Rolling Motion

Rolling motion combines two kinds of motion at the same time:

• translational motion — the object's centre of mass moves forward
• rotational motion — the object rotates about its centre

A bicycle wheel, car tire, bowling ball, rolling cylinder, and soccer ball can all undergo rolling motion.

For an object rolling without slipping, translational speed and angular speed are connected by:

v = rω

where:

• v = speed of the centre of mass in m/s
• r = radius of the rolling object in m
• ω = angular velocity in rad/s

This relationship is one of the most important equations for rolling motion.

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Rolling Without Slipping

An object rolls without slipping when the point touching the ground is instantaneously at rest relative to the ground.

As the wheel completes one revolution, it moves forward by a distance equal to its circumference:

d = 2πr

During the same revolution, its angular displacement is:

Δθ = 2π rad

Therefore:

d = rΔθ

More generally:

x = rθ

Differentiating this relationship gives:

v = rω

and:

a = rα

for the tangential acceleration associated with rolling.

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Translational and Rotational Motion Together

Consider a wheel rolling along a road.

The centre of the wheel moves forward with speed v.

At the same time, the wheel rotates with angular velocity ω.

These motions are linked:

v = rω

If the wheel rolls faster:

• v increases
• ω increases

If the radius is larger, the same angular velocity produces a greater translational speed.

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Worked Example: Finding Linear Speed

A wheel has:

r = 0.40 m

ω = 15 rad/s

Using:

v = rω

v = (0.40)(15)

v = 6.0 m/s

The centre of the wheel moves forward at 6.0 m/s.

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Finding Angular Velocity

Rearrange:

ω = v/r

A bicycle wheel has:

r = 0.35 m

and the bicycle travels at:

v = 7.0 m/s

Then:

ω = 7.0/0.35

ω = 20 rad/s

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Effect of Wheel Radius

Suppose two wheels roll at the same translational speed.

Wheel A:

r = 0.20 m

Wheel B:

r = 0.50 m

If:

v = 10 m/s

then for Wheel A:

ω = 10/0.20

ω = 50 rad/s

For Wheel B:

ω = 10/0.50

ω = 20 rad/s

The smaller wheel must rotate faster to maintain the same translational speed.

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Revolutions and Distance

Each complete revolution moves a rolling wheel forward by one circumference:

distance per revolution = 2πr

Therefore:

d = N(2πr)

where N is the number of revolutions.

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Worked Example: Distance Travelled

A wheel has:

r = 0.30 m

and completes:

N = 20 revolutions

Circumference:

2πr = 2π(0.30)

≈ 1.885 m

Distance:

d = 20(1.885)

d ≈ 37.7 m

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Finding the Number of Revolutions

Rearrange:

N = d/(2πr)

Suppose a bicycle travels:

d = 500 m

with wheels of radius:

r = 0.35 m

Then:

N = 500/[2π(0.35)]

N ≈ 227

The wheels complete approximately:

227 revolutions

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Rotational Frequency

If a wheel completes f revolutions per second:

ω = 2πf

Since:

v = rω

then:

v = 2πrf

This gives another useful relationship between rolling speed and rotational frequency.

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Worked Frequency Example

A wheel has:

r = 0.25 m

and rotates at:

f = 8.0 Hz

Then:

v = 2πrf

v = 2π(0.25)(8)

v ≈ 12.6 m/s

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Converting rpm to Rolling Speed

Rotating wheels are often described using revolutions per minute.

Suppose a wheel rotates at:

600 rpm

First convert to revolutions per second:

f = 600/60

f = 10 Hz

Then:

ω = 2πf

ω = 20π

ω ≈ 62.8 rad/s

For a wheel with:

r = 0.20 m

v = rω

v = (0.20)(62.8)

v ≈ 12.6 m/s

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Speeds of Different Points on a Rolling Wheel

An interesting feature of rolling motion is that different points on the wheel have different speeds relative to the ground.

The centre moves at:

v

The rotational speed of the rim relative to the centre is also:

rω = v

when rolling without slipping.

These motions combine differently at different locations.

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Speed at the Top of the Wheel

At the top of the wheel, translational and rotational velocities point in the same direction.

Therefore:

v_top = v + rω

Since:

rω = v

then:

v_top = 2v

The top of a rolling wheel moves twice as fast as its centre relative to the ground.

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Speed at the Centre

The centre of the wheel has only the translational velocity:

v_centre = v

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Speed at the Bottom

At the contact point, the rotational velocity is opposite the translational velocity.

Therefore:

v_bottom = v − rω

For rolling without slipping:

rω = v

so:

v_bottom = 0

The point touching the ground is instantaneously stationary relative to the ground.

This does not mean that the wheel itself stops.

A moment later, that point has moved away from the ground and another point becomes the contact point.

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Example: Speeds Around a Wheel

A wheel rolls at:

v = 8.0 m/s

Then:

Top:

v_top = 16 m/s

Centre:

v_centre = 8.0 m/s

Bottom:

v_bottom = 0 m/s

These values are all measured relative to the ground.

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The Path of a Point on the Rim

A point on the edge of a rolling wheel follows a curved path called a cycloid.

The point:

• briefly stops at the ground
• accelerates upward and forward
• reaches maximum speed at the top
• moves downward and forward
• returns to zero speed at the next contact point

This explains why a rolling wheel is more complicated than simply rotating or simply translating.

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Acceleration in Rolling Motion

If a wheel speeds up while rolling without slipping:

a = rα

where:

• a = translational acceleration of the centre
• α = angular acceleration

For a fixed radius:

greater α → greater a

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Worked Acceleration Example

A wheel has:

r = 0.50 m

and:

α = 6.0 rad/s²

Then:

a = rα

a = (0.50)(6.0)

a = 3.0 m/s²

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Rolling Down an Incline

When a round object rolls down a slope, gravity causes its centre of mass to accelerate downhill while the object also gains angular velocity.

The object's gravitational potential energy becomes:

• translational kinetic energy
• rotational kinetic energy

For ideal rolling:

mgh = 1/2 mv² + 1/2 Iω²

Using:

ω = v/r

we obtain:

mgh = 1/2 mv² + 1/2 I(v²/r²)

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Why Shape Matters

Moment of inertia depends on mass distribution.

Common examples include:

Objects with more mass concentrated farther from the axis have larger moments of inertia.

That affects how gravitational energy is divided between translational and rotational motion.

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General Speed Equation Down a Hill

Start with:

mgh = 1/2 mv² + 1/2 I(v²/r²)

Factor:

mgh = 1/2 v²(m + I/r²)

Therefore:

v² = 2mgh/(m + I/r²)

or:

v = √[2mgh/(m + I/r²)]

This equation can be used for any rigid object rolling without slipping when I is known.

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Solid Sphere Rolling Down a Hill

For a solid sphere:

I = 2/5 mr²

Substitute:

mgh = 1/2 mv² + 1/2(2/5 mr²)(v²/r²)

mgh = 1/2 mv² + 1/5 mv²

mgh = 7/10 mv²

Therefore:

v = √(10gh/7)

Notice that mass and radius cancel.

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Worked Sphere Example

A solid sphere rolls from rest through a vertical height:

h = 2.5 m

Using:

g = 9.8 m/s²

v = √[(10)(9.8)(2.5)/7]

v = √35

v ≈ 5.92 m/s

If:

r = 0.20 m

then:

ω = v/r

ω = 5.92/0.20

ω ≈ 29.6 rad/s

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Solid Cylinder Rolling Down a Hill

For a solid cylinder:

I = 1/2 mr²

Energy conservation gives:

mgh = 1/2 mv² + 1/4 mv²

mgh = 3/4 mv²

Therefore:

v = √(4gh/3)

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Thin Ring Rolling Down a Hill

For a thin ring:

I = mr²

Then:

mgh = 1/2 mv² + 1/2 mv²

mgh = mv²

Therefore:

v = √(gh)

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Which Object Rolls Fastest?

For objects released from rest at the same height:

Solid sphere:

v = √(10gh/7)

Solid cylinder:

v = √(4gh/3)

Thin ring:

v = √(gh)

Therefore:

solid sphere → fastest

solid cylinder → next

thin ring → slowest

The object with the smaller rotational inertia relative to mr² uses a smaller fraction of its energy for rotation and gains more translational speed.

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Mass Does Not Determine the Winner

In the ideal rolling equations, mass cancels.

A 2 kg solid sphere and a 20 kg solid sphere released from the same height reach the same final speed, provided:

• both roll without slipping
• air resistance is negligible
• rolling resistance is negligible
• they have the same shape and mass distribution

Mass alone does not determine which object reaches the bottom first.

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The Role of Friction

Friction plays an essential role in rolling motion.

Without sufficient friction, an object may:

• slide instead of roll
• spin without moving properly
• fail to satisfy v = rω

For many rolling situations, static friction provides the torque required for rotational acceleration.

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Static Friction in Rolling

For rolling without slipping, the surfaces at the contact point do not slide relative to each other.

Therefore, the friction is generally static friction.

Static friction can produce torque without necessarily dissipating mechanical energy.

This is an important distinction.

It is incorrect to assume that:

friction always means mechanical energy is lost.

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Why Friction Is Needed on an Incline

Consider a solid ball released from rest on a slope.

Gravity acts through the centre of mass.

Therefore, gravity by itself produces no torque about the centre.

Static friction at the contact point produces the torque needed to make the ball rotate.

The ball therefore:

• accelerates downhill
• angularly accelerates
• rolls without slipping

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Direction of Friction

The direction of static friction is determined by the tendency to slip, not simply by the direction the object is moving.

For a passive object rolling down an incline from rest, static friction commonly acts up the slope.

Why?

Without friction, gravity would cause the object to slide downhill without enough rotation.

The contact point would tend to slip downhill.

Static friction acts uphill to oppose that tendency and provides the torque needed for rotation.

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Friction and Torque

For a rolling object on an incline:

τ = fr

where f is the static friction force.

Newton's Second Law for rotation gives:

fr = Iα

Since:

a = rα

then:

α = a/r

Therefore:

fr = I(a/r)

so:

f = Ia/r²

This shows directly how friction produces angular acceleration.

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Translational Dynamics on an Incline

For an object rolling downhill:

mg sin θ − f = ma

while rotationally:

fr = Iα

and:

a = rα

These equations can be combined to find the acceleration.

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Acceleration of a Rolling Object

Substituting:

f = Ia/r²

into:

mg sin θ − f = ma

gives:

mg sin θ − Ia/r² = ma

Therefore:

a = g sin θ / [1 + I/(mr²)]

This equation shows how moment of inertia affects acceleration.

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Solid Sphere on an Incline

For a solid sphere:

I = 2/5 mr²

Therefore:

a = g sin θ / (1 + 2/5)

a = g sin θ /(7/5)

a = 5/7 g sin θ

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Solid Cylinder on an Incline

For a solid cylinder:

I = 1/2 mr²

Therefore:

a = g sin θ /(1 + 1/2)

a = 2/3 g sin θ

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Thin Ring on an Incline

For a thin ring:

I = mr²

Therefore:

a = g sin θ /(1 + 1)

a = 1/2 g sin θ

Again:

sphere accelerates fastest

cylinder next

ring slowest

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Worked Incline Example

A solid cylinder rolls without slipping down a:

30°

incline.

Its acceleration is:

a = 2/3 g sin θ

a = 2/3(9.8)(sin 30°)

a = 2/3(9.8)(0.5)

a ≈ 3.27 m/s²

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What Happens If Friction Is Too Small?

Static friction has a maximum value:

f_s ≤ μ_sN

If the friction required for rolling exceeds the maximum available static friction, the object begins to slip.

Once slipping occurs:

v ≠ rω

and the simple rolling-without-slipping relationships no longer apply.

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Rolling with Slipping

A wheel can both translate and rotate while slipping.

There are two common possibilities.

Too Much Translation

If:

v > rω

the wheel is moving forward faster than its rotation corresponds to.

The contact point tends to slide forward relative to the ground.

Too Much Rotation

If:

v < rω

the wheel is spinning too quickly for its forward speed.

The contact point tends to slide backward relative to the ground.

Kinetic friction acts to oppose the relative sliding.

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Car Accelerating

When a driven car wheel accelerates, the engine applies torque to the wheel.

The tire tends to push the road backward.

The road exerts static friction forward on the tire.

That forward friction force accelerates the vehicle.

This is an important example showing that friction does not always oppose the motion of an object's centre of mass.

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Car Braking

During braking, the brakes apply torque that reduces the wheel's angular velocity.

If the tire continues rolling without slipping, static friction helps produce the required translational deceleration.

If the brakes are strong enough to lock the wheel:

ω = 0

while the vehicle still has:

v > 0

The tire then skids.

Kinetic friction acts between the tire and road.

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Anti-Lock Braking Systems

An anti-lock braking system (ABS) helps prevent wheels from locking during hard braking.

Keeping the wheels near rolling conditions allows the tires to maintain better control and steering ability than when they are fully skidding.

Rolling and friction therefore have important safety applications in transportation.

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Rolling Resistance

Even when a wheel rolls without obvious slipping, real systems can lose mechanical energy.

Tires and surfaces deform slightly.

Bearings experience friction.

Air resistance acts on the vehicle.

These effects create rolling resistance.

Therefore, real rolling motion is not perfectly energy conserving.

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Energy Losses

In real rolling systems, mechanical energy can be transferred into:

• thermal energy
• sound
• deformation
• air motion

A more realistic energy equation may be:

E_initial = E_final + E_losses

For example:

mgh = 1/2 mv² + 1/2 Iω² + E_losses

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Worked Example with Energy Loss

A rolling object begins with:

500 J

of gravitational potential energy.

During its descent:

80 J

is transferred into thermal energy and other forms.

At the bottom:

E_mechanical = 500 − 80

E_mechanical = 420 J

Therefore:

1/2 mv² + 1/2 Iω² = 420 J

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Rolling Wheels on Vehicles

Vehicle wheels demonstrate the connection between linear and angular motion.

For rolling without slipping:

v_vehicle = rω_wheel

If the vehicle speed increases, the wheels must rotate faster.

If a vehicle uses smaller wheels at the same road speed, those wheels must rotate at a greater angular velocity.

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Worked Vehicle Example

A car travels at:

v = 25 m/s

Its tires have radius:

r = 0.32 m

Angular velocity:

ω = v/r

ω = 25/0.32

ω ≈ 78.1 rad/s

Convert to revolutions per second:

f = ω/(2π)

f ≈ 12.4 Hz

Convert to rpm:

rpm = 12.4 × 60

rpm ≈ 746 rpm

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Bicycle Wheels

A bicycle travelling at a particular speed provides another straightforward rolling system.

Suppose:

v = 9.0 m/s

r = 0.35 m

Then:

ω = 9.0/0.35

ω ≈ 25.7 rad/s

The frequency is:

f = ω/(2π)

f ≈ 4.09 Hz

Therefore:

rpm ≈ 245 rpm

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Rolling and Gears

Rolling-type contact also occurs between gears.

At the point where two ideal gears mesh, their tangential speeds match:

r₁ω₁ = r₂ω₂

Therefore:

ω₁/ω₂ = r₂/r₁

A smaller gear rotates faster than a larger gear.

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Pulley and Belt Systems

For an ideal belt that does not slip:

v_belt = r₁ω₁ = r₂ω₂

This is essentially the same no-slip relationship used for rolling wheels.

If one pulley is larger, it rotates more slowly for the same belt speed.

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Yo-Yos and Spools

Rolling principles also appear in:

• yo-yos
• spools
• reels
• winches

These systems can be more complicated because forces may act at different radii.

The direction of friction may also depend on how and where the force is applied.

The basic principles remain:

• translation follows ΣF = ma
• rotation follows Στ = Iα
• no-slip motion provides a relationship between linear and angular quantities

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Rolling and Angular Momentum

A rolling object has angular momentum associated with its rotation.

About its centre of mass:

L = Iω

Since:

ω = v/r

we can also write:

L = Iv/r

The amount of rotational angular momentum therefore depends on:

• speed
• radius
• moment of inertia

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Rolling and Total Kinetic Energy

For rolling without slipping:

E_total = 1/2 mv² + 1/2 Iω²

Since:

ω = v/r

then:

E_total = 1/2 mv² + 1/2 I(v²/r²)

Factor:

E_total = 1/2 v²(m + I/r²)

This equation makes it easy to compare rolling objects with different moments of inertia.

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Complete Rolling Problem

A solid cylinder has:

m = 8.0 kg

r = 0.25 m

It rolls without slipping at:

v = 5.0 m/s

Angular Velocity

ω = v/r

ω = 5.0/0.25

ω = 20 rad/s

Moment of Inertia

For a solid cylinder:

I = 1/2 mr²

I = 1/2(8)(0.25²)

I = 4(0.0625)

I = 0.25 kg·m²

Translational Kinetic Energy

E_trans = 1/2 mv²

E_trans = 1/2(8)(5²)

E_trans = 100 J

Rotational Kinetic Energy

E_rot = 1/2 Iω²

E_rot = 1/2(0.25)(20²)

E_rot = 50 J

Total Kinetic Energy

E_total = 100 + 50

E_total = 150 J

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Complete Incline Problem

A solid sphere rolls without slipping from rest down a slope with vertical height:

h = 4.0 m

Radius:

r = 0.20 m

Find its speed and angular velocity at the bottom.

For a solid sphere:

I = 2/5 mr²

Energy conservation gives:

v = √(10gh/7)

Substitute:

v = √[(10)(9.8)(4.0)/7]

v = √56

v ≈ 7.48 m/s

Then:

ω = v/r

ω = 7.48/0.20

ω ≈ 37.4 rad/s

Notice that the object's mass was not needed.

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Complete Dynamics Problem

A solid sphere rolls without slipping down a:

25°

incline.

For a solid sphere:

a = 5/7 g sin θ

Therefore:

a = 5/7(9.8)(sin 25°)

a ≈ 2.96 m/s²

If the sphere starts from rest and travels:

d = 6.0 m

use:

v² = v₀² + 2ad

v² = 2(2.96)(6)

v² ≈ 35.5

v ≈ 5.96 m/s

If:

r = 0.15 m

then:

ω = v/r

ω = 5.96/0.15

ω ≈ 39.7 rad/s

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Solving Rolling Motion Problems

A useful first question is:

Is the object rolling without slipping?

If yes, use:

v = rω

and, when appropriate:

a = rα

Then identify the type of problem.

Speed and Rotation

Use:

v = rω

Revolutions and Distance

Use:

d = 2πrN

Rolling Energy

Use:

E_total = 1/2 mv² + 1/2 Iω²

Rolling Down a Hill

Use:

mgh = 1/2 mv² + 1/2 Iω²

Dynamics on an Incline

Use:

ΣF = ma

Στ = Iα

and:

a = rα

Real Systems with Losses

Include energy transferred to:

• heat
• sound
• deformation

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Common Mistakes

Using v = rω when the object is slipping

This equation applies specifically to rolling without slipping.

Forgetting translation

A rolling object is not merely rotating. Its centre of mass is also moving.

Forgetting rotation

A rolling object's kinetic energy is not just 1/2 mv².

Assuming the contact point moves at v

For pure rolling, the contact point is instantaneously stationary relative to the ground.

Thinking every point on the wheel has the same speed

Relative to the ground, the top moves at 2v, the centre at v, and the contact point at zero.

Assuming friction always acts backward

Static friction acts against the tendency of surfaces to slip. Its direction depends on the situation.

Assuming friction always wastes energy

Static friction in ideal rolling does not necessarily dissipate mechanical energy.

Ignoring moment of inertia

Mass distribution affects rolling acceleration and energy.

Assuming heavier objects always roll faster

For ideal objects of the same shape, mass cancels from the rolling equations.

Using degrees in rotational relationships

Angular quantities used with equations such as x = rθ should use radians.

Forgetting to convert rpm

Convert rpm into rad/s when using v = rω.

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Key Terms

Rolling motion — motion involving simultaneous translation of an object's centre of mass and rotation about its centre.

Rolling without slipping — rolling condition in which the contact point is instantaneously stationary relative to the surface.

Translational motion — movement of an object's centre of mass from one position to another.

Rotational motion — motion in which an object turns around an axis.

Angular velocity (ω) — rate of change of angular position.

Translational velocity (v) — velocity of the object's centre of mass.

Angular acceleration (α) — rate of change of angular velocity.

Translational acceleration (a) — rate of change of translational velocity.

Moment of inertia (I) — measure of how an object's mass is distributed relative to its rotation axis.

Static friction — friction acting between surfaces that are not sliding relative to each other.

Kinetic friction — friction acting when surfaces slide relative to each other.

Rolling resistance — resistance to rolling caused by effects such as deformation, bearing friction, and surface interactions.

Contact point — point where a rolling object touches the surface.

Cycloid — curved path followed by a point on the rim of an ideal rolling wheel.

No-slip condition — relationship v = rω connecting translational and rotational motion.

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Key Takeaways

• Rolling motion combines translation and rotation.
• For rolling without slipping, v = rω.
• The corresponding acceleration relationship is a = rα when the no-slip condition applies.
• One complete revolution moves a wheel forward by 2πr.
• The smaller the wheel, the greater its angular velocity must be at the same translational speed.
• Relative to the ground, the centre of a rolling wheel moves at v.
• The top of the wheel moves instantaneously at 2v.
• The contact point moves instantaneously at 0 m/s during pure rolling.
• Rolling objects possess both translational and rotational kinetic energy.
• Their total kinetic energy is 1/2 mv² + 1/2 Iω².
• Moment of inertia affects rolling acceleration and energy distribution.
• A solid sphere rolls down an ideal incline faster than a solid cylinder or thin ring released under comparable conditions.
• Mass alone does not determine rolling acceleration for geometrically similar ideal objects.
• Static friction can provide the torque required for rolling.
• Static friction does not necessarily dissipate mechanical energy.
• If there is insufficient static friction, the object slips.
• During slipping, v ≠ rω.
• Real rolling systems lose useful mechanical energy through rolling resistance, deformation, bearing friction, and air resistance.
• Rolling principles apply to wheels, vehicles, bicycles, gears, pulleys, balls, cylinders, spools, and many engineering systems.

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Check Your Understanding

1. Describe what is meant by rolling without slipping.
2. A wheel of radius 0.40 m rotates at 12 rad/s while rolling without slipping. Calculate its translational speed.
3. A bicycle travels at 8.0 m/s on wheels of radius 0.32 m. Calculate the angular velocity of the wheels.
4. A wheel of radius 0.25 m completes 50 revolutions. Calculate the distance it travels.
5. A car travels at 20 m/s on tires with radius 0.30 m. Calculate the angular velocity of the tires and their approximate rotational speed in rpm.
6. A wheel rolls without slipping at 6.0 m/s. Determine the instantaneous speeds relative to the ground of the top, centre, and bottom points of the wheel.
7. Explain why the contact point of a wheel can have an instantaneous speed of zero even though the wheel is moving forward.
8. A wheel has radius 0.50 m and angular acceleration 4.0 rad/s². Calculate the translational acceleration if it rolls without slipping.
9. Explain why static friction is often necessary for an object to roll down an incline without slipping.
10. A solid cylinder rolls without slipping from rest through a vertical height of 3.0 m. Calculate its speed at the bottom.
11. A solid sphere and a thin ring have the same mass and radius and are released from rest at the top of the same incline. Which reaches the bottom first? Explain using moment of inertia and energy.
12. A 6.0 kg solid cylinder of radius 0.30 m rolls without slipping at 5.0 m/s. Calculate its angular velocity, moment of inertia, translational kinetic energy, rotational kinetic energy, and total kinetic energy.
13. A solid sphere rolls down a 30° incline without slipping. Calculate its translational acceleration using a = 5/7 g sin θ.
14. A wheel is moving forward at 10 m/s, but its rim speed rω is only 7.0 m/s. Is it rolling without slipping? Describe the type of slipping occurring and the direction kinetic friction would tend to act on the wheel.
15. A 10 kg solid sphere of radius 0.25 m starts from rest and rolls without slipping down a hill with a vertical drop of 5.0 m. During the descent, 60 J of mechanical energy is transferred into thermal energy and other forms. Calculate the sphere's final translational speed and angular velocity.