Uniform circular motion

Watch the second hand of a clock, the tip of a ceiling fan blade, or a stone whirled on a string, and you are watching the same quiet paradox play out again and again: something moving at a perfectly steady speed, yet accelerating the entire time. Ask a beginner whether a car going around a circular track at a constant 60 km/h is accelerating, and the instinctive answer is often "no — the speedometer never moves." That instinct is wrong, and untangling why is the whole point of uniform circular motion (UCM), the closing topic of the CBSE Class 11 chapter "Motion in a Plane." The resolution hinges on a fact vectors force you to confront early: velocity is not just a number, it is a number and a direction. A particle can keep the number fixed and still have velocity change, purely because the direction keeps swinging around. Any change in velocity, however it happens, is an acceleration — so a body tracing a circle at constant speed is accelerating at every single instant, even though its speedometer reading never twitches. That acceleration always points one way: straight in, toward the centre of the circle. It is called centripetal acceleration, from Latin roots meaning "centre-seeking," and its size is fixed by just two numbers, the speed and the radius. This article builds that idea from a simple vector-diagram argument, pins down the formula $a_c = v^2/r = \omega^2 r$ , connects angular speed $\omega$ to the everyday language of period and frequency, and works through two full numerical examples — a whirled stone and a spinning fan blade — building the vocabulary and formula fluency needed before the cause of this acceleration, centripetal force, is tackled in the next chapter, Laws of Motion.

Key takeaways

A particle moving at constant speed around a circle is still accelerating, because its velocity's direction keeps changing even though its magnitude does not.
This acceleration is called centripetal acceleration, it always points radially inward, toward the centre of the circle, and its magnitude is $a_c = \dfrac{v^2}{r}$ .
Angular speed $\omega$ (rad/s) measures how fast the angle is swept; it relates to linear speed by $v = \omega r$ , giving the equivalent form $a_c = \omega^2 r$ .
The period $T$ is the time for one full revolution and the frequency $f = 1/T$ counts revolutions per second; angular speed connects to both via $\omega = \dfrac{2\pi}{T} = 2\pi f$ .
"Uniform" refers only to constant speed — the acceleration vector itself is not constant, because its direction rotates continuously along with the particle.
This chapter covers only the kinematics of UCM (the acceleration produced by direction change); the force that produces it, $F_c = mv^2/r$ , and topics like banking of roads belong to the later chapter Laws of Motion.
Key idea: in uniform circular motion the speed never changes but the direction of velocity changes every instant, producing a centre-pointing centripetal acceleration $a_c = v^2/r = \omega^2 r$ whose magnitude stays fixed even as its direction continuously rotates.

The core insight

A particle moving at constant speed $v$ around a circle of radius $r$ still accelerates, because its velocity's direction rotates continuously; that acceleration always points toward the centre with magnitude $a_c = v^2/r = \omega^2 r$ , where $\omega = v/r = 2\pi/T = 2\pi f$ .

Where it fits in the syllabus

Uniform circular motion is the final headline topic of CBSE Class 11 Physics, chapter Motion in a Plane, immediately following projectile motion, and it is explicitly named in the prescribed 2025-26 scope with the formulas $a_c = v^2/r = \omega^2 r$ and $\omega = 2\pi/T$ . A student is expected to explain why constant-speed circular motion is still accelerated motion, derive the direction and magnitude of centripetal acceleration from a vector diagram, and convert fluently between linear speed, angular speed, period, and frequency. Non-uniform circular motion, banking of roads, and the force that causes circular motion (centripetal force, $F_c = mv^2/r$ ) are explicitly reserved for the later chapter Laws of Motion and are kept out of this section.

The physical picture

Set up a circle of radius $r$ centred at the origin, and let a particle travel around it at constant speed $v$ . At every point of the path, the velocity vector is tangent to the circle — it points along the direction of motion, perpendicular to the radius drawn to that point. As the particle moves from one point to a nearby point, the radius vector rotates by some small angle, and because velocity is always perpendicular to the radius, the velocity vector rotates by that exact same angle. The velocity's length never changes (speed is constant), but its direction sweeps around continuously, just like the radius does. A rotating vector of fixed length is exactly a changing vector, so there is a genuine acceleration here even without any change in speed. To find its direction, picture the velocity vectors at two nearby instants, both drawn from the same starting point, both the same length $v$ , but tilted by a small angle relative to each other. The vector connecting their tips — the change in velocity, $\Delta \vec v$ — points inward, roughly perpendicular to both, toward the centre of the circle. As the time interval shrinks toward zero, this stays true at every instant: the instantaneous acceleration points exactly toward the centre. That is why it is called centripetal acceleration, and its magnitude works out to:

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

where: $v$ is the constant speed of the particle (m/s) and $r$ is the radius of the circular path (m); $a_c$ comes out in m/s². Physically, this says the acceleration grows with the square of the speed (double the speed, and you need four times the inward acceleration to keep curving on the same circle) and shrinks as the circle gets larger (a gentler curve needs less inward pull for the same speed). Motion around a circle is also naturally described by the angle swept rather than the arc length travelled. The angular speed $\omega$ is the rate at which this angle changes, and it connects to the linear speed through the radius:

v = \omega r

where: $\omega$ is measured in radians per second (rad/s), and $r$ is the same radius as before. Substituting this into the centripetal acceleration formula gives an equally useful alternate form, expressed purely in terms of $\omega$ and $r$ :

a_c = \omega^2 r

Finally, two familiar everyday quantities complete the toolkit. The period $T$ is the time taken for one complete trip around the circle, and the frequency $f = 1/T$ is the number of complete trips per second. Since one full revolution sweeps through an angle of $2\pi$ radians in time $T$ , angular speed is:

\omega = \frac{2\pi}{T} = 2\pi f

Figure

A particle at two nearby positions on a circle with its velocity vectors drawn tangent to the circle at each position, both vectors the same length, and a separate small vector diagram showing the change in velocity pointing radially inward toward the centre of the circle

It is worth being precise about what "uniform" describes here: only the speed is constant, not the acceleration as a vector. The magnitude $a_c = v^2/r$ does stay fixed throughout the motion, but its direction keeps rotating along with the radius vector, always pointing from the particle inward toward the centre. This is genuinely different from the constant-acceleration motion studied earlier in this chapter (as in projectile motion), where both the magnitude and the direction of the acceleration vector stayed completely fixed throughout. Note too that this article develops only the kinematics of UCM — the changing-direction geometry that produces $a_c$ — and deliberately stops short of asking what physical agent (tension, friction, gravity) supplies the force $F_c = mv^2/r$ that causes this acceleration; that dynamics question, along with banking of roads, belongs to the chapter Laws of Motion.

Derivation ACentripetal acceleration from the geometry of a rotating velocity vector

The goal is to derive the magnitude $a_c = v^2/r$ using nothing beyond similar triangles and the definition of angular speed — no calculus beyond the idea of a very small angle, and no forces at all, since this is purely a statement about how the velocity vector's direction must change if the particle stays on a circle of fixed radius at fixed speed. The key trick is to notice that the triangle formed by the two position vectors and the chord between them is similar to the triangle formed by the two velocity vectors and the change in velocity between them, because both triangles are formed by rotating a fixed-length vector through the same small angle.

Step 1. Set up two nearby instants. Let the particle be at point $P_1$ at time $t$ and at nearby point $P_2$ at time $t+\Delta t$ , having swept a small angle $\Delta\theta$ around the centre in that time. The two radius vectors $\vec r_1$ and $\vec r_2$ both have length $r$ and differ in direction by $\Delta\theta$ . Because velocity is always perpendicular to the radius (tangent to the circle), the two velocity vectors $\vec v_1$ and $\vec v_2$ both have length $v$ and also differ in direction by exactly the same angle $\Delta\theta$ .

Step 2. Compare the two isosceles triangles. One triangle has two sides of length $r$ (the radius vectors) enclosing angle $\Delta\theta$ , with the chord $|\Delta \vec r\,|$ as its third side. The other has two sides of length $v$ (the velocity vectors) enclosing the same angle $\Delta\theta$ , with $|\Delta \vec v\,|$ as its third side. Since both triangles are isosceles with the same enclosed angle, they are similar, which gives the proportion $\dfrac{|\Delta \vec v\,|}{v} = \dfrac{|\Delta \vec r\,|}{r}$ .

Step 3. Take the small-angle limit. For very small $\Delta\theta$ , the chord length $|\Delta \vec r\,|$ becomes essentially equal to the arc length travelled, $v\,\Delta t$ . Substituting into the proportion from Step 2 gives $|\Delta \vec v\,| = \dfrac{v}{r}\,|\Delta \vec r\,| \approx \dfrac{v}{r}(v\,\Delta t) = \dfrac{v^2}{r}\,\Delta t$ .

Step 4. Divide by $\Delta t$ and take the limit. The magnitude of the instantaneous acceleration is $a_c = \lim_{\Delta t \to 0} \dfrac{|\Delta \vec v\,|}{\Delta t} = \dfrac{v^2}{r}$ . For the direction, as $\Delta\theta \to 0$ the chord $\Delta \vec v$ (which connects the tip of $\vec v_1$ to the tip of $\vec v_2$ ) becomes perpendicular to the velocity vectors themselves — and since the velocity is tangent to the circle, "perpendicular to velocity" at that point means pointing exactly along the radius, toward the centre. This confirms both the magnitude $a_c = v^2/r$ and the centripetal (centre-seeking) direction. Substituting $v = \omega r$ gives the equivalent form $a_c = \omega^2 r$ .

Worked examples

Board level - stone on a string

A stone tied to a string is whirled in a horizontal circle of radius 2 m at a constant speed of 4 m/s. Find (a) its angular speed, (b) the time period, and (c) its centripetal acceleration.

Step 1. Find angular speed using $v = \omega r$ . Rearranging, $\omega = \dfrac{v}{r} = \dfrac{4}{2} = 2\ \text{rad/s}$ .

Step 2. Find the time period using $\omega = 2\pi/T$ . Rearranging, $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{2} = \pi \approx 3.14\ \text{s}$ .

Step 3. Find centripetal acceleration directly from $a_c = v^2/r$ : $a_c = \dfrac{4^2}{2} = \dfrac{16}{2} = 8\ \text{m/s}^2$ .

Step 4. Cross-check with the alternate form $a_c = \omega^2 r$ : $a_c = (2)^2(2) = 8\ \text{m/s}^2$ — the two routes agree, confirming the arithmetic.

Final answers: $\omega = 2\ \text{rad/s}$ , $T \approx 3.14\ \text{s}$ , $a_c = 8\ \text{m/s}^2$ .

Competitive level - fan blade tip

A ceiling fan blade tip moves in a circle of radius 0.3 m, completing 300 revolutions per minute. Find (a) the frequency in Hz, (b) the angular speed, (c) the linear speed of the tip, and (d) the centripetal acceleration, and compare its size to $g = 9.8\ \text{m/s}^2$ .

Step 1. Convert the given rate to frequency in Hz. 300 revolutions per minute means $f = \dfrac{300}{60} = 5\ \text{Hz}$ (revolutions per second).

Step 2. Find angular speed using $\omega = 2\pi f$ : $\omega = 2\pi(5) = 10\pi \approx 31.4\ \text{rad/s}$ .

Step 3. Find the linear speed of the tip using $v = \omega r$ : $v = 31.4 \times 0.3 \approx 9.42\ \text{m/s}$ .

Step 4. Find centripetal acceleration using $a_c = \omega^2 r$ : $a_c = (31.4)^2(0.3) \approx 986.0 \times 0.3 \approx 295.8\ \text{m/s}^2$ .

Step 5. Compare to $g$ : $\dfrac{a_c}{g} = \dfrac{295.8}{9.8} \approx 30$ , so the blade tip experiences roughly 30 times the acceleration due to gravity, purely from spinning at a modest 300 rpm over a small 0.3 m radius — a reminder of how sensitive $a_c = \omega^2 r$ is to $\omega$ , since it depends on the square of the angular speed.

Final answers: $f = 5\ \text{Hz}$ , $\omega \approx 31.4\ \text{rad/s}$ , $v \approx 9.42\ \text{m/s}$ , $a_c \approx 295.8\ \text{m/s}^2 \approx 30g$ . Timing note: in a board exam this whole problem is worth splitting into small labelled parts, each finished in under a minute; the fastest route is always $f \to \omega \to v \to a_c$ in that order — never try to jump straight to $a_c$ from rpm without first converting to $\omega$ in rad/s, since a slip in units there wastes far more time than the actual arithmetic.

Common traps

Believing "uniform" circular motion means zero acceleration because speed is constant — acceleration exists because the direction of velocity changes continuously; only the magnitude (speed) stays unchanged.
Thinking centripetal acceleration points along the direction of motion — it always points radially inward, toward the centre, perpendicular to the velocity at every instant, never forward or backward along the path.
Confusing angular speed $\omega$ (rad/s, how fast the angle changes) with linear speed $v$ (m/s, how fast the arc length changes) — they are related by $v = \omega r$ but are different quantities with different units, and mixing them up (or forgetting to convert rpm to rad/s) is one of the most common scoring errors.
Forgetting to convert revolutions per minute to Hz (divide by 60) before using $\omega = 2\pi f$ — plugging rpm directly into the formula gives an answer sixty times too large.
Trying to bring in centripetal force, tension, friction, or banking angles here — those are dynamics topics that belong to the next chapter, Laws of Motion; this chapter only develops the kinematic result $a_c = v^2/r = \omega^2 r$ , with no reference to what causes it.

You should be able to…

Explain, using a vector-diagram argument, why a particle in uniform circular motion is accelerating despite having constant speed.
State the direction (radially inward) and derive the magnitude ( $a_c = v^2/r = \omega^2 r$ ) of centripetal acceleration from first principles.
Convert fluently between angular speed $\omega$ , period $T$ , and frequency $f$ , using $\omega = 2\pi/T = 2\pi f$ , including converting rpm to Hz first.
Relate linear speed to angular speed via $v = \omega r$ , and use it to move between the two forms of the centripetal acceleration formula.
Compute centripetal acceleration for a body given any two of speed, radius, angular speed, or period, and sanity-check an answer using both formula forms.
Recognise when a problem is asking only about UCM kinematics versus when it is (wrongly, for this chapter) asking about centripetal force or banking, which belong to Laws of Motion.

Ask the tutor: Curious why the vector diagram in the derivation above still works even as the two points get infinitesimally close together — ask the AI tutor to walk through the limiting argument one more time with a concrete small angle.

If the speed never changes in uniform circular motion, where does the energy for the acceleration come from?

No energy is needed for this particular acceleration, because the force producing it is always perpendicular to the velocity. A force perpendicular to motion does zero work (recall $W = Fd\cos\theta$ , and here $\theta = 90$ degrees), so it changes the direction of velocity without changing its magnitude, meaning kinetic energy stays exactly constant throughout. That is precisely why the speed can stay fixed even though a nonzero acceleration is present the whole time.

Why is it called centripetal and not centrifugal, if the stone actually seems to pull outward on the string?

The felt "outward pull" you notice while spinning something is really your own body's inertia trying to keep moving in a straight line, resisted by the string pulling you inward — there is no real outward force acting on the stone itself in an inertial frame. The actual acceleration of the stone, and the actual net force on it, both point inward, toward the centre, which is exactly what "centripetal" (centre-seeking) describes. The apparent outward "centrifugal" push is a fictitious effect that only shows up if you insist on analysing things from the rotating stone's own point of view, not from the ground.

Is angular speed the same thing as angular velocity, and does the direction of omega matter here?

For this chapter's purposes, treat $\omega$ simply as a scalar rate, in radians per second, describing how fast the angle is being swept — that is all that is needed to compute $a_c = \omega^2 r$ or convert between $\omega$ , $T$ , and $f$ . Angular velocity as a full vector, with a direction given by the right-hand rule along the rotation axis, becomes important later when angular momentum and torque are introduced in System of Particles and Rotational Motion, but it is not required, and not tested, at this stage.

Why does the same fan spinning at a constant rpm give a much bigger acceleration near the tip than near the hub?

Because $a_c = \omega^2 r$ : every point on a rigid spinning fan blade shares the same angular speed $\omega$ , since the whole blade turns together, but each point has a different radius $r$ from the axis. A point near the hub has small $r$ and therefore small $a_c$ , while a point near the tip has large $r$ and therefore much larger $a_c$ , growing in direct proportion to distance from the axis even though every point completes a revolution in the exact same time.