the outset.

Worked example

A stone tied to a string is whirled in a horizontal circle of radius $0.8\ \text{m}$ , completing one revolution every $0.4\ \text{s}$ . Find its speed and its centripetal acceleration.

Step 1. Identify the given quantities and the target formulas. Here $r = 0.8\ \text{m}$ and $T = 0.4\ \text{s}$ ; speed follows from $v = 2\pi r/T$ , and then centripetal acceleration follows from $a_c = v^2/r$ .

Step 2. Compute the speed. $v = \dfrac{2\pi r}{T} = \dfrac{2\pi(0.8)}{0.4} = \dfrac{5.0265}{0.4} \approx 12.57\ \text{m/s}$ .

Step 3. Compute the centripetal acceleration using that speed. $a_c = \dfrac{v^2}{r} = \dfrac{(12.57)^2}{0.8} = \dfrac{158.0}{0.8} \approx 197.5\ \text{m/s}^2$ .

Step 4. State the final answer with a sanity check on scale. The stone moves at about $12.57\ \text{m/s}$ and experiences a centripetal acceleration of about $197.5\ \text{m/s}^2$ , directed toward the centre of the circle — roughly twenty times the acceleration due to gravity, which is why even a short, fast-spinning string exerts a large inward pull.

Competitive stretch

A satellite orbits Earth in a circle of radius $7000\ \text{km}$ at a constant speed of $7.5\ \text{km/s}$ . Converting to SI units, $r = 7\times10^6\ \text{m}$ and $v = 7.5\times10^3\ \text{m/s}$ , so

a_c = \frac{v^2}{r} = \frac{(7.5\times10^3)^2}{7\times10^6} = \frac{5.625\times10^7}{7\times10^6} \approx 8.04\ \text{m/s}^2.

At every instant this acceleration points from the satellite straight toward Earth's centre — so directly above the North Pole, it points straight down — even though the satellite's speed stays a constant $7.5\ \text{km/s}$ throughout the orbit. Notice the two examples span nine orders of magnitude in radius yet use the identical formula: the physics of "constant speed, ever-turning direction" is scale-free.

Common traps

Believing constant speed means zero acceleration. Speed and velocity are not the same thing; in uniform circular motion speed is constant but velocity (and hence acceleration) is not, because direction keeps changing.
Assuming centripetal acceleration points outward. This is the common but incorrect intuition from feeling "pushed outward" in a turning vehicle — the acceleration itself always points inward, toward the centre.
Confusing period $T$ with frequency $f$ . They are reciprocals, $f = 1/T$ ; swapping them inverts every subsequent calculation of speed.
Mixing units before applying $a_c = v^2/r$ . Radius quoted in km alongside speed in m/s (or vice versa) is a frequent source of errors of several orders of magnitude — convert everything to SI first.

Ask the tutor: if "accelerating at constant speed" still feels counterintuitive, ask the AI tutor to walk through the velocity-vector triangle with a few more numeric examples.

If the acceleration is always changing direction, how can we say uniform circular motion has a constant acceleration formula

The magnitude $a_c = v^2/r$ stays constant throughout the motion because both $v$ and $r$ are fixed — only the direction of the acceleration vector keeps changing, always pointing toward the (also moving-with-the-body-relative, but geometrically fixed) centre. So it is correct to say the size of the acceleration never changes even though the acceleration vector itself is different at every point on the circle.

Why does a component of acceleration perpendicular to velocity not change the speed

Speed is the length of the velocity vector, and a tiny rotation of a vector, to first order, changes its direction without changing its length — the same reason turning a clock's second hand does not change how long the hand is. Only a component of acceleration parallel to velocity would stretch or shrink the vector and change the speed. In uniform circular motion the acceleration is purely perpendicular, so it can only steer the motion, never speed it up or slow it down.

Is this the full treatment of circular motion, or is there more coming later

This article deliberately stops at kinematics: constant speed, the geometric derivation of $a_c = v^2/r$ , and the direction of the acceleration. What is not covered here — the force that actually causes this acceleration (centripetal force), angular velocity $\omega$ and angular acceleration $\alpha$ , and applications like banking of roads — belongs to the Laws of Motion unit that follows, where circular motion is revisited with forces included.

Uniform circular motion

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads