Physics · Rolling motion

Rolling without slipping

For rolling without slipping on stationary ground, centre-of-mass speed equals rolling radius times angular speed.

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Formula

v_{CM} = R ω

$v_{\mathrm{CM}}$ in $\mathrm{m\,s^{-1}}$ ; R in m; $\omega$ in $\mathrm{rad\,s^{-1}}$ .

Variables & units

Variables & units
Symbol	Meaning	Unit
$vCM$	Nonnegative centre-of-mass speed relative to stationary ground	m/s
$R$	Constant rolling radius of the rigid circular body	m
$ω$	Nonnegative angular speed \|ωz\|	rad/s

LaTeX

v_{\mathrm{CM}}=R\omega

QUANTA STUDYquanta-study.de

Wheel of radius 0.280 m rotating clockwise, with positive z out of the page. In the ground frame, centre velocity is 1.680 m/s rightward; the top arrow is twice as long. Bottom contact velocity is zero. Below, equal translation and rotation arrows cancel. — On stationary ground, rotation cancels translation at contact. With R = 0.280 m, ωz = −6 rad/s and vCM = 1.680 m/s, the top point moves at 3.360 m/s; contact velocity is vC = 0. This does not imply aC = 0.

Rolling without slipping calculator

Find the speed, angular velocity or radius of a wheel rolling without slipping, on stationary ground or on a conveyor belt.

Calculation model

Find

Positive rolling radius R

Distance from the wheel centre to the tread, e.g. 35 cm for a 28-inch bicycle tyre.

Velocity of the wheel centre v (right +)

Velocity of the wheel centre, positive to the right, e.g. 18 km/h for a bicycle.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for ω, substitute, evaluate

1. Rearrange for ω
$\omega_z=-\frac{v_{\mathrm{CM},x}}R$
2. Substitute values (in SI units)
- R = 35 cm = 0.35 m
- vCM = 18 km/h = 5 m/s
$-\frac{5}{0.35}$
3. Result
ω = -14.29 rad/s

Result: Angular velocity ω (anticlockwise +)

ω = -14.29 rad/s

With R = 35 cm and vCM = 18 km/h, ω = -14.29 rad/s.

Quick tips

Sign check on stationary ground: a wheel rolling to the right turns clockwise, so ω is negative. The size follows from v = R · |ω|. On a moving belt the rule does not always hold.
On stationary ground the contact point is momentarily at rest, and the top of the rim moves twice as fast as the wheel centre.

Assumptions and limits

Rigid wheel on a straight surface: x points right, z out of the image, positive ωz is counterclockwise. No slip requires vCM,x+Rωz=Vsurface. This kinematic condition does not determine friction, contact acceleration or drive dynamics. The surface is stationary: Vsurface=0.

Derivation & background

The short form $v_{\mathrm{CM}}=R\omega$ uses nonnegative magnitudes on stationary ground. With x rightward and z out of the plane, the signed relation is $v_{\mathrm{CM},x}=-R\omega_z$ . Generally, $v_{C,x}=v_{\mathrm{CM},x}+R\omega_z$ must match surface velocity. Static friction can maintain this constraint but is zero in ideal uniform rolling without further horizontal forces. Zero contact velocity means neither zero acceleration nor permanent attachment of that material point.

Exam blueprint

Validity range

Rigid circular body with constant rolling radius on a straight surface; no tangential slip. $v_{\mathrm{CM}}=R\omega$ uses nonnegative speeds and stationary ground. A moving surface needs the relative contact constraint; any required static friction must be available but can also be zero.

Derivation steps

Without slipping, contact velocity matches surface velocity; translation and rotation add at contact.

1Choose x rightward, z out of the plane and a rigid circular body with $R>0$ .
2Each material point has $\vec v_P=\vec v_{\mathrm{CM}}+\vec\omega\times\vec r_{P/\mathrm{CM}}$ . Positive $\omega_z$ means counterclockwise.
3At bottom contact $\vec r=(0,-R,0)$ , giving the rotational x contribution $+R\omega_z$ .
4Without slipping, $v_{C,x}=v_{\mathrm{CM},x}+R\omega_z=V_{\mathrm{surface}}$ .
5Stationary ground has $V_{\mathrm{surface}}=0$ , so $v_{\mathrm{CM},x}=-R\omega_z$ .
6Taking magnitudes gives $v_{\mathrm{CM}}=R|\omega_z|$ ; the upper short form calls this nonnegative magnitude $\omega$ .
7On a moving belt, $v_{\mathrm{CM},x}-V_{\mathrm{belt}}=-R\omega_z$ ; contact velocity vanishes only relative to the belt.

Rearrangements

Angular velocity

|\omega_z|=\frac{v_{\mathrm{CM}}}R

Magnitudes on stationary ground, no slip and positive radius. Use the signed contact equation for direction.

Task variant

A rigid wheel of $R=0.30\,\mathrm m$ rolls without slipping on stationary ground at angular speed $4.0\,\mathrm{rad\,s^{-1}}$ . Find centre-of-mass speed.

Approach: $v_{\mathrm{CM}}=R|\omega_z|$ .
Substitution gives $v_{\mathrm{CM}}=1.20\,\mathrm{m\,s^{-1}}$ .

Check: $\frac{v_{\mathrm{CM}}}R=4.0\,\mathrm{rad\,s^{-1}}$ ; magnitude alone does not specify direction.

Common mistakes

Equating contact and centre velocity, or using the stationary-ground rule on a moving belt.

Check $v_{C,x}=v_{\mathrm{CM},x}+R\omega_z=V_{\mathrm{surface}}$ . Contact is at rest only instantaneously; its acceleration need not vanish.

Exam context

Connect rolling wheels, cylinders and spheres in motion and energy problems.

Formula cluster

Rolling motion

Related relations: rotationsenergie-starrer-koerper, drehimpuls-starrer-koerper, winkelgeschwindigkeit-kreisfrequenz.

Erot = ½Iω²

Rotational kinetic energy of a rigid body

L = Iω

Angular momentum of a rigid body

ω = 2πf; v = ω·r

Angular Velocity and Angular Frequency

Worked example

A wheel with $R=0.30\,\mathrm m$ and $\omega=4.0\,\mathrm{rad\,s^{-1}}$ rolls without slipping at $v_{\mathrm{CM}}=1.20\,\mathrm{m\,s^{-1}}$ .

Applications

Connect rolling wheels, cylinders and spheres in motion and energy problems.

How do you calculate with Rolling without slipping?

Here is how to work through a typical Rolling without slipping (vCM = Rω) task step by step:

1
Task
A rigid wheel of $R=0.30\,\mathrm m$ rolls without slipping on stationary ground at angular speed $4.0\,\mathrm{rad\,s^{-1}}$ . Find centre-of-mass speed.
Solution path
1. Approach: $v_{\mathrm{CM}}=R|\omega_z|$ .
2. Substitution gives $v_{\mathrm{CM}}=1.20\,\mathrm{m\,s^{-1}}$ .
Check: $\frac{v_{\mathrm{CM}}}R=4.0\,\mathrm{rad\,s^{-1}}$ ; magnitude alone does not specify direction.

Understand and apply the equation

In rolling without slipping, rotation matches centre-of-mass motion relative to the surface. For a circular body on stationary ground this gives the familiar magnitude relation $v_{\mathrm{CM}}=R|\omega_z|$ . With signed quantities or on a conveyor belt, contact-point velocity reveals whether the condition actually holds.

What does “without slipping” mean for a wheel?

A wheel moves forward while rotating. At the ground the two velocity contributions can cancel: the material point touching it is momentarily at rest. This is the physical basis of the rolling constraint. If the surface itself moves, its velocity must be included. Check contact motion before calculating forces or energy.[1]

Prior knowledge: velocity, radius and rotation direction. The centre of mass is the mass-weighted centre; for a uniform wheel it lies at the geometric centre.

The surface sets contact-point velocity

v_{\mathrm{CM}}=R|\omega_z|

The upper short form uses $\omega$ as nonnegative angular speed. Here $v_{\mathrm{CM}}$ is speed relative to stationary ground.[1]

A rigid circular body can translate and rotate simultaneously. Without slipping, the material point instantaneously at contact has the same tangential velocity as the surface. On stationary ground its velocity is zero. Other body points have different velocities: the contact is neither a permanently fixed material point nor the centre of mass.[2]

Velocity from bottom to top

QUANTA STUDYquanta-study.de

Stationary ground,

R=0.280\,\mathrm{m}

,

v_{\mathrm{CM},x}=1.680\,\mathrm{m\,s^{-1}}

and

\omega_z=-6\,\mathrm{rad\,s^{-1}}

. Horizontal velocity is zero at the bottom, 1.680

\mathrm{m\,s^{-1}}

at the centre and 3.360

\mathrm{m\,s^{-1}}

at the top. y is height relative to the centre; lateral points also have a vertical component.[2]

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Add translation and rotation at contact

\mathbf v_P=\mathbf v_{\mathrm{CM}}+\boldsymbol\omega\times\mathbf r_{P/\mathrm{CM}}

A body point velocity combines centre-of-mass translation and rotation about the centre of mass.[2]

Choose x to the right and z out of the diagram. Positive $\omega_z$ means counterclockwise. At the bottom contact the relative position vector is $(0,-R,0)$ , so the cross product gives x component $+R\omega_z$ . Contact speed is therefore $v_{\mathrm{CM},x}+R\omega_z$ . Setting it to zero on stationary ground gives $v_{\mathrm{CM},x}=-R\omega_z$ . Taking magnitudes then gives $v_{\mathrm{CM}}=R|\omega_z|$ .[2]

\begin{aligned}v_{C,x}&=v_{\mathrm{CM},x}+R\omega_z=V_{\mathrm{surface}}\\v_{\mathrm{CM},x}-V_{\mathrm{surface}}&=-R\omega_z\end{aligned}

This contact equation also applies to a straight moving surface; all velocities must use the same frame.[2]

Rearrange magnitudes and directions carefully

\begin{aligned}|\omega_z|&=\frac{v_{\mathrm{CM}}}{R}\\R&=\frac{v_{\mathrm{CM}}}{|\omega_z|}\\\omega_z&=-\frac{v_{\mathrm{CM},x}-V_{\mathrm{surface}}}{R}\end{aligned}

$R>0$ ; rearranging for radius additionally requires $|\omega_z|$ >0. The first two forms assume stationary ground.

Radius in m times angular speed in $\mathrm{rad\,s^{-1}}$ gives $\mathrm{m\,s^{-1}}$ . With frequency in revolutions per second, $v=2\pi Rf$ , not $Rf$ . Inferring radius from $\frac v\omega$ is meaningful only without slipping; otherwise the discrepancy is not a new geometric radius. Tyre deformation, a noncircular contour or changing contact may require a separately defined effective rolling radius.[1]

Why the upper rim is faster

\begin{aligned}v_{\mathrm{top},x}&=v_{\mathrm{CM},x}-R\omega_z=2v_{\mathrm{CM},x}\\v_{C,x}&=0\end{aligned}

On stationary ground, rotation reinforces translation at the top and cancels it at the bottom.[2]

The upper equation is an instantaneous velocity statement. The material point now at the ground leaves that location immediately afterwards. In uniform rolling, centre-of-mass acceleration and $\alpha$ vanish, yet this contact material point has upward centripetal acceleration $R\omega^2$ . Thus $v_C=0$ does not imply $a_C=0$ . Accelerated motion also includes tangential rotational acceleration.[2]

a_{\mathrm{CM},x}=-R\alpha_z

For sustained rolling on a straight stationary surface with constant R, differentiation gives the centre-of-mass acceleration condition. It does not describe contact material point acceleration.[1]

Stationary ground and moving belt

A wheel rolls uniformly to the right

Given: A rigid wheel of $R=0.280\,\mathrm{m}$ rolls without slipping on stationary ground. $v_{\mathrm{CM},x}=+1.680\,\mathrm{m\,s^{-1}}$ stays constant for 4.00 s. +z points out of the plane.
Find: Find $\omega_z$ , rim velocities, distance and contact-point acceleration.

Approach: The stationary support requires zero contact velocity. $v_{\mathrm{Kontakt},x}=v_{\mathrm{CM},x}+R\omega_z=0$ . The signed contact condition determines rotation direction.
$\omega_z=-\frac{1.680}{0.280}=-6.00\,\mathrm{rad/s}$
Add rotational contributions to translation at top and bottom.
$\begin{aligned}v_{\mathrm{top},x}&=3.360\,\mathrm{m/s}\\v_{C,x}&=0\end{aligned}$
The centre of mass travels a distance; angular travel counts rotation magnitude.
$\begin{aligned}s&=1.680(4)=6.720\,\mathrm m\\|\Delta\theta|&=6(4)=24.0\,\mathrm{rad}\end{aligned}$
At $\alpha=0$ , centripetal acceleration of the bottom material point remains.
$a_{C,y}=R\omega_z^2=0.280(36)=10.08\,\mathrm{m/s^2}$

Result: $\omega_z=-6.00\,\mathrm{rad\,s^{-1}}$ , clockwise. Top speed is 3.360 $\mathrm{m\,s^{-1}}$ , bottom speed instantaneously zero; travel is 6.720 m and angular travel 24.0 rad. The bottom material point accelerates upward at 10.08 $\mathrm{m\,s^{-2}}$ .

Check: Independently, $s=R|\Delta\theta|=0.280\,\mathrm m\cdot24.0=6.720\,\mathrm m$ . Rotation rate is $\frac{|\omega_z|}{2\pi}\cdot60\,\mathrm{s/min}\approx57.296\,\mathrm{min^{-1}}$ , and the number of turns is $\frac{24}{2\pi}\approx3.820$ . At a lateral rim point, two velocity contributions of 1.680 $\mathrm{m\,s^{-1}}$ are perpendicular; the magnitude is $\sqrt2\cdot1.680\,\mathrm{m/s}\approx2.376\,\mathrm{m/s}$ .

The same principle on a conveyor belt

Given: A wheel of $R=0.200\,\mathrm{m}$ contacts a straight belt moving at Vbelt=+0.500 $\mathrm{m\,s^{-1}}$ . Its centre moves at +1.100 $\mathrm{m\,s^{-1}}$ in the laboratory; no slipping and both speeds constant.
Find: Find $\omega_z$ and laboratory velocities of contact and upper rim.

Approach: For a moving support equate contact velocity to belt velocity. $v_{\mathrm{Kontakt},x}=v_{\mathrm{CM},x}+R\omega_z=v_{\mathrm{Band}}$ . Subtract surface speed from centre speed.
$\omega_z=-\frac{1.100-0.500}{0.200}=-3.00\,\mathrm{rad/s}$
In the laboratory contact must move with the belt.
$v_{C,x}=1.100+0.200(-3)=0.500\,\mathrm{m/s}$
At the top the rotational contribution has opposite sign.
$v_{\mathrm{top},x}=1.100-0.200(-3)=1.700\,\mathrm{m/s}$

Result: In the laboratory, $\omega_z=-3.00\,\mathrm{rad\,s^{-1}}$ , $v_{C,x}=+0.500\,\mathrm{m\,s^{-1}}$ and $v_{\mathrm{top},x}=+1.700\,\mathrm{m\,s^{-1}}$ . The stationary-ground relation would be wrong here: $1.100\ne0.600$ .

Check: In the uniformly co-moving belt frame, centre, contact and upper velocities are 0.600, 0 and 1.200 m/s. There $v_{\mathrm{CM},\mathrm{rel}}=R|\omega_z|$ and $v_{\mathrm{top},\mathrm{rel}}=2v_{\mathrm{CM},\mathrm{rel}}$ hold again. Angular velocity is unchanged by this pure uniform translation of the reference frame.

A rolling constraint is not a force balance

Static friction can enforce rolling when a tangential force is needed. Required force must satisfy $|F_{\mathrm{static}}|\le\mu_sN$ . Its direction follows dynamics and impending relative motion, not a universal rule opposing centre-of-mass travel. An ideal body can roll uniformly on horizontal ground with no other horizontal forces even when static friction is zero.[1][2]

K=\frac12 Mv_{\mathrm{CM}}^2+\frac12 I_{\mathrm{CM}}\omega^2=\frac12(I_{\mathrm{CM}}+MR^2)\omega^2

The last equality applies to rolling on stationary ground. Moment of inertia about the contact axis already includes both energy contributions.[2]

Do not add $\frac12Mv_{\mathrm{CM}}^2$ again to the contact-axis form; that would count translation twice. On a moving belt, laboratory kinetic energy still uses the separate centre-of-mass form with laboratory velocities. The rolling constraint does not replace translational or rotational dynamics and alone determines neither drive force nor acceleration.[2]

Static and kinetic friction must match actual contact motion: the static limit discussed here can permit rolling without slipping, whereas slip requires a different friction model. The rolling condition alone gives neither the required static force nor its direction.[1]

Rotational energy of a rigid body is only one of the two terms in this section’s centre-of-mass expression. Use it for intrinsic rotation about the centre of mass and keep translation separate; the contact-axis expression on stationary ground already includes both contributions.[1]

Five contact checks

Calculate · Foundation

A circular wheel of $R=0.150\,\mathrm{m}$ rolls without slipping on stationary ground; $|\omega_z|=8.00\,\mathrm{rad\,s^{-1}}$ . Find centre-of-mass speed.

Show hint

The magnitude form needs no additional direction specification here.

Learning goal: Find centre speed from rolling condition.

Show solution and calculation

Approach: On stationary ground use $v_{\mathrm{CM}}=R|\omega_z|$ .
Substitution gives $v_{\mathrm{CM}}=0.150\cdot8.00\,\mathrm{m\,s^{-1}}=1.20\,\mathrm{m\,s^{-1}}$ .

Check: $\frac{v_{\mathrm{CM}}}R=8.00\,\mathrm{rad\,s^{-1}}$ . Magnitude alone does not determine rotation direction.

v_{\mathrm{CM}}=1.20\,\mathrm{m/s}

Calculate · Application

On stationary ground $R=0.120\,\mathrm{m}$ and $\omega_z=+5.00\,\mathrm{rad\,s^{-1}}$ , counterclockwise. Find $v_{\mathrm{CM},x}$ and $v_{\mathrm{top},x}$ without slipping.

Show hint

Use signed relation $v_{\mathrm{CM},x}=-R\omega_z$ .

Learning goal: Relate rotation sign to point velocities.

Show solution and calculation

Approach: With x rightward and z out of the page, use $v_{\mathrm{CM},x}=-R\omega_z$ ; the top has $v_{\mathrm{top},x}=2v_{\mathrm{CM},x}$ .
Hence $v_{\mathrm{CM},x}=-0.600\,\mathrm{m\,s^{-1}}$ and $v_{\mathrm{top},x}=-1.20\,\mathrm{m\,s^{-1}}$ .
The wheel rolls left, with positive centre-of-mass speed magnitude.

Check: at the bottom, $v_{C,x}=-0.600+0.120\cdot5.00=0\,\mathrm{m\,s^{-1}}$ .

\begin{aligned}v_{\mathrm{CM},x}&=-0.600\,\mathrm{m/s}\\v_{\mathrm{top},x}&=-1.20\,\mathrm{m/s}\end{aligned}

Calculate · Application

A wheel on stationary ground has $R=0.250\,\mathrm{m}$ , $v_{\mathrm{CM},x}=+2.00\,\mathrm{m\,s^{-1}}$ and $\omega_z=-6.00\,\mathrm{rad\,s^{-1}}$ . Check rolling and give required $\omega_z$ without slipping.

Show hint

First calculate contact velocity in the ground frame.

Learning goal: Check slipping and find required rotation.

Show solution and calculation

Approach: Check $v_{C,x}=v_{\mathrm{CM},x}+R\omega_z$ ; stationary ground requires zero for rolling.
Substitution gives $v_{C,x}=2.00+0.250(-6.00)=+0.500\,\mathrm{m\,s^{-1}}$ , so contact slides right.
Without slipping, $\omega_z=-\frac{v_{\mathrm{CM},x}}R=-8.00\,\mathrm{rad\,s^{-1}}$ is required.

Check: that value cancels the two contact contributions.

\begin{aligned}v_{C,x}&=+0.500\,\mathrm{m/s}\\\omega_{z,\mathrm{no\ slip}}&=-8.00\,\mathrm{rad/s}\end{aligned}

Interpret · Application

A wheel of $R=0.20\,\mathrm{m}$ moves at $v_{\mathrm{CM},x}=+1.0\,\mathrm{m\,s^{-1}}$ in the lab on a belt at $v_{\mathrm{belt}}$ =+2.0 $\mathrm{m\,s^{-1}}$ . +z points out of the page and +x right. What $\omega_z$ permits rolling without slipping? Which way does it rotate?

Show hint

Contact velocity must equal belt velocity.

Learning goal: Check rolling relative to a moving belt.

Show solution and calculation

Approach: At bottom contact the rotational contribution in x is R times $\omega_z$ .
Equating total contact velocity to 2 $\mathrm{m\,s^{-1}}$ gives $\omega_z=+5\,\mathrm{rad\,s^{-1}}$ , counterclockwise.
Although the centre moves right, rotation is positive here.
The familiar clockwise rule assumes stationary ground.

\begin{aligned}v_{C,x}&=v_{\mathrm{CM},x}+R\omega_z=v_{\mathrm{belt}}\\{}\omega_z&=\frac{2.0-1.0}{0.20}=+5\,\mathrm{rad\,s^{-1}}\end{aligned}

[1]

Error analysis · Application

A wheel of $R=0.20\,\mathrm{m}$ rolls uniformly on stationary ground at $|\omega|=4\,\mathrm{rad\,s^{-1}}$ . Someone infers from zero contact velocity that the material point currently there also has zero acceleration and stays on the ground. Check both claims.

Show hint

Instantaneous velocity and its time derivative are different.

Learning goal: Avoid treating instantaneous rest as a stationary material point.

Show solution and calculation

Approach: Uniform centre motion gives zero centre acceleration, but each rim point has rotational acceleration toward the centre.
At the bottom it points upward with magnitude 3.2 $\mathrm{m\,s^{-2}}$ .
That point subsequently leaves contact while another material point touches the ground.
Instantaneous rest therefore implies neither zero acceleration nor permanent attachment at the same ground location.

a_C=R\omega^2=0.20\cdot4^2=3.2\,\mathrm{m\,s^{-2}}

[1]

When a different contact model is needed

The simple relation assumes a rigid circular body of radius R and planar contact on a straight surface. With slipping, tangential contact velocities differ. Deformation or rolling resistance needs additional contact and energy assumptions; three-dimensional rolling on curved surfaces needs suitable vector geometry. The example with $R=0.30\,\mathrm m$ and angular speed 4 $\mathrm{rad\,s^{-1}}$ remains correct on stationary ground: $v_{\mathrm{CM}}=1.2\,\mathrm{m\,s^{-1}}$ .[1][2]

What to remember

Always check contact velocity relative to the surface. On stationary ground, rotation cancels centre motion at the bottom and reinforces it at the top. On a belt, include belt motion. The constraint relates velocities; required friction, acceleration and energy need additional balances.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Linked relevant learning transitions; preserved existing derivations, worked examples, exercises and figures in full.
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Quanta. Rolling without slipping. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rollbedingung-ohne-rutschen

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Frequently asked questions about Rolling without slipping

What does Rolling without slipping describe physically?+

The rolling condition means that contact instantaneously has no slip relative to the surface. For a rigid circular body on stationary ground, nonnegative speeds satisfy $v_{\mathrm{CM}}=R|\omega_z|$ . The upper short form labels $|\omega_z|$ as ω. With x rightward and z out of the plane, the signed equation instead is $v_{\mathrm{CM},x}=-R\omega_z$ . This minus sign connects rightward travel with clockwise rotation. The contact material point is at rest only instantaneously, not permanently fixed. Other body points move differently: translation and rotation add at the top. The constraint alone does not specify which force causes or maintains the motion.

How can I check a calculation involving Rolling without slipping?+

Calculate contact velocity by adding translation and rotation. With R=0.280 m and $v_{\mathrm{CM},x}$ =+1.680 m/s on stationary ground, ωz=−6.00 rad/s. Thus $v_{C,x}=1.680+0.280(-6.00)=0\,\mathrm{m\,s^{-1}}$ . At the top, the rotational contribution reverses sign, giving $v_{\mathrm{top},x}$ =3.360 m/s. A distance check after 4.00 s gives s=6.720 m and angular travel 24.0 rad; radius times angular travel also gives 6.720 m. Despite instantaneous zero contact velocity, that material point has upward acceleration $R\omega^2=10.08\,\mathrm{m\,s^{-2}}$ in uniform rolling. Do not confuse contact velocity with contact material point acceleration; they describe different aspects of motion.

What assumptions and limits apply to Rolling without slipping?+

The simple radius relation assumes a rigid circular body, constant rolling radius and no tangential slip. Magnitude form $v_{\mathrm{CM}}=R\omega$ uses centre speed relative to stationary ground. On a moving belt, contact velocity must match belt speed: $v_{\mathrm{CM},x}+R\omega_z=V_{\mathrm{belt}}$ . Static friction can maintain this condition; required force must not exceed $\mu_sN$ . However, ideal uniform rolling with no other horizontal forces needs no nonzero static friction. Deformation, rolling resistance or a noncircular contour require additional models. Zero contact velocity does not make that material point a permanently fixed rotation axis or automatically justify every torque balance about it.

How do I check units and scale for Rolling without slipping?+

R in m and ω in rad/s give v in m/s. Frequency f in revolutions per second needs factor 2π: $v=2\pi Rf$ on stationary ground. First identify whether values are magnitudes or signed components. Positive R cannot make nonnegative speeds negative; direction follows the signed contact equation. With R=0.250 m, $v_{\mathrm{CM},x}$ =+2.00 m/s and ωz=−6.00 rad/s, contact velocity is +0.500 m/s, indicating slip. Rearranging $\frac v\omega$ would not provide a valid geometric radius measurement here. A simple scaling check doubles centre speed when R doubles at the same angular speed, provided the rolling model continues to hold.

When should I combine Rolling without slipping with another relation?+

For acceleration on straight stationary ground with constant R, differentiate $v_{\mathrm{CM},x}=-R\omega_z$ to obtain $a_{\mathrm{CM},x}=-R\alpha_z$ . Dynamics additionally needs $\sum F_x=Ma_{\mathrm{CM},x}$ and torque balance about the centre of mass. For energy use $K=\frac12Mv_{\mathrm{CM}}^2+\frac12I_{\mathrm{CM}}\omega^2$ . On stationary ground this equals $\frac12I_{\mathrm{contact}}\omega^2$ , where $I_{\mathrm{contact}}=I_{\mathrm{CM}}+MR^2$ . Do not add translation energy again to the contact-axis form, or the same contribution is counted twice. On a conveyor belt the separate laboratory-energy expression with laboratory velocities remains reliable; the simple contact-axis form no longer applies unchanged. Sliding requires actual friction dynamics and must not be forced to satisfy the rolling constraint.

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F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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Rolling without slipping

Variables & units

Rolling without slipping calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Rolling motion

Worked example

Applications

How do you calculate with Rolling without slipping?

What does “without slipping” mean for a wheel?

The surface sets contact-point velocity

Velocity from bottom to top

Add translation and rotation at contact

Rearrange magnitudes and directions carefully

Why the upper rim is faster

Stationary ground and moving belt

A wheel rolls uniformly to the right

The same principle on a conveyor belt

A rolling constraint is not a force balance

Five contact checks

A circular wheel of R=0.150 mR=0.150\,\mathrm{m}R=0.150m rolls without slipping on stationary ground; ∣ωz∣=8.00 rad s−1|\omega_z|=8.00\,\mathrm{rad\,s^{-1}}∣ωz​∣=8.00rads−1. Find centre-of-mass speed.

On stationary ground R=0.120 mR=0.120\,\mathrm{m}R=0.120m and ωz=+5.00 rad s−1\omega_z=+5.00\,\mathrm{rad\,s^{-1}}ωz​=+5.00rads−1, counterclockwise. Find vCM,xv_{\mathrm{CM},x}vCM,x​ and vtop,xv_{\mathrm{top},x}vtop,x​ without slipping.

When a different contact model is needed

What to remember

Sources for this explanation

Frequently asked questions about Rolling without slipping

Subject references and further reading

More formulas: Mechanics

Retain Rolling without slipping for exams

A circular wheel of $R=0.150\,\mathrm{m}$ rolls without slipping on stationary ground; $|\omega_z|=8.00\,\mathrm{rad\,s^{-1}}$ . Find centre-of-mass speed.

On stationary ground $R=0.120\,\mathrm{m}$ and $\omega_z=+5.00\,\mathrm{rad\,s^{-1}}$ , counterclockwise. Find $v_{\mathrm{CM},x}$ and $v_{\mathrm{top},x}$ without slipping.