---
title: Rolling Motion
module: Rotation
moduleNumber: 5
lessonNumber: 3
order: 503
summary: >
  A rolling wheel is doing two things at once — translating and spinning — but the
  no-slip condition $v_{cm}=R\omega$ locks them together, and that single
  constraint is what makes rolling tractable. We use it to split the kinetic energy
  into $\tfrac12Mv_{cm}^2+\tfrac12I\omega^2$, find how fast a cylinder reaches the
  bottom of an incline, and show why the contact point is instantaneously at rest.
  The static friction that enforces rolling does no work; we track its direction
  from the tendency to slip, and mark exactly where the model breaks once the
  required friction exceeds $\mu_sN$.
topics: [Rotation]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 9 — Rotation; §9-6"
---

## Rolling constraint and rigid-body kinematics

A rigid wheel rolling without slipping on a fixed surface has an instantaneous
contact point at rest relative to the surface. The centre-of-mass speed and
angular speed satisfy

$$
v_{\rm cm}=R\omega,
\qquad a_{\rm cm}=R\alpha.
$$

The relation is a kinematic constraint, not an independent force law. It fails
when the contact surfaces slide. A material point on the rim is not permanently
stationary: it is the contact point only for one instant and subsequently moves
around the wheel.

$$
% caption: Rolling superposes translation of the centre at $v_{cm}$ with rotation about the centre at angular speed $\omega$. The velocities add to $2v_{cm}$ at the top rim and cancel to zero at the instantaneous contact point.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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$$

**Force and torque derivation on an incline.**

Resolve forces along an incline of angle $\theta$, taking downhill as positive.
The centre-of-mass equation and the torque equation about the centre are

$$
Mg\sin\theta-f_s=Ma,
\qquad f_sR=I\alpha.
$$

No-slip rolling imposes $a=R\alpha$. Elimination of $f_s$ and $\alpha$ gives

$$
a=\frac{g\sin\theta}{1+I/(MR^2)},
\qquad f_s=\frac{I}{R^2}\,a.
$$

The solved friction is uphill for a body released from rest, even though the body
moves downhill. Friction opposes the relative motion that contact would have in
the absence of friction, not necessarily the direction of centre motion. If the
required $f_s$ exceeds $\mu_sN$, the assumed rolling solution is impossible and
the problem must be reformulated with kinetic friction.

**Energy derivation and consistency check.**

The same acceleration follows from energy. Substitution of $\omega=v/R$ into
rolling kinetic energy gives

$$
K=\frac12Mv^2\left(1+\frac{I}{MR^2}\right).
$$

After a downhill distance $s$, the vertical drop is $s\sin\theta$. Differentiating
the energy equation with respect to time gives the force-torque result. Agreement
checks that static friction is doing no work at the fixed contact surface
while still transferring energy between translational and rotational motion.

Friction direction follows from the force-torque equations. The energy equation
sets the speed relation without determining that direction. Gravitational-energy
partition appears directly in the energy equation. Both methods describe the same
motion from complementary perspectives.

## Slipping, curved paths, and contact limits

When $v\ne R\omega$, the relative contact velocity determines kinetic-friction
direction. A wheel dropped with excessive spin has contact moving backward relative
to the ground; kinetic friction acts forward, increasing $v$ and decreasing
$\omega$. A wheel with excessive translation has contact moving forward; friction
acts backward, decreasing $v$ and increasing rotation. In both cases, the rolling
state is an attractor only when friction and geometry permit it.

Mechanical energy is not conserved during sliding. The kinetic-friction work is
negative because contact surfaces move relative to each other. Linear momentum of
the wheel alone is not conserved because ground exerts an external impulse. Angular
momentum about the centre changes because friction has a torque. These statements
avoid an incorrect attempt to conserve both translational and rotational kinetic
energy through a slipping transient.

**Rolling up and down curved surfaces.**

On a smooth curved track, the rolling constraint can remain valid if static
friction supplies the required tangential torque. Normal force contributes no
torque about the centre when it acts along the radius to contact. Energy gives the
centre speed as a function of height, while radial force balance determines whether
the body remains in contact. A body can lose contact when required normal force
would become negative; rolling kinematics then no longer describes its subsequent
free flight.

The instantaneous-centre construction calculates velocities on a curved path, but
the contact point changes continually. Torque equations should normally be
written about the centre of mass or another fixed inertial point. Treating the
instantaneous contact point as a permanent pivot produces incorrect acceleration
and force conclusions.

> **Worked example.** A solid cylinder has $I=MR^2/2$ and rolls from rest down a
> $30^\circ$ incline. Its acceleration is
>
> $$
> a=\frac{g\sin30^\circ}{1+1/2}=3.27\ \mathrm{m\,s^{-2}}.
> $$
>
> The required static friction is $f_s=Ma/2$, directed uphill, so the minimum
> coefficient of static friction is $f_s/(Mg\cos30^\circ)=0.192$. A smaller
> available coefficient lets the cylinder slide, and this rolling acceleration no
> longer applies.

**Experimental checks and units.**

The rolling constraint can be tested with video by tracking centre displacement
and rim angle. A line fit of $x$ against $\theta$ has slope $R$ in no-slip motion.
The residual pattern distinguishes random tracking error from systematic slip.
Wheel radius should be measured at the contact surface, including a tire's loaded
effective radius when deformation is appreciable.

The dimensionless ratio $I/(MR^2)$ carries the shape dependence of the
acceleration and is independent of overall scale for geometrically similar rigid
bodies. The model assumes rigid shape, a fixed surface, and static contact. Tire deformation,
rolling resistance, bearing friction, and air drag produce energy loss outside
the ideal result.

## Energy, effective inertia, and instantaneous axes

The ratio $\beta=I/(MR^2)$ collects all shape dependence for a rigid body rolling
on a fixed incline. The acceleration and energy fractions are

$$
\frac{a}{g\sin\theta}=\frac1{1+\beta},
\qquad
\frac{K_{rot}}{K_{total}}=\frac{\beta}{1+\beta}.
$$

Moving mass outward increases $I$ and reduces downhill acceleration. This result
is exploited in flywheels, where high rim inertia stores rotational energy, and in
rolling competitions, where a compact mass distribution reaches the bottom first.
The comparison requires equal radius and equal centre-of-mass height change. A
larger-radius body can have a different path and contact geometry, so it is not
automatically comparable by inertia ratio alone.

The parallel-axis theorem must be used with care. Rolling energy contains inertia
about the centre of mass, because rotational velocity is taken relative to that
centre. An inertia about the contact point can describe instantaneous kinetic
energy, but the contact point is not fixed through time. Mixing these two axes in
one energy equation double counts or omits translational energy.

**Instantaneous-axis energy interpretation.**

For no-slip rolling, kinetic energy can alternatively be written about the
instantaneous contact point:

$$
K=\frac12I_P\omega^2,
\qquad I_P=I_{cm}+MR^2.
$$

Substitution gives the sum of translation and centre-of-mass rotation.
This is an energy identity at one instant. It does not authorize writing
$\sum\tau_P=I_P\alpha$ as though the contact point were a fixed inertial pivot.
The acceleration of that point is generally nonzero, and the standard fixed-axis
torque equation has additional terms for a moving reference point.

Treat the contact as an instantaneous centre to obtain the velocity of any point
in a speed calculation. Force problems require the moving-point dynamics
explicitly. Centre-of-mass force and torque equations remain the reliable method
for acceleration.

## Resistance and traction

Real wheels lose energy even without macroscopic slip. Tire or wheel deformation
shifts the normal-force resultant ahead of the geometrical contact point, producing
a resisting torque. Bearing friction and internal material hysteresis add further
losses. A common simplified model uses a rolling-resistance force opposite motion,
but its magnitude can depend on speed, load, pressure, temperature, and surface.

Rolling resistance differs from kinetic friction. It can act while the contact
point is instantaneously at rest, because deformation creates a distributed
contact region and internal energy loss. The ideal static-friction statement "does
no work" refers to rigid bodies and fixed contact surfaces; it does not imply that
real rolling motion is lossless.

**Constraint failure and contact limits.**

The inequality $|f_s|\le\mu_sN$ is a feasibility condition. On an incline,
$N=Mg\cos\theta$ only when no additional forces have normal components. A rope,
aerodynamic force, or curved track changes $N$ and therefore the maximum available
static friction. Solving the ideal rolling equations before checking this
inequality is efficient; assuming the inequality is satisfied without checking
can give an unphysical solution.

If contact is lost, neither normal force nor friction acts. The body then follows
translation under gravity while retaining angular velocity in the absence of
external torque about its centre. The relation $v=R\omega$ generally ceases to
hold during flight. On recontact, a collision and possible slipping transient can
change both speeds abruptly.

## Signs, velocity fields, and powered rolling

One sign convention can assign downhill translation positive and clockwise
rotation positive for a body descending a slope. The constraint then has either
$v=R\omega$ or $v=-R\omega$ depending on the axes used in the diagram. The
physical requirement is consistency through velocity, acceleration, and torque
equations. Replacing a signed constraint by an unsigned magnitude relation midway
through a calculation is a frequent source of reversed friction direction.

The most reliable check is geometric: identify the velocity of the contact point
relative to the surface. If the hypothesized no-friction motion would make it slide
downhill, static friction must point uphill; if it would slide uphill, friction
must point downhill. This reasoning works for rolling starts, braking, powered
wheels, and moving belts.

**Rigid-body velocity derivation.**

The velocity of point $P$ in a planar rigid body is

$$
\vec v_P=\vec v_{cm}+\vec\omega\times\vec r_{P/cm}.
$$

For the bottom point of a wheel, $\vec r_{P/cm}$ has magnitude $R$ downward.
With the sign of $\vec\omega$ chosen for forward rolling, the rotational
term is backward and has magnitude $R\omega$. The no-slip condition is therefore
the vector equality that makes their sum zero. At the top point, the rotational
term is forward, giving twice the centre speed. At side points, the two velocity
components are perpendicular and their magnitude is $\sqrt2v_{cm}$.

Differentiation gives point acceleration:

$$
\vec a_P=\vec a_{cm}+\vec\alpha\times\vec r_{P/cm}
+\vec\omega\times(\vec\omega\times\vec r_{P/cm}).
$$

The final centripetal term explains why the contact point can have zero velocity
but nonzero acceleration. An instantaneous-centre diagram determines velocities;
the full rigid-body relation is required for acceleration.

**External torque and powered rolling.**

An applied axle torque $\tau_d$ changes the force balance. On level ground with
forward positive,

$$
f_s=Ma,
\qquad \tau_d-f_sR=I\alpha,
\qquad a=R\alpha.
$$

Elimination gives

$$
a=\frac{\tau_d/R}{M+I/R^2},
\qquad f_s=\frac{M\tau_d/R}{M+I/R^2}.
$$

Static friction points forward for a driven wheel because the drive torque would
otherwise make the contact surface slip backward. A wheel pulled at its axle is
different: the applied force can accelerate translation directly, and friction may
point backward to provide the torque needed for rolling. The direction cannot be
deduced from motion direction alone.

**Contact mechanics and modelling limits.**

The phrase "without slipping" idealizes contact as a point with sufficient static
friction. Real contact has finite area. A pneumatic tire deforms, a railroad wheel
has microscopic creep, and a soft ball changes shape. The effective rolling radius
can differ from unloaded geometric radius. The rigid-body constraint remains valid
when these deviations are small relative to experimental resolution, but they
become important in traction, braking, and energy-loss measurements.

The coefficient of static friction is not the friction force divided by normal
force in every rolling problem. It sets an upper bound. A freely rolling wheel on
level ground at constant speed can have essentially zero static friction in the
ideal model even if the surface has a large coefficient. The available friction is
then not being used because no torque is required to change angular speed.

> **Worked example.** A solid cylinder of mass $4.0\ \mathrm{kg}$ and radius
> $0.20\ \mathrm m$ is driven on level ground by an axle torque $6.0\ \mathrm{N\,m}$.
> Its inertia is $I=MR^2/2=0.080\ \mathrm{kg\,m^2}$, so
>
> $$
> a=\frac{6.0/0.20}{4.0+0.080/(0.20)^2}=5.0\ \mathrm{m\,s^{-2}}.
> $$
>
> Static friction is $Ma=20\ \mathrm N$ forward, needing a coefficient
> $20/(4.0g)=0.51$. A surface below that cannot sustain the no-slip acceleration
> and the wheel spins instead: drive torque, wheel inertia, and surface traction
> all set the realistic acceleration limit.

> **Worked example.** A solid sphere ($I=\tfrac25MR^2$) starts rolling uphill at
> speed $v_0$. Energy conservation gives
>
> $$
> Mgh=\frac12Mv_0^2+\frac12\left(\frac25MR^2\right)
> \frac{v_0^2}{R^2}=\frac{7}{10}Mv_0^2,
> \qquad h=\frac{7v_0^2}{10g}.
> $$
>
> A sliding point mass would rise only $v_0^2/(2g)$, so the rolling sphere climbs
> higher at the same centre speed because its initial rotation carries energy too.
> The comparison holds the initial translation speed fixed, not the total energy;
> which quantity is held fixed must be stated in any rolling comparison.

## Energy release and moving surfaces

The formula for downhill speed can be rearranged to identify how much height is
needed to reach a target centre speed:

$$
h=\frac{v^2}{2g}\left(1+\frac{I}{MR^2}\right).
$$

At fixed target speed, a hoop requires twice the height required by a sliding
point mass, while a solid sphere requires $7/5$ of that height. In laboratory
design, the form uses the directly controlled height and photogate-measured final
speed. The relation predicts the ideal result before frictional losses are
estimated from measured discrepancies.

If a rolling object is released from the same height but has a displaced centre of
mass, its potential-energy change is $Mg\Delta y_{cm}$, not necessarily $Mgh$
based on geometric radius. An asymmetric wheel can also have a varying inertia
about the instantaneous orientation. The simple constant-$I$ formula assumes a
body with fixed centre-of-mass distance from the axis and a surface path whose
height relation has been stated.

**Rolling on a moving surface.**

The no-slip condition is relative to the surface. On a conveyor belt moving at
speed $u$, the contact-point ground-frame velocity is $u$, giving

$$
v_{cm}-R\omega=u
$$

for one common sign convention. The contact point is therefore not at rest in the
laboratory frame, and static friction can do work on the wheel in that frame. A
wheel placed on a moving belt may acquire both translation and rotation while
remaining non-slipping after the transient. The claim that static friction does no
work is restricted to a fixed surface and cannot be applied unchanged here.

**Rolling versus pure rotation.**

A wheel mounted on a frictionless fixed axle has $v_{cm}=0$ and can rotate with
nonzero $\omega$. A sliding block has $\omega=0$ and can translate with nonzero
$v_{cm}$. Rolling is the constrained combination of both motions. The same rigid
body can occupy all three regimes depending on supports and contact conditions.
Separating the motions before applying the constraint prevents the mistake of
assigning $v=R\omega$ to every rotating wheel.

The centre-of-mass kinetic energy and rotational kinetic energy have different
origins but are coupled by contact in no-slip rolling. A torque-free wheel in flight
retains its rotation while its centre follows projectile motion; on recontact,
static or kinetic friction determines whether the speeds satisfy the rolling
constraint. This sequence tests whether a solution applies the constraint only
while its physical contact condition exists.

**Measurement uncertainty.**

An experimental estimate of $I/(MR^2)$ from incline acceleration uses

$$
\frac{I}{MR^2}=\frac{g\sin\theta}{a}-1.
$$

Small error in acceleration is amplified when $a$ is measured from short travel
times, so longer tracks and multiple trials improve the estimate. Angle uncertainty
also enters through $\sin\theta$. A comparison among shapes should control radius,
surface, release method, and mass distribution; otherwise observed arrival order
cannot be attributed solely to dimensionless inertia.

The resulting model remains experimentally falsifiable and mechanically coherent.

A spinning wheel placed on a rough surface can initially slip. Kinetic friction
both accelerates its centre and reduces angular speed until $v=R\omega$ holds.
Mechanical energy decreases because the contact point slides during this transient,
even though linear and angular momentum relations remain applicable.

A full rolling trajectory combines the distance constraint, energy equation, and
incline acceleration. The sequence runs from release through constant
acceleration to a known distance, where speed and angular speed follow from the
same rolling relation.

At an instant of no-slip rolling, the contact point is an instantaneous centre of
rotation for velocity calculation. The centre then has speed $R\omega$, but this
construction must not be used blindly for acceleration because the instantaneous
centre changes from moment to moment.

$$
% caption: At each instant the contact point acts as a centre of rotation: every point's velocity is perpendicular to its line from the contact and proportional to that distance, so the centre moves at $v=R\omega$ and the top at $2v=2R\omega$.
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$$

Static friction is the only force creating torque about the centre on a body rolling
down an ideal incline. The torque equation and translation equation are solved
together; substituting the rolling constraint gives the acceleration.

$$
% caption: On an incline the only torque about the centre comes from static friction $f_s$ at the rim; with lever arm $R$ it supplies $f_sR=I\alpha$. The weight $Mg$ acts through the centre and exerts no torque there.
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$$

Shape comparisons are organized by the dimensionless inertia ratio. The incline
acceleration formula takes the tabulated values as input and yields a numerical race
order.

| Shape | $I/(MR^2)$ | $a/(g\sin\theta)$ |
| --- | --- | --- |
| solid sphere | $2/5$ | $5/7$ |
| solid cylinder | $1/2$ | $2/3$ |
| hoop | $1$ | $1/2$ |

Static friction changes direction when the required angular acceleration reverses.
A wheel driven at its axle tends to spin the contact point backward relative to the
ground, so friction acts forward and accelerates the centre. A braking torque has
the opposite tendency and can make friction point backward.

When the rolling constraint is violated, contact has nonzero relative speed and
kinetic friction replaces static friction. Mechanical energy then decreases because
friction acts through a nonzero sliding displacement.

$$
% caption: When translation and spin are mismatched, $v_{cm}\neq R\omega$: the contact slides at $v_{slip}=v_{cm}-R\omega$, so kinetic friction acts against that sliding until rolling is re-established.
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\end{tikzpicture}
$$

A rolling body sent uphill converts both translation and rotation into gravitational
potential energy. At the turn, both $v$ and $\omega$ vanish if rolling persists.

$$
% caption: A body rolling up an incline trades both translational and rotational kinetic energy for height; under the no-slip constraint $v$ and $\omega$ reach zero together at the turning point.
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$$

The no-slip condition can be tested by tracking a marked rim point and the centre.
Over a time interval, centre displacement should equal arc length $R\Delta\theta$.
The slope of a graph of centre displacement against rotation angle estimates radius
and indicates systematic slipping through departure from a straight line.

The velocity of a rim point is the vector sum of centre translation and rotation
relative to the centre. At the top, both point forward. At the bottom, they cancel.
The corresponding accelerations do not cancel at contact; rolling contact has zero
velocity but generally nonzero acceleration.

## Inclined rolling and energy checks

A body of mass $M$ and centre-of-mass inertia $I$ has rolling energy

$$
K=\frac12Mv_{\rm cm}^2+\frac12I\omega^2.
$$

Rolling down an incline through height $h$ gives

$$
Mgh=\frac12Mv^2\left(1+\frac{I}{MR^2}\right),
\qquad
v=\sqrt{\frac{2gh}{1+I/(MR^2)}}.
$$

Bodies with smaller $I/(MR^2)$ have larger translational speed at the same height.
A solid sphere has ratio $2/5$, a solid cylinder $1/2$, and a hoop $1$.

$$
% caption: Released from the same height, a solid sphere, solid cylinder, and hoop accelerate as $a=g\sin\theta/(1+I/MR^2)$; the smaller inertia ratio reaches the bottom first.
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$$

Static friction supplies the torque required to increase angular speed. On an
incline, it acts uphill for a body released from rest. It does no work on a fixed
surface because the instantaneous contact point has zero displacement. The force
is nevertheless essential; without it, the rolling constraint cannot be maintained.

The contact-point condition follows from vector addition. The rim velocity relative
to the centre has magnitude $R\omega$ and is tangent to the rim. At bottom it
opposes centre translation exactly; at top it reinforces it.

Energy partition depends only on the dimensionless inertia ratio
$\beta=I/(MR^2)$. The rotational fraction is $\beta/(1+\beta)$ and translational
fraction is $1/(1+\beta)$, which isolates the shape dependence.

Force and torque equations determine static-friction direction. With positive
downhill direction, $Mg\sin\theta-f_s=Ma$ and $f_sR=I\alpha=Ia/R$. A body
released from rest requires positive angular acceleration, so friction points
uphill. This conclusion changes for a driven or braking wheel.

The race result follows from one parameter, $I/(MR^2)$, when mass, radius, and
height change are equal. It is not caused by a heavier body exerting greater
downhill force; mass cancels from the acceleration equation.

Differentiating the no-slip condition requires a fixed surface and a consistent
choice of signs. The centre-of-mass acceleration and angular acceleration have
equal magnitudes after multiplication by radius, but they point in different
geometric directions in a planar diagram.

Without friction, a body initially sliding down an incline has no torque about its
centre from gravity or the normal force. Its angular speed remains constant while
translation accelerates, so the rolling constraint cannot develop from rest.

**The rolling constraint as a relative-velocity condition.**

The no-slip condition belongs to the two surfaces at their point of contact. A
material point on the rim of the wheel has the same instantaneous velocity as the
material point on the surface that it touches. On a stationary horizontal floor,
the centre has forward velocity and the rotational contribution at the bottom rim
point has equal backward velocity. Their vector sum is zero. With a signed
coordinate chosen so that forward translation and clockwise rotation are positive,
the scalar condition is $v_{\rm cm}=R\omega$. The sign convention must be stated;
using an anticlockwise-positive angular coordinate changes the corresponding
algebraic sign without changing the physical contact condition.

This is a condition on velocity, not a statement that the contact material has
zero acceleration or stays at one location on the floor. A different rim particle
reaches contact a moment later. The contact point on the wheel has nonzero
acceleration in general, and it cannot automatically be used as a fixed pivot in
Newton's second-law equations. For a wheel moving right on level ground, the top
rim point has speed $2v_{\rm cm}$ and the bottom point has zero speed only at that
instant. The velocity field follows by adding the centre-of-mass translation to
the rotational velocity about the centre.

Integrating the velocity condition gives the displacement form
$s=R\theta$ when the radius is constant and the surface is fixed. It links an
encoder reading to travelled distance and gives the acceleration relation
$a_{\rm cm}=R\alpha$ while rolling continues. A changing radius, an elastic tire,
or any slip invalidates this simple conversion. An odometer based on wheel turns
therefore measures distance accurately only when the effective rolling radius is
known and the tire does not slide relative to the road.

The condition changes on a moving surface. If a belt moves right at speed $u$,
the bottom point of a rolling wheel must have ground-frame velocity $u$, not zero.
The appropriate scalar equation is $v_{\rm cm}-R\omega=u$ under the same sign
choice. Conveyor-belt problems and rolling on a translating vehicle require this
relative-velocity condition. Friction can be static even though the contact point is
moving in the laboratory frame; static friction requires zero relative velocity,
not zero velocity in a particular reference frame.

**Energy partition and the inertia ratio.**

In a rigid body rolling without slip, the translational and rotational kinetic
energies are tied to the same centre speed. Write the moment of inertia as
$I=\beta MR^2$, where the dimensionless number $\beta$ records how far the mass
lies from the rotation axis. Substitution of $\omega=v_{\rm cm}/R$ gives total
kinetic energy $K=(1/2)Mv_{\rm cm}^2(1+\beta)$. The translational fraction is
$1/(1+\beta)$ and the rotational fraction is $\beta/(1+\beta)$. A hoop has
$\beta=1$, so its kinetic energy is divided equally. A uniform solid cylinder has
$\beta=1/2$, so one third is rotational. A uniform solid sphere has
$\beta=2/5$, so its rotational share is smaller still.

Release from height $h$ on a fixed track gives
$Mgh=(1/2)Mv_{\rm cm}^2(1+\beta)$ when rolling resistance and air drag are
negligible. Thus $v_{\rm cm}^2=2gh/(1+\beta)$. For a drop of
$0.80\ \mathrm{m}$, a uniform cylinder has speed
$3.23\ \mathrm{m\,s^{-1}}$ at the bottom, whereas a hoop has speed
$2.80\ \mathrm{m\,s^{-1}}$. The common mass cancels, and the radius cancels
when the bodies have the same shape ratio. The difference arises from the energy
needed to establish angular speed, not from a difference in gravitational force.

On a stationary rigid surface, ideal static friction does no mechanical work on
the rolling body because its point of application has zero instantaneous velocity
in the ground frame. Static friction can nevertheless be essential: it supplies
the torque that changes angular speed and enforces the kinematic constraint. The
energy equation remains valid because no energy is dissipated at the contact. A
moving belt is different. Its contact point has nonzero laboratory-frame velocity,
so static friction can transfer energy between belt and wheel without slip.

Energy accounting also separates rolling from sliding. When a wheel skids,
kinetic friction acts at a contact with nonzero relative speed and converts
mechanical energy into internal energy. Applying the no-slip energy formula
during that interval overestimates the final speed. The calculation sequence
therefore identifies the contact regime first, imposes the rolling constraint only
for the no-slip stage, and uses an energy loss term whenever kinetic friction is
present. The same sequence applies to a wheel that begins in a skid and later
settles into pure rolling.

$$
% caption: The ratio $\beta=I/MR^2$ fixes how rolling kinetic energy splits: a solid cylinder keeps one third in rotation, a hoop one half. A larger rotational share means a slower centre at the same height.
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**Static friction follows the impending slip.**

Static friction has no universal direction for a rolling wheel. Its direction is
set by the relative motion that would occur at contact if friction were removed.
A body released from rest on an incline has its centre accelerated downhill by
the slope but has no torque about the centre. The bottom rim would therefore tend
to slide downhill relative to the track. Static friction acts uphill, producing
the rotational acceleration needed for rolling. The uphill friction force does
not oppose the centre-of-mass motion; it opposes the local sliding tendency.

For a body with $I=\beta MR^2$ rolling down an incline of angle $\theta$, the
force and torque equations give $a=g\sin\theta/(1+\beta)$ and
$f_s=Mg\sin\theta\,\beta/(1+\beta)$ directed uphill. A $2.0\ \mathrm{kg}$ uniform
cylinder ($\beta=1/2$) on a $20^\circ$ incline needs a static-friction magnitude
of $2.24\ \mathrm{N}$, so contact stays static only if
$\mu_s\geq f_s/(Mg\cos\theta)=0.121$. This is a threshold, not an equality: the
friction equals $\mu_sN$ only at impending slip, not throughout rolling.

Axle driving reverses the usual intuition. A motor torque that tries to spin a
wheel clockwise on level ground would make its bottom rim move backward relative
to the road if translation did not respond. The road then exerts forward static
friction on the wheel. That forward force accelerates the centre of mass while
the motor supplies the rotation. During braking, a torque that reduces the
clockwise spin can make the contact tend to slide forward; static friction then
points backward. A wheel moving right does not determine the friction direction
by itself. The applied torque, the slope, and the current spin rate must all be
included.

A direct diagnostic is to solve the translational and rotational equations using
an assumed positive friction direction. A negative result means the physical
friction direction is opposite to the assumed arrow. The procedure avoids changing
signs halfway through an equation set. If the required magnitude exceeds
$\mu_sN$, static friction cannot enforce $v_{\rm cm}=R\omega$; the wheel slips
and kinetic friction points against the actual relative motion at contact. The
rolling constraint must then be removed until a new no-slip state is established.

$$
% caption: Static-friction direction follows the tendency to slip. A wheel released on an incline needs uphill friction; an axle-driven wheel needs forward friction to couple motor torque to translation; a braking wheel needs backward friction.
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An extended incline calculation begins by separating the force balance from the
torque balance. Take the positive tangential direction down the slope and let a
rigid body start from rest. Its weight component is positive, while the static
friction force is uphill and therefore negative in the translational equation.
A body with moment of inertia $I=\beta MR^2$ has the centre-of-mass equation
$Ma=Mg\sin\theta-f_s$. About the centre, the only torque along the rotation axis
comes from static friction, giving $f_sR=I\alpha$ when the positive angular
direction matches the rolling sense. The rolling condition imposes
$a=R\alpha$. The three statements describe different parts of the mechanics and
should not be replaced by a single memorized acceleration formula.

Substitution gives $f_s=\beta Ma$ and then
$a=g\sin\theta/(1+\beta)$. The corresponding friction magnitude is
$f_s=Mg\sin\theta\,\beta/(1+\beta)$, directed uphill. The sign matters: an
uphill contact force reduces the translational acceleration below the frictionless
value while generating the rotation required for no-slip motion. A reader who
places friction downhill in this release problem obtains a torque in the wrong
direction and predicts a wheel that translates and spins incompatibly. The force
does not slow the total conversion of gravitational potential energy; it changes
the partition between translation and rotation.

> **Worked example.** A $2.4\ \mathrm{kg}$ uniform solid cylinder of radius
> $0.18\ \mathrm{m}$ is released on an $18^\circ$ incline. With $\beta=1/2$, the
> acceleration is $a=g\sin18^\circ/(1+\beta)=2.02\ \mathrm{m\,s^{-2}}$ and the
> required friction is $2.43\ \mathrm{N}$ uphill. A five-metre run from rest takes
> $t=\sqrt{2s/a}=2.22\ \mathrm{s}$ and produces a centre speed
> $4.49\ \mathrm{m\,s^{-1}}$ and an angular speed
> $\omega=v/R=24.9\ \mathrm{rad\,s^{-1}}$. The radius sets the final angular speed
> but not the acceleration, because for geometrically similar bodies a larger
> radius raises both the rim speed at a given $\omega$ and the moment of inertia by
> the same squared length scale.
>
> Two checks confirm the result. The vertical drop over the path is
> $5\sin18^\circ=1.55\ \mathrm{m}$, and energy gives
> $v^2=2gh/(1+\beta)=20.2\ \mathrm{m^2\,s^{-2}}$, matching the kinematic
> $v^2=2as$. The no-slip requirement also sets a minimum coefficient of static
> friction, $\mu_s\geq f_s/(Mg\cos18^\circ)=0.108$; a coefficient below this cannot
> maintain rolling.

The equations change at a contact transition. Each contact state has its own
constraint, force law, and energy account.

| contact state | relations used in the calculation | observable or inequality that selects the state | energy account on a fixed track |
|---|---|---|---|
| pure rolling | $v_{\rm cm}=R\omega$, $a=R\alpha$, and $\lvert f_s\rvert\leq\mu_sN$ | $v_{\rm slip}=v_{\rm cm}-R\omega=0$ within measurement uncertainty | Static friction has zero instantaneous power at the stationary contact. |
| kinetic slip | Solve translation and rotation independently; $f_k$ opposes $v_{\rm slip}$. | $v_{\rm slip}\ne0$ after the static-friction requirement is exceeded | Kinetic friction converts mechanical energy to internal energy. |
| loss of contact | $N=0$ and contact friction vanishes; gravity supplies the centre-of-mass acceleration. | On a circular crest, $N=M(g-v^2/\rho)$ reaches zero. | In the absence of air torque, spin kinetic energy remains constant during flight. |
| rolling resistance with no slip | Add $\tau_{rr}$, or its equivalent $F_{rr}=\tau_{rr}/R$, to the mechanical model. | A fitted energy loss persists even while $v_{\rm slip}=0$. | For nearly constant resistance, $\Delta E_{\rm mech}\simeq-F_{rr}d$. |

The first row gives the incline result above. The second and third rows
remove the no-slip constraint because there is either sliding contact or no
contact. The final row retains the constraint while introducing a torque that
represents deformation or bearing loss; it does not turn static friction into a
dissipative force.

Rolling resistance is a separate physical effect from ideal static friction. A
real tire or wheel deforms near the contact region. The normal-pressure resultant
is displaced slightly from the geometric contact line, producing a resisting
torque opposite to rotation. Bearings and internal material hysteresis can add
further resisting torques. A convenient low-speed model writes the resistance as
a torque $\tau_{rr}$ or as an equivalent force $F_{rr}=\tau_{rr}/R$ opposite to
the rolling direction. This force removes mechanical energy over distance even
when the wheel does not slide. It is often approximately proportional to normal
load over a limited speed range, but its coefficient is not the coefficient of
static friction.

On level ground with no drive torque, a rolling-resistance force produces both a
translation deceleration and a rotational deceleration. The detailed division
depends on where the contact resultant acts and on bearing torque; treating the
resistance as a single force through the centre misses its rotational effect.
An energy measurement is usually cleaner: over distance $d$, the mechanical
energy change is approximately $-F_{rr}d$ when resistance is nearly constant.
By contrast, ideal static friction at a stationary contact has zero instantaneous
power on the wheel. It may be nonzero in the force and torque equations, yet it
does not by itself reduce the total mechanical energy. Confusing these two roles
leads to the false claim that every friction force must dissipate energy.

Finite static friction sets the first contact-limit test. The rolling solution
requires the force found from the coupled equations to satisfy
$|f_s|\leq\mu_sN$. The normal force is not always $Mg\cos\theta$. On a curved
surface it also depends on the required normal acceleration, and a changing
normal load changes the available static-friction bound. A wheel can therefore
begin a trajectory in pure rolling and later slip even with a constant material
coefficient. The comparison must be repeated wherever the slope, curvature,
speed, drive torque, or normal load changes substantially.

Loss of normal contact is a separate failure condition. For a wheel rolling over
a convex circular crest, let $\rho$ be the radius of curvature of the centre path.
At the highest point, the inward direction is downward and the normal-force
balance gives $N=M(g-v^2/\rho)$. Contact requires $N\geq0$; the limiting speed is
$v=\sqrt{g\rho}$. Above that speed the track would need to pull downward on the
wheel, which an ordinary unilateral contact cannot do. The wheel then leaves the
surface and follows projectile motion until another contact occurs. The rolling
constraint, the contact torque, and the static-friction calculation all end at
the instant of separation.

A finite contact patch also has practical limits before complete separation. A
large drive or braking torque shifts the pressure distribution across the patch.
If the resultant reaches an edge, the rigid point-contact approximation no longer
describes the load distribution reliably. Pneumatic tires can deform, rock, or
develop partial slip while the centre still follows the surface. High-speed
experiments therefore monitor normal force, slip ratio, and wheel angular speed
together. A match between $v_{\rm cm}$ and $R\omega$ alone does not prove that
the ideal static-contact model remains valid under a rapidly changing load.

Once slip or separation begins, the earlier constraint cannot be imposed as an
extra equation. During slip, kinetic friction is set by the direction of relative
motion and the translational and rotational accelerations must be solved
independently. During separation, there is neither normal force nor contact
friction, so gravity determines the centre trajectory while angular momentum
about the centre remains constant if air torque is negligible. A correct solution
marks the transition time, matches velocity and angular speed continuously, and
applies the rolling formula only before the transition.

An equivalent-force resistance model extends the incline calculation when contact
remains no-slip. Use a generalized resisting force
$F_{rr}$ opposite to increasing path length, representing the rolling-resistance
torque divided by radius together with any approximately constant bearing loss.
The combined translational and rotational equations then give
$(M+I/R^2)a=Mg\sin\theta-F_{rr}$. For the solid cylinder considered above, the
effective inertial coefficient is $M+I/R^2=1.5M=3.6\ \mathrm{kg}$. If the
equivalent resistance is $0.45\ \mathrm{N}$, the acceleration on the same
$18^\circ$ slope becomes $1.90\ \mathrm{m\,s^{-2}}$, rather than
$2.02\ \mathrm{m\,s^{-2}}$. The reduction is modest because the downhill weight
component is about $7.27\ \mathrm{N}$, but its accumulated energy loss grows
linearly with distance.

This compact model has a defined scope. It predicts centre acceleration and total
energy loss, but it does not determine the pressure distribution in the contact
patch. A separate torque model is needed when the location of the normal resultant
or a bearing torque matters. The distinction is relevant for a bicycle coasting
downhill: aerodynamic drag grows strongly with speed, rolling resistance changes
more slowly, and bearing loss may be nearly constant over a narrow speed range.
Combining all three into one constant friction force can fit a short data segment
while failing outside that range. Each loss mechanism should be identified before
a coefficient is extrapolated.

Contact failure can also be detected from measured kinematics before separation
is visually obvious. Define a signed slip speed as the tangential velocity of the
wheel material at contact relative to the surface. On fixed level ground it is
$v_{\rm slip}=v_{\rm cm}-R\omega$ under the adopted convention. Pure rolling
requires this quantity to remain zero within measurement uncertainty. A sustained
nonzero value identifies slip; a rapidly fluctuating value can indicate tire
deformation, sensor timing error, or intermittent contact. Normal-load data adds
the second check: a near-zero load makes any friction-based rolling prediction
fragile because the available static-friction bound also approaches zero.