JEE Mains Physics · Motion in a Plane

Uniform Circular Motion: Angular Speed and Acceleration

A body moving round a circle of radius r at angular speed ω has speed v = ωr and an acceleration v²/r = ω²r pointing at the centre, even when its speed is constant; if the speed changes as well, a tangential acceleration dv/dt adds to it at right angles.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Nineteen PYQs, sixteen of them multiple choice, and none from 2026. Seven compare the distance travelled round a circle with the displacement, or the average velocity with the speed. Twelve use the centripetal acceleration: a value from a period or an rpm, its direction as a vector, how the speed must change with the radius, and three where the speed itself changes so a tangential acceleration joins in. Three of the seven distance questions come with a figure; read the angle turned off it first.

Concept 1 of 2: Distance, displacement and average velocity on a circle

Going round a circle, the distance travelled is the length of the arc, but the displacement is the straight chord from start to finish. After a full turn the displacement is zero; after half a turn it is the diameter. Average speed uses the arc, average velocity uses the chord, so on a circle the average velocity is always smaller than the speed.

Definition

Angle turned $\theta = \omega t$ (in radians); arc $s = r\theta$ .
Displacement $= 2r\sin\dfrac{\theta}{2}$ : $r\sqrt{2}$ for a quarter or three quarters of a turn, $2r$ for half a turn, 0 for a whole one.
Average speed $= v$ ; average velocity $= \dfrac{2r\sin(\theta/2)}{t}$ .
$\dfrac{v}{v_{avg}} = \dfrac{\theta}{2\sin(\theta/2)}$ .
Clock hands: the second hand turns once a minute, the minute hand once an hour, the hour hand once in 12 hours.

Arc and chord

s = r\theta \qquad |\Delta\vec{r}| = 2r\sin\frac{\theta}{2} \qquad \frac{v}{v_{avg}} = \frac{\theta}{2\sin(\theta/2)}

Worked example

A particle moves at constant speed on a circle of radius 6 m and turns through

60^{\circ}

in 2 s. Find the distance travelled, the displacement, the average speed and the average velocity.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q6Moderate

Example 1 · Motion in a Plane · Uniform Circular Motion: Angular Speed and Acceleration

A particle is moving with constant speed in a circular path. When the particle turns by an angle

90^{\circ}

, the ratio of instantaneous velocity to its average velocity is

\pi:x\sqrt{2}

. The value of

x

will be

Average velocity from the arc

Average velocity is displacement over time, and the displacement is the chord 2r sin(θ/2). Dividing the arc length by the time gives the average speed instead.

Degrees in s = rθ

The arc length rθ needs θ in radians. Ninety degrees is π/2, so a quarter turn of radius r is πr/2 long, not 90r.

Drill 6 more on distance, displacement and average velocity on a circle

Concept 2 of 2: Centripetal and tangential acceleration

At constant speed the velocity still changes direction all the time, and that change needs an acceleration pointing at the centre: v²/r, or ω²r. Because it is always at right angles to the velocity, it turns the motion without speeding it up, and the force behind it does no work. If the speed is also changing, a second acceleration dv/dt acts along the velocity; the two are at right angles, so they add by Pythagoras.

Definition

$\omega = \dfrac{2\pi}{T} = 2\pi f$ ; for N rpm, $\omega = \dfrac{2\pi N}{60}$ . Speed $v = \omega r$ .
Centripetal acceleration $a_c = \dfrac{v^{2}}{r} = \omega^{2}r$ , towards the centre; as a vector $\vec{a} = -\omega^{2}\vec{r}$ .
In uniform circular motion the force is at right angles to the velocity and to the momentum, so it does no work and the speed stays constant.
Same mass and same centripetal force: $v \propto \sqrt{r}$ . A central force $F \propto 1/R^{n}$ gives $T \propto R^{(n+1)/2}$ .
Changing speed: tangential acceleration $a_t = \dfrac{dv}{dt}$ ; total $a = \sqrt{a_t^{2} + a_c^{2}}$ . Only the tangential force does work: power $P = F_t v$ .
$a_t = a_c$ at every instant: $v\dfrac{dv}{ds} = \dfrac{v^{2}}{R}$ gives $v = v_0e^{s/R}$ , and the time to turn through θ is $\dfrac{R}{v_0}(1 - e^{-\theta})$ .

Circular motion

v = \omega r \qquad a_c = \frac{v^{2}}{r} = \omega^{2}r \qquad a = \sqrt{a_t^{2} + a_c^{2}},\ a_t = \frac{dv}{dt}

Worked example

A stone on a string goes round a horizontal circle of radius 0.5 m at 120 rpm. Find its angular speed, its speed and its centripetal acceleration.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 29 January 2023 · Q108Moderate

Example 2 · Motion in a Plane · Uniform Circular Motion: Angular Speed and Acceleration

An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at

x = + 2\text{ }m

, its velocity is

- 4\widehat{\jmath}\text{ }m/s

. The object's velocity (v) and acceleration

(a)

x = - 2\text{ }m

will be

Constant speed, zero acceleration

In uniform circular motion the speed is constant but the direction is not, so the acceleration is v²/r towards the centre. Only straight-line motion at constant speed has zero acceleration.

Forgetting to convert rpm

ω in rad/s is 2π × (revolutions per second). For N rpm divide by 60 first: 120 rpm is 4π rad/s, not 240π.

Adding the two accelerations directly

Tangential and centripetal accelerations are at right angles. Their total is √(a_t² + a_c²), never a_t + a_c.

Drill 11 more on centripetal and tangential acceleration

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Distance, displacement and average velocity on a circle
Arc and chord
$s = r\theta \qquad |\Delta\vec{r}| = 2r\sin\frac{\theta}{2} \qquad \frac{v}{v_{avg}} = \frac{\theta}{2\sin(\theta/2)}$
Centripetal and tangential acceleration
Circular motion
$v = \omega r \qquad a_c = \frac{v^{2}}{r} = \omega^{2}r \qquad a = \sqrt{a_t^{2} + a_c^{2}},\ a_t = \frac{dv}{dt}$

Watch out for (5)

Average velocity from the arc
→ Distance, displacement and average velocity on a circle
Degrees in s = rθ
→ Distance, displacement and average velocity on a circle
Constant speed, zero acceleration
→ Centripetal and tangential acceleration
Forgetting to convert rpm
→ Centripetal and tangential acceleration
Adding the two accelerations directly
→ Centripetal and tangential acceleration

Test yourself on Motion in a Plane

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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