In short

Any motion of a rigid body can be split into translation of the centre of mass plus rotation about the centre of mass. The velocity of any point P on the body is \vec{v}_P = \vec{v}_{\text{cm}} + \vec{\omega} \times \vec{r}', where \vec{r}' is P's position relative to the centre of mass. At any instant, there exists a single point — the instantaneous centre of rotation — about which the entire body is purely rotating.

Paint a dot on the tyre of your bicycle and watch it as you ride down a straight road. The dot does not travel in a straight line. It rises, sweeps forward, dips down to kiss the road, and lifts again — tracing a looping arch that mathematicians call a cycloid. The hub of the wheel, meanwhile, moves in a perfectly straight line at constant speed.

How does a single wheel produce two such different motions at the same time? The answer is the central idea of this article: every rigid body motion, no matter how complicated, is just two simple motions added together — translation and rotation. The hub translates forward. Every other point on the wheel translates forward and rotates around the hub. Add those two contributions, and the dot on the rim traces its elegant cycloid.

This decomposition is not a trick that works only for wheels. It works for every rigid body — a spinning cricket ball in flight, an autorickshaw wheel on a potholed road, a ceiling fan wobbling on a loose mount, a gymnast somersaulting through the air. Once you learn to see general motion as translation plus rotation, even the most complicated rigid body movement becomes two problems you already know how to solve.

Two simple motions

Before combining anything, make sure you are comfortable with the two ingredients separately.

Pure translation

Slide a book across a table without spinning it. Every point on the book — every corner, every edge — moves with the same velocity in the same direction. If the centre of the book moves 10 cm to the right, the top-left corner also moves 10 cm to the right. That is pure translation: zero rotation, every point shares the same velocity \vec{v}_{\text{cm}}.

Pure rotation

Now pin one corner of the book to the table and spin it. Every point on the book moves in a circle around that pin. Points far from the pin move fast; points near the pin move slowly; the pin itself does not move at all. That is pure rotation about a fixed axis: zero translation of the axis, every point has a velocity that depends on how far it is from the axis and in what direction.

The key question

What happens when you slide the book and spin it at the same time? Every point's velocity is the sum of two parts: the translation velocity (the same for every point) and the rotational velocity (different for each point, depending on its distance and direction from the centre). That sum is general motion — and it describes everything from a rolling wheel to a tumbling satellite.

Decomposition of general motion into translation plus rotation Three discs side by side. The first shows pure translation — all velocity arrows point right and are equal. The second shows pure rotation — arrows are tangent to the circle, clockwise. The third shows their sum: the top arrow is large and rightward, the bottom arrow is zero, and the side arrow points diagonally. Translation + Rotation = v = 0 General motion
Any rigid body motion is translation (every point has the same velocity) plus rotation (velocities depend on position). At one special point — the bottom of the third disc — the two contributions cancel exactly, giving zero velocity.

Why the centre of mass is the natural reference point

You could decompose general motion around any reference point — a corner, an edge, an arbitrary nail driven through the body. But the centre of mass (CM) is the natural choice because Newton's second law governs it directly:

\vec{F}_{\text{net}} = M\,\vec{a}_{\text{cm}}

The CM moves as if the entire mass M were concentrated there and all external forces acted on that single point. Once you know the CM's translational motion, the remaining motion of the body is purely rotational about the CM. This is not an approximation — it is an exact decomposition that follows from the definition of the centre of mass.

Deriving the velocity of any point

Take a rigid body moving in any manner. Let \vec{r}_{\text{cm}} be the position of the centre of mass, and \vec{r}_P be the position of some arbitrary point P on the body.

Step 1. Write the position of P as the CM position plus the offset.

\vec{r}_P = \vec{r}_{\text{cm}} + \vec{r}' \tag{1}

Why: \vec{r}' is the position of P relative to the CM. This splits the position into "where is the CM?" and "where is P relative to the CM?"

Step 2. Differentiate both sides with respect to time.

\vec{v}_P = \vec{v}_{\text{cm}} + \frac{d\vec{r}'}{dt} \tag{2}

Why: velocity is the time derivative of position. The derivative of the sum is the sum of the derivatives.

Step 3. Recognise what \frac{d\vec{r}'}{dt} is.

In the CM frame — a reference frame attached to and moving with the centre of mass — the body is purely rotating. The CM is at rest in this frame, and every other point moves in a circle around it. For pure rotation with angular velocity \vec{\omega}, the velocity of a point at position \vec{r}' from the axis is:

\frac{d\vec{r}'}{dt} = \vec{\omega} \times \vec{r}' \tag{3}

Why: this is the standard result for circular motion. The cross product gives a vector perpendicular to both \vec{\omega} and \vec{r}', with magnitude \omega\,r'\sin\theta — exactly the tangential speed of a point revolving at distance r'\sin\theta from the rotation axis.

Step 4. Substitute equation (3) into equation (2).

\boxed{\vec{v}_P = \vec{v}_{\text{cm}} + \vec{\omega} \times \vec{r}'} \tag{4}

Why: combine the two pieces. The velocity of any point on the body equals the translational velocity of the CM plus the rotational velocity about the CM. This single equation describes all rigid body motion.

Velocity of any point on a rigid body

For a rigid body with centre-of-mass velocity \vec{v}_{\text{cm}} and angular velocity \vec{\omega}, the velocity of any point P on the body is:

\vec{v}_P = \vec{v}_{\text{cm}} + \vec{\omega} \times \vec{r}'

where \vec{r}' = \vec{r}_P - \vec{r}_{\text{cm}} is the position vector of P relative to the centre of mass.

Read this equation in pieces. \vec{v}_{\text{cm}} is the velocity every point would have if the body were only translating — it is the same for all points. \vec{\omega} \times \vec{r}' is the extra velocity due to rotation — it varies from point to point, depending on where P sits relative to the CM. The full velocity \vec{v}_P is the vector sum of these two contributions. Points far from the CM have large rotational contributions; the CM itself has \vec{r}' = 0, so its velocity is just \vec{v}_{\text{cm}}.

Pure rotation versus general planar motion

With equation (4) in hand, you can classify rigid body motions cleanly:

Motion type \vec{v}_{\text{cm}} \vec{\omega} What it looks like
Pure translation \neq 0 = 0 Every point moves with the same velocity. A sliding block on ice.
Pure rotation about a fixed axis = 0 \neq 0 One point (on the axis) is stationary; all others move in circles. A ceiling fan.
General planar motion \neq 0 \neq 0 Translation and rotation happen simultaneously. A rolling wheel, a sliding ladder, a spinning coin skidding across a table.

The first two are special cases of equation (4). General motion — where both \vec{v}_{\text{cm}} and \vec{\omega} are nonzero — is the rule, not the exception. Most moving objects in the real world both translate and rotate.

Rolling — where translation meets rotation

The most important example of general motion is a wheel rolling without slipping on a flat surface. The centre of the wheel translates forward at velocity v_{\text{cm}}. The wheel rotates about its centre at angular velocity \omega. The rolling condition links the two:

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

Why: in one full rotation (2\pi/\omega seconds), the wheel covers a distance equal to its circumference (2\pi R). So v_{\text{cm}} = 2\pi R / (2\pi/\omega) = R\omega. Equivalently, the contact point has zero velocity — the translation and rotation cancel exactly there.

Watch what happens to a dot on the rim. The animation below tracks two points: the centre (which moves in a straight line) and a dot on the rim (which traces a cycloid).

Animated: point on a rolling wheel traces a cycloid A wheel of radius 1 m rolls to the right at 2 m/s. The centre moves in a straight line at constant height. A red dot on the rim traces a cycloid — looping up and touching the ground once per rotation. ground
The centre (dark) moves in a straight line. The dot on the rim (red) traces a cycloid — rising to twice the wheel's height, then dipping down to touch the ground. Click replay to watch again.

The cycloid dramatically illustrates the decomposition. The centre has only the translational velocity — steady and horizontal. The rim point has translation plus rotation, which sometimes adds (at the top, where the dot is moving fastest) and sometimes cancels (at the bottom, where the dot is momentarily stationary).

How speed varies around the rim

For a wheel rolling without slipping (v_{\text{cm}} = R\omega), the speed of a point on the rim depends on its angular position. Let \theta be the angle measured from the contact point (bottom of the wheel), going counterclockwise.

Step 1. Write the position of the rim point relative to the centre.

x' = R\sin\theta, \quad y' = -R\cos\theta

Why: at \theta = 0 (contact point), x' = 0 and y' = -R — directly below the centre. At \theta = \pi (top), x' = 0 and y' = R — directly above. This parameterises the rim correctly.

Step 2. Apply equation (4). The CM moves rightward at v_{\text{cm}}. The wheel rotates clockwise (angular velocity \vec{\omega} = -\omega\hat{k} in the counterclockwise-positive convention). The rotational velocity is:

v_{Px} = v_{\text{cm}} - \omega R\cos\theta = v_{\text{cm}}(1 - \cos\theta)
v_{Py} = -\omega R\sin\theta = -v_{\text{cm}}\sin\theta

Why: the cross product \vec{\omega} \times \vec{r}' for clockwise rotation adds a rightward component -\omega R\cos\theta and a downward component -\omega R\sin\theta. With v_{\text{cm}} = R\omega, everything simplifies.

Step 3. Compute the speed.

|\vec{v}_P| = v_{\text{cm}}\sqrt{(1 - \cos\theta)^2 + \sin^2\theta} = v_{\text{cm}}\sqrt{2 - 2\cos\theta}

Using the identity 1 - \cos\theta = 2\sin^2(\theta/2):

\boxed{|\vec{v}_P| = 2v_{\text{cm}}\left|\sin\frac{\theta}{2}\right|} \tag{5}

Why: the trigonometric identity converts the square root into a clean sine function. The speed varies smoothly from 0 at the contact point to 2v_{\text{cm}} at the top.

Check the special cases:

Drag the point in the figure below to see how the speed changes as you move around the wheel.

Interactive: speed of a rim point versus angular position on a rolling wheel A curve showing speed divided by v_cm equals 2 times the sine of theta over 2, as theta goes from 0 to 2 pi. Speed is zero at the contact point and reaches 2 v_cm at the top of the wheel. θ (angle from contact point) speed / v_cm 0 π/2 π 3π/2 2π 0 1 2 drag the red point along the axis
Drag the red point to change the angle $\theta$. At the contact point ($\theta = 0$), speed is zero. At the top ($\theta = 180°$), speed reaches $2v_{\text{cm}}$. The arch-shaped curve is $2\sin(\theta/2)$.

The instantaneous centre of rotation

Here is a remarkable fact: for any planar motion where \omega \neq 0, there is exactly one point in the plane with zero velocity at that instant. This is the instantaneous centre of rotation (ICR). At that instant, the entire body behaves as if it were purely rotating about the ICR.

To find the ICR, set \vec{v}_P = 0 in equation (4):

\vec{v}_{\text{cm}} + \vec{\omega} \times \vec{r}'_{\text{ICR}} = 0

This has a unique solution for \vec{r}'_{\text{ICR}} (the position of the ICR relative to the CM) whenever \omega \neq 0.

Finding the ICR geometrically

If you know the velocity direction of two points on the body, draw the perpendicular to each velocity vector at each point. The ICR is where those two perpendiculars intersect — because the velocity at any point is perpendicular to the line joining that point to the centre of rotation.

For a rolling wheel, the contact point has zero velocity. So the ICR is the contact point itself. Every point on the wheel is instantaneously rotating about the contact point. Verify this: the top of the wheel is at distance 2R from the contact point, so its speed is \omega \times 2R = 2v_{\text{cm}}. The centre is at distance R, so its speed is \omega \times R = v_{\text{cm}}. Both match equation (5).

For pure translation (\omega = 0), every point has the same velocity and no perpendicular construction can converge. The ICR is at infinity — which makes sense, because a body translating is "rotating about a point infinitely far away" with an infinitely small angular velocity. In the limit, \omega \to 0 and r_{\text{ICR}} \to \infty, with \omega \times r_{\text{ICR}} staying finite and equal to v_{\text{cm}}.

The ICR is instantaneous — it generally moves over time. For a rolling wheel, the ICR is always at the contact point, but which physical point on the rim serves as the contact point changes continuously. The ICR is a geometric concept, not a physical marker pinned to the body.

Worked examples

Example 1: Velocity of points on an autorickshaw wheel

An autorickshaw moves at 36 km/h (= 10 m/s) along a straight road. Each rear wheel has radius R = 0.25 m. The wheels roll without slipping. Find the velocity of (a) the topmost point, (b) the contact point, and (c) the point on the front edge of the wheel, level with the axle.

Velocity vectors on a rolling autorickshaw wheel A circle representing the wheel. At the top, a rightward arrow labelled 20 m/s. At the centre, a shorter rightward arrow labelled 10 m/s. At the contact point, a label says v = 0. At the front (level with centre), a diagonal arrow pointing down-right is labelled 10√2 m/s. A curved arrow inside indicates clockwise rotation. O ω v_cm = 10 m/s T 20 m/s C: v = 0 F 10√2 m/s R = 0.25 m v_cm = 10 m/s ω = 40 rad/s direction of motion →
The top of the wheel moves at 20 m/s — twice the autorickshaw's speed. The contact point is momentarily at rest. The front point moves diagonally at $10\sqrt{2}$ m/s.

Step 1. Find the angular velocity.

\omega = \frac{v_{\text{cm}}}{R} = \frac{10}{0.25} = 40 \text{ rad/s (clockwise)}

Why: rolling without slipping means v_{\text{cm}} = R\omega. This is the kinematic constraint that ties translation to rotation.

Step 2. Topmost point T. Here \theta = \pi (directly opposite the contact point).

v_T = 2v_{\text{cm}}\sin\left(\frac{\pi}{2}\right) = 2 \times 10 \times 1 = 20 \text{ m/s, directed rightward}

Why: at the top, both the translational velocity (v_{\text{cm}} rightward) and the rotational velocity (R\omega rightward) point in the same direction. They add: 10 + 10 = 20 m/s.

Step 3. Contact point C. Here \theta = 0.

v_C = 2v_{\text{cm}}\sin(0) = 0

Why: at the contact point, the translational velocity (v_{\text{cm}} rightward) and the rotational velocity (R\omega leftward) point in opposite directions and cancel exactly. This is the rolling condition in action.

Step 4. Front point F (level with the axle, on the leading edge). Here \theta = \pi/2.

v_{Fx} = v_{\text{cm}}(1 - \cos\,\tfrac{\pi}{2}) = 10(1 - 0) = 10 \text{ m/s (rightward)}
v_{Fy} = -v_{\text{cm}}\sin\,\tfrac{\pi}{2} = -10 \text{ m/s (downward)}
|v_F| = \sqrt{10^2 + 10^2} = 10\sqrt{2} \approx 14.1 \text{ m/s}

Why: at this point, the rotational velocity is directed straight downward (R\omega = 10 m/s), perpendicular to the translational velocity (v_{\text{cm}} = 10 m/s, rightward). The resultant is the diagonal at 45° below horizontal.

Result: v_T = 20 m/s (= 72 km/h), v_C = 0, v_F = 10\sqrt{2} \approx 14.1 m/s at 45° below horizontal.

What this shows: Different points on the same wheel have dramatically different speeds. The top of the tyre moves at twice the speed of the autorickshaw, while the bottom is momentarily at rest. If mud flings off the tyre, it is the top — the fastest-moving part — that sends it flying the farthest.

Example 2: The sliding ladder — finding the instantaneous centre of rotation

A painter's ladder of length 5 m leans against a smooth wall. Its lower end B rests on the floor 3 m from the wall and slides outward at 2 m/s. Find (a) the velocity of the upper end A, (b) the angular velocity of the ladder, and (c) the location of the instantaneous centre of rotation.

Sliding ladder with instantaneous centre of rotation A ladder leans from a wall to the floor. The lower end B at (3, 0) has a rightward velocity arrow. The upper end A at (0, 4) has a downward velocity arrow. Dashed perpendiculars from each velocity meet at the ICR at (3, 4). The distances from ICR to A (3 m) and ICR to B (4 m) are labelled. wall A B 2 m/s 1.5 m/s ICR (3, 4) 3 m 4 m 3 m 4 m
The perpendicular to each velocity (dashed lines) meet at the ICR — the point (3, 4). At this instant, the entire ladder rotates about this point.

Step 1. Find the height of the upper end.

The ladder has length L = 5 m. With B at 3 m from the wall: y_A = \sqrt{L^2 - x_B^2} = \sqrt{25 - 9} = 4 m.

Why: the Pythagorean theorem. The 3-4-5 right triangle gives clean numbers.

Step 2. Find the velocity of end A. The ladder is rigid, so x_B^2 + y_A^2 = 25 at all times. Differentiate:

2x_B\,\dot{x}_B + 2y_A\,\dot{y}_A = 0
\dot{y}_A = -\frac{x_B\,\dot{x}_B}{y_A} = -\frac{3 \times 2}{4} = -1.5 \text{ m/s}

Why: the negative sign means A moves downward — as the bottom slides out, the top slides down. The constraint of constant length forces this relationship.

Result (a): End A moves downward at 1.5 m/s.

Step 3. Locate the ICR. End B moves horizontally (rightward), so the perpendicular to its velocity is a vertical line through B: x = 3. End A moves vertically (downward), so the perpendicular to its velocity is a horizontal line through A: y = 4. These perpendiculars meet at (3, 4) — that is the ICR.

Why: every point on the body instantaneously moves perpendicular to the line joining it to the ICR. So the ICR lies along the perpendicular to each point's velocity.

Step 4. Find the angular velocity using the ICR.

The distance from the ICR to B is \sqrt{(3-3)^2 + (4-0)^2} = 4 m. Since B moves at 2 m/s and is purely rotating about the ICR at this instant:

\omega = \frac{v_B}{r_{B \to \text{ICR}}} = \frac{2}{4} = 0.5 \text{ rad/s}

Verify: the distance from the ICR to A is \sqrt{(3-0)^2 + (4-4)^2} = 3 m. So v_A = \omega \times 3 = 0.5 \times 3 = 1.5 m/s. ✓

Why: this confirms the ICR is correct. Both endpoints give the same angular velocity, as they must for a rigid body.

Result: (a) v_A = 1.5 m/s downward. (b) \omega = 0.5 rad/s. (c) The ICR is at coordinates (3, 4) — directly above B and to the right of A.

What this shows: The ICR is not on the ladder itself — it sits at the corner of a rectangle formed by the wall, floor, and the two endpoints. The geometric construction (perpendiculars to velocity) gives the ICR cleanly, and from it you can find the velocity of any point on the ladder.

Common confusions

If you came here to understand how translation and rotation combine, apply equation (4) to rolling wheels and sliding ladders, and find instantaneous centres of rotation, you have what you need. What follows is for readers who want the formal theorem and its three-dimensional generalisation.

Chasles' theorem

The decomposition you have been using has a name: Chasles' theorem (also written Chasles's theorem). In two dimensions, it states:

Any displacement of a rigid body in a plane is equivalent to a translation of any chosen reference point plus a rotation about that reference point.

In three dimensions, the theorem is stronger and more elegant:

Any displacement of a rigid body in three-dimensional space is equivalent to a screw motion — a translation along an axis combined with a rotation about that same axis.

A screw motion is exactly what it sounds like: the body advances along a line (the screw axis) while simultaneously rotating around it, like a bolt turning into a nut. Pure translation is a screw with zero rotation angle. Pure rotation is a screw with zero translation distance. Every other rigid body motion is a helical combination of both.

Why the ICR exists (for \omega \neq 0)

For planar motion, set \vec{v}_P = 0 in equation (4):

\vec{v}_{\text{cm}} + \vec{\omega} \times \vec{r}'_{\text{ICR}} = 0

With \vec{\omega} = \omega\hat{k} and \vec{v}_{\text{cm}} = (v_x, v_y), the cross product gives:

\vec{\omega} \times \vec{r}'_{\text{ICR}} = \omega\hat{k} \times (a\hat{i} + b\hat{j}) = \omega\,a\,\hat{j} - \omega\,b\,\hat{i} = (-\omega b, \omega a)

Setting the sum to zero: v_x - \omega b = 0 and v_y + \omega a = 0, giving:

a = -\frac{v_y}{\omega}, \quad b = \frac{v_x}{\omega}

As long as \omega \neq 0, this has a unique solution — there is exactly one point with zero velocity. When \omega = 0, the equations have no finite solution, confirming that the ICR moves to infinity for pure translation.

Rolling as a preview of what comes next

Rolling without slipping is the gateway to the deepest ideas in rotational mechanics. The constraint v_{\text{cm}} = R\omega ties translational and rotational motion together, and that coupling produces conservation laws (angular momentum about the contact point) and energy relations (kinetic energy has both translational and rotational contributions) that the next few articles will develop. The moment of inertia — the rotational analogue of mass — determines how the body's kinetic energy splits between translation and rotation. The parallel-axis theorem connects the moment of inertia about the CM to the moment of inertia about the contact point. All of these ideas rest on the decomposition you learned in this article: general motion = translation + rotation.

Where this leads next