Angular motion variables

Watch a stone whirled on a string, or a point on a spinning fan blade, and you are watching two motions at once: a linear one (a definite speed, in metres per second) and an angular one (a definite rate of turning, in radians per second). Physics gives circular motion its own vocabulary — angular displacement $\theta$ , angular velocity $\omega$ , angular acceleration $\alpha$ — precisely because these angular quantities are the same for every point on a rotating body, however far it sits from the axis, while the linear speed of each point is different. Once you know how $\theta$ , $\omega$ and $\alpha$ connect back to the familiar linear quantities, a whole family of circular-motion problems — fans, wheels, orbiting stones, cars on curves — collapses into a handful of short, exact formulas. The central result this article builds toward is centripetal acceleration, $a_c = v^2/r = \omega^2 r$ , and the equally important idea that the force producing it, centripetal force, is not some new kind of force but simply a name for whichever real force is doing the pulling.

The core insight

Angular displacement, angular velocity and angular acceleration describe circular motion by the angle turned rather than the distance covered, and they connect to linear speed and centripetal acceleration through $v = r\omega$ and $a_c = v^2/r = \omega^2 r$ .

The core idea

Angular displacement $\theta$ is the angle, in radians, swept by the radius joining a particle to the centre of its circular path. If the particle covers an arc length $s$ on a circle of radius $r$ ,

\theta = \frac{s}{r}

where: $s$ is the arc length (m), $r$ is the radius (m), and $\theta$ is in radians — this formula is only valid with angles in radians, since a radian is defined precisely as the angle for which arc length equals radius.

Angular velocity $\omega$ is the rate at which this angle changes,

\omega = \frac{d\theta}{dt}, \qquad \omega = \frac{2\pi}{T} = 2\pi f

where: $T$ is the period of one revolution (s) and $f$ is the frequency of revolution (Hz); $\omega$ carries SI unit $\text{rad/s}$ . Angular acceleration $\alpha$ is the rate at which $\omega$ itself changes,

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

with SI unit $\text{rad/s}^2$ . In uniform circular motion the speed never changes, so $\alpha = 0$ throughout — even though the particle is still accelerating in another sense, because the direction of its velocity is constantly turning. Every point on a rotating rigid body shares the same $\theta$ , $\omega$ and $\alpha$ , but a point farther from the axis moves faster in the ordinary sense. The bridge between the two languages is the radius itself:

v = r\omega, \qquad a_t = r\alpha

where: $v$ is linear (tangential) speed, $a_t$ is tangential acceleration (the rate of change of speed along the path, relevant only when motion is non-uniform), and $r$ is the distance from the axis. Double the radius at the same $\omega$ and the linear speed doubles too — this is why the outer teeth of a gear or the tip of a fan blade move far faster than a point near the hub, even though every point completes the circle in the same time.

Figure

A particle on a circle showing the angle theta swept between two positions, with velocity vectors tangent to the circle at each position and the centripetal acceleration vector pointing inward toward the centre

Centripetal acceleration. Because the velocity vector keeps rotating even while its length stays fixed, a particle in uniform circular motion is always accelerating, and this acceleration always points toward the centre. Its magnitude is

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

and Newton's second law then gives the centripetal force, $F_c = ma_c = mv^2/r = m\omega^2 r$ — not a new, independent force, but simply the name for whatever real force (tension, friction, gravity, normal reaction) happens to point toward the centre and supply this acceleration.

Derivation ADeriving centripetal acceleration from the changing velocity vector

The goal is to show, using only the geometry of a circle and the definition of average acceleration, that a particle moving at constant speed $v$ around a circle of radius $r$ has an acceleration of magnitude $v^2/r$ directed toward the centre — without appealing to calculus, just a small-angle chord construction that a Class 11 student can draw and measure.

Step 1. Set up two nearby positions. Let the particle be at point A at time $t$ , with velocity $\vec{v}_A$ tangent to the circle, and at a nearby point B a short time $\Delta t$ later, having swept a small angle $\Delta\theta$ , with velocity $\vec{v}_B$ tangent at B. Because the motion is uniform, $|\vec{v}_A| = |\vec{v}_B| = v$ — only the direction of the velocity has changed, by the same small angle $\Delta\theta$ that the radius itself swept.

Step 2. Find the change in velocity. Draw $\vec{v}_A$ and $\vec{v}_B$ as two arrows of equal length $v$ , tail to tail, separated by the small angle $\Delta\theta$ between them (this is always true for tangent vectors on a circle: the angle between the velocities equals the angle between the radii). The vector $\Delta\vec{v} = \vec{v}_B - \vec{v}_A$ joins the tip of $\vec{v}_A$ to the tip of $\vec{v}_B$ , forming an isosceles triangle with two sides of length $v$ and included angle $\Delta\theta$ . For small $\Delta\theta$ (measured in radians), the chord length of this thin triangle is closely approximated by $|\Delta\vec{v}| \approx v\,\Delta\theta$ .

Step 3. Divide by the time elapsed. The magnitude of the average acceleration over the interval is $\bar{a} = |\Delta\vec{v}|/\Delta t \approx v\,\Delta\theta/\Delta t$ . As $\Delta t \to 0$ , the ratio $\Delta\theta/\Delta t$ becomes exactly the angular velocity $\omega$ , so the instantaneous acceleration has magnitude $a = v\omega$ . Using $v = r\omega$ to eliminate $\omega$ , this becomes $a = v(v/r) = v^2/r$ , or equivalently $a = \omega^2 r$ after eliminating $v$ instead — the two familiar forms of the centripetal acceleration formula.

Step 4. Fix the direction. As $\Delta\theta \to 0$ , the isosceles triangle of Step 2 becomes very thin, and the vector $\Delta\vec{v}$ becomes perpendicular to $\vec{v}_A$ itself. Since $\vec{v}_A$ is tangent to the circle at A, a direction perpendicular to it at A points either straight toward the centre or straight away from it; a careful look at the geometry (B lags slightly behind A around the curve) shows $\Delta\vec{v}$ always points inward. So the acceleration, however small the interval, is directed toward the centre of the circle — hence "centripetal," centre-seeking.

Worked example

A particle moves in a circle of radius $0.5\ \text{m}$ , completing $10$ revolutions in $5\ \text{s}$ at constant speed. Find its angular velocity, linear speed and centripetal acceleration; then find the centripetal force needed to keep a $0.3\ \text{kg}$ mass moving this way, and name what could physically supply it.

Step 1. Find the period. Ten revolutions in $5\ \text{s}$ means one revolution takes $T = 5/10 = 0.5\ \text{s}$ .

Step 2. Find the angular velocity. $\omega = 2\pi/T = 2\pi/0.5 = 4\pi \approx 12.57\ \text{rad/s}$ .

Step 3. Find the linear speed. $v = r\omega = 0.5 \times 12.57 \approx 6.28\ \text{m/s}$ .

Step 4. Find the centripetal acceleration two ways, as a check. Using $a_c = \omega^2 r$ : $a_c = (12.57)^2 \times 0.5 \approx 158.0 \times 0.5 = 79.0\ \text{m/s}^2$ . Using $a_c = v^2/r$ : $a_c = (6.28)^2/0.5 \approx 39.4/0.5 = 78.9\ \text{m/s}^2$ — the two agree within rounding, confirming $v = r\omega$ was used consistently.

Step 5. Find the centripetal force on a $0.3\ \text{kg}$ mass moving this way. $F_c = ma_c = 0.3 \times 79.0 \approx 23.7\ \text{N}$ . Answer: $\omega \approx 12.57\ \text{rad/s}$ , $v \approx 6.28\ \text{m/s}$ , $a_c \approx 79.0\ \text{m/s}^2$ , and $F_c \approx 23.7\ \text{N}$ , which would physically be supplied by the tension in a string if the mass were a stone being whirled, or by friction if it were, say, a puck constrained to this circular path on a rough surface — the "centripetal force" is just the name for that real, inward-pointing force, not something added on top of it.

Common traps

Treating "centripetal force" as a separate force to draw on a free-body diagram alongside tension, gravity or friction — it is the name for the net inward force already supplied by these, not an extra one.
Forgetting that in uniform circular motion $\alpha = 0$ (constant $\omega$ ), even though the particle is still accelerating centripetally because its velocity's direction keeps changing.
Mixing degrees and radians when substituting into $v = r\omega$ , $s = r\theta$ or $a_t = r\alpha$ — these relations hold only when the angle is in radians.
Using $a_c = v^2/r$ and $a_c = \omega^2 r$ inconsistently instead of checking with $v = r\omega$ ; either form gives the identical answer, so pick whichever the given data makes easier.

Ask the tutor: Want to practice converting between linear and angular quantities, or work through more examples of identifying what physically supplies the centripetal force in different setups?

Why do we bother with angular quantities like omega when we already have linear speed v?

Because omega describes the whole rotating body at once. Every point on a spinning wheel, disc or fan blade shares the same angular velocity and angular acceleration regardless of how far it sits from the axis, while the linear speed $v = r\omega$ is different for every point. It is far simpler to say "the wheel turns at $\omega = 10\ \text{rad/s}$ " than to separately track the linear speed of every point on it, and $\omega$ plugs directly into $v = r\omega$ whenever a specific point's speed is actually needed.

If the speed is constant in uniform circular motion, why is the particle accelerating at all?

Because acceleration measures the rate of change of the velocity vector, and a vector changes if either its magnitude or its direction changes. In uniform circular motion the magnitude (speed) stays fixed, but the direction of velocity is continuously rotating to stay tangent to the circle. That continuous turning is itself a change in velocity, so the particle is accelerating even at constant speed — and this acceleration points toward the centre, which is exactly the centripetal acceleration $a_c = v^2/r$ .

Is centripetal force the same thing as centrifugal force?

No, and this is a very common mix-up. Centripetal force is a real, inward-pointing force (or combination of forces) that an outside observer sees acting on the object to keep it curving — it is what tension, gravity or friction is actually doing. Centrifugal force is not a real force at all; it is the outward "push" you feel only if you are riding along inside the rotating frame yourself, a bookkeeping effect of analysing motion from a non-inertial, rotating viewpoint. In this article, working entirely from the outside (an inertial frame), only centripetal force appears.

Angular motion variables

The core idea

Worked example

Common traps

Before this

Where this leads