In short

Angular velocity \vec{\omega} is a vector that points along the axis of rotation, with its direction given by the right-hand rule. The velocity of any point on a rotating body is \vec{v} = \vec{\omega} \times \vec{r}, where \vec{r} is the position vector from the axis. For two objects rotating at angular velocities \omega_1 and \omega_2, the relative angular velocity is \omega_{\text{rel}} = \omega_1 - \omega_2 (same-sense rotation) or \omega_1 + \omega_2 (opposite-sense). In meshing gears, the no-slip condition gives \omega_1 r_1 = \omega_2 r_2.

Watch the second hand of a clock. It sweeps around the dial at a steady rate — one full revolution every 60 seconds. Now watch the minute hand. It also sweeps around, but much slower — one revolution every 60 minutes. Both hands rotate about the same axis (the centre pin), yet if you sat on the tip of the minute hand and watched the second hand, it would appear to race ahead of you, lapping you once every 61.02 seconds instead of every 60. That difference in angular speeds — the angular velocity of one object as seen from another rotating object — is relative angular velocity.

But before you can subtract angular velocities, you need to know something surprising: angular velocity is not just a number. It is a vector. It has a direction — and that direction matters.

Angular velocity is a vector

When a bicycle wheel spins, every point on the rim traces a circle. The rate of rotation — how many radians the wheel turns through per second — is the angular speed \omega. So far, this is a scalar: just a number with units of rad/s.

But ask yourself: which way does the wheel spin? If you look at the wheel from the left side, it turns clockwise. Look at it from the right side, it turns anticlockwise. Same physical rotation, two different descriptions. To avoid this ambiguity, you need a direction that captures the rotation unambiguously. That direction is along the axis of rotation — the axle of the wheel.

The right-hand rule

Curl the fingers of your right hand in the direction the wheel is spinning. Your thumb points in the direction of the angular velocity vector \vec{\omega}.

Right-hand rule for angular velocity direction A wheel rotates anticlockwise when viewed from the right. Curled fingers show the rotation direction. The thumb points to the right, indicating the direction of the angular velocity vector along the axis. ω⃗ wheel (viewed from the side) axis of rotation rotation
Curl your right hand in the direction of rotation (anticlockwise as viewed from the right). Your thumb points to the right — that is the direction of $\vec{\omega}$. The magnitude $|\vec{\omega}|$ equals the angular speed in rad/s.

For a potter's wheel spinning anticlockwise when viewed from above, the right-hand rule tells you \vec{\omega} points upward (out of the wheel). For the front wheel of a bicycle moving forward, the wheel spins so the top moves forward and the bottom moves backward — curl your right hand in that direction, and \vec{\omega} points to the left (from the rider's perspective).

Why does \vec{\omega} point along the axis and not somewhere in the plane of rotation? Because the axis is the only direction that every part of the rotating body agrees on. Points on the rim move in constantly changing directions — but the axis is fixed, and it uniquely describes the plane in which everything rotates.

The sign convention

For rotation confined to a plane (most problems in 2D kinematics), you can simplify. Choose a positive direction for \vec{\omega} — typically, anticlockwise is positive when viewed from the standard orientation. Then \omega > 0 means anticlockwise, and \omega < 0 means clockwise. The vector machinery still works behind the scenes, but you only need to track the sign.

The velocity of a point on a rotating body: \vec{v} = \vec{\omega} \times \vec{r}

Here is the central result of rotational kinematics. If a rigid body rotates about a fixed axis with angular velocity \vec{\omega}, then the velocity of any point P on the body is:

\boxed{\vec{v} = \vec{\omega} \times \vec{r}}

where \vec{r} is the position vector of P measured from any point on the axis of rotation.

This is not a definition pulled from the sky. You can derive it from the scalar relation v = r\omega and the requirement that \vec{v} is perpendicular to both \vec{\omega} and \vec{r}.

Deriving \vec{v} = \vec{\omega} \times \vec{r}

Consider a point P on a rigid body rotating about a fixed axis. Let \vec{r} be the position vector of P from a point O on the axis.

Step 1. The point P traces a circle. The radius of this circle is the perpendicular distance from P to the axis, which is r \sin\theta, where \theta is the angle between \vec{\omega} (along the axis) and \vec{r}.

Why: the component of \vec{r} along the axis does not contribute to circular motion — only the perpendicular component r_\perp = r\sin\theta determines the radius of the circle P traces.

Step 2. The speed of P is:

v = \omega \cdot r_\perp = \omega\, r \sin\theta

Why: this is the scalar relation v = \omega R applied to the circle of radius R = r\sin\theta that P actually traces.

Step 3. The direction of \vec{v} is tangent to the circle — perpendicular to both \vec{\omega} and \vec{r}.

Why: P moves along the circumference of its circle at every instant. This tangent direction is perpendicular to the radius of the circle (which is the perpendicular component of \vec{r}) and perpendicular to the axis (the direction of \vec{\omega}).

Step 4. Combine magnitude and direction. The cross product \vec{\omega} \times \vec{r} has:

Why: the cross product is the unique vector operation that produces a result with magnitude ab\sin\theta and direction perpendicular to both input vectors. Since \vec{v} has exactly these properties, \vec{v} = \vec{\omega} \times \vec{r}.

Therefore:

\vec{v} = \vec{\omega} \times \vec{r}

This formula is extraordinarily powerful. Give me the angular velocity vector and the position of any point on the body, and I can tell you the velocity of that point — magnitude and direction — in one line.

Velocity of a point on a rotating body as the cross product of omega and r A rotating body with axis vertical. Point P is at position r from the axis. The velocity v is perpendicular to both omega and r, pointing tangentially out of the page. axis ω⃗ O r⃗ P v⃗ = ω⃗ × r⃗ θ r sin θ
Point P sits at position $\vec{r}$ from a point O on the axis. The angular velocity $\vec{\omega}$ points along the axis (upward, by the right-hand rule). The velocity $\vec{v} = \vec{\omega} \times \vec{r}$ is tangent to the circle P traces, perpendicular to both $\vec{\omega}$ and $\vec{r}$. The speed is $v = \omega r\sin\theta$.

Connecting linear and angular quantities

The cross-product relationship \vec{v} = \vec{\omega} \times \vec{r} is the vector version of v = r\omega. But you can extend this to acceleration as well, linking every linear quantity to its angular counterpart.

Linear quantity Angular quantity Relation
Displacement s Angular displacement \theta s = r\theta
Velocity v Angular velocity \omega v = r\omega (or \vec{v} = \vec{\omega} \times \vec{r})
Tangential acceleration a_t Angular acceleration \alpha a_t = r\alpha
Centripetal acceleration a_c Angular velocity \omega a_c = r\omega^2

Why: all these relations follow from the same geometry — a point at distance r from the axis traces a circle of circumference 2\pi r. Every radian of rotation sweeps r metres of arc. So the conversion factor between angular and linear is always r.

For uniform rotation (\alpha = 0, constant \omega), the angular displacement in time t is \theta = \omega t, and the arc length swept is s = r\omega t = vt. For uniformly accelerated rotation, the angular kinematic equations mirror the linear ones exactly:

\omega = \omega_0 + \alpha t
\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2
\omega^2 = \omega_0^2 + 2\alpha\theta

Why: these are mathematically identical to v = u + at, s = ut + \frac{1}{2}at^2, and v^2 = u^2 + 2as — just with \theta replacing s, \omega replacing v, and \alpha replacing a. The structure of constant-acceleration kinematics does not depend on whether the motion is along a line or around a circle.

Relative angular velocity

When two objects rotate about the same axis, each with its own angular velocity, the angular velocity of one as seen by the other is their relative angular velocity.

Think of two runners on a circular track. If runner A goes around at \omega_A and runner B at \omega_B, both anticlockwise, then from A's perspective, B appears to rotate at \omega_B - \omega_A. If they run in opposite senses, B appears to go at \omega_B + \omega_A.

Same axis, same sense

\omega_{\text{rel}} = \omega_1 - \omega_2

Why: both objects rotate in the same direction. In the frame of object 2, you subtract its rotation from everything — so object 1 appears to rotate at the difference.

Same axis, opposite sense

\omega_{\text{rel}} = \omega_1 + \omega_2

Why: when they rotate in opposite directions, from the frame of one, the other sweeps through angles even faster — their rates add.

The general vector form

When the rotations are about different axes, or when you need the full three-dimensional treatment, relative angular velocity is simply the vector difference:

\vec{\omega}_{\text{rel}} = \vec{\omega}_1 - \vec{\omega}_2

Why: angular velocity vectors add and subtract just like any other vectors. If \vec{\omega}_1 and \vec{\omega}_2 point in the same direction, the difference is smaller in magnitude (they partially cancel). If they point in opposite directions, the difference is larger (they reinforce).

The clock hands — a classic relative angular velocity problem

The minute hand of a clock completes one revolution (2\pi rad) in 60 minutes. The hour hand completes one revolution in 12 hours (720 minutes). Both rotate clockwise (taking clockwise as negative by convention, but their magnitudes are what matter for the relative rate).

\omega_{\text{min}} = \frac{2\pi}{60} = \frac{\pi}{30} \text{ rad/min}
\omega_{\text{hr}} = \frac{2\pi}{720} = \frac{\pi}{360} \text{ rad/min}

Both hands rotate in the same sense (clockwise), so the relative angular velocity of the minute hand with respect to the hour hand is:

\omega_{\text{rel}} = \omega_{\text{min}} - \omega_{\text{hr}} = \frac{\pi}{30} - \frac{\pi}{360} = \frac{12\pi - \pi}{360} = \frac{11\pi}{360} \text{ rad/min}

Why: the minute hand is faster, so from the hour hand's point of view, the minute hand races ahead at this rate. In one day, the minute hand laps the hour hand 22 times — which is why the hands overlap 22 times in 24 hours (not 24, because the hour hand is also moving).

Applications: gears, pulleys, and rolling constraints

The cross-product formula \vec{v} = \vec{\omega} \times \vec{r} and the relative angular velocity concept power some of the most important mechanical systems. Here are the key applications.

Meshing gears — the no-slip condition

When two gears mesh, their teeth interlock at the point of contact. At that point, the teeth of both gears have the same linear velocity — if they did not, the teeth would either separate (losing contact) or jam (attempting to occupy the same space). This is the no-slip condition.

Two meshing gears with the no-slip condition Gear A (smaller, radius r1) and Gear B (larger, radius r2) mesh at a contact point. The linear velocity at the contact point is the same for both gears: v = omega_1 times r_1 = omega_2 times r_2. The gears rotate in opposite directions. A r₁ ω₁ B r₂ ω₂ contact v₁ v₂ No-slip: ω₁r₁ = ω₂r₂
Gear A (smaller, radius $r_1$) drives gear B (larger, radius $r_2$). They rotate in opposite directions. At the point of contact, the linear velocity is the same for both: $v_1 = v_2$, giving $\omega_1 r_1 = \omega_2 r_2$.

For gear A (radius r_1, angular velocity \omega_1) and gear B (radius r_2, angular velocity \omega_2), the linear velocity at the contact point is:

v_{\text{contact}} = \omega_1 r_1 = \omega_2 r_2

Why: each gear's contact point moves at v = \omega r (the scalar form of \vec{v} = \vec{\omega} \times \vec{r} for a point on the rim). Since the teeth mesh without slipping, these speeds must be equal.

This gives the gear ratio:

\frac{\omega_1}{\omega_2} = \frac{r_2}{r_1} = \frac{N_2}{N_1}

where N_1 and N_2 are the number of teeth on each gear (since teeth are uniformly spaced, the number of teeth is proportional to the radius).

Why: the smaller gear spins faster. If gear B has twice the radius of gear A, gear A must spin twice as fast to keep up at the contact point. This is why a bicycle's small chainring makes the rear wheel spin faster — you are trading force for speed.

Note that meshing gears always rotate in opposite directions. If gear A turns anticlockwise, gear B turns clockwise at the contact point. This is because the teeth push each other — the top of gear A's teeth push upward, and the corresponding teeth on gear B are pushed upward too, which means B rotates in the opposite sense.

Pulleys connected by a belt or chain

A bicycle chain connects a large front chainring to a smaller rear sprocket. The chain does not stretch or slip (ideally), so every point on the chain moves at the same linear speed. This is the same no-slip condition as meshing gears:

\omega_{\text{front}} \cdot r_{\text{front}} = \omega_{\text{rear}} \cdot r_{\text{rear}}

The difference from gears: pulleys connected by a belt or chain rotate in the same direction (both turn the way the chain moves). Gears mesh and reverse direction; belts and chains transmit and preserve direction.

Rolling without slipping — a preview

When a wheel rolls on a surface without slipping, the contact point between the wheel and the ground is momentarily at rest. This means the velocity from rotation exactly cancels the forward translational velocity at the contact point:

v_{\text{cm}} = \omega R

where v_{\text{cm}} is the velocity of the wheel's centre and R is the radius. This is the rolling constraint — a powerful condition that you will use extensively in problems involving rolling bodies. The full treatment, including the forces and energy of rolling, comes in the article on rolling motion.

Worked examples

Example 1: Two meshing gears in a machine

In a workshop lathe, gear A has 30 teeth (radius r_1 = 6 cm) and rotates at 120 rpm. It directly meshes with gear B, which has 60 teeth (radius r_2 = 12 cm). Find: (a) the angular velocity of gear B, and (b) the linear velocity at the point where the teeth mesh.

Two meshing gears: gear A (30 teeth) drives gear B (60 teeth) Gear A (30 teeth, radius 6 cm, 120 rpm) on the left meshes directly with gear B (60 teeth, radius 12 cm) on the right. They rotate in opposite directions. The linear velocity at the contact point is the same for both. A 30 teeth, 120 rpm r₁ = 6 cm r₁ ω₁ B 60 teeth, r₂ = 12 cm r₂ ω₂ contact v v
Gear A (30 teeth, $r_1 = 6$ cm, 120 rpm) meshes directly with gear B (60 teeth, $r_2 = 12$ cm). They rotate in opposite directions. At the contact point, both teeth move at the same linear speed $v$.

Step 1. Convert the angular velocity of gear A to rad/s.

\omega_1 = 120 \text{ rpm} = 120 \times \frac{2\pi}{60} = 4\pi \text{ rad/s} \approx 12.57 \text{ rad/s}

Why: 1 revolution = 2\pi radians, 1 minute = 60 seconds. Multiplying rpm by 2\pi/60 converts directly to rad/s.

Step 2. Apply the no-slip condition to find \omega_2.

At the contact point, the teeth of both gears move at the same linear speed:

\omega_1 r_1 = \omega_2 r_2
\omega_2 = \omega_1 \times \frac{r_1}{r_2} = 4\pi \times \frac{6}{12} = 2\pi \text{ rad/s} \approx 6.28 \text{ rad/s}

Why: the linear velocity at the rim of each gear is v = \omega r. Since the teeth mesh without slipping, v must be the same for both gears. The larger gear (twice the radius) turns at half the angular speed.

This corresponds to 60 rpm — exactly half the input speed. You can also confirm this using the tooth ratio: \omega_2 = \omega_1 \times N_1/N_2 = 120 \times 30/60 = 60 rpm. The tooth ratio and radius ratio give the same answer because teeth are uniformly spaced, so N \propto r.

Step 3. Find the linear velocity at the contact point.

v = \omega_1 r_1 = 4\pi \times 0.06 = 0.24\pi \approx 0.754 \text{ m/s}

Verify: v = \omega_2 r_2 = 2\pi \times 0.12 = 0.24\pi \approx 0.754 m/s. The values match — the no-slip condition is satisfied.

Why: both calculations give the same v because the no-slip condition forces \omega_1 r_1 = \omega_2 r_2. This is the speed at which the gear teeth slide past the contact point — about 0.75 m/s, roughly walking pace.

Result: Gear B rotates at 2\pi rad/s (60 rpm), in the opposite direction to gear A. The linear velocity at the point of contact is 0.24\pi \approx 0.754 m/s.

What this shows: The no-slip condition \omega_1 r_1 = \omega_2 r_2 determines everything. The larger gear turns slower by exactly the radius ratio (or equivalently, the tooth ratio). The smaller gear is the "driver" — it spins fast and the larger gear follows at reduced speed but with greater torque, which is how a lathe's gear train steps down the motor speed to the cutting speed.

Example 2: Clock hands — when do they overlap after 12:00?

The minute hand and hour hand of a clock are both at the 12 o'clock position at exactly 12:00. (a) Find the relative angular velocity of the minute hand with respect to the hour hand. (b) At what time do they next overlap?

Clock showing minute and hour hands at 12:00 and their angular velocities A clock face with both hands at 12. The minute hand has angular velocity omega_min = 2pi/60 rad/min. The hour hand has angular velocity omega_hr = 2pi/720 rad/min. Both move clockwise. 12 3 6 9 minute hour clockwise ω_min = 2π/60 rad/min ω_hr = 2π/720 rad/min ω_rel = ω_min − ω_hr = 11π/360 rad/min
At 12:00, both hands coincide. The minute hand moves at $\omega_{\text{min}} = 2\pi/60$ rad/min, the hour hand at $\omega_{\text{hr}} = 2\pi/720$ rad/min. Both move clockwise. The relative angular velocity is $11\pi/360$ rad/min.

(a) Relative angular velocity

Step 1. Write the angular velocities.

The minute hand completes one revolution (2\pi radians) in 60 minutes:

\omega_{\text{min}} = \frac{2\pi}{60} = \frac{\pi}{30} \text{ rad/min}

The hour hand completes one revolution in 12 hours = 720 minutes:

\omega_{\text{hr}} = \frac{2\pi}{720} = \frac{\pi}{360} \text{ rad/min}

Why: angular velocity is the angle swept per unit time. Both hands rotate steadily, so \omega = \Delta\theta / \Delta t = 2\pi / T, where T is the period.

Step 2. Both hands rotate in the same direction (clockwise), so the relative angular velocity of the minute hand with respect to the hour hand is:

\omega_{\text{rel}} = \omega_{\text{min}} - \omega_{\text{hr}} = \frac{\pi}{30} - \frac{\pi}{360}
= \frac{12\pi - \pi}{360} = \frac{11\pi}{360} \text{ rad/min}

Why: the minute hand is faster, so it gains on the hour hand. The rate of gain is the difference. Finding a common denominator: \pi/30 = 12\pi/360.

(b) Time to next overlap

Step 3. The hands overlap again when the minute hand has gained exactly one full revolution (2\pi radians) on the hour hand.

\omega_{\text{rel}} \times t = 2\pi
t = \frac{2\pi}{\omega_{\text{rel}}} = \frac{2\pi}{11\pi/360} = \frac{2\pi \times 360}{11\pi} = \frac{720}{11} \text{ min}
t = 65.\overline{45} \text{ min} \approx 65 \text{ min } 27 \text{ s}

Why: the minute hand must sweep 2\pi more than the hour hand for them to coincide again. At the relative rate of 11\pi/360 rad/min, this takes 720/11 minutes.

Result: The relative angular velocity of the minute hand with respect to the hour hand is \frac{11\pi}{360} rad/min \approx 0.0960 rad/min. They next overlap at approximately 1:05:27 (65 minutes and 27 seconds after 12:00).

What this shows: Relative angular velocity tells you how fast one rotating object gains on another. The factor of \frac{11}{12} that appears (since \omega_{\text{rel}}/\omega_{\text{min}} = 11/12) is why the hands overlap 11 times in every 12-hour cycle, not 12 times. The hour hand's own motion "steals" one overlap per cycle.

Common confusions

If you came here to understand angular velocity as a vector, the \vec{v} = \vec{\omega} \times \vec{r} formula, and the gear ratio, you have what you need. What follows is for readers preparing for JEE Advanced who want the formal vector treatment and the angular velocity addition theorem.

Angular velocity as an axial vector (pseudovector)

Angular velocity \vec{\omega} is not an ordinary ("polar") vector like displacement or force. It is an axial vector (also called a pseudovector). The difference: under a mirror reflection, a polar vector reverses direction, but an axial vector does not.

Think of a spinning top viewed in a mirror. The spin direction appears to reverse (clockwise becomes anticlockwise), but if you apply the right-hand rule in the mirror world, the right-hand rule itself reverses (because your right hand becomes a left hand in the mirror). The net result: \vec{\omega} defined by the right-hand rule in the real world and \vec{\omega} defined by the right-hand rule in the mirror world are the same vector. This means \vec{\omega} does not flip under reflection — it is a pseudovector.

Why this matters: when you take cross products of polar and axial vectors, you need to keep track of which kind each is. \vec{v} = \vec{\omega} \times \vec{r} is a cross product of a pseudovector (\vec{\omega}) and a polar vector (\vec{r}), giving a polar vector (\vec{v}) — consistent, since velocity is indeed a true vector.

The angular velocity addition theorem

When body B rotates relative to body A with angular velocity \vec{\omega}_{B/A}, and body A rotates relative to the ground with \vec{\omega}_{A}, then the angular velocity of B relative to the ground is:

\vec{\omega}_B = \vec{\omega}_A + \vec{\omega}_{B/A}

This is a vector addition — the angular velocities add as vectors, including direction. This theorem is critical for problems with multiple rotating frames: a gyroscope mounted on a rotating platform, or the Earth spinning on its axis while orbiting the Sun.

For the Earth: \vec{\omega}_{\text{spin}} = 2\pi/(24 \text{ hr}) about the polar axis (from south to north), and \vec{\omega}_{\text{orbit}} = 2\pi/(365.25 \text{ days}) about the normal to the ecliptic plane. These vectors are not parallel (the Earth's axis is tilted at 23.4° to the ecliptic normal), so the total angular velocity is their vector sum — not just the sum of their magnitudes.

Acceleration of a point on a rotating body

Differentiating \vec{v} = \vec{\omega} \times \vec{r} with respect to time:

\vec{a} = \frac{d\vec{v}}{dt} = \frac{d\vec{\omega}}{dt} \times \vec{r} + \vec{\omega} \times \frac{d\vec{r}}{dt}
\vec{a} = \vec{\alpha} \times \vec{r} + \vec{\omega} \times \vec{v}
\vec{a} = \vec{\alpha} \times \vec{r} + \vec{\omega} \times (\vec{\omega} \times \vec{r})

Why: the product rule applies to cross products just as it does to ordinary products. The first term \vec{\alpha} \times \vec{r} is the tangential acceleration (due to changing \omega). The second term \vec{\omega} \times (\vec{\omega} \times \vec{r}) is the centripetal acceleration (pointing toward the axis). Together, they give the complete acceleration of any point on a rotating body.

The centripetal term \vec{\omega} \times (\vec{\omega} \times \vec{r}) points radially inward and has magnitude \omega^2 r\sin\theta — which is the familiar \omega^2 R where R = r\sin\theta is the perpendicular distance from the axis. The tangential term \vec{\alpha} \times \vec{r} is tangent to the circle and has magnitude \alpha r\sin\theta = \alpha R.

Instantaneous axis of rotation

Any rigid body motion (at any instant) can be decomposed into a translation plus a rotation about some axis. For a rolling wheel, the instantaneous axis of rotation passes through the contact point with the ground — not through the axle. Every point on the wheel instantaneously rotates about this contact point. The point at the very bottom of the wheel has zero velocity (it is on the axis), the centre has velocity v = \omega R (at distance R from the contact point), and the top has velocity 2v = 2\omega R (at distance 2R). This elegant picture — where all velocities radiate from a single point — simplifies many rolling problems.

Where this leads next