Physics
Kinematics
Circular Motion

Kinematics · JEE & NEET Physics

Circular Motion: notes and previous year questions

Why moving round a circle at a steady speed still needs an acceleration, and how angles, angular velocity and centripetal acceleration describe it.

15 JEE Main questions (2012–2026)
6 NEET questions (2001–2024)

Circular Motion in short

Moving round a circle at constant speed is accelerated motion, because the direction of the velocity changes.
The velocity is tangent to the circle; the centripetal acceleration points to the centre.
In radians, θ = s/r; a full turn is 2π rad = 360°, and 1 rad ≈ 57.3°.
Angular velocity ω = dθ/dt is the same for every point of a turning body; the linear speed is v = rω.

1Always turning, always accelerating

Whirl a ball on a string at a steady speed. The size of its velocity (the speed) stays the same, but its direction keeps changing. A change of velocity is an acceleration, so the ball accelerates even at constant speed.

The velocity is always tangent to the circle. If the string breaks, the ball flies off along the tangent.
The acceleration of uniform circular motion always points to the centre.

2Angles in radians

The angular displacement $\theta$ is the angle swept by the radius. In radians, it is the arc length divided by the radius. An arc exactly one radius long makes an angle of 1 rad ≈ 57.3°.

s = r\theta

θ in radians.

2\pi \text{ rad} = 360°

A full turn: the circumference 2πr holds 2π ≈ 6.28 radii. So π rad = 180° and π/2 rad = 90°.

Degrees to radians: multiply by $\pi/180$ .
Radians to degrees: multiply by $180/\pi$ . Example: $1.5 \times 57.3° \approx 86°$ .

Example: a wheel of radius 0.5 m turning through 4 rad moves a point on its rim $s = 0.5 \times 4 = 2$ m.

3Angular velocity, period and frequency

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

Angular velocity, in rad/s. Its direction is along the axis of rotation (right-hand rule).

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

Angular acceleration, in rad/s².

Two riders on the same merry-go-round share the same $\omega$ , but the outer one covers more distance, so moves faster. To go from angular to linear quantities, multiply by the radius:

Linear	Angular	Relation
Arc length $s$	$\theta$	$s = r\theta$
Speed $v$	$\omega$	$v = r\omega$
Tangential acceleration $a_t$	$\alpha$	$a_t = r\alpha$

In uniform circular motion the speed and $\omega$ are constant. The period $T$ is the time for one turn and the frequency $f$ is the number of turns per second:

T = \frac{2\pi}{\omega} = \frac{2\pi r}{v}

f = \frac{1}{T} = \frac{\omega}{2\pi}

In hertz (Hz). A fan making 3 turns per second: f = 3 Hz, ω = 6π ≈ 18.8 rad/s.

4Centripetal acceleration

Take the velocities at two nearby points: same length, different directions. Placed tail to tail, their difference $\Delta\vec{v}$ points toward the centre. The resulting acceleration is called centripetal (centre-seeking):

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

5Centripetal and centrifugal

	Centripetal	Centrifugal
What it is	the real net inward force	a pseudo force
Seen from	an inertial frame (the ground)	the rotating frame
Direction	toward the centre	away from the centre
Size	$mv^2/r$	$m\omega^2 r$

The centripetal force is not a new kind of force: it is whatever real force points inward, such as the tension in a string, the friction on a car's tyres on a flat bend, or gravity for a satellite.

6Speeding up on a circle

In non-uniform circular motion the speed changes too, and the acceleration has two perpendicular parts:

Centripetal $a_c = v^2/r$ , toward the centre: it changes the direction.
Tangential $a_t = dv/dt = r\alpha$ , along the path: it changes the speed.

a = \sqrt{a_c^2 + a_t^2}

\tan\phi = \frac{a_t}{a_c}

φ is the angle between the total acceleration and the radius.

For a constant angular acceleration, the equations of motion carry over directly:

Straight line	Rotation
$v = u + at$	$\omega = \omega_0 + \alpha t$
$s = ut + \tfrac{1}{2}at^2$	$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$
$v^2 = u^2 + 2as$	$\omega^2 = \omega_0^2 + 2\alpha\theta$

Example: a wheel starting from rest with $\alpha = 2\ \text{rad/s}^2$ reaches $\omega = 10$ rad/s after 5 s, having turned $\theta = \tfrac{1}{2}(2)(25) = 25$ rad. A stone on a 0.5 m string making 2 turns per second has $\omega = 4\pi$ rad/s and $a_c = 16\pi^2 \times 0.5 \approx 80\ \text{m/s}^2$ .

Summary

Key ideas

Moving round a circle at constant speed is accelerated motion, because the direction of the velocity changes.
The velocity is tangent to the circle; the centripetal acceleration points to the centre.
In radians, θ = s/r; a full turn is 2π rad = 360°, and 1 rad ≈ 57.3°.
Angular velocity ω = dθ/dt is the same for every point of a turning body; the linear speed is v = rω.
Period T = 2π/ω and frequency f = 1/T.
Centripetal acceleration is v²/r = ω²r: doubling the speed quadruples it.
In uniform circular motion the acceleration is perpendicular to the velocity, so the speed stays constant.
The centripetal force is the real net inward force (tension, friction, gravity); the centrifugal force is a pseudo force of the rotating frame.
If the inward force vanishes, the object flies off along the tangent.
When the speed changes, a tangential acceleration a_t = rα adds to the centripetal one, and the total is √(a_c² + a_t²).
With constant α, the angular equations of motion mirror the straight-line ones.

Every equation

Arc length: $s = r\theta$
Full turn: $2\pi \text{ rad} = 360°$
Angular velocity: $\omega = d\theta/dt$
Angular acceleration: $\alpha = d\omega/dt$
Linear speed: $v = r\omega$
Tangential acceleration: $a_t = r\alpha$
Period: $T = \frac{2\pi}{\omega} = \frac{2\pi r}{v}$
Frequency: $f = \frac{1}{T} = \frac{\omega}{2\pi}$
Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2 r = v\omega$
Centripetal force: $F = \frac{mv^2}{r}$
Centrifugal (pseudo) force: $F = m\omega^2 r$
Total acceleration: $a = \sqrt{a_c^2 + a_t^2}$
Its angle with the radius: $\tan\phi = a_t / a_c$
Angular velocity: $\omega = \omega_0 + \alpha t$
Angle turned: $\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$
Without time: $\omega^2 = \omega_0^2 + 2\alpha\theta$

Previous year questions with solutions

Real JEE and NEET questions on circular motion. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A particle is rotating in a circular path and at any instant its motion can be described as

$\theta =\frac{5t^{4}}{40}-\frac{t^{3}}{3}$ .

The angular acceleration of the particle after 10 seconds is _________ rad/s².

A150
B120
C130
D170

Show answer and solution

Answer: Option C

Tidy the first term before differentiating anything: $\dfrac{5t^{4}}{40} = \dfrac{t^{4}}{8}$ . Then $\omega = \dfrac{d\theta}{dt} = \dfrac{t^{3}}{2} - t^{2}$ , and $\alpha = \dfrac{d\omega}{dt} = \dfrac{3t^{2}}{2} - 2t = 1.5t^{2} - 2t$ . At $t = 10$ that is $150 - 20 = 130\ \mathrm{rad/s^{2}}$ . The $150$ on offer is what you get by dropping the second term — and note that $\alpha$ grows with $t$ here, so the three angular equations are shut out and the definitions are the only way in.

Q2NEET 2024One correct option

A bob is whirled in a horizontal circle by means of a string at an initial speed of $10 \mathrm{rpm}$ . If the tension in the string is quadrupled while keeping the radius constant, the new speed is:

A20 rpm
B40 rpm
C5 rpm
D10 rpm

Show answer and solution

Answer: Option A

The tension is the inward pull, so $T = \dfrac{mv^{2}}{r}$ with $m$ and $r$ both held fixed. Quadrupling it quadruples $v^{2}$ and so doubles $v$ , giving $2 \times 10 = 20\ \mathrm{rpm}$ . Working in revolutions per minute rather than $\mathrm{m/s}$ changes nothing, because at a fixed radius the two are proportional and the factor of $2$ passes straight through.

Q3JEE Main 2024One correct option

A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $9 m$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $m/s^{2}$ ) :

A $4{\pi}^{2}{\mathrm{ms}}^{-2}$
B $16{\pi}^{2}{\mathrm{ms}}^{-2}$
C $57600{\pi}^{2}{\mathrm{ms}}^{-2}$
DZero

Show answer and solution

Answer: Option B

Three minutes is $180\ \mathrm{s}$ , so $120$ turns give $f = \dfrac{120}{180} = \dfrac{2}{3}\ \mathrm{s^{-1}}$ and $\omega = 2\pi f = \dfrac{4\pi}{3}\ \mathrm{rad/s}$ . Then $a = \omega^{2}r = \dfrac{16\pi^{2}}{9} \times 9 = 16\pi^{2}\ \mathrm{m/s^{2}}$ . The monkey sits still on the shoulder, so it travels the same track on the same radius as the man — which is the whole reason it is in the question, and why Zero is wrong. Still with respect to the cyclist is not still with respect to the ground.

Practice questions, easy to hard

Three questions from the circular motion practice ladder: one easy, one medium, one hard.

Q4One correct option

The same idea over a shorter stretch of the circle.

A body moves at a steady $5\ \mathrm{m/s}$ round a circle. What is the magnitude of the change in its velocity over a quarter of a revolution?

A $5\ \mathrm{m/s}$
B $10\ \mathrm{m/s}$
Czero
D $5\sqrt{2}\ \mathrm{m/s}$

Show answer and solution

Answer: Option D

A quarter turn is $\theta = 90^{\circ}$ , so $|\Delta\vec{v}| = 2 \times 5 \times \sin 45^{\circ} = \dfrac{10}{\sqrt{2}} = 5\sqrt{2}\ \mathrm{m/s}$ , about $7.07$ . The two velocities are equal in length and perpendicular, so the change is the diagonal of a square of side $5$ — the same sum done with a picture. Zero belongs to a complete revolution, not a quarter of one.

Q5Numerical answer

Read $F = \dfrac{mv^{2}}{r}$ backwards. The radius and the mass stay put, so the force and the square of the speed rise and fall together.

The pull on a whirling bob is made $9$ times larger on the same string. By what factor does its speed go up?

Show answer and solution

Answer: 3

$F \propto v^{2}$ , so $v \propto \sqrt{F}$ and the speed rises by $\sqrt{9} = 3$ . Since the string length has not changed, the number of turns per minute goes up by that same factor of $3$ — $\omega = v/r$ with $r$ fixed makes the two proportional.