Rolling Without Slipping

In short

A wheel rolls without slipping when its centre moves forward at exactly v_{\text{cm}} = R\omega, where R is the radius and \omega the angular speed. Under this condition, the contact point has instantaneous velocity zero, the top of the wheel moves at 2v_{\text{cm}}, and every other point moves somewhere in between. A rolling wheel is identical, at every instant, to a wheel rotating purely about the contact point — an instantaneous axis that keeps sliding along the ground as the wheel advances.

Take a coin and a sheet of paper. Place the paper on your desk. Now roll the coin — not slide, roll — across the paper so it leaves a clean track. Nothing unusual, right? You have done it a hundred times.

Now think about what a single point on the edge of that coin is doing. At one instant it is at the top of the coin, zipping forward. An instant later it is at the side, moving diagonally down. Then at the bottom — pressed against the paper. And here is the strange part: at the instant it touches the paper, that point is not moving at all. If it were, the coin would be skidding, not rolling. The point pressed against the paper is momentarily at rest, while the same point half a turn later is moving faster than the centre of the coin.

This is what makes rolling subtle. A rolling wheel is not just "translation plus rotation" in some vague sense — there is a precise relationship between the two, and if that relationship is off by even a little, the wheel either skids forward or its tyres screech backward. This article nails down that relationship and then uses it to work out what every point on the wheel is doing at every instant.

The rolling condition

Watch a cycle tyre roll forward without slipping. The tyre's centre moves forward at some speed v_{\text{cm}}. At the same time, the tyre is spinning about its centre at some angular speed \omega. These two numbers are not independent. If the wheel spins too fast compared to how quickly it is moving forward, the tyre would spin against the road — that is what happens when you accelerate too hard on a wet road and the wheels screech. If the wheel spins too slowly, the tyre would drag forward across the road — that is what happens when you slam the brakes and lock the wheels and they skid.

Pure rolling is the Goldilocks condition between these two failure modes. To pin it down, imagine marking a single point on the tyre. In one full rotation, the tyre turns through an angle of 2\pi radians. During that rotation, the wheel must move forward by exactly one circumference — because the tread that was on the road at the start unrolls onto the road as the wheel turns, and then the same patch of tread touches the road again when the wheel has advanced by one circumference.

A wheel of radius $R$ that completes one full rotation advances forward by its circumference $2\pi R$. Two rotations advance it by $4\pi R$. This is the rolling condition written geometrically — the tread length equals the ground distance.

One rotation corresponds to the wheel moving forward by 2\pi R (the circumference). Write this as a rate:

\text{distance per rotation} = 2\pi R

\text{rotations per second} = \frac{\omega}{2\pi}

Why: angular speed \omega is measured in radians per second. One full rotation is 2\pi radians, so \omega/(2\pi) is the number of rotations per second.

Multiply the two to get the forward speed of the centre:

v_{\text{cm}} = \underbrace{2\pi R}_{\text{metres per rotation}} \times \underbrace{\frac{\omega}{2\pi}}_{\text{rotations per second}} = R\omega

Why: the 2\pi factors cancel. Every 2\pi radians of rotation corresponds to 2\pi R of forward motion. The ratio — metres per radian of turning — is exactly R.

This is the rolling condition, and it is the single most important equation in this article.

Rolling without slipping

A rigid body of radius R rolls without slipping when its centre-of-mass velocity v_{\text{cm}} and its angular velocity \omega about the centre satisfy

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

If v_{\text{cm}} > R\omega, the body slides forward across the surface (think of a car skidding on ice). If v_{\text{cm}} < R\omega, the body spins in place or the surface moves backward under it (think of a drive wheel spinning on wet mud without gripping).

The rolling condition also applies to accelerations. Differentiate both sides with respect to time:

\frac{dv_{\text{cm}}}{dt} = R\frac{d\omega}{dt}

a_{\text{cm}} = R\alpha

Why: the derivative of v_{\text{cm}} is the linear acceleration a_{\text{cm}} of the centre; the derivative of \omega is the angular acceleration \alpha. Since R is a constant (the radius does not change), it just passes through the derivative. The rolling condition holds at every instant, so its rate of change holds too.

This acceleration form is what you use when the wheel is speeding up or slowing down — say, a rolling sphere on an incline or a car accelerating from rest.

The point of contact is momentarily at rest

Now here is the claim that trips up almost everyone learning rolling for the first time: the point of the wheel that is touching the ground has zero velocity at that instant. Not zero in some average sense. Not "slow". Literally zero.

Read that again. The car is moving at 60 km/h. The tyre is spinning. But the patch of rubber that happens to be pressed against the road at this exact moment — it is stationary. If you could somehow freeze time and reach in, that bit of rubber would not be moving.

Why? Think about what it would mean if it were not zero. If the contact patch moved, there would be relative motion between the tyre and the road. That is exactly what "slipping" means. Rolling without slipping, by definition, requires zero relative velocity between the tyre and the road at the contact point. Since the road is stationary, the contact point on the tyre must also be stationary — for that instant.

This is weird because the tyre is clearly not stationary as a whole. The centre is cruising along at v_{\text{cm}}. The top of the tyre is whizzing. Every part is moving except the one part that touches the road, and that one part changes every millisecond as the wheel turns.

You can prove it from the velocities directly. The velocity of any point P on a rolling wheel is the sum of two contributions:

Translation: every point moves with the centre at velocity \vec{v}_{\text{cm}} (pointing forward along the road).
Rotation about the centre: each point, in addition, moves around the centre at angular speed \omega, with tangential speed R\omega at the rim.

At the contact point — the bottom of the wheel — the rotation contributes a tangential velocity of magnitude R\omega pointing backward (a point at the bottom of a rotating wheel, rotating in the sense a rolling wheel rotates, moves to the rear). The translation contributes v_{\text{cm}} pointing forward.

The two contributions point in opposite directions. Their sum is:

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

Apply the rolling condition v_{\text{cm}} = R\omega:

v_{\text{contact}} = R\omega - R\omega = 0

Why: the rolling condition is precisely the statement that the forward translational velocity of the centre equals the backward tangential velocity from the rotation, at the contact point. The two cancel out exactly. This cancellation is what makes rolling without slipping possible.

Velocity of every point on a rolling wheel

Now extend that argument to every point on the wheel, not just the contact point. Set up coordinates: the centre of the wheel is at position \vec{r}_{\text{cm}}, and a point P on the wheel sits at position \vec{r} relative to the centre. The wheel rotates with angular velocity \omega (taken to be in the positive sense — counterclockwise when viewed from the standard orientation where the wheel rolls to the right).

The velocity of P relative to the ground is:

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

Why: this is the standard decomposition of rigid-body motion. The first term \vec{v}_{\text{cm}} is the velocity P would have if the wheel were only translating. The second term \vec{\omega} \times \vec{r} is the velocity P would have from pure rotation about the centre. Since rolling is translation plus rotation, the total velocity is their sum.

Take four points around the wheel — bottom, top, front, back — and apply this formula to each. Let the wheel roll to the right, so \vec{v}_{\text{cm}} = v_{\text{cm}} \hat{x} (horizontal, pointing right). The angular velocity \vec{\omega} points into the page (using the right-hand rule for a wheel rolling to the right: your fingers curl in the direction of rotation, your thumb points into the page).

Velocity vectors at four points on a wheel rolling to the right. The centre C moves at $v$. The top moves at $2v$ (twice as fast as the centre). The contact point at the bottom is instantaneously at rest. The front and back points move at $v\sqrt{2}$, angled 45° from the ground.

Now compute each of the four special points. Write v = v_{\text{cm}} = R\omega for short.

Top of the wheel (\vec{r} = R\hat{y}, pointing up from the centre):

\vec{\omega} \times \vec{r} = (-\omega \hat{z}) \times (R\hat{y}) = -\omega R(\hat{z} \times \hat{y}) = -\omega R(-\hat{x}) = R\omega\, \hat{x}

\vec{v}_{\text{top}} = v_{\text{cm}}\hat{x} + R\omega\, \hat{x} = (v_{\text{cm}} + R\omega)\hat{x} = 2v_{\text{cm}}\, \hat{x}

Why: both contributions point forward. The translation carries the whole wheel forward at v_{\text{cm}}; the rotation also carries the top of the wheel forward (a point at the top of a wheel rotating in this sense moves forward). They add, giving 2v_{\text{cm}}. The top of a rolling wheel moves at twice the speed of its centre — which is why you should never try to stick your hand on the top of a spinning car tyre.

Bottom of the wheel (\vec{r} = -R\hat{y}, pointing down from the centre):

\vec{\omega} \times \vec{r} = (-\omega \hat{z}) \times (-R\hat{y}) = \omega R(\hat{z} \times \hat{y}) = -R\omega\, \hat{x}

\vec{v}_{\text{bottom}} = v_{\text{cm}}\hat{x} - R\omega\, \hat{x} = (v_{\text{cm}} - R\omega)\hat{x} = 0

Why: rotation carries the bottom of the wheel backward (opposite to the top), so the two contributions exactly cancel under the rolling condition. The contact point has zero velocity, as promised.

Front of the wheel (\vec{r} = R\hat{x}, pointing forward from the centre):

\vec{\omega} \times \vec{r} = (-\omega \hat{z}) \times (R\hat{x}) = -\omega R(\hat{z} \times \hat{x}) = -R\omega\, \hat{y}

\vec{v}_{\text{front}} = v_{\text{cm}}\hat{x} - R\omega\, \hat{y} = v\hat{x} - v\hat{y}

The magnitude is |\vec{v}_{\text{front}}| = \sqrt{v^2 + v^2} = v\sqrt{2}, pointing downward and forward at 45°.

Why: the front of the wheel is moving forward because of translation and moving downward because of the rotation (the leading edge of a forward-rolling wheel is descending toward the ground). The two combine into a diagonal velocity of magnitude v\sqrt{2}.

Back of the wheel (\vec{r} = -R\hat{x}):

\vec{v}_{\text{back}} = v\hat{x} + v\hat{y}

Magnitude v\sqrt{2}, pointing upward and forward at 45°. The back of the wheel is climbing up as it rolls forward, which is why the trailing edge of a moving tyre always looks like it is being scooped up off the road.

Summary:

Point	Velocity	Magnitude	Direction
Top	2v\, \hat{x}	2v	Forward (fastest)
Centre	v\, \hat{x}	v	Forward
Front	v\hat{x} - v\hat{y}	v\sqrt{2}	Forward-down 45°
Back	v\hat{x} + v\hat{y}	v\sqrt{2}	Forward-up 45°
Contact	0	0	— (stationary)

The velocity of any point P at angle \phi (measured from the upward vertical, anticlockwise) on the rim has magnitude

|\vec{v}_P| = 2v\sin(\phi/2) = 2v_{\text{cm}} \left|\sin\frac{\phi}{2}\right|

where \phi = 0 at the top (giving 0… wait, that's the top, which should be 2v). The convention needs care: measure \phi from the contact point instead.

Measure \phi from the contact point, going around the rim. The distance from the contact point to any point P on the rim is 2R\sin(\phi/2) — a standard chord formula. Since the rolling wheel is equivalent to pure rotation about the contact point (the next section proves this), the speed of P is \omega times this distance:

|\vec{v}_P| = \omega \cdot 2R\sin(\phi/2) = 2R\omega \sin(\phi/2) = 2v_{\text{cm}}\sin(\phi/2)

At \phi = 0 (contact point): |\vec{v}_P| = 0. ✓ At \phi = \pi (top of wheel): |\vec{v}_P| = 2v_{\text{cm}} \sin(\pi/2) = 2v_{\text{cm}}. ✓ At \phi = \pi/2 (front of wheel): |\vec{v}_P| = 2v_{\text{cm}} \sin(\pi/4) = v_{\text{cm}}\sqrt{2}. ✓

All three special values match what you just computed the long way. The chord formula is exact.

Rolling as pure rotation about the contact point

The fact that the contact point has zero velocity suggests a powerful reframing. If there is a point on the wheel that is not moving at a given instant, you can think of the wheel as rotating about that point — just for that instant. This is the idea of an instantaneous axis of rotation.

Here is the claim: at any instant, a wheel rolling without slipping is kinematically identical to a wheel that is purely rotating (no translation at all) about its current contact point, at the same angular velocity \omega.

Why is that true? Two descriptions are kinematically identical if they assign the same velocity to every point of the wheel. The "translation + rotation about centre" description gives the velocities you just computed. Compute the same velocities using "pure rotation about the contact point" and check that they match.

Take the contact point as the axis. A point P on the wheel at position \vec{r}' relative to the contact point rotates about it with angular velocity \omega. Its velocity is:

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

Why: for pure rotation about a fixed point, every point moves with velocity \omega \times (position from that point). This is the standard formula for rotation, with no translation term.

Check the four special points:

Contact point (\vec{r}' = 0): \vec{v} = 0. ✓
Centre (\vec{r}' = R\hat{y}, pointing from contact to centre):

\vec{v} = (-\omega\hat{z}) \times (R\hat{y}) = R\omega\,\hat{x} = v_{\text{cm}}\hat{x}. \quad \checkmark

Top (\vec{r}' = 2R\hat{y}, pointing from contact to top):

\vec{v} = (-\omega\hat{z}) \times (2R\hat{y}) = 2R\omega\,\hat{x} = 2v_{\text{cm}}\hat{x}. \quad \checkmark

Front (\vec{r}' = R\hat{x} + R\hat{y}, pointing from contact to front):

\vec{v} = (-\omega\hat{z}) \times (R\hat{x} + R\hat{y}) = -R\omega\hat{y} + R\omega\hat{x} = v\hat{x} - v\hat{y}. \quad \checkmark

Every velocity matches exactly. The two descriptions are the same kinematically.

A rolling wheel seen as pure rotation about its contact point. Each concentric circle around the contact point traces a line of equal speed — points on the same circle move at the same rate. The top is the farthest point from the axis and moves the fastest ($2v$). The centre is at distance $R$ and moves at $v$. The contact point itself is at the axis, with zero speed.

This reframing is extraordinarily useful. When you need the speed of any point on a rolling body, you do not have to add two vectors. Just measure the distance d from that point to the current contact point, and the speed is \omega d. No cross products, no axis directions.

The catch — the thing that makes this an instantaneous axis rather than a fixed axis — is that the contact point is not a fixed point on the wheel. It is always a different material point. The bit of tread touching the road right now will be on its way up in a millisecond, replaced by a new bit of tread that was just trailing behind. The axis is fixed in space (it is always "wherever the wheel touches the road") but not fixed on the wheel itself.

This is why rotation about the contact point is kinematically equivalent to rolling only at one instant. The next instant, the contact point has moved forward, and you have a new instantaneous axis, also at the new contact point, at the same angular speed \omega. Over time, the instantaneous axis slides along the ground — but at any single instant, treating it as a fixed axis gives you the right velocities.

Watch the rolling condition in action

The simulation below tracks a single marked point on the rim of a rolling wheel. Watch the trace it carves out — a curve called a cycloid. The marked point rises from the ground (where it is momentarily stationary), arcs over the top (where it moves at 2v), and returns to the ground for the next contact. Its forward motion is never constant; it slows to a halt at each touch.

Watch the marked point on the rim (red) trace a cycloid as the wheel rolls at 3 m/s. The point touches the ground, rises, reaches the top where it moves fastest, and comes back down. The wheel centre (dark) moves in a perfectly straight horizontal line. The sharp cusps at each ground contact are the moments when the rim point has zero velocity. Click replay to watch again.

Every cusp in the red curve — the sharp points at the ground — is a moment of zero velocity. Between cusps, the point sweeps through the air. The horizontal distance between cusps is 2\pi R, the wheel's circumference, because one cusp happens every full rotation. You are watching the rolling condition in action.

Worked examples

Example 1: The auto-rickshaw tyre

An auto-rickshaw accelerates from rest. Its rear tyre has a radius of 0.22 m. When the auto is moving at 36 km/h, find (a) the angular speed of the tyre, (b) the speed of the topmost point of the tyre, (c) the speed of a point on the rim that is level with the centre, at the leading edge.

The tyre rolling at 10 m/s. The top zips forward at 20 m/s. The front leading point moves diagonally down and forward at $10\sqrt{2}$ m/s. The contact patch is momentarily at rest.

Step 1. Convert the centre-of-mass speed to SI units.

v_{\text{cm}} = 36 \text{ km/h} = 36 \times \frac{1000}{3600} \text{ m/s} = 10 \text{ m/s}

Why: the rolling condition uses SI units throughout. 1 km/h is 1000/3600 m/s — a factor of 5/18 — so 36 km/h is exactly 10 m/s, a clean number.

Step 2. Apply the rolling condition to get \omega.

\omega = \frac{v_{\text{cm}}}{R} = \frac{10 \text{ m/s}}{0.22 \text{ m}} = 45.5 \text{ rad/s}

Why: solving v_{\text{cm}} = R\omega for \omega gives \omega = v_{\text{cm}}/R. The units check: m/s divided by m gives 1/s, which is rad/s (the radian is dimensionless).

Step 3. Speed of the topmost point.

The top of the wheel moves at twice the centre's speed:

v_{\text{top}} = 2v_{\text{cm}} = 20 \text{ m/s}

Why: at the top, the translational and rotational velocity contributions both point forward and are both equal to v_{\text{cm}}. They add.

Step 4. Speed of the leading rim point (at the same height as the centre, at the front of the wheel).

Using the decomposition: translation gives v_{\text{cm}} = 10 m/s forward. The rotation about the centre gives R\omega = 10 m/s tangentially — at the front of the wheel, tangential means downward. Combining:

|\vec{v}_{\text{front}}| = \sqrt{v_{\text{cm}}^2 + (R\omega)^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \approx 14.1 \text{ m/s}

Why: the two contributions are perpendicular (one horizontal, one vertical), so their magnitudes combine by Pythagoras. The velocity is directed 45° below the horizontal, pointing forward and down.

Result: \omega \approx 45.5 rad/s; v_{\text{top}} = 20 m/s; v_{\text{front}} \approx 14.1 m/s at 45° below horizontal. The contact point has speed 0.

What this shows: the same rolling wheel has points moving at every speed from 0 (contact) through v (centre) to v\sqrt{2} (sides) up to 2v (top). The angular speed is a single number that determines all of them, via v = R\omega and the instantaneous-axis picture.

Example 2: Mud flying off a rolling tyre

A motorcycle rides through a shallow puddle at 54 km/h. Its wheel has a radius of 0.30 m. A small glob of mud stuck to the rim just at the top of the wheel breaks loose and flies forward, behaving like a projectile after release. Neglect air resistance. (a) What is the initial speed of the mud glob? (b) If the top of the wheel is 0.60 m above the ground, how far forward of its release point does the mud land?

A glob of mud leaves the top of the wheel moving horizontally at $2v_{\text{cm}} = 30$ m/s. Gravity pulls it down; it lands about 3.3 m ahead of where it was released.

Step 1. Find v_{\text{cm}} in m/s.

v_{\text{cm}} = 54 \text{ km/h} \times \frac{5}{18} = 15 \text{ m/s}

Step 2. The mud is at the top of the rolling wheel when it breaks loose. Its velocity at that instant is the velocity of the topmost point of the wheel.

v_{\text{mud, initial}} = 2v_{\text{cm}} = 30 \text{ m/s}

The mud moves horizontally (forward) at this speed.

Why: the top of a rolling wheel moves at 2v, purely horizontal. Once the mud detaches, there is no tangential acceleration from the wheel — the mud keeps whatever velocity it had at the moment of release, then is acted on by gravity alone.

Step 3. Find the time of flight. Treat the mud as a projectile launched horizontally from height h = 0.60 m.

h = \tfrac{1}{2} g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.60}{9.8}} = \sqrt{0.1224} \approx 0.350 \text{ s}

Why: horizontal launch means the initial vertical velocity is zero. The mud falls freely under gravity; its fall time depends only on the height and g, not on the horizontal speed.

Step 4. Horizontal distance travelled during the fall.

x = v_{\text{mud, initial}} \times t = 30 \times 0.350 \approx 10.5 \text{ m}

But wait — the wheel has also moved forward during that 0.350 s, at v_{\text{cm}} = 15 m/s:

x_{\text{wheel}} = 15 \times 0.350 = 5.25 \text{ m}

So relative to the release point on the ground (not on the moving wheel), the mud lands at 10.5 m. Relative to the wheel, it lands 10.5 - 5.25 = 5.25 m ahead of the wheel — which matches the intuition that mud shed from the top flies well ahead of the vehicle.

Why: the ground-frame distance is just horizontal speed times time, because gravity only changes the vertical velocity. The wheel-frame distance is smaller because the wheel also moved during the fall.

Result: the mud leaves at 30 m/s, flies for 0.35 s, and lands about 10.5 m ahead of its launch point on the ground.

What this shows: the top of a rolling wheel really does move at 2v_{\text{cm}}, not v_{\text{cm}}. This is why mudflaps exist — without them, mud flung from the top of the tyre would sail forward at twice the vehicle's speed and shower everything in front. The mudflap blocks the upward-backward trajectory of mud thrown off the back of the wheel, where mud is still attached longer because that is the rising edge.

Common confusions

"The contact point has low velocity, not zero." No — it is exactly zero at the instant of contact. If it were not zero, the tyre would be slipping. The reason this feels strange is that every other point on the wheel is moving, so it is tempting to assume the bottom point is moving too. But the whole idea of rolling without slipping is defined by this zero.
"The top of the wheel moves at v_{\text{cm}}, the same as the centre." No — the top moves at 2v_{\text{cm}}. This catches almost everyone because the wheel is a rigid body and "it all moves together" feels right. But the rigid-body motion at any instant is rotation about the contact point, and the top is 2R from the contact point while the centre is only R from it. Distance-to-axis times \omega gives speed; the top's speed is doubled.
"Rolling requires friction." It does on an incline or when the wheel is accelerating — static friction provides the torque needed to spin the wheel up or down. On a flat road at constant speed, however, a rolling wheel needs no friction at all; it just coasts. Rolling resistance (from tyre deformation) is a separate, smaller effect.
"If v_{\text{cm}} = R\omega always, then a wheel can never be on ice." It can — on ice, the wheel does slip. The rolling condition is a statement about what counts as rolling; it is not a universal law. A wheel on ice with the engine pushing too hard has R\omega > v_{\text{cm}}: the wheel spins faster than the car moves. The mismatch is the slipping. Pure rolling is a special case, not the rule.
"Angular velocity is the same at every point on the wheel." Yes! Angular velocity is a property of the whole rigid body, not of any particular point. Every point on the wheel rotates at the same \omega. It is the linear velocities that differ from point to point — precisely because they are proportional to distance from the axis, and different points are at different distances.
"The centre of the wheel traces a cycloid too." No — the centre moves in a straight horizontal line, at constant speed v_{\text{cm}}. Only points on the rim trace cycloids. Points between the rim and the centre trace curtate cycloids (no cusps, just a wavy line). Points beyond the rim (impossible for a rigid wheel, but imagine a flange) trace prolate cycloids that loop backward.

If you understand v = R\omega and the instantaneous axis picture, you have what you need for every standard rolling problem. What follows is for readers who want the full cycloid equations, the deeper geometry of rolling curves, and the connection to a body rolling on a curved surface.

The cycloid equations

A point on the rim of a wheel of radius R that rolls without slipping along the x-axis at angular speed \omega traces a cycloid. Its position as a function of time is:

x(t) = R(\omega t - \sin(\omega t))

y(t) = R(1 - \cos(\omega t))

where t = 0 corresponds to the marked point being at the origin (touching the ground).

Deriving these. The centre of the wheel moves at v_{\text{cm}} = R\omega, so at time t the centre is at (R\omega t,\, R). Relative to the centre, the marked point rotates at angular speed \omega. If at t = 0 it is directly below the centre (at position (0, -R) relative to the centre), then at time t it is at position (-R\sin(\omega t),\, -R\cos(\omega t)) relative to the centre — the minus signs capture that the wheel rotates clockwise when rolling to the right, so the marked point swings backward and upward as the wheel turns.

Adding the centre's position and the relative position:

x(t) = R\omega t - R\sin(\omega t) = R(\omega t - \sin(\omega t))

y(t) = R + (-R\cos(\omega t)) = R(1 - \cos(\omega t))

Check the zeros of the velocity. The marked point's velocity components are:

\dot{x}(t) = R\omega(1 - \cos(\omega t)), \qquad \dot{y}(t) = R\omega \sin(\omega t)

At \omega t = 0, 2\pi, 4\pi, \ldots (whenever the marked point touches the ground), both \dot{x} and \dot{y} are zero. The speed vanishes exactly at the ground contacts, which is the instantaneous-axis result from before, written out as a curve.

Rolling on a curved surface

Everything above assumed the wheel rolls on a flat surface. If the wheel of radius R rolls without slipping on the inside of a larger circular track of radius R_0, the centre of the wheel moves along a circle of radius R_0 - R, and a point on the rim traces a hypocycloid — the same zero-velocity-at-ground result holds, but now the "ground" is curved.

The rolling condition on a curve has a slightly richer form: the arc length traced on the outside surface equals the arc length traced on the wheel's rim. If you parametrise each path by arc length s, then ds_{\text{track}} = ds_{\text{wheel}} — which in terms of angular parameters becomes

R_0\, d\phi_{\text{track}} = R\, d\theta_{\text{wheel}}

where \phi_{\text{track}} is the angle subtended by the wheel's contact point at the track's centre and \theta_{\text{wheel}} is the angle through which the wheel has rotated.

A special case: a wheel of radius R rolling inside a track of radius 2R produces a hypocycloid that is a straight line — a classical result known as the Cardan circles theorem. Every point on the rim oscillates back and forth along a diameter of the outer track. You can demonstrate this at home with a small circular biscuit-tin lid rolling inside a larger one.

The instantaneous axis for any rigid body

The "instantaneous axis = contact point" result generalises beyond wheels. For any rigid body undergoing planar motion, there is always a point in the plane (possibly not on the body itself) whose velocity is zero at a given instant. This point is the instantaneous centre of zero velocity, and the body's motion at that instant is equivalent to pure rotation about it.

For a rolling wheel, this point is the contact point — on the body. For a ladder sliding down a wall (top sliding down, bottom sliding out along the floor), the instantaneous centre is at the intersection of the wall and the floor extended perpendicular to the ladder's endpoint velocities — not on the ladder at all.

This is the foundation of the kinematics of machine linkages and is used heavily in mechanical engineering to analyse the motion of pistons, cranks, and connecting rods. In every case, the trick is the same: locate the point that is not moving this instant, then treat the whole body as rotating about it.

A subtle point about rolling on an incline

When a wheel rolls without slipping down an incline, there is still a single contact point with zero velocity relative to the incline surface. The instantaneous axis is perpendicular to the incline at the contact point — but the wheel's centre is not moving horizontally; it is moving parallel to the slope. Everything about the rolling-without-slipping kinematics still holds, with one subtle change: v_{\text{cm}} now refers to the speed of the centre along the incline, not along the ground. The condition v_{\text{cm}} = R\omega is the same; the direction of v_{\text{cm}} is what has changed.

This generalisation is what Rolling on Inclines and Advanced Problems builds on to find the acceleration of a sphere rolling down a ramp.

Where this leads next

Energy in Rolling Motion — the kinetic energy of a rolling body splits into translational and rotational parts; the rolling condition links them into a single expression.
Rolling on Inclines and Advanced Problems — what acceleration a rolling sphere picks up on a ramp, and why the one with the smallest moment of inertia wins.
Angular Momentum — the rotational analogue of momentum, which a rolling wheel carries in addition to its linear momentum.
Torque — the turning effect of a force, and the quantity that determines how rolling motion speeds up or slows down.
Moment of Inertia — the rotational analogue of mass; the number I/(MR^2) that determines which round body wins a downhill race.