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.

Inclined Planes and Rolling Objects

When an object rolls down an inclined plane, its gravitational potential energy is transformed into two forms of kinetic energy:

• translational kinetic energy as its centre of mass moves down the slope
• rotational kinetic energy as the object spins

For an object rolling without slipping:

v = rω

and:

a = rα

These relationships connect the object's translational and rotational motion.

The object's moment of inertia determines how its available energy is divided between translation and rotation. As a result, objects with the same mass and radius but different shapes can accelerate differently down the same incline.

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Forces on a Rolling Object

Consider a round object rolling down an incline at angle θ.

The important forces are:

• weight mg, acting vertically downward
• normal force N, perpendicular to the surface
• static friction f, acting along the surface

The component of gravity parallel to the incline is:

mg sin θ

The perpendicular component is:

mg cos θ

If there is no acceleration perpendicular to the surface:

N = mg cos θ

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Why Gravity Makes the Object Move Downhill

The component:

mg sin θ

pulls the centre of mass down the slope.

Newton's Second Law along the slope can be written:

mg sin θ − f = ma

for a passive object released from rest and rolling without slipping.

The friction term appears because static friction commonly acts uphill in this situation.

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Why Friction Is Needed for Rotation

Gravity acts through the object's centre of mass.

Therefore, gravity produces no torque about the centre.

The normal force also acts along a line through the centre for an ideal circular object.

Static friction at the contact point provides the torque:

τ = fr

This torque produces angular acceleration:

τ = Iα

Therefore:

fr = Iα

For rolling without slipping:

a = rα

so:

α = a/r

Substitute:

fr = I(a/r)

Therefore:

f = Ia/r²

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Deriving the Acceleration Down an Incline

Start with translational motion:

mg sin θ − f = ma

Substitute:

f = Ia/r²

giving:

mg sin θ − Ia/r² = ma

Factor out a:

mg sin θ = a(m + I/r²)

Therefore:

a = mg sin θ/(m + I/r²)

Dividing by m:

a = g sin θ/[1 + I/(mr²)]

This is one of the most important equations for rolling objects on inclined planes.

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What the Acceleration Equation Tells Us

The acceleration depends on:

• gravitational field strength g
• incline angle θ
• moment of inertia relative to mr²

A larger moment of inertia generally produces a smaller translational acceleration.

More of the object's response to gravity goes into increasing its rotational motion.

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Moment of Inertia and Shape

Different shapes distribute their mass differently.

A thin ring has all of its mass relatively far from its axis.

A solid sphere has more of its mass closer to the axis.

Therefore, the ring has a larger moment of inertia relative to mr².

---

Acceleration of a Solid Sphere

For a solid sphere:

I = 2/5 mr²

Substitute into:

a = g sin θ/[1 + I/(mr²)]

Then:

a = g sin θ/(1 + 2/5)

a = g sin θ/(7/5)

Therefore:

a = 5/7 g sin θ

---

Acceleration of a Solid Cylinder

For a solid cylinder:

I = 1/2 mr²

Therefore:

a = g sin θ/(1 + 1/2)

a = g sin θ/(3/2)

a = 2/3 g sin θ

---

Acceleration of a Thin Spherical Shell

For a thin spherical shell:

I = 2/3 mr²

Therefore:

a = g sin θ/(1 + 2/3)

a = 3/5 g sin θ

---

Acceleration of a Thin Ring

For a thin ring:

I = mr²

Therefore:

a = g sin θ/(1 + 1)

a = 1/2 g sin θ

---

Comparing the Objects

For the same incline:

Solid sphere:

a = 5/7 g sin θ ≈ 0.714g sin θ

Solid cylinder:

a = 2/3 g sin θ ≈ 0.667g sin θ

Thin spherical shell:

a = 3/5 g sin θ = 0.600g sin θ

Thin ring:

a = 1/2 g sin θ = 0.500g sin θ

Therefore:

solid sphere → greatest acceleration

solid cylinder → next

thin spherical shell → next

thin ring → smallest acceleration

---

The Rolling Race

Imagine releasing a solid sphere, solid cylinder, spherical shell, and thin ring simultaneously from rest at the top of the same incline.

Assuming they all roll without slipping:

1. solid sphere arrives first
2. solid cylinder arrives second
3. thin spherical shell arrives third
4. thin ring arrives last

The result depends on mass distribution, not simply mass.

---

Why the Sphere Wins

The solid sphere has the smallest value of:

I/(mr²)

among these objects.

It therefore requires a smaller fraction of its energy for rotation and gains translational speed more quickly.

The ring has a larger moment of inertia because its mass is concentrated farther from the axis.

More of its energy is associated with rotation, leaving less for translational motion at a given height loss.

---

Does Mass Matter?

Notice that mass cancels from:

a = g sin θ/[1 + I/(mr²)]

when I has the form:

I = kmr²

Then:

a = g sin θ/(1 + k)

Therefore, for geometrically similar ideal objects, acceleration does not depend on mass.

A 1 kg solid sphere and a 20 kg solid sphere have the same ideal rolling acceleration down the same incline.

---

Does Radius Matter?

Radius also cancels when:

I = kmr²

because:

I/(mr²) = k

Therefore, ideal rolling acceleration for geometrically similar objects does not depend directly on radius either.

A large solid sphere and a small solid sphere can have the same translational acceleration down the same incline.

Their angular accelerations will differ because:

α = a/r

The smaller sphere has the larger angular acceleration.

---

Worked Acceleration Example

A solid sphere rolls down a:

30°

incline.

For a solid sphere:

a = 5/7 g sin θ

Using:

g = 9.8 m/s²

and:

sin 30° = 0.5

a = 5/7(9.8)(0.5)

a = 3.5 m/s²

---

Comparing with a Sliding Object

If a frictionless block slides down the same 30° incline:

a = g sin θ

a = (9.8)(0.5)

a = 4.9 m/s²

The rolling sphere accelerates at only:

3.5 m/s²

because some of the gravitational effect produces rotational acceleration.

---

Cylinder on the Same Incline

For a solid cylinder:

a = 2/3 g sin θ

a = 2/3(9.8)(0.5)

a ≈ 3.27 m/s²

Therefore:

• sliding block: 4.90 m/s²
• solid sphere: 3.50 m/s²
• solid cylinder: 3.27 m/s²

---

Ring on the Same Incline

For a thin ring:

a = 1/2 g sin θ

a = 1/2(9.8)(0.5)

a = 2.45 m/s²

The ring has the smallest acceleration of these objects.

---

Energy Conservation on an Incline

Rolling problems can also be solved using conservation of energy.

Suppose an object begins from rest at vertical height h.

Initial energy:

E_i = mgh

At the bottom:

E_f = 1/2 mv² + 1/2 Iω²

Therefore:

mgh = 1/2 mv² + 1/2 Iω²

For rolling without slipping:

ω = v/r

so:

mgh = 1/2 mv² + 1/2 I(v²/r²)

---

General Final-Speed Equation

Starting with:

mgh = 1/2 mv² + 1/2 I(v²/r²)

factor out v²:

mgh = 1/2 v²(m + I/r²)

Multiply by 2:

2mgh = v²(m + I/r²)

Therefore:

v² = 2mgh/(m + I/r²)

Divide the numerator and denominator by m:

v² = 2gh/[1 + I/(mr²)]

Therefore:

v = √{2gh/[1 + I/(mr²)]}

This equation gives the speed after descending through a vertical height h.

---

Final Speed of a Solid Sphere

For a solid sphere:

I/(mr²) = 2/5

Therefore:

v² = 2gh/(1 + 2/5)

v² = 2gh/(7/5)

v² = 10gh/7

so:

v = √(10gh/7)

---

Final Speed of a Solid Cylinder

For a solid cylinder:

I/(mr²) = 1/2

Therefore:

v² = 2gh/(3/2)

v² = 4gh/3

so:

v = √(4gh/3)

---

Final Speed of a Thin Ring

For a thin ring:

I/(mr²) = 1

Therefore:

v² = 2gh/2

v² = gh

so:

v = √(gh)

---

Comparing Final Speeds

For the same vertical drop h:

The solid sphere again has the greatest translational speed.

---

Worked Energy Example

A solid cylinder rolls from rest down an incline through a vertical height:

h = 3.0 m

Its final speed is:

v = √(4gh/3)

v = √[(4)(9.8)(3.0)/3]

v = √39.2

v ≈ 6.26 m/s

If its radius is:

r = 0.25 m

then:

ω = v/r

ω = 6.26/0.25

ω ≈ 25.0 rad/s

---

Checking the Energy

Suppose the cylinder has:

m = 8.0 kg

Initial gravitational energy:

E_p = mgh

E_p = (8)(9.8)(3)

E_p = 235.2 J

Translational kinetic energy:

E_trans = 1/2 mv²

E_trans = 1/2(8)(6.26²)

E_trans ≈ 156.8 J

For a solid cylinder:

I = 1/2 mr²

Rotational kinetic energy for ideal rolling becomes:

E_rot = 1/4 mv²

Therefore:

E_rot ≈ 78.4 J

Total:

156.8 + 78.4 = 235.2 J

Energy is conserved.

---

How Energy Is Divided

The fraction of kinetic energy that becomes rotational energy depends on the object's moment of inertia.

For a solid cylinder:

E_trans = 1/2 mv²

E_rot = 1/4 mv²

Therefore, rotational energy is half as large as translational energy.

For a thin ring:

E_trans = 1/2 mv²

E_rot = 1/2 mv²

The energy is divided equally between translation and rotation.

---

Energy Distribution in a Solid Sphere

For a solid sphere:

I = 2/5 mr²

Therefore:

E_rot = 1/2(2/5 mr²)(v²/r²)

E_rot = 1/5 mv²

while:

E_trans = 1/2 mv²

The sphere puts a smaller proportion of its total kinetic energy into rotation than the ring does.

This helps explain why the sphere reaches a greater translational speed.

---

The Role of Static Friction

It may seem strange that friction is needed for rolling but mechanical energy can still be conserved.

For ideal rolling without slipping, the point of contact is instantaneously stationary relative to the surface.

Therefore, static friction does not normally dissipate mechanical energy at the contact point.

It provides the torque required for rotation without necessarily transforming mechanical energy into thermal energy.

---

Direction of Static Friction

For a passive round object released from rest on an incline, friction usually acts up the slope.

Without friction, gravity would accelerate the centre of mass downhill but would not provide enough torque about the centre to establish rolling.

The contact point would tend to slide downhill.

Static friction acts uphill and produces the torque needed for rotation.

---

Calculating the Friction Force

We found:

f = Ia/r²

If:

I = kmr²

then:

f = kma

Since:

a = g sin θ/(1 + k)

we obtain:

f = [k/(1 + k)]mg sin θ

for the ideal passive object rolling down the incline.

---

Friction on a Solid Sphere

For a solid sphere:

k = 2/5

Therefore:

f = [(2/5)/(7/5)]mg sin θ

f = 2/7 mg sin θ

The friction force acts uphill.

---

Friction on a Solid Cylinder

For a solid cylinder:

k = 1/2

Therefore:

f = [(1/2)/(3/2)]mg sin θ

f = 1/3 mg sin θ

---

Friction on a Thin Ring

For a thin ring:

k = 1

Therefore:

f = 1/2 mg sin θ

The ring requires a relatively large static-friction force to maintain rolling.

---

Is There Enough Friction?

Rolling without slipping requires the needed friction to remain below the maximum static friction:

f_required ≤ μ_sN

Since:

N = mg cos θ

the available maximum static friction is:

f_s,max = μ_smg cos θ

If the required friction exceeds this value, the object begins to slip.

---

Minimum Coefficient of Friction

For an object with:

I = kmr²

the required friction is:

f = [k/(1 + k)]mg sin θ

For no slipping:

[k/(1 + k)]mg sin θ ≤ μ_smg cos θ

Cancel mg:

[k/(1 + k)]sin θ ≤ μ_s cos θ

Therefore:

μ_s ≥ [k/(1 + k)]tan θ

This gives the minimum coefficient of static friction required for pure rolling.

---

Worked Friction Example

A solid cylinder rolls down a:

30°

incline.

For a solid cylinder:

k = 1/2

Therefore:

μ_s,min = [0.5/(1.5)]tan 30°

μ_s,min = 1/3(0.577)

μ_s,min ≈ 0.192

If the coefficient of static friction is greater than about 0.192, the cylinder can roll without slipping under these ideal conditions.

---

What Happens When the Surface Is Too Slippery?

If there is insufficient static friction:

• the object slides as it rotates
• v ≠ rω
• translational and rotational motion are no longer locked together
• kinetic friction may act
• mechanical energy may be transformed into thermal energy

The simple rolling-without-slipping equations cannot then be used without modification.

---

Steeper Inclines

From:

a = g sin θ/(1 + k)

a steeper incline produces a larger acceleration because sin θ increases.

However, the static friction required to maintain rolling also increases.

The minimum coefficient:

μ_s,min = [k/(1 + k)]tan θ

increases as the incline becomes steeper.

Therefore, an object that rolls without slipping on a gentle slope may begin to slip on a sufficiently steep slope.

---

Incline Length and Vertical Height

Energy calculations depend primarily on the vertical height change, not the length or angle of the path.

If an object drops through vertical height h:

ΔE_p = mgh

Two ramps with different angles but the same vertical drop provide the same gravitational potential-energy decrease.

However, the steeper ramp generally produces greater acceleration and a shorter travel time.

---

Finding Vertical Height from Ramp Length

If the ramp has length d and angle θ:

h = d sin θ

Therefore:

mgh = mgd sin θ

This can be useful when a problem gives ramp length instead of vertical height.

---

Worked Ramp Example

A solid sphere rolls from rest down a:

5.0 m

long ramp inclined at:

20°

Vertical height:

h = d sin θ

h = 5.0 sin 20°

h ≈ 1.71 m

For a solid sphere:

v = √(10gh/7)

v = √[(10)(9.8)(1.71)/7]

v ≈ 4.89 m/s

---

Solving the Same Problem with Kinematics

For a solid sphere:

a = 5/7 g sin θ

a = 5/7(9.8)(sin 20°)

a ≈ 2.39 m/s²

Using:

v² = v₀² + 2ad

and v₀ = 0:

v² = 2(2.39)(5.0)

v² ≈ 23.9

v ≈ 4.89 m/s

The energy and dynamics methods agree.

---

Energy Method vs Dynamics Method

Use the energy method when you need:

• final speed
• energy distribution
• effect of vertical height
• energy losses

Use the dynamics method when you need:

• acceleration
• friction force
• time
• forces and torques

Both methods describe the same physical system.

---

Time Down an Incline

If acceleration is constant and the object begins from rest:

d = 1/2 at²

Therefore:

t = √(2d/a)

Since objects with smaller I/(mr²) have greater acceleration, they also take less time to travel the same incline.

---

Worked Time Comparison

A solid sphere and thin ring start from rest on a:

4.0 m

long incline at:

30°

Solid Sphere

a = 5/7(9.8)(0.5)

a = 3.50 m/s²

t = √(8/3.50)

t ≈ 1.51 s

Thin Ring

a = 1/2(9.8)(0.5)

a = 2.45 m/s²

t = √(8/2.45)

t ≈ 1.81 s

The solid sphere reaches the bottom first.

---

Objects with the Same Mass and Radius

Suppose a ring and disk both have:

m = 5 kg

r = 0.30 m

Their moments of inertia are:

Ring:

I = mr²

I = (5)(0.30²)

I = 0.45 kg·m²

Disk:

I = 1/2 mr²

I = 0.225 kg·m²

The ring has twice the moment of inertia.

Therefore, it accelerates more slowly down the incline.

---

Mass Distribution Is More Important Than Total Mass

Two objects can have the same:

• total mass
• radius
• starting height

and still behave differently.

The important difference is where the mass is located relative to the axis.

Mass farther from the axis increases moment of inertia.

This makes the object harder to angularly accelerate.

---

Hollow and Solid Objects

A hollow object generally has more of its mass farther from its centre than a comparable solid object.

Therefore:

hollow object → larger I

and usually:

larger I → smaller rolling acceleration

For example, a solid sphere reaches the bottom before a thin spherical shell.

---

Real Rolling Objects

Real objects do not behave perfectly.

Energy can be transferred into:

• deformation
• thermal energy
• sound
• bearing friction
• air resistance

A real tire continually deforms where it contacts the road.

This contributes to rolling resistance.

---

Energy Conservation with Losses

If mechanical energy is transferred into other forms:

E_initial = E_final + E_losses

For an object rolling downhill:

mgh = 1/2 mv² + 1/2 Iω² + E_losses

This means the object's final speed will be lower than predicted by the ideal model.

---

Worked Problem with Energy Loss

A solid cylinder has:

m = 10 kg

and rolls from rest through:

h = 4.0 m

During the motion:

70 J

is transferred into thermal energy and other forms.

Initial gravitational energy:

E_i = mgh

E_i = (10)(9.8)(4)

E_i = 392 J

Mechanical energy remaining:

392 − 70 = 322 J

For a solid cylinder:

E_total = 3/4 mv²

Therefore:

322 = 3/4(10)v²

322 = 7.5v²

v² = 42.93

v ≈ 6.55 m/s

Without the energy loss:

v = √(4gh/3)

v ≈ 7.23 m/s

The losses reduce the final speed.

---

Vehicle on a Slope

Vehicle wheels also combine translation and rotation.

When a vehicle rolls downhill:

• gravitational potential energy decreases
• translational kinetic energy can increase
• rotational kinetic energy of the wheels can increase

Braking may transform some of this mechanical energy into thermal energy.

Regenerative braking can instead recover part of the energy as electrical energy.

---

Bicycle on a Hill

A bicycle and rider descending a hill involve several rotating components:

• front wheel
• rear wheel
• gears
• chain components
• pedals in some situations

A more complete energy model is:

mgh = 1/2 Mv² + Σ(1/2 Iω²) + E_losses

where the summation includes the rotating components.

This shows why rotating mass can matter in vehicle and bicycle performance.

---

Rolling Objects in Sports

Inclined rolling motion appears in:

• bowling
• golf
• skateboarding
• cycling
• wheelchair sports
• ball sports
• ramps and tracks

Understanding rolling motion helps explain how:

• surface friction
• ball construction
• mass distribution
• slope angle

affect motion.

---

Complete Dynamics Problem

A solid cylinder of mass:

m = 6.0 kg

and radius:

r = 0.20 m

rolls without slipping down a:

25°

incline.

Acceleration

For a solid cylinder:

a = 2/3 g sin θ

a = 2/3(9.8)(sin 25°)

a ≈ 2.76 m/s²

Friction

For a solid cylinder:

f = 1/3 mg sin θ

f = 1/3(6)(9.8)(sin 25°)

f ≈ 8.28 N

The friction acts uphill.

Angular Acceleration

α = a/r

α = 2.76/0.20

α ≈ 13.8 rad/s²

This single example connects:

• force
• torque
• translational acceleration
• angular acceleration

---

Complete Energy Problem

A solid sphere has:

m = 4.0 kg

r = 0.15 m

It starts from rest and rolls without slipping down a slope through:

h = 2.5 m

Initial Energy

E_i = mgh

E_i = (4)(9.8)(2.5)

E_i = 98 J

Final Speed

For a solid sphere:

v = √(10gh/7)

v = √[(10)(9.8)(2.5)/7]

v ≈ 5.92 m/s

Angular Velocity

ω = v/r

ω = 5.92/0.15

ω ≈ 39.5 rad/s

Translational Energy

E_trans = 1/2 mv²

E_trans = 1/2(4)(5.92²)

E_trans ≈ 70 J

Rotational Energy

For a solid sphere:

E_rot = 1/5 mv²

E_rot = 1/5(4)(5.92²)

E_rot ≈ 28 J

Energy Check

70 + 28 = 98 J

The initial gravitational potential energy has become translational and rotational kinetic energy.

---

Complete Comparison Problem

A solid sphere and thin ring start from rest at the same height:

h = 5.0 m

Solid Sphere

v = √(10gh/7)

v = √[(10)(9.8)(5)/7]

v ≈ 8.37 m/s

Thin Ring

v = √(gh)

v = √[(9.8)(5)]

v = 7.00 m/s

The sphere reaches the bottom with the greater translational speed because its smaller relative moment of inertia means a smaller proportion of its energy is stored in rotation.

---

Problem-Solving Strategy

First identify the object's shape and moment of inertia.

Then determine whether it is rolling without slipping.

If acceleration is required, use:

a = g sin θ/[1 + I/(mr²)]

If final speed is required and energy is conserved, use:

mgh = 1/2 mv² + 1/2 Iω²

with:

ω = v/r

If the ramp length is given, find height using:

h = d sin θ

If friction is required, use:

mg sin θ − f = ma

and:

fr = Iα

with:

a = rα

If energy losses occur, include them explicitly:

E_initial = E_final + E_losses

Finally, check whether your answer makes physical sense:

• larger I should usually mean smaller translational acceleration
• steeper slope should mean greater acceleration
• final mechanical energy cannot exceed the available input energy without an additional energy source

---

Common Mistakes

Treating a rolling object like a sliding block

A rolling object has rotational motion as well as translation.

Using only 1/2 mv²

Rolling objects also have rotational kinetic energy.

Forgetting the moment of inertia

Different shapes accelerate differently because their mass distributions differ.

Assuming heavier objects roll faster

For geometrically similar ideal objects, mass cancels.

Assuming larger objects roll faster

Radius also cancels from ideal acceleration for geometrically similar shapes.

Thinking friction must act downhill

For a passive object rolling down an incline, static friction normally acts uphill.

Thinking static friction always wastes energy

Ideal static friction can provide torque without dissipating mechanical energy.

Using v = rω when the object is slipping

This relationship requires rolling without slipping.

Ignoring the vertical height

Energy depends on the vertical change in height, not simply the ramp length.

Using the wrong moment-of-inertia equation

Always identify the object's shape before calculating.

Assuming mechanical energy is always conserved

Real systems can transfer energy into thermal energy, sound, and deformation.

---

Key Terms

Inclined plane — sloping surface that makes an angle with the horizontal.

Rolling without slipping — motion in which v = rω and the contact point is instantaneously stationary relative to the surface.

Moment of inertia — measure of how mass is distributed relative to a rotation axis.

Translational kinetic energy — kinetic energy associated with movement of the centre of mass.

Rotational kinetic energy — kinetic energy associated with rotation.

Static friction — friction between surfaces that are not sliding relative to one another.

Rolling resistance — resistance caused by deformation, friction and other real effects during rolling.

Angular acceleration — rate of change of angular velocity.

Translational acceleration — rate of change of the velocity of the centre of mass.

Gravitational potential energy — energy associated with position in a gravitational field.

Energy conservation — principle stating that total energy cannot be created or destroyed.

Mass distribution — arrangement of an object's mass relative to its rotation axis.

Coefficient of static friction — quantity describing the maximum static friction available between two surfaces relative to the normal force.

---

Key Takeaways

• Objects rolling down inclines undergo both translation and rotation.
• Rolling without slipping requires v = rω and a = rα.
• Gravity provides the energy that produces both translational and rotational motion.
• Static friction commonly provides the torque required for rolling.
• For a passive object released down an incline, static friction generally acts uphill.
• The acceleration of a rolling object is a = g sin θ/[1 + I/(mr²)].
• Moment of inertia determines how easily an object gains angular velocity.
• Larger I/(mr²) produces smaller translational acceleration.
• Mass distribution matters more than total mass when comparing ideal rolling shapes.
• For geometrically similar ideal objects, mass and radius cancel from the acceleration equation.
• A solid sphere accelerates faster than a solid cylinder, spherical shell, or thin ring on the same ideal incline.
• Energy conservation gives mgh = 1/2 mv² + 1/2 Iω².
• Rolling objects divide their kinetic energy between translation and rotation.
• Static friction in ideal rolling does not necessarily reduce mechanical energy.
• Insufficient static friction causes slipping.
• Steeper slopes require more friction to maintain pure rolling.
• Real rolling systems experience losses through deformation, heat, sound, bearing friction, and air resistance.
• Both energy methods and force-and-torque methods can solve inclined rolling problems.

---

Check Your Understanding

1. Identify the three main forces acting on an ideal object rolling down an inclined plane.
2. Explain why gravity alone does not provide the torque about the centre needed for a round object to begin rolling from rest.
3. A solid sphere rolls without slipping down a 30° incline. Calculate its acceleration.
4. A solid cylinder rolls down the same 30° incline. Calculate its acceleration and compare it with the solid sphere.
5. Calculate the acceleration of a thin ring rolling without slipping down a 30° incline.
6. A solid sphere, solid cylinder, thin spherical shell and thin ring are released simultaneously from rest on the same incline. Predict their finishing order and explain your reasoning using moment of inertia.
7. Explain why increasing the mass of a geometrically similar rolling object does not change its ideal acceleration down an incline.
8. A solid cylinder starts from rest and rolls through a vertical height of 4.0 m. Calculate its speed at the bottom.
9. The cylinder in Question 8 has a radius of 0.20 m. Calculate its angular velocity at the bottom.
10. A solid sphere rolls down a 6.0 m long incline at 25°. Calculate the vertical height lost and its final speed if it starts from rest.
11. A 5.0 kg solid cylinder rolls without slipping down a 20° incline. Calculate its translational acceleration, angular acceleration if its radius is 0.25 m, and the static friction force acting on it.
12. Explain why static friction can act on a rolling object without reducing its mechanical energy.
13. A solid cylinder rolls down a 35° incline. Determine the minimum coefficient of static friction required to prevent slipping.
14. A 10 kg solid cylinder rolls from rest through a vertical height of 3.0 m. During the descent, 50 J is transferred into thermal energy and sound. Calculate its final translational speed.
15. A solid sphere and thin ring, each with the same mass and radius, roll from rest down an incline with a vertical drop of 5.0 m. Calculate the final translational speed of each. Then explain why their speeds differ even though they started with the same gravitational potential energy per unit mass.

Gyroscopes and Rotational Stability

A gyroscope is a rapidly rotating object whose spinning motion gives it a strong tendency to maintain the direction of its rotation axis.

A simple gyroscope can be made from a spinning wheel or disk mounted so that its axis is free to change direction. When the wheel is not spinning, gravity can easily make it tip. When it spins rapidly, its behaviour changes dramatically because it possesses angular momentum.

For a rigid object rotating about a fixed principal axis:

L = Iω

where:

• L = angular momentum, measured in kg·m²/s
• I = moment of inertia, measured in kg·m²
• ω = angular velocity, measured in rad/s

The greater the angular momentum, the more difficult it is for an external torque to rapidly change the direction of the rotation axis.

---

Angular Momentum and Stability

Angular momentum is a vector quantity. It has both magnitude and direction.

For a spinning wheel, the angular momentum vector points along the rotation axis.

Its direction can be determined using the right-hand rule:

• curl the fingers of your right hand in the direction of rotation
• your thumb points in the direction of angular momentum

The spinning wheel tends to maintain this angular momentum unless an external torque acts.

This tendency produces what we commonly call rotational stability.

---

What Rotational Stability Means

Rotational stability does not mean that a spinning object is impossible to move.

It means that changing the direction of its angular momentum requires a torque acting over time.

The relationship is:

Στ_external = ΔL/Δt

or, in instantaneous form:

Στ_external = dL/dt

Therefore, an external torque changes angular momentum.

If the external torque is small, the direction of a large angular momentum changes relatively slowly.

---

Why a Fast-Spinning Gyroscope Is Stable

Consider two identical wheels.

One is spinning slowly.

The other is spinning rapidly.

Since:

L = Iω

the faster wheel has greater angular momentum.

For the same external torque, the faster-spinning wheel experiences a smaller rate of directional change relative to the size of its angular momentum.

This gives it greater rotational stability.

---

Factors Affecting Gyroscopic Stability

For a simple rigid rotor:

L = Iω

Therefore, rotational stability generally increases when:

• moment of inertia I increases
• angular velocity ω increases

Moment of inertia can be increased by placing more mass farther from the rotation axis.

Angular momentum can be increased by spinning the rotor faster.

A large, rapidly rotating wheel can therefore have substantial rotational stability.

---

Effect of Angular Velocity

Suppose a gyroscope has:

I = 0.20 kg·m²

At:

ω = 20 rad/s

its angular momentum is:

L = Iω

L = (0.20)(20)

L = 4.0 kg·m²/s

If its angular velocity increases to:

ω = 60 rad/s

then:

L = (0.20)(60)

L = 12 kg·m²/s

Tripling the angular velocity triples the angular momentum.

The faster-spinning gyroscope therefore has greater resistance to rapid changes in the direction of its axis under the same applied torque.

---

Effect of Moment of Inertia

Suppose two rotors spin at the same angular velocity:

ω = 30 rad/s

Rotor A:

I = 0.10 kg·m²

Rotor B:

I = 0.40 kg·m²

For Rotor A:

L = (0.10)(30)

L = 3.0 kg·m²/s

For Rotor B:

L = (0.40)(30)

L = 12 kg·m²/s

Rotor B has four times the angular momentum.

Its larger moment of inertia gives it greater angular momentum at the same angular velocity.

---

Mass Distribution Matters

Moment of inertia depends not only on mass but also on where the mass is located.

For a point mass:

I = mr²

The r² term means that moving mass farther from the axis can greatly increase moment of inertia.

This is why a rotor with substantial mass near its outer edge can have a large angular momentum.

---

Torque Changes Angular Momentum

A torque does not necessarily make a spinning gyroscope simply fall over.

Instead, torque changes its angular momentum.

Remember:

Στ = dL/dt

Because angular momentum is a vector, torque can change:

• the magnitude of L
• the direction of L
• or both

In many gyroscope situations, gravity mainly changes the direction of the angular momentum vector.

This produces precession.

---

What Is Precession?

Precession is the gradual turning of the rotation axis of a spinning object.

Instead of simply falling downward under gravity, a spinning gyroscope can have its axis move sideways around a vertical direction.

The tip of the rotation axis traces approximately a circular path.

This behaviour can seem surprising because the gravitational force acts downward, yet the gyroscope's axis moves mainly sideways.

Angular momentum explains why.

---

A Gyroscope Supported at One End

Imagine a spinning wheel with one end of its axle supported.

Gravity acts downward on the wheel's centre of mass.

The support point acts as a pivot.

Because the line of action of gravity does not pass through the pivot, gravity produces a torque.

The magnitude of this torque is approximately:

τ = rmg

when the axle is horizontal, where r is the horizontal distance from the pivot to the centre of mass.

This torque changes the direction of the angular momentum.

The result is precession.

---

Why the Gyroscope Does Not Immediately Fall

If the wheel were not spinning, gravity would cause it to rotate downward about the support.

When it is spinning rapidly, however, it already has a large angular momentum vector along its axle.

The gravitational torque continuously changes the direction of this vector.

Instead of immediately rotating downward, the axis moves around the pivot.

This sideways turning is precession.

---

A Vector View of Precession

Suppose the angular momentum initially points horizontally.

During a short time Δt, gravity produces a torque.

The resulting change in angular momentum is:

ΔL = τΔt

This small ΔL points in the direction of the torque.

The new angular momentum is the vector sum of:

L_new = L_old + ΔL

If ΔL is mostly perpendicular to L, the magnitude of L changes little, but its direction changes.

Repeated continuously, this produces precession.

---

Precession Rate

For a simple rapidly spinning gyroscope under gravity, the precession angular speed can be approximated by:

Ω = τ/L

Since:

L = Iω

then:

Ω = τ/(Iω)

For a gyroscope supported at one end with its axle approximately horizontal:

τ = mgr

so:

Ω = mgr/(Iω)

where:

• Ω = precession angular speed
• m = mass
• g = gravitational field strength
• r = distance from pivot to centre of mass
• I = moment of inertia of the spinning rotor
• ω = rotor angular velocity

This equation is an approximation for steady precession under appropriate conditions, but it reveals the important relationships clearly.

---

Faster Spin Means Slower Precession

Since:

Ω = τ/L

a larger angular momentum produces a smaller precession rate for the same torque.

Therefore:

greater ω → greater L → slower precession

This explains why a rapidly spinning gyroscope can appear very stable.

As it slows, its precession generally becomes more noticeable.

---

Worked Precession Example

A gyroscope has:

m = 1.5 kg

r = 0.20 m

I = 0.060 kg·m²

ω = 100 rad/s

First calculate angular momentum:

L = Iω

L = (0.060)(100)

L = 6.0 kg·m²/s

Gravitational torque:

τ = mgr

τ = (1.5)(9.8)(0.20)

τ = 2.94 N·m

Precession rate:

Ω = τ/L

Ω = 2.94/6.0

Ω = 0.49 rad/s

---

What Happens as the Gyroscope Slows?

Real gyroscopes experience:

• bearing friction
• air resistance
• internal deformation

These processes reduce the rotor's angular speed.

As ω decreases:

L = Iω decreases

For a similar gravitational torque:

Ω = τ/L

therefore increases.

The gyroscope often precesses faster and becomes less stable as it slows.

Eventually, it may wobble strongly and fall.

---

Precession Is Not the Same as Spin

A gyroscope can have several motions occurring simultaneously.

Spin — rapid rotation of the rotor around its own axis.

Precession — slower rotation of the spin axis around another direction.

Nutation — small oscillations or wobbling of the axis that can occur in addition to precession.

These motions should not be confused.

---

Nutation

A gyroscope does not always settle immediately into smooth precession.

Its axis may bob or wobble as it moves.

This motion is called nutation.

Nutation can occur when:

• the gyroscope is released suddenly
• its initial motion does not match steady precession
• external disturbances act on it

Friction may gradually reduce the wobbling.

---

Spinning Tops

A spinning top is a familiar example of gyroscopic behaviour.

When spinning rapidly:

• angular momentum is large
• the top remains relatively upright
• its axis precesses slowly

As friction reduces its spin:

• angular momentum decreases
• precession becomes more noticeable
• wobbling increases
• the top eventually falls

---

Why a Non-Spinning Top Falls

A stationary top has no spin angular momentum:

L = 0

Gravity produces a torque about the contact point.

With no large angular momentum already present, the top simply begins rotating in the direction produced by the gravitational torque and falls.

A rapidly spinning top responds differently because its existing angular momentum must be redirected.

---

Bicycle Wheels and Rotational Stability

A spinning bicycle wheel has angular momentum.

As wheel speed increases:

ω increases

and therefore:

L increases

This can contribute to the behaviour and stability of a moving bicycle.

However, bicycle stability is more complicated than a simple gyroscope.

Real bicycle stability also involves:

• steering geometry
• rider control
• centre of mass
• tire-ground forces
• forward speed

The gyroscopic effect of the wheels is one contributor, not the complete explanation.

---

Motorcycles

Motorcycle wheels also possess substantial angular momentum.

At higher speeds, changing the orientation of the rotating wheels requires torque.

This contributes to the characteristic handling of motorcycles.

Riders intentionally use steering and body movement to change the direction of the vehicle and its rotating wheels.

---

Aircraft Gyroscopes

Gyroscopes have historically been important in aircraft instruments.

They can provide stable references for measuring orientation.

Applications include instruments that indicate:

• aircraft attitude
• heading
• turning motion

Modern aircraft increasingly use electronic and inertial sensors rather than relying only on large mechanical spinning gyroscopes.

---

Ships

Gyroscopes can be used for navigation and stabilization on ships.

A gyrocompass uses rotational principles and Earth's rotation to provide a directional reference.

Gyroscopic stabilization systems can also produce torques that help oppose unwanted rolling motion.

---

Spacecraft

Spacecraft cannot always rely on aerodynamic surfaces for orientation because there may be essentially no surrounding air.

Rotational systems can therefore be used to control spacecraft orientation.

Examples include:

• reaction wheels
• momentum wheels
• control moment gyroscopes

---

Reaction Wheels

A reaction wheel is a motor-driven flywheel inside a spacecraft.

Suppose the spacecraft and wheel initially have zero total angular momentum.

If the motor accelerates the wheel in one direction, conservation of angular momentum causes the spacecraft body to rotate in the opposite direction.

Symbolically:

L_spacecraft + L_wheel = constant

If the initial total angular momentum is zero:

L_spacecraft = −L_wheel

This allows spacecraft orientation to be changed without pushing against the surrounding environment.

---

Worked Reaction-Wheel Example

A spacecraft body has:

I_s = 800 kg·m²

A reaction wheel changes its angular momentum by:

+40 kg·m²/s

Ignoring external torques, the spacecraft must gain:

−40 kg·m²/s

of angular momentum.

Therefore:

L_s = I_sω_s

−40 = (800)ω_s

ω_s = −0.050 rad/s

The negative sign indicates that the spacecraft rotates opposite to the reaction wheel's angular momentum change.

---

Control Moment Gyroscopes

A control moment gyroscope (CMG) uses a rapidly spinning rotor.

Instead of mainly changing the rotor's spin speed, the system changes the direction of its angular momentum.

Changing that direction requires torque.

By controlling this process, the spacecraft experiences an equal and opposite torque.

CMGs can produce substantial control torques and are useful for orienting large spacecraft.

---

Gyroscopes in Phones

Modern smartphones commonly use tiny MEMS gyroscopes rather than visible spinning wheels.

MEMS stands for microelectromechanical systems.

These sensors detect rotational motion using tiny vibrating structures.

They help devices determine:

• angular velocity
• orientation changes
• screen rotation
• motion for games
• camera stabilization

The operating mechanism differs from a traditional spinning-wheel gyroscope, but the device performs the same general job of sensing rotational motion.

---

Navigation Systems

Gyroscopes are important components of inertial navigation systems.

These systems may combine:

• gyroscopes
• accelerometers
• computers

Gyroscopes measure rotational motion while accelerometers measure linear acceleration.

Together, these measurements can help estimate:

• orientation
• velocity
• position

Such systems are useful when external navigation signals are unavailable.

---

Camera Stabilization

Gyroscopic sensors can detect unwanted camera rotation.

A stabilization system can then compensate using:

• moving lens elements
• sensor movement
• motorized gimbals
• electronic correction

This helps produce smoother images and video.

---

Drones

Drones use gyroscopic sensors as part of their flight-control systems.

The sensors detect changes in orientation.

The control system can then adjust motor speeds to correct:

• roll
• pitch
• yaw

This allows rapid automatic stabilization.

---

Rotational Stability in Sports

Angular momentum also helps explain stability in many sports.

Examples include:

• spinning balls
• bicycle and motorcycle racing
• gymnastics
• diving
• figure skating
• discus throwing

A rotating athlete or object can maintain a relatively stable rotation axis when its angular momentum is large and external torques are small.

---

Spinning Balls

A spinning ball has angular momentum along its rotation axis.

External torques can change the direction of this angular momentum.

However, a rapidly spinning ball may maintain its spin orientation more effectively than a slowly rotating one.

This contributes to the stable rotation seen in objects such as:

• footballs
• rugby balls
• thrown discs
• spinning projectiles

Aerodynamic forces may also strongly influence their motion.

---

Stability Does Not Mean No Change

A common misconception is that angular momentum prevents the axis from changing direction.

It does not.

Instead:

torque causes angular momentum to change

A large angular momentum means that a particular torque produces a relatively small directional change over a short time.

Therefore, the axis changes more slowly.

---

Angular Impulse

If a torque acts for time Δt:

ΔL = τΔt

This is the rotational equivalent of linear impulse:

Δp = FΔt

Suppose a gyroscope has:

L = 20 kg·m²/s

and a perpendicular torque of:

2.0 N·m

acts for:

0.10 s

Then:

ΔL = (2.0)(0.10)

ΔL = 0.20 kg·m²/s

This change is small compared with the original angular momentum.

The axis therefore changes direction only slightly.

---

Comparing Two Gyroscopes

Gyroscope A:

L = 5 kg·m²/s

Gyroscope B:

L = 50 kg·m²/s

Suppose the same perpendicular torque acts on each for the same time.

They receive the same ΔL.

However, ΔL is a much larger fraction of Gyroscope A's original angular momentum.

Therefore, Gyroscope A experiences the larger change in orientation.

Gyroscope B is more rotationally stable.

---

Conservation of Angular Momentum

If:

Στ_external = 0

then:

L_total = constant

This is the principle of conservation of angular momentum.

A freely rotating object in space with negligible external torque tends to maintain its angular momentum vector.

That means both the magnitude and direction of total angular momentum remain constant.

This is the fundamental reason angular momentum is so important for rotational stability.

---

Gyroscopic Stability in Space

In space, external torques can be extremely small.

A spinning spacecraft can therefore maintain a relatively fixed angular momentum direction for long periods.

This principle has historically been used for spin stabilization.

A spacecraft may be deliberately rotated so that its angular momentum helps maintain its orientation.

---

Spin-Stabilized Spacecraft

Suppose an entire spacecraft rotates.

Its angular momentum is:

L = Iω

If external torque is very small:

L ≈ constant

The spacecraft's spin axis therefore tends to remain pointed in approximately the same inertial direction.

This can provide passive stability without continuous active control.

---

Earth as a Rotating System

Earth also has enormous angular momentum because:

• it has enormous moment of inertia
• it rotates continuously

Its rotation axis is therefore relatively stable.

However, external gravitational torques from the Moon and Sun cause Earth's axis to undergo very slow precession.

This is a large-scale example of the same basic physics seen in a gyroscope.

---

Precession of Earth's Axis

Earth is not a perfect sphere. It has a slight equatorial bulge.

The gravitational effects of the Moon and Sun on this bulge produce torque.

This causes Earth's rotation axis to slowly change direction.

The axis traces out a very slow precessional motion over thousands of years.

The physical principle is the same:

external torque → change in direction of angular momentum

---

Energy and Gyroscopes

A spinning gyroscope also possesses rotational kinetic energy:

E_rot = 1/2 Iω²

Angular momentum is:

L = Iω

These quantities are related but are not the same.

A gyroscope can have both:

• substantial rotational energy
• substantial angular momentum

Angular momentum is the quantity most directly connected to directional rotational stability.

---

Angular Momentum vs Rotational Energy

Increasing ω increases both L and E_rot.

However:

L ∝ ω

while:

E_rot ∝ ω²

Doubling angular velocity doubles angular momentum but quadruples rotational kinetic energy.

---

Friction and Gyroscope Motion

Real gyroscopes eventually slow because of friction and air resistance.

Energy is transferred from rotational kinetic energy into:

• thermal energy
• sound
• air motion

As ω decreases:

• rotational energy decreases
• angular momentum decreases
• rotational stability decreases
• precession can become more rapid
• wobbling may increase

Eventually the gyroscope can no longer maintain its characteristic stable motion.

---

Complete Gyroscope Problem

A gyroscope rotor has:

I = 0.080 kg·m²

ω = 150 rad/s

Its mass is:

m = 2.0 kg

and its centre of mass is:

r = 0.15 m

from the pivot.

Angular Momentum

L = Iω

L = (0.080)(150)

L = 12 kg·m²/s

Gravitational Torque

With the axle horizontal:

τ = mgr

τ = (2.0)(9.8)(0.15)

τ = 2.94 N·m

Approximate Precession Rate

Ω = τ/L

Ω = 2.94/12

Ω ≈ 0.245 rad/s

Precession Period

The time for one complete precession is:

T = 2π/Ω

T = 2π/0.245

T ≈ 25.6 s

---

What If the Spin Speed Halves?

Suppose friction reduces:

ω = 150 rad/s

to:

ω = 75 rad/s

Angular momentum becomes:

L = (0.080)(75)

L = 6.0 kg·m²/s

For approximately the same gravitational torque:

Ω = 2.94/6.0

Ω = 0.49 rad/s

The angular momentum has halved and the approximate precession rate has doubled.

This demonstrates why a slowing gyroscope tends to precess more rapidly.

---

Analyzing Gyroscopic Systems

When examining a gyroscope, first identify the rotation axis and direction of angular momentum.

Then determine whether an external torque acts.

Ask:

• What force creates the torque?
• Where is the pivot?
• In which direction is L?
• In which direction is τ?
• Is the torque changing the magnitude of L, its direction, or both?
• Is the rotor speeding up or slowing down?
• How will changing I or ω affect stability?

This approach is often more useful than trying to think only in terms of forces.

---

Common Mistakes

Thinking a gyroscope has no tendency to fall because gravity disappears

Gravity still acts. Its torque changes the angular momentum and produces precession.

Thinking angular momentum is a scalar

Angular momentum has both magnitude and direction.

Confusing spin with precession

Spin is rotation about the rotor's own axis. Precession is the slower motion of that axis.

Thinking torque always increases rotational speed

Torque can change the direction of angular momentum without greatly changing its magnitude.

Assuming a gyroscope is perfectly stable

External torques can always change its angular momentum.

Thinking faster spin causes faster gravitational precession

For a simple rapidly spinning gyroscope under the same torque, greater angular momentum generally means slower precession.

Confusing angular momentum and rotational kinetic energy

L = Iω, while E_rot = 1/2 Iω².

Assuming all modern gyroscopes contain spinning wheels

Many electronic devices use MEMS gyroscopes with microscopic vibrating structures.

Claiming gyroscopic effects completely explain bicycle stability

Bicycle stability also involves steering geometry, rider control, tire forces, and other effects.

---

Key Terms

Gyroscope — device or rotating system that uses rotational motion to maintain or measure orientation.

Rotational stability — tendency of a rotating object with angular momentum to resist rapid changes in the direction of its rotation axis.

Angular momentum (L) — vector quantity describing rotational motion; for a simple rigid rotor about a fixed principal axis, L = Iω.

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

Angular velocity (ω) — rate and direction of rotation.

Torque (τ) — turning effect that changes angular momentum.

Precession — gradual turning of the rotation axis of a spinning object due to an external torque.

Precession angular speed (Ω) — rate at which the rotation axis precesses.

Nutation — oscillation or wobbling of a rotating object's axis superimposed on precession.

Right-hand rule — method for determining the direction of angular velocity and angular momentum.

Angular impulse — product of torque and time, equal to the change in angular momentum.

Reaction wheel — rotating wheel used to control spacecraft orientation by exchanging angular momentum with the spacecraft.

Control moment gyroscope — device that produces control torque by changing the direction of a spinning rotor's angular momentum.

MEMS gyroscope — microscopic electromechanical sensor used to detect rotational motion.

Spin stabilization — use of angular momentum from rotation to help maintain an object's orientation.

---

Key Takeaways

• Gyroscopes demonstrate the effects of angular momentum and torque.
• For a simple rigid rotor, angular momentum is L = Iω.
• Angular momentum is a vector directed along the rotation axis.
• A rapidly spinning object can have substantial rotational stability.
• Increasing moment of inertia increases angular momentum when angular velocity is unchanged.
• Increasing angular velocity increases angular momentum when moment of inertia is unchanged.
• External torque changes angular momentum.
• The fundamental relationship is Στ_external = ΔL/Δt.
• Torque can change the direction of angular momentum without greatly changing its magnitude.
• This directional change produces gyroscopic precession.
• Precession is different from the rotor's spin.
• For simple steady gravitational precession, Ω ≈ τ/L.
• Greater angular momentum generally produces slower precession for the same torque.
• As a gyroscope slows because of friction, its angular momentum decreases and its behaviour becomes less stable.
• Nutation is the additional wobbling motion that can accompany precession.
• Angular momentum conservation explains why freely rotating objects tend to maintain their orientation when external torque is negligible.
• Gyroscopic principles are used in spacecraft, navigation, aircraft systems, ships, vehicles and stabilization systems.
• Reaction wheels control spacecraft by exchanging angular momentum internally.
• Modern phones and drones commonly use MEMS gyroscopes rather than large spinning rotors.
• Earth's slow axial precession is a large-scale example of torque changing the direction of angular momentum.

---

Check Your Understanding

1. What is a gyroscope, and what property of rotational motion gives it stability?
2. A rotor has I = 0.15 kg·m² and rotates at 80 rad/s. Calculate its angular momentum.
3. The angular velocity of the rotor in Question 2 doubles. What happens to its angular momentum?
4. Two gyroscopes rotate at the same angular velocity. One has twice the moment of inertia of the other. Compare their angular momenta and rotational stability.
5. Explain how the right-hand rule can be used to determine the direction of a gyroscope's angular momentum.
6. What does an external torque do to angular momentum?
7. Explain qualitatively why a spinning gyroscope supported at one end precesses rather than simply falling straight downward.
8. Distinguish between spin, precession, and nutation.
9. A gyroscope has angular momentum L = 15 kg·m²/s and experiences a perpendicular torque of 3.0 N·m. Estimate its precession angular speed using Ω = τ/L.
10. A gyroscope has I = 0.050 kg·m² and initially rotates at 200 rad/s. Friction reduces its angular velocity to 100 rad/s. Calculate its initial and final angular momentum and predict how this change affects its rotational stability.
11. Explain why increasing a rotor's angular velocity generally increases its gyroscopic stability.
12. Explain how a reaction wheel can rotate a spacecraft even though the spacecraft has nothing external to push against.
13. Compare a traditional mechanical gyroscope with a MEMS gyroscope used in a smartphone.
14. A gyroscope rotor has I = 0.10 kg·m², ω = 120 rad/s, mass 2.5 kg, and its centre of mass is 0.20 m from the pivot. With the axle horizontal, calculate its angular momentum, gravitational torque, and approximate precession angular speed.
15. A spinning gyroscope gradually slows because of friction. Explain the complete sequence of changes involving rotational kinetic energy, angular momentum, precession, stability, and eventual motion of the gyroscope.

Rotating Machinery and Vehicles

Rotational motion is central to the operation of engines, electric motors, turbines, generators, wheels, gears, transmissions, pumps, fans, and many transportation systems.

A machine often begins with one form of energy and uses rotating components to transfer, transform, and control that energy.

For example, in an electric vehicle:

electrical energy → motor rotation → drivetrain rotation → wheel rotation → vehicle motion

Understanding these systems requires several ideas from rotational mechanics:

• torque
• angular velocity
• angular acceleration
• moment of inertia
• rotational kinetic energy
• angular momentum
• work and power
• gears
• rolling motion
• friction
• energy conservation

---

Rotation in Machines

A rotating machine usually contains one or more components turning about an axis.

Examples include:

• shafts
• gears
• wheels
• pulleys
• turbines
• flywheels
• rotors
• crankshafts
• propellers
• fans

The rotational motion of these components can transmit energy from one part of a machine to another.

For a rotating object:

ω = Δθ/Δt

and:

α = Δω/Δt

The rotational equivalent of Newton's Second Law is:

Στ = Iα

This equation connects the torque applied to a machine with its resulting angular acceleration.

---

Torque in Machinery

Torque is the turning effect of a force.

For a force acting at distance r from an axis:

τ = rF sin θ

For a perpendicular force:

τ = rF

A machine can increase torque by:

• increasing the applied force
• increasing the distance from the axis
• using gears or other mechanical systems

Torque is particularly important when machinery must:

• start a heavy load
• accelerate a vehicle
• lift an object
• turn a cutting tool
• overcome friction
• climb a hill

---

Worked Torque Example

A motor applies a tangential force of:

F = 400 N

at a radius:

r = 0.15 m

Torque:

τ = rF

τ = (0.15)(400)

τ = 60 N·m

If opposing friction produces:

10 N·m

of torque, the net torque is:

τ_net = 60 − 10

τ_net = 50 N·m

---

Torque and Angular Acceleration

Newton's Second Law for rotation gives:

α = τ_net/I

Suppose the rotating component has:

I = 5.0 kg·m²

and:

τ_net = 50 N·m

Then:

α = 50/5

α = 10 rad/s²

A smaller moment of inertia would produce a greater angular acceleration for the same torque.

---

Moment of Inertia in Machine Design

Moment of inertia measures how strongly an object resists changes in rotational motion.

It depends on:

• total mass
• shape
• mass distribution
• location of the rotation axis

Mass farther from the axis produces a larger moment of inertia.

For example:

Thin ring:

I = MR²

Solid disk:

I = 1/2 MR²

Solid sphere:

I = 2/5 MR²

Machine designers must consider moment of inertia whenever rotating components need to speed up or slow down quickly.

---

High and Low Moment of Inertia

A low moment of inertia can be useful when a component must:

• accelerate quickly
• decelerate quickly
• change speed frequently
• respond rapidly to controls

A high moment of inertia can be useful when a component should:

• maintain steady rotation
• store rotational energy
• smooth variations in speed

Flywheels are deliberately designed with significant moment of inertia for this reason.

---

Wheels

The wheel is one of the most important applications of rotational mechanics.

A wheel converts between:

• rotational motion
• translational motion

For rolling without slipping:

v = rω

and:

a = rα

These relationships connect the rotation of a vehicle's wheels with the vehicle's forward motion.

---

Worked Wheel Example

A car wheel has:

r = 0.32 m

and rotates at:

ω = 75 rad/s

Vehicle speed:

v = rω

v = (0.32)(75)

v = 24 m/s

Converting to km/h:

24 × 3.6 = 86.4 km/h

---

Why Wheels Reduce Energy Losses

Rolling motion usually involves much less resistance than dragging an object across a surface.

Instead of surfaces continuously sliding against one another, a properly rolling wheel has an instantaneous contact point that is stationary relative to the road.

This is why wheels are so effective for transportation.

Real wheels still experience:

• tire deformation
• bearing friction
• air resistance
• surface deformation

These effects create rolling resistance and other energy losses.

---

Friction and Vehicle Wheels

Friction between tires and the road is essential.

When a driven wheel accelerates, the tire pushes backward on the road.

The road exerts a forward static-friction force on the tire.

This force accelerates the vehicle.

Without sufficient friction, the wheel spins:

rω > v

and the tire slips relative to the road.

---

Wheel Torque and Driving Force

If torque τ is delivered to a wheel of radius r, the corresponding tangential force is approximately related by:

F = τ/r

For example, if:

τ = 600 N·m

and:

r = 0.30 m

then:

F = 600/0.30

F = 2000 N

This helps explain why large wheel torque is valuable when accelerating a vehicle.

---

Gears

Gears transfer rotational motion and torque between shafts.

When two gears mesh:

• their edges have the same tangential speed at the contact point in an ideal system
• they rotate in opposite directions
• their angular velocities depend on their radii or numbers of teeth

For two gears:

r₁ω₁ = r₂ω₂

Therefore:

ω₁/ω₂ = r₂/r₁

---

Gear Ratio

Gear ratio can be expressed using the number of teeth:

gear ratio = N_output/N_input

where:

• N_output = number of teeth on the output gear
• N_input = number of teeth on the input gear

A larger output gear rotates more slowly but can provide greater torque.

---

Worked Gear Example

A driving gear has:

20 teeth

and turns a gear with:

60 teeth.

The gear ratio is:

60/20 = 3

If the input rotates at:

900 rpm

the output rotates at:

900/3

= 300 rpm

Ideally, the output torque increases by a factor of 3.

If input torque is:

40 N·m

then:

τ_output = 3(40)

τ_output = 120 N·m

---

Speed-Torque Trade-Off

Gears do not create energy.

They trade:

angular speed ↔ torque

In an ideal system:

P_input = P_output

Since:

P = τω

we have:

τ_inω_in = τ_outω_out

Therefore, increasing torque requires a corresponding decrease in angular speed if power remains constant.

---

Why Vehicles Need Gears

A vehicle operates under very different conditions.

Starting from rest requires:

• high wheel torque
• relatively low wheel speed

Cruising at high speed requires:

• greater wheel speed
• less torque multiplication

A transmission allows the motor or engine to operate effectively while the wheel requirements change.

---

Low Gears

Low gears provide:

• high torque multiplication
• lower output speed

They are useful for:

• starting from rest
• climbing steep hills
• towing
• carrying heavy loads
• rapid low-speed acceleration

---

High Gears

Higher gears provide:

• lower torque multiplication
• greater wheel speed for a given engine speed

They are useful during:

• cruising
• high-speed travel
• efficient operation

A gearbox therefore allows the drivetrain to match the machine's power source to the required motion.

---

Bicycle Gears

Bicycles provide a visible example of gear ratios.

A smaller front chainring combined with a larger rear sprocket makes pedalling easier when climbing.

The rider turns the pedals more times for each wheel revolution, but less torque is required from the rider for a given wheel-driving demand.

A larger front chainring and smaller rear sprocket allows greater wheel rotation per pedal revolution but requires more pedal force under comparable conditions.

---

Engines

An internal-combustion engine converts chemical energy from fuel into mechanical energy.

A simplified pathway is:

chemical energy → thermal/internal energy → piston motion → crankshaft rotation

The pistons move approximately back and forth.

Connecting rods transfer their motion to the crankshaft.

The crankshaft converts this reciprocating motion into rotational motion.

---

Crankshaft Torque

Gas pressure acting on a piston creates force.

The connecting rod transfers force to the crankshaft.

Because this force acts at a distance from the crankshaft axis, it creates torque.

The torque varies during the engine cycle because:

• gas pressure changes
• crank angle changes
• the effective lever arm changes

A real engine therefore does not produce perfectly constant torque.

---

Flywheels in Engines

A flywheel can help smooth these torque variations.

Its rotational kinetic energy is:

E_rot = 1/2 Iω²

During parts of the cycle when the engine produces excess torque, the flywheel gains rotational energy.

During other parts, the flywheel releases some of that energy.

This helps maintain a more nearly steady crankshaft speed.

---

Worked Flywheel Example

A flywheel has:

I = 6.0 kg·m²

and rotates at:

ω = 100 rad/s

Its rotational energy is:

E_rot = 1/2(6)(100²)

E_rot = 30,000 J

If its angular speed increases to:

110 rad/s

then:

E_rot = 1/2(6)(110²)

E_rot = 36,300 J

Increase:

36,300 − 30,000

= 6300 J

The flywheel has stored an additional 6300 J of rotational energy.

---

Electric Motors

Electric motors convert:

electrical energy → rotational mechanical energy

They are used in:

• electric vehicles
• trains
• elevators
• pumps
• fans
• industrial machinery
• robots
• appliances

Motor performance is often described using:

• torque
• angular speed
• power
• efficiency

---

Motor Power

Rotational power is:

P = τω

Suppose an electric motor produces:

τ = 150 N·m

at:

ω = 200 rad/s

Then:

P = (150)(200)

P = 30,000 W

or:

30 kW

---

Why Torque and Power Are Different

Torque describes the turning effect.

Power describes the rate of energy transfer.

A machine can produce high torque at low speed while transferring relatively little power.

For example:

τ = 500 N·m

ω = 2 rad/s

P = (500)(2)

P = 1000 W

Another machine might produce less torque but much greater power because it rotates much faster.

---

Converting rpm to Angular Velocity

Machine speeds are commonly given in revolutions per minute.

Convert rpm to rad/s using:

ω = rpm × 2π/60

For example:

3000 rpm

gives:

ω = 3000 × 2π/60

ω ≈ 314 rad/s

If torque is:

100 N·m

then:

P = (100)(314)

P ≈ 31.4 kW

---

Torque Curves and Power Curves

The torque produced by a motor or engine may change with rotational speed.

Since:

P = τω

power also changes with speed.

Maximum torque and maximum power therefore do not necessarily occur at the same rotational speed.

This is important when selecting:

• motors
• engines
• transmissions
• operating speeds

for engineering applications.

---

Vehicle Drivetrains

A drivetrain transfers rotational energy from the engine or motor to the wheels.

A simplified conventional drivetrain may include:

engine → clutch/torque converter → transmission → driveshaft → differential → axles → wheels

An electric drivetrain may be simpler:

battery → inverter → electric motor → reduction gear → differential → wheels

---

The Differential

When a vehicle turns, the outside wheel travels farther than the inside wheel.

Therefore, the two driven wheels need different angular velocities.

A differential allows the wheels to rotate at different speeds while still transmitting torque to them.

Without this ability, tires could scrub or slip significantly during turns.

---

Turning Vehicle Example

Suppose a vehicle turns so that:

• inner wheel path radius = 8 m
• outer wheel path radius = 10 m

If both wheels complete the turn in the same time, the outer wheel must travel farther.

Therefore:

v_outer > v_inner

For equal-size wheels:

ω_outer > ω_inner

The differential allows this difference.

---

Rotational Energy in Vehicle Wheels

A moving vehicle has translational kinetic energy:

E_trans = 1/2 Mv²

Each wheel also has rotational kinetic energy:

E_rot = 1/2 Iω²

Therefore, some of the energy supplied by the engine or motor goes into rotating the wheels.

---

Worked Vehicle-Wheel Example

A wheel has:

I = 1.5 kg·m²

r = 0.30 m

The vehicle travels at:

v = 20 m/s

For rolling without slipping:

ω = v/r

ω = 20/0.30

ω ≈ 66.7 rad/s

Rotational kinetic energy:

E_rot = 1/2(1.5)(66.7²)

E_rot ≈ 3337 J

For four identical wheels:

4E_rot ≈ 13.3 kJ

This energy must be supplied when the vehicle accelerates.

---

Why Lightweight Wheels Can Matter

Reducing wheel mass can reduce:

• translational mass
• moment of inertia

Therefore, less energy may be required to accelerate both the vehicle and its wheels.

Mass near the wheel's outer edge has a particularly strong effect because moment of inertia depends strongly on distance from the axis.

---

Braking

Braking requires reducing both:

• translational kinetic energy
• rotational kinetic energy

Conventional friction brakes transform much of this mechanical energy into thermal energy.

The brake pads exert forces on a rotating disc.

These forces produce a torque opposite the direction of rotation.

This negative torque reduces angular velocity.

---

Braking Work

For constant braking torque:

W = τΔθ

If torque opposes the angular displacement, the work is negative.

Suppose:

τ = −200 N·m

and the brake disc rotates through:

Δθ = 40 rad

Then:

W = (−200)(40)

W = −8000 J

The disc's rotational mechanical energy decreases by 8000 J.

---

Regenerative Braking

Electric vehicles can recover some mechanical energy during braking.

The motor operates as a generator.

A simplified energy pathway is:

vehicle kinetic energy → wheel rotation → motor/generator → electrical energy → battery

Some energy is still transferred into:

• heat
• electrical resistance
• mechanical friction
• other losses

Regenerative braking therefore cannot recover all the vehicle's kinetic energy.

---

Trains

Trains make extensive use of rotational mechanics.

Important rotating components include:

• traction motors
• axles
• wheels
• gears
• generators
• braking systems

Electric trains can use regenerative braking to return electrical energy to the supply system or store it, depending on the system design.

---

Aircraft

Aircraft contain many rotating systems, including:

• propellers
• turbines
• compressors
• generators
• pumps
• wheels

A propeller converts rotational mechanical energy into motion of air, producing thrust.

A turbine extracts energy from moving gases and transfers it to a rotating shaft.

---

Jet Engines

In a turbine engine, major rotating components may include:

• compressor
• turbine
• shafts

The turbine extracts energy from hot expanding gases.

Some of that rotational energy drives the compressor.

The system continuously transfers energy through rotating machinery at very high angular speeds.

---

Turbines and Generators

A turbine extracts energy from a moving fluid.

Examples include:

• steam turbines
• wind turbines
• hydroelectric turbines
• gas turbines

The turbine rotates a shaft.

The shaft can drive a generator.

Energy transformation may be represented as:

fluid energy → rotational mechanical energy → electrical energy

---

Wind Turbines

Wind turbines have very large blades rotating relatively slowly.

The rotor can produce substantial torque.

A generator then converts mechanical rotational energy into electrical energy.

Depending on turbine design, a gearbox may be used to alter the rotational speed delivered to the generator.

---

Hydroelectric Systems

Water falling through a height loses gravitational potential energy.

The moving water turns a turbine.

The turbine drives a generator.

The energy pathway is approximately:

gravitational potential energy → kinetic energy of water → rotational energy → electrical energy

This is conservation of energy applied to a large engineering system.

---

Pumps

A pump does essentially the reverse of a turbine.

A motor supplies rotational mechanical energy.

The rotating pump transfers energy to a fluid.

This can increase:

• fluid pressure
• fluid speed
• gravitational potential energy

Pumps are used in:

• water systems
• vehicles
• heating and cooling
• industrial processing
• medical equipment

---

Fans and Propellers

A rotating fan or propeller transfers energy to the surrounding fluid.

The blades exert forces on air or water.

The fluid exerts equal and opposite forces on the blades.

This produces:

• airflow
• thrust
• pressure differences

The motor must continuously provide torque because fluid resistance opposes the rotation.

---

Bearings

Rotating shafts need support.

Bearings allow shafts to rotate while reducing unwanted friction.

Common designs include:

• ball bearings
• roller bearings
• plain bearings

Reducing bearing friction improves:

• efficiency
• component life
• temperature control

---

Lubrication

Lubricants reduce direct contact between moving surfaces.

This can reduce:

• friction
• heating
• wear

Lubrication is particularly important in:

• engines
• gearboxes
• bearings
• turbines
• industrial machinery

Without adequate lubrication, friction can cause rapid energy loss and mechanical damage.

---

Efficiency of Rotating Machines

No real rotating machine is 100% efficient.

Efficiency is:

efficiency = useful output energy/input energy × 100%

or:

efficiency = useful output power/input power × 100%

Energy may be transferred into less useful forms through:

• friction
• electrical resistance
• sound
• vibration
• deformation
• air resistance

---

Worked Efficiency Example

A motor receives:

P_input = 50 kW

and produces:

P_output = 43 kW

Efficiency:

efficiency = 43/50 × 100%

efficiency = 86%

Power transferred into other forms:

50 − 43

= 7 kW

---

Gearbox Efficiency

Real gearboxes also lose some mechanical energy.

Suppose:

P_input = 20 kW

and gearbox efficiency is:

92%

Then:

P_output = 0.92(20)

P_output = 18.4 kW

If the output shaft rotates at:

ω = 80 rad/s

then:

τ = P/ω

τ = 18,400/80

τ = 230 N·m

---

Energy Flow Through a Machine

A useful engineering approach is to follow the energy through the entire system.

Consider an electric vehicle:

battery chemical energy

↓

electrical energy

↓

motor rotational energy

↓

drivetrain rotational energy

↓

wheel rotation

↓

vehicle translational kinetic energy

At each stage, some energy may be transferred into less useful forms.

This allows engineers to identify where efficiency improvements can be made.

---

Mechanical Power Through a Shaft

A rotating shaft transfers mechanical power according to:

P = τω

This means the same amount of power can be transmitted using different combinations of torque and angular velocity.

For example:

System A:

τ = 100 N·m

ω = 200 rad/s

P = 20 kW

System B:

τ = 400 N·m

ω = 50 rad/s

P = 20 kW

Both transmit the same power.

---

Why Some Machines Use High Rotational Speed

For a given power:

τ = P/ω

Increasing angular velocity allows the same power to be transmitted with less torque.

This can sometimes allow:

• smaller shafts
• smaller motors
• more compact machinery

However, high rotational speed also introduces challenges involving:

• vibration
• bearing loads
• friction
• heating
• structural stresses
• safety

Engineering always involves trade-offs.

---

Centripetal Effects in Rotating Machinery

Parts of a rotating machine continually accelerate toward the rotation axis.

Centripetal acceleration is:

a_c = rω²

The required inward force is:

F_c = mrω²

Because ω is squared, forces can become extremely large at high rotational speeds.

This is a major safety consideration in:

• turbines
• flywheels
• engines
• centrifuges
• grinding wheels

---

Worked High-Speed Rotor Example

A:

m = 0.50 kg

component rotates at:

r = 0.20 m

with:

ω = 300 rad/s

Centripetal force:

F_c = mrω²

F_c = (0.50)(0.20)(300²)

F_c = 9000 N

Even a relatively small mass can experience a very large force at high angular velocity.

---

Balancing Rotating Machinery

If mass is distributed unevenly around a rotating shaft, the centre of mass may not lie exactly on the rotation axis.

This can cause:

• vibration
• noise
• bearing wear
• structural stress

Rotating components are therefore carefully balanced.

Vehicle wheels are a familiar example. Small balancing masses may be attached to a wheel so that its mass distribution is more uniform around the axis.

---

Resonance and Vibration

Rotating machinery can produce periodic forces.

If these forces match a natural frequency of the machine or structure, large vibrations may develop.

This is called resonance.

Engineers must consider rotational speed, stiffness, damping, and structural design to avoid dangerous vibration.

---

Angular Momentum in Machinery

A rotating component has angular momentum:

L = Iω

A large flywheel, turbine, or wheel can therefore possess substantial angular momentum.

Changing its rotation requires torque:

Στ = ΔL/Δt

This means rapidly changing the orientation or rotational speed of heavy rotating machinery can require large forces and torques.

---

Gyroscopic Effects in Vehicles

Rotating wheels possess angular momentum.

When a motorcycle, bicycle, aircraft propeller, or other rotating system changes orientation, its angular momentum vector must also change.

This can produce gyroscopic effects.

These effects can influence:

• steering
• handling
• stability
• structural forces

However, vehicle stability usually depends on many factors, not gyroscopic effects alone.

---

Engineering Trade-Offs

Machine design rarely involves maximizing only one quantity.

Engineers may need to balance:

• torque
• angular speed
• power
• efficiency
• moment of inertia
• mass
• strength
• friction
• heat generation
• cost
• reliability
• safety

For example, a large flywheel can store more rotational energy but may:

• increase mass
• increase structural loads
• require stronger bearings
• increase safety risks at high speed

---

Designing a Vehicle Wheel

A vehicle wheel must satisfy competing requirements.

It should be:

• strong enough to support loads
• light enough to reduce unnecessary mass
• stiff enough to maintain shape
• capable of transmitting torque
• able to withstand braking forces
• resistant to fatigue
• balanced for smooth rotation

Its radius also affects:

• angular velocity
• required torque
• gearing
• tire behaviour

Rotational mechanics therefore directly influences wheel design.

---

Designing a Gearbox

A gearbox must provide suitable combinations of:

• input speed
• output speed
• input torque
• output torque

It must also handle:

• friction
• heating
• lubrication
• gear-tooth forces
• wear
• efficiency

A theoretically ideal gear ratio is only the beginning of the engineering problem.

---

Complete Drivetrain Problem

An electric motor produces:

τ_motor = 120 N·m

at:

ω_motor = 300 rad/s

Motor Power

P_motor = τω

P_motor = (120)(300)

P_motor = 36,000 W = 36 kW

The motor drives a:

4:1 reduction gear

Ideally, output angular velocity becomes:

ω_output = 300/4

ω_output = 75 rad/s

Ideal output torque:

τ_output = 4(120)

τ_output = 480 N·m

Suppose the gearbox is:

90% efficient

Actual output power:

P_output = 0.90(36,000)

P_output = 32,400 W

Actual output torque:

τ_output = P_output/ω_output

τ_output = 32,400/75

τ_output = 432 N·m

If the wheel radius is:

r = 0.30 m

the corresponding tangential force is approximately:

F = τ/r

F = 432/0.30

F = 1440 N

This example connects:

• torque
• angular velocity
• power
• gear ratio
• efficiency
• wheel force

---

Complete Vehicle-Speed Problem

Suppose the output shaft directly drives wheels with:

r = 0.30 m

and:

ω = 75 rad/s

Vehicle speed:

v = rω

v = (0.30)(75)

v = 22.5 m/s

Convert to km/h:

22.5 × 3.6

= 81 km/h

The drivetrain therefore converts the motor's high-speed rotation into lower-speed, higher-torque wheel rotation.

---

Complete Energy Problem

A vehicle of mass:

M = 1200 kg

accelerates from rest to:

v = 20 m/s

Each of its four wheels has:

I = 1.2 kg·m²

and radius:

r = 0.30 m

Translational Energy

E_trans = 1/2 Mv²

E_trans = 1/2(1200)(20²)

E_trans = 240,000 J

Wheel Angular Velocity

ω = v/r

ω = 20/0.30

ω ≈ 66.7 rad/s

Rotational Energy per Wheel

E_rot = 1/2 Iω²

E_rot = 1/2(1.2)(66.7²)

E_rot ≈ 2669 J

For four wheels:

E_rot,total ≈ 10,676 J

Total Kinetic Energy

E_total ≈ 240,000 + 10,676

E_total ≈ 250,676 J

The motor must supply energy for both translation and rotation, in addition to overcoming losses.

---

Complete Braking Problem

Suppose the same vehicle must stop.

Approximately:

250.7 kJ

of kinetic energy must be removed.

If regenerative braking recovers:

60%

then:

E_recovered = 0.60(250.7)

≈ 150.4 kJ

The remaining:

250.7 − 150.4

≈ 100.3 kJ

must be transferred into other forms such as thermal energy, aerodynamic effects, rolling resistance, and electrical losses.

---

Analyzing an Engineering System

When analyzing rotating machinery, use a systematic approach.

First identify the energy source.

Then identify the rotating components.

Determine:

• torque
• angular speed
• gear ratios
• moment of inertia
• energy transfers
• power
• efficiency
• frictional losses

Finally, determine the useful output.

A useful chain is:

energy input → torque → rotation → transmission → useful output

Then identify where energy is transferred into less useful forms.

---

Choosing the Right Equation

For torque:

τ = rF sin θ

For rotational dynamics:

Στ = Iα

For rotational speed:

ω = Δθ/Δt

For rotational kinetic energy:

E_rot = 1/2 Iω²

For rotational power:

P = τω

For rotational work:

W = τΔθ

For angular momentum:

L = Iω

For rolling wheels:

v = rω

For tangential acceleration:

a = rα

For centripetal acceleration:

a_c = rω²

For efficiency:

efficiency = useful output/input × 100%

Together, these equations provide a toolkit for analyzing many real machines.

---

Common Mistakes

Confusing torque with power

Torque describes turning effect. Power describes the rate of energy transfer.

Thinking gears create energy

Gears trade torque for angular speed. They do not create power.

Ignoring gearbox efficiency

Real gears lose some energy through friction, vibration, and heating.

Using rpm directly in P = τω

Convert rpm to rad/s first.

Forgetting rotational kinetic energy

Vehicle wheels and other rotating components store energy as they spin.

Ignoring moment of inertia

Two components with equal mass can respond very differently if their mass distributions differ.

Assuming friction is always undesirable

Tire-road friction is essential for acceleration, braking, and steering.

Assuming friction never wastes energy

Bearing friction, sliding friction, and deformation can transfer useful mechanical energy into thermal energy.

Forgetting the no-slip condition

For an ideal rolling wheel, v = rω.

Ignoring centripetal forces at high speed

High angular velocity can produce enormous internal forces.

Assuming a more powerful machine always produces more torque

Power depends on both torque and angular velocity.

Ignoring system-level energy losses

A machine may have several stages, each with its own efficiency.

---

Key Terms

Rotating machinery — machines containing components that rotate about an axis to transfer or transform energy.

Drivetrain — collection of components transferring mechanical power from an engine or motor to the wheels.

Torque — turning effect of a force.

Angular velocity — rate at which angular position changes.

Angular acceleration — rate of change of angular velocity.

Moment of inertia — measure of resistance to angular acceleration based on mass distribution.

Rotational kinetic energy — energy associated with rotation, E_rot = 1/2 Iω².

Rotational power — rate of rotational energy transfer, P = τω.

Gear ratio — relationship between input and output gears that determines changes in speed and torque.

Transmission — system of gears or other mechanisms used to control the relationship between motor or engine speed and wheel speed.

Differential — mechanism allowing driven wheels to rotate at different speeds while transmitting torque.

Flywheel — rotating component designed to store rotational energy and smooth variations in speed.

Crankshaft — rotating shaft that converts reciprocating piston motion into rotation in many engines.

Bearing — component that supports rotating shafts while reducing friction.

Rolling resistance — resistance to rolling caused by deformation and other real effects.

Regenerative braking — process that converts some vehicle mechanical energy back into electrical energy during braking.

Efficiency — proportion of input energy or power converted into useful output.

Balancing — adjustment of mass distribution around a rotation axis to reduce vibration.

---

Key Takeaways

• Rotational mechanics is fundamental to engines, motors, vehicles, turbines and industrial machines.
• Torque produces angular acceleration according to Στ = Iα.
• Moment of inertia determines how strongly a rotating component resists changes in rotational speed.
• Wheels connect rotational and translational motion through v = rω.
• Tire-road friction allows vehicles to accelerate, brake and steer.
• Wheel torque can produce a tangential driving force.
• Gears trade angular speed for torque.
• Ideal gears conserve mechanical power rather than creating energy.
• Rotational power is P = τω.
• Low gears provide greater torque multiplication and are useful for starting and climbing.
• Higher gears allow greater vehicle speed for a given engine or motor speed.
• Internal-combustion engines convert piston motion into crankshaft rotation.
• Electric motors convert electrical energy directly into rotational mechanical energy.
• Flywheels store rotational kinetic energy and can smooth variations in machine speed.
• Vehicle wheels possess rotational kinetic energy in addition to the vehicle's translational kinetic energy.
• Conventional braking transforms mechanical energy mainly into thermal energy.
• Regenerative braking can recover some vehicle kinetic energy.
• Differentials allow driven wheels to rotate at different speeds during turns.
• Bearings and lubrication reduce friction and improve efficiency.
• High-speed rotating machinery can experience very large centripetal forces.
• Rotating components must be carefully balanced to control vibration.
• Engineering design requires trade-offs among torque, speed, power, mass, efficiency, strength, reliability, and safety.
• Rotational mechanics provides a framework for analyzing complete engineering systems rather than only isolated rotating objects.

---

Check Your Understanding

1. Identify four examples of rotating components found in machines or vehicles.
2. A motor applies a tangential force of 500 N at a radius of 0.12 m. Calculate the torque.
3. A machine experiences a net torque of 80 N·m and has a moment of inertia of 10 kg·m². Calculate its angular acceleration.
4. A vehicle wheel has radius 0.35 m and rotates at 60 rad/s. Calculate the vehicle's speed if the wheel rolls without slipping.
5. A 20-tooth gear drives an 80-tooth gear. Determine the gear ratio. If the input rotates at 1200 rpm, calculate the output speed.
6. Explain why increasing output torque with a gearbox normally causes output angular speed to decrease.
7. A motor produces 200 N·m of torque at 150 rad/s. Calculate its mechanical power.
8. A motor rotates at 2400 rpm while producing 100 N·m of torque. Convert the angular speed to rad/s and calculate the power.
9. Explain why vehicles use low gears when starting from rest or climbing steep hills.
10. A flywheel has I = 8.0 kg·m² and rotates at 120 rad/s. Calculate its rotational kinetic energy.
11. A gearbox receives 40 kW and is 90% efficient. Calculate its output power. If the output shaft rotates at 100 rad/s, calculate its output torque.
12. Explain the purpose of a differential when a vehicle turns.
13. A 1400 kg vehicle travels at 25 m/s. Each of its four wheels has I = 1.5 kg·m² and radius 0.32 m. Calculate the vehicle's translational kinetic energy, angular velocity of each wheel, total rotational kinetic energy of the wheels, and total kinetic energy.
14. A rotating machine contains a 0.40 kg component located 0.25 m from the axis and rotating at 400 rad/s. Calculate the centripetal force acting on the component and explain why high-speed machinery requires careful engineering.
15. An electric motor produces 150 N·m at 320 rad/s and drives the wheels through a 4:1 reduction gearbox that is 92% efficient. The wheels have radius 0.30 m. Calculate the motor power, gearbox output speed, gearbox output power, output torque, approximate tangential driving force at the wheels, and vehicle speed if the wheels roll without slipping.

Advanced Applications of Rigid Body Mechanics

Rigid body mechanics brings together the major ideas of translational and rotational motion to analyze objects whose size, shape, and mass distribution matter.

In an ideal rigid body, the distances between particles remain constant even when the object moves or rotates. Real materials deform slightly, but the rigid-body model is extremely useful for analyzing:

• vehicles and wheels
• gears and transmissions
• pulleys and winches
• cranes
• turbines and generators
• flywheels
• robots
• bridges and structures
• rolling objects
• rotating spacecraft

Advanced problems often require several physics principles at the same time rather than a single equation.

---

Translation and Rotation Together

A rigid body can undergo:

• pure translation
• pure rotation
• combined translation and rotation

A rolling wheel is a familiar example of combined motion.

Its centre of mass translates while the wheel rotates around its centre.

For rolling without slipping:

v = rω

and:

a = rα

These relationships connect the linear and rotational descriptions of the same system.

---

Translational and Rotational Equations

Many equations in rotational mechanics have direct translational equivalents.

Recognizing these parallels helps organize complex problems.

---

The Centre of Mass

The centre of mass is the point at which the mass of an object can be considered concentrated when analyzing its translational motion.

For the centre of mass:

ΣF_external = Ma_CM

This equation applies even when the object is also rotating.

The rotational motion can then be analyzed separately using torque.

For rotation about a suitable axis:

Στ = Iα

This separation of translation and rotation is one of the most powerful ideas in rigid-body mechanics.

---

Forces and Torques

A force can produce:

• translational acceleration
• rotational acceleration
• both

Whether a force produces torque depends on where and how it acts.

Torque magnitude is:

τ = rF sin θ

or:

τ = Fd

where d is the perpendicular distance from the axis to the force's line of action.

A force acting through the centre of mass may accelerate the object without producing torque about the centre.

A force acting away from the centre can produce both translation and rotation.

---

Worked Force-and-Torque Example

A:

20 kg

solid disk of radius:

0.50 m

experiences a tangential force of:

60 N

at its rim.

Ignoring other horizontal forces:

Translational acceleration:

a = F/m

a = 60/20

a = 3.0 m/s²

For a solid disk:

I = 1/2 mr²

I = 1/2(20)(0.50²)

I = 2.5 kg·m²

Torque:

τ = rF

τ = (0.50)(60)

τ = 30 N·m

Angular acceleration:

α = τ/I

α = 30/2.5

α = 12 rad/s²

The same force produces both translation and rotation.

---

Moment of Inertia in Complex Systems

Moment of inertia is the rotational equivalent of mass, but unlike ordinary mass it depends strongly on mass distribution.

For several point masses:

I = Σmr²

For a rigid object, the appropriate moment of inertia depends on:

• total mass
• shape
• rotation axis
• distribution of mass

Moving mass farther from the axis increases I.

This affects:

• angular acceleration
• rotational kinetic energy
• angular momentum
• rolling behaviour

---

Parallel-Axis Theorem

Sometimes an object rotates about an axis that does not pass through its centre of mass.

The parallel-axis theorem can be used:

I = I_CM + Md²

where:

• I_CM = moment of inertia about a parallel axis through the centre of mass
• M = total mass
• d = distance between the axes

This is useful for doors, rods, pendulums, machinery, and structural components.

---

Worked Parallel-Axis Example

A uniform rod has:

M = 6.0 kg

length:

ℓ = 2.0 m

About its centre:

I_CM = 1/12 Mℓ²

I_CM = 1/12(6)(2²)

I_CM = 2.0 kg·m²

For an axis through one end:

d = ℓ/2 = 1.0 m

Using the parallel-axis theorem:

I = 2.0 + (6)(1.0²)

I = 8.0 kg·m²

This agrees with the standard rod-about-one-end result:

I = 1/3 Mℓ²

---

Conservation of Energy in Rigid Bodies

Complex rigid-body systems can often be solved efficiently using energy conservation.

For an ideal system:

E_initial = E_final

Possible energy stores include:

• gravitational potential energy
• elastic potential energy
• translational kinetic energy
• rotational kinetic energy

A general mechanical-energy equation might be:

mgh + 1/2 kx² = 1/2 mv² + 1/2 Iω²

Not every term will appear in every problem.

---

Rolling Objects

For an object rolling without slipping:

v = rω

Total kinetic energy is:

E_k = 1/2 mv² + 1/2 Iω²

Substituting:

ω = v/r

gives:

E_k = 1/2 mv² + 1/2 I(v²/r²)

This shows why moment of inertia affects the translational speed of rolling objects.

---

Rolling Down an Incline

For an object starting from rest and descending through vertical height h:

mgh = 1/2 mv² + 1/2 Iω²

Using:

ω = v/r

gives:

v = √{2gh/[1 + I/(mr²)]}

Objects with smaller I/(mr²) reach greater translational speeds.

For ideal objects:

solid sphere → solid cylinder → thin ring

from fastest to slowest.

---

Force-and-Torque Method for Rolling

The same problem can be solved using dynamics.

Along the incline:

mg sin θ − f = ma

About the centre:

fr = Iα

and:

a = rα

Combining these equations gives:

a = g sin θ/[1 + I/(mr²)]

This illustrates an important feature of advanced mechanics:

different methods can describe the same physical system and should give consistent results.

---

Choosing Energy or Dynamics

The energy method is often easiest when the problem asks for:

• final speed
• energy transformations
• maximum height
• energy losses

The force-and-torque method is often better when the problem asks for:

• acceleration
• friction
• tension
• normal force
• torque

Strong problem solving involves choosing the method that provides the required quantity most directly.

---

Pulley Systems

Real pulleys can have significant moment of inertia.

A massive pulley therefore stores rotational kinetic energy.

Suppose two masses m₁ and m₂ are connected by a rope over a pulley.

If the rope does not slip:

a = rα

The tensions on the two sides of the pulley are generally different because their difference must produce torque.

For the pulley:

(T₂ − T₁)r = Iα

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Advanced Pulley Example

Consider:

m₁ = 4.0 kg

m₂ = 6.0 kg

pulley radius:

r = 0.20 m

pulley moment of inertia:

I = 0.30 kg·m²

Assume m₂ moves downward.

For m₁:

T₁ − m₁g = m₁a

For m₂:

m₂g − T₂ = m₂a

For the pulley:

(T₂ − T₁)r = Iα

Since:

α = a/r

then:

T₂ − T₁ = Ia/r²

Combining the equations gives:

a = (m₂ − m₁)g/[m₁ + m₂ + I/r²]

Calculate:

I/r² = 0.30/0.20²

= 7.5 kg

Therefore:

a = (6 − 4)(9.8)/(4 + 6 + 7.5)

a = 19.6/17.5

a = 1.12 m/s²

---

Why a Massive Pulley Reduces Acceleration

If the pulley had negligible moment of inertia:

a = (6 − 4)(9.8)/(6 + 4)

a = 1.96 m/s²

With the massive pulley:

a = 1.12 m/s²

Some of the gravitational effect must accelerate the pulley rotationally.

The system therefore accelerates less.

---

Solving the Pulley with Energy

Suppose the masses move distance h.

The net decrease in gravitational potential energy is:

(m₂ − m₁)gh

This becomes:

• translational kinetic energy of m₁
• translational kinetic energy of m₂
• rotational kinetic energy of the pulley

Therefore:

(m₂ − m₁)gh = 1/2 m₁v² + 1/2 m₂v² + 1/2 Iω²

with:

ω = v/r

Again, energy and dynamics describe the same system.

---

Angular Momentum in Complex Systems

Angular momentum is conserved when the net external torque about the chosen axis is zero:

L_initial = L_final

For several objects:

ΣL_initial = ΣL_final

For simple rigid rotors:

L = Iω

Angular momentum conservation is especially useful for:

• rotational collisions
• spacecraft
• spinning platforms
• flywheels
• athletes
• coupled disks

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Rotational Collision Example

A disk has:

I₁ = 3.0 kg·m²

ω₁ = 8.0 rad/s

A second disk has:

I₂ = 5.0 kg·m²

and is initially stationary.

The disks are brought together and rotate as one.

Angular momentum conservation:

I₁ω₁ = (I₁ + I₂)ω_f

(3)(8) = (3 + 5)ω_f

24 = 8ω_f

ω_f = 3.0 rad/s

---

Energy in the Rotational Collision

Initial energy:

E_i = 1/2(3)(8²)

E_i = 96 J

Final energy:

E_f = 1/2(8)(3²)

E_f = 36 J

Difference:

96 − 36 = 60 J

Angular momentum is conserved, but rotational kinetic energy is not.

The 60 J is transferred into forms such as:

• thermal energy
• sound
• deformation

This demonstrates why different conservation laws must not be confused.

---

Angular Momentum and Changing Shape

If external torque is negligible:

I_iω_i = I_fω_f

A person on a rotating platform can change angular velocity by changing mass distribution.

If I decreases:

ω increases

If I increases:

ω decreases

However, rotational kinetic energy does not necessarily remain constant because the person may perform internal work while changing body position.

---

Impulse and Angular Impulse

Linear impulse changes linear momentum:

FΔt = Δp

Angular impulse changes angular momentum:

τΔt = ΔL

These relationships are useful when forces or torques act strongly for short periods.

Examples include:

• impacts
• braking
• sports
• machine clutches
• collisions
• sudden loads

---

Work and Energy in Rotating Systems

For constant torque:

W = τΔθ

The rotational work-energy theorem is:

W_net = ΔE_rot

Therefore:

W_net = 1/2 Iω_f² − 1/2 Iω_i²

This can sometimes replace a longer torque-and-angular-acceleration calculation.

---

Rotational Power

Power measures the rate of energy transfer.

For rotation:

P = τω

This equation is extremely important in engineering.

A motor producing large torque at low angular velocity can have the same power as a motor producing smaller torque at high angular velocity.

---

Worked Machine-Power Example

A motor produces:

τ = 250 N·m

at:

ω = 160 rad/s

Power:

P = τω

P = (250)(160)

P = 40,000 W

or:

40 kW

If the motor operates for:

30 s

at this constant power, energy transferred is:

E = Pt

E = (40,000)(30)

E = 1.20 × 10⁶ J

or:

1.20 MJ

---

Gears and Conservation of Energy

Ideal gears change torque and angular speed while conserving power.

For an ideal gear system:

P_in = P_out

Therefore:

τ_inω_in = τ_outω_out

If angular speed decreases, torque increases.

This is why gear reduction is useful when large output torque is needed.

---

Gearbox Example

A motor provides:

τ_in = 80 N·m

ω_in = 300 rad/s

Input power:

P_in = (80)(300)

P_in = 24 kW

A gearbox reduces angular speed to:

ω_out = 75 rad/s

For an ideal gearbox:

τ_out = P/ω_out

τ_out = 24,000/75

τ_out = 320 N·m

The angular speed is quartered while torque is quadrupled.

---

Real Gearboxes

Real gearboxes are not perfectly efficient.

Suppose the previous gearbox is:

90% efficient.

Then:

P_out = 0.90(24,000)

P_out = 21,600 W

Output torque:

τ_out = 21,600/75

τ_out = 288 N·m

Energy is transferred into less useful forms through:

• friction
• vibration
• sound
• heating

---

Vehicles as Rigid-Body Systems

A vehicle provides an excellent example of integrated rigid-body mechanics.

Its motion involves:

• translation of the vehicle
• rotation of wheels
• motor or engine rotation
• gears
• driveshafts
• friction
• braking
• energy transfer

A complete model may require both translational and rotational kinetic energy.

---

Vehicle Energy

For a vehicle of mass M moving at speed v:

E_vehicle = 1/2 Mv²

Each wheel also stores:

E_wheel = 1/2 Iω²

Therefore:

E_total = 1/2 Mv² + Σ(1/2 Iω²)

For rolling wheels:

ω = v/r

Ignoring wheel rotation underestimates the total energy required to accelerate the vehicle.

---

Vehicle Acceleration and Wheel Inertia

Suppose four identical wheels each have moment of inertia I and radius r.

Their combined rotational kinetic energy is:

4(1/2 Iω²)

Using:

ω = v/r

gives:

E_rot,wheels = 2Iv²/r²

This behaves as though the vehicle has additional effective inertia during acceleration.

Reducing wheel moment of inertia can therefore improve responsiveness.

---

Braking Systems

When a vehicle brakes, both translational and rotational kinetic energy must decrease.

Friction brakes convert much of this mechanical energy into thermal energy.

Regenerative braking converts part of it into electrical energy.

A complete energy equation might be:

E_k,initial = E_battery + E_thermal + other losses

Total energy remains conserved.

---

Regenerative Braking Example

A vehicle has total kinetic energy:

600 kJ

before braking.

Its regenerative system recovers:

65%

of this energy.

Recovered energy:

E_recovered = 0.65(600)

E_recovered = 390 kJ

Energy transferred into other forms:

600 − 390

= 210 kJ

---

Flywheel Energy Storage

Flywheels store energy as rotational kinetic energy:

E_rot = 1/2 Iω²

They can be used to:

• smooth power fluctuations
• recover braking energy
• provide short bursts of power
• stabilize machinery

Because energy depends on ω², increasing rotational speed can greatly increase stored energy.

However, high-speed flywheels require strong materials and careful containment.

---

Engineering Safety and High-Speed Rotation

A rotating mass requires centripetal force:

F_c = mrω²

Because the force depends on ω², very high rotational speeds can produce enormous stresses.

Potential failures include:

• cracking
• blade separation
• bearing failure
• shaft failure
• catastrophic rotor breakup

Engineering designs therefore include substantial safety margins.

---

Turbines

Turbines transfer energy from moving fluids into rotational mechanical energy.

Examples include:

• wind turbines
• steam turbines
• hydroelectric turbines
• gas turbines

A turbine's rotating shaft may drive:

• generators
• compressors
• pumps
• machinery

Analysis can involve torque, power, angular momentum, energy conservation, and efficiency.

---

Turbine Power Example

A turbine shaft rotates at:

ω = 120 rad/s

while delivering:

τ = 5000 N·m

Mechanical power:

P = τω

P = (5000)(120)

P = 600,000 W

or:

600 kW

If the generator is:

94% efficient

then electrical output power is:

P_electric = 0.94(600)

P_electric = 564 kW

---

Cranes and Rotational Equilibrium

Rigid-body mechanics also applies when objects are not accelerating.

For static equilibrium:

ΣF = 0

and:

Στ = 0

A crane must satisfy both conditions when holding a stationary load.

Counterweights create torque opposing the load torque.

---

Crane Example

A:

5000 kg

load hangs:

12 m

from the crane's pivot.

Load torque:

τ_load = rmg

τ_load = (12)(5000)(9.8)

τ_load = 588,000 N·m

A counterweight is positioned:

6.0 m

on the opposite side.

For equilibrium:

τ_counter = τ_load

6m_cg = 588,000

m_c = 588,000/[6(9.8)]

m_c = 10,000 kg

This simplified model neglects the mass of the crane structure itself.

---

Tipping Stability

A rigid object remains stable when the vertical line through its centre of mass falls within its base of support.

If the line moves beyond the edge of the base, gravity produces a torque that tends to tip the object.

This principle is important for:

• cranes
• trucks
• forklifts
• buildings
• robots
• construction equipment

---

Vehicle Rollover

A vehicle with a high centre of mass is generally more susceptible to rollover.

During a turn, the required lateral forces create moments that can shift loads between the wheels.

Factors affecting rollover risk include:

• centre-of-mass height
• track width
• speed
• turning radius
• suspension
• tire forces

Rigid-body mechanics helps engineers evaluate these factors.

---

Robotics

Robots combine many rigid-body systems:

• rotating joints
• links
• gears
• motors
• arms
• wheels

A robotic arm must overcome torque caused by:

• its own weight
• payload weight
• angular acceleration
• external forces

As an arm extends, the torque required at its base can increase substantially because the lever arm increases.

---

Robotic Arm Example

A robotic arm holds:

m = 20 kg

at a horizontal distance:

r = 1.5 m

from its shoulder joint.

The gravitational torque is:

τ = rmg

τ = (1.5)(20)(9.8)

τ = 294 N·m

The motor must produce at least this much opposing torque merely to hold the load stationary, ignoring the arm's own mass.

More torque is required to accelerate the arm upward.

---

Spacecraft Attitude Control

Rigid-body mechanics also governs spacecraft rotation.

A spacecraft can change orientation using:

• reaction wheels
• control moment gyroscopes
• thrusters

Reaction wheels rely on conservation of angular momentum.

If an internal wheel gains angular momentum in one direction, the spacecraft body gains angular momentum in the opposite direction.

L_total = constant

when external torque is negligible.

---

Reaction-Wheel Example

A spacecraft body has:

I_s = 1000 kg·m²

A reaction wheel changes its angular momentum by:

+50 kg·m²/s

The spacecraft body must change by:

−50 kg·m²/s

Therefore:

Δω_s = ΔL/I

Δω_s = −50/1000

Δω_s = −0.050 rad/s

The spacecraft rotates opposite to the wheel's change in angular momentum.

---

Gyroscopic Effects

Rapidly spinning rotors can possess large angular momentum:

L = Iω

An external torque changes that angular momentum:

τ = ΔL/Δt

If the torque mainly changes the direction of L, the rotor can undergo precession.

Gyroscopic effects are important in:

• spacecraft
• navigation systems
• motorcycles
• rotating machinery

---

Conservation Laws in Complex Systems

Three major conservation principles are especially useful in rigid-body mechanics.

Energy conservation

Use when tracking transformations between potential, translational, rotational, thermal, and other energy stores.

Linear momentum conservation

Use when the net external impulse on a system is negligible.

Angular momentum conservation

Use when the net external torque about the chosen axis is negligible.

The correct conservation law depends on the system and conditions.

---

Conservation Laws Can Apply Differently

A system can conserve one quantity without conserving another.

For example, during an inelastic rotational collision:

• angular momentum may be conserved
• rotational kinetic energy may decrease

During braking:

• total energy is conserved
• mechanical energy decreases
• angular momentum of the vehicle alone is not conserved because external torque acts

Always identify the system and external interactions before applying a conservation law.

---

Choosing the System Boundary

The system boundary determines which forces, torques, and energy transfers are considered internal or external.

For example, consider a braking wheel.

If the system is only the wheel, friction transfers energy out of the wheel.

If the system includes the wheel and brake assembly, the energy remains within the system but changes primarily into thermal energy.

Choosing the system carefully makes conservation equations much clearer.

---

Multi-Stage Engineering Systems

Many real systems involve several stages.

Consider:

motor → gearbox → shaft → wheel → vehicle

Each stage may change:

• angular speed
• torque
• energy
• power

Ideally:

P_in = P_out

but real systems have efficiency losses.

If several components have efficiencies η₁, η₂ and η₃:

η_total = η₁η₂η₃

when efficiencies are written as decimals.

---

Worked Multi-Stage Efficiency Example

A motor has efficiency:

η₁ = 0.90

A gearbox has:

η₂ = 0.92

A drivetrain has:

η₃ = 0.95

Overall efficiency:

η_total = (0.90)(0.92)(0.95)

η_total ≈ 0.787

or:

78.7%

If electrical input power is:

100 kW

useful mechanical output is approximately:

78.7 kW

---

Engineering Optimization

Real engineering is rarely about maximizing only one variable.

A designer may need to balance:

• mass
• moment of inertia
• strength
• torque
• speed
• power
• efficiency
• stability
• friction
• heat
• reliability
• cost
• safety

For example, increasing flywheel mass can increase stored energy but also increases structural loads and system mass.

Increasing rotational speed can dramatically increase stored energy but also increases mechanical stress.

---

Dimensional and Unit Checks

Advanced problems should always include a unit check.

Examples:

Torque:

N·m

Angular momentum:

kg·m²/s

Moment of inertia:

kg·m²

Angular velocity:

rad/s

Angular acceleration:

rad/s²

Energy:

J

Power:

W

Correct units provide a useful check that the correct relationships have been used.

---

Communicating a Physics Solution

A strong physics solution should show more than a numerical answer.

A clear solution normally includes:

• known quantities
• unknown quantity
• relevant physical principle
• equation
• substitution
• calculation
• units
• direction where relevant
• interpretation

For complex problems, a short diagram can also clarify:

• forces
• distances
• axes
• directions
• energy changes

---

Using Appropriate Terminology

Instead of writing:

"The wheel makes the car move."

A more precise explanation is:

"The motor supplies torque to the driven wheel. Static friction between the tire and road produces the external horizontal force that accelerates the vehicle."

Instead of:

"The ring is slower because it is harder to spin."

A stronger explanation is:

"The ring has a larger moment of inertia relative to mr², so a greater fraction of its kinetic energy is rotational and its translational acceleration down the incline is smaller."

Precise terminology demonstrates understanding.

---

Complete Integrated Engineering Problem

An electric motor delivers:

τ_motor = 100 N·m

at:

ω_motor = 400 rad/s

It drives a wheel through a:

5:1 reduction gearbox

with efficiency:

90%

The wheel radius is:

r = 0.35 m

Motor Power

P_motor = τω

P_motor = (100)(400)

P_motor = 40,000 W

Gearbox Output Speed

ω_out = 400/5

ω_out = 80 rad/s

Output Power

P_out = 0.90(40,000)

P_out = 36,000 W

Output Torque

τ_out = P_out/ω_out

τ_out = 36,000/80

τ_out = 450 N·m

Approximate Driving Force

F = τ/r

F = 450/0.35

F ≈ 1286 N

Vehicle Speed

For rolling without slipping:

v = rω

v = (0.35)(80)

v = 28 m/s

or:

100.8 km/h

This one system uses:

• torque
• angular velocity
• power
• gears
• efficiency
• rolling motion
• translational force

---

Complete Conservation Problem

A:

5.0 kg

solid cylinder of radius:

0.30 m

rolls from rest down a ramp through a vertical height:

4.0 m

and then moves up another ramp.

Assume no energy losses.

For a solid cylinder:

I = 1/2 mr²

At the bottom:

mgh = 1/2 mv² + 1/2 Iω²

Using:

ω = v/r

gives:

mgh = 1/2 mv² + 1/4 mv²

mgh = 3/4 mv²

Therefore:

v² = 4gh/3

v² = 4(9.8)(4)/3

v² ≈ 52.27

v ≈ 7.23 m/s

As it rolls up the second ramp, translational and rotational kinetic energy transform back into gravitational potential energy.

With no losses, it reaches the same vertical height:

4.0 m

This illustrates conservation of mechanical energy in combined translation and rotation.

---

Complete Problem with Energy Loss

Suppose the same cylinder loses:

40 J

of mechanical energy before reaching the second ramp.

Initial energy:

E_i = mgh

E_i = (5)(9.8)(4)

E_i = 196 J

Remaining mechanical energy:

196 − 40 = 156 J

At maximum height on the second ramp:

mgh₂ = 156

h₂ = 156/[5(9.8)]

h₂ ≈ 3.18 m

Energy loss reduces the maximum height.

---

Complete Angular Momentum and Energy Problem

A person stands on a rotating platform.

Initially:

I_i = 6.0 kg·m²

ω_i = 4.0 rad/s

The person pulls masses inward so that:

I_f = 3.0 kg·m²

With negligible external torque:

I_iω_i = I_fω_f

(6)(4) = (3)ω_f

ω_f = 8.0 rad/s

Initial rotational energy:

E_i = 1/2(6)(4²)

E_i = 48 J

Final rotational energy:

E_f = 1/2(3)(8²)

E_f = 96 J

Angular momentum is conserved, but rotational kinetic energy increases.

The additional 48 J comes from internal work performed by the person.

This example demonstrates why conservation of angular momentum does not automatically imply conservation of mechanical energy.

---

Complete Static-Equilibrium Problem

A horizontal beam has length:

6.0 m

and mass:

100 kg

It is supported at its left end.

A cable provides an upward force at the right end.

A:

200 kg

load hangs:

4.0 m

from the pivot.

The beam's own weight acts at its centre:

3.0 m

from the pivot.

Taking torques about the left support:

F_cable(6) − (100)(9.8)(3) − (200)(9.8)(4) = 0

6F_cable = 2940 + 7840

6F_cable = 10,780

F_cable ≈ 1797 N

Choosing the support as the pivot eliminates the unknown support force from the torque equation.

This is an important strategy in equilibrium problems.

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Evaluating a Real Engineering Design

When evaluating an engineering system, ask several questions.

Motion: Is the system translating, rotating, or both?

Forces: What external forces act?

Torques: Which forces produce torque, and about which axis?

Inertia: How is mass distributed?

Energy: What energy enters, leaves, or changes form?

Momentum: Are linear or angular momentum conservation useful?

Friction: Is friction useful, harmful, or both?

Efficiency: Where is energy transferred into less useful forms?

Stability: Could the system tip, vibrate, slip, or become unstable?

Safety: What stresses and loads occur?

This approach connects classroom physics to real engineering analysis.

---

Common Mistakes

Treating translation and rotation as unrelated

Many rigid bodies undergo both simultaneously.

Forgetting rotational kinetic energy

A rolling or rotating component can store significant energy.

Using F = ma when torque is also required

An off-centre force can produce both translational and angular acceleration.

Using τ = Iα without identifying the correct axis

Moment of inertia depends on the chosen axis.

Assuming tension is the same on both sides of a massive pulley

Different tensions are required to produce net torque on the pulley.

Applying angular momentum conservation when external torque is significant

Always examine external torques first.

Assuming angular momentum conservation means kinetic energy conservation

The two conservation laws have different conditions.

Ignoring system boundaries

Whether a force, torque, or energy transfer is external depends on the chosen system.

Ignoring efficiency

Real engineering systems transfer some energy into heat, sound, vibration, and deformation.

Forgetting v = rω

This relationship is essential for rolling without slipping.

Using rpm as though it were rad/s

Convert rotational speed before using equations such as P = τω.

Ignoring units and directions

Advanced solutions should communicate both magnitude and physical meaning.

---

Key Terms

Rigid body — idealized object whose shape and internal distances remain constant during motion.

Centre of mass — point representing the average position of an object's mass for translational analysis.

Torque — rotational effect of a force about an axis.

Moment of inertia — measure of resistance to angular acceleration determined by mass distribution and axis.

Parallel-axis theorem — relationship I = I_CM + Md² used to find moment of inertia about an axis parallel to one through the centre of mass.

Rolling without slipping — combined translation and rotation satisfying v = rω.

Rotational kinetic energy — energy associated with rotation, E_rot = 1/2 Iω².

Angular momentum — rotational momentum; for a simple rigid rotor, L = Iω.

Angular impulse — change in angular momentum produced by torque acting over time.

Rotational power — rate of rotational energy transfer, P = τω.

Static equilibrium — condition in which both net force and net torque are zero.

Conservation law — principle stating that a physical quantity remains constant under specified conditions.

System boundary — chosen division between a physical system and its surroundings.

Efficiency — ratio of useful output energy or power to input energy or power.

Drivetrain — system that transfers mechanical power from a motor or engine to driven components.

Reaction wheel — rotating device used to control spacecraft orientation through angular momentum exchange.

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Key Takeaways

• Rigid-body mechanics integrates translational and rotational physics.
• The centre of mass follows ΣF = Ma_CM.
• Rotational motion follows Στ = Iα about an appropriate axis.
• A single force can produce translation, rotation, or both.
• Moment of inertia depends on both mass distribution and rotation axis.
• The parallel-axis theorem extends moment-of-inertia calculations to displaced axes.
• Rolling objects possess both translational and rotational kinetic energy.
• Rolling without slipping connects motion through v = rω and a = rα.
• Massive pulleys require different rope tensions because a net torque is needed to rotate them.
• Energy methods can simplify systems containing several moving and rotating components.
• Angular momentum is conserved only when net external torque is zero.
• Angular momentum can be conserved even when rotational kinetic energy changes.
• Rotational work is related to torque and angular displacement.
• Rotational power is P = τω.
• Gears trade angular speed for torque rather than creating energy.
• Real engineering systems must account for efficiency and energy losses.
• Static equilibrium requires both ΣF = 0 and Στ = 0.
• Engineering applications often require several physical principles simultaneously.
• Choosing the correct system boundary is essential when applying conservation laws.
• Clear physics communication requires equations, units, terminology, assumptions, and interpretation.

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Check Your Understanding

1. Explain the difference between pure translation, pure rotation, and combined translation and rotation.
2. A 12 kg solid disk of radius 0.40 m experiences a tangential force of 50 N at its rim. Calculate its translational acceleration, torque about its centre, moment of inertia, and angular acceleration.
3. A uniform rod has mass 8.0 kg and length 3.0 m. Use the parallel-axis theorem to determine its moment of inertia about one end.
4. Explain why a massive pulley causes an Atwood machine to accelerate less than an ideal massless pulley.
5. Two masses of 3.0 kg and 5.0 kg are connected over a pulley with radius 0.20 m and moment of inertia 0.20 kg·m². Calculate the acceleration of the system.
6. A solid cylinder rolls from rest through a vertical height of 5.0 m. Calculate its translational speed and angular velocity at the bottom if its radius is 0.25 m.
7. Explain how the force-and-torque method and the energy method can both be used to analyze a rolling object.
8. A motor produces 180 N·m of torque at 200 rad/s. Calculate its mechanical power and the energy transferred during 45 s of operation.
9. A gearbox receives 30 kW at 300 rad/s and reduces the output speed to 75 rad/s. If its efficiency is 92%, calculate the output power and output torque.
10. Two disks have I₁ = 4.0 kg·m² and I₂ = 6.0 kg·m². Disk 1 initially rotates at 15 rad/s, while Disk 2 is stationary. They couple together. Calculate their final angular velocity, initial rotational kinetic energy, final rotational kinetic energy, and energy transferred into other forms.
11. Explain why angular momentum can be conserved while rotational kinetic energy changes.
12. A crane holds a 3000 kg load 10 m from its pivot. A counterweight is positioned 5.0 m on the opposite side. Ignoring the crane's own mass, determine the counterweight required for rotational equilibrium.
13. A vehicle has mass 1500 kg and travels at 20 m/s. Each of its four wheels has I = 1.4 kg·m² and radius 0.32 m. Calculate the vehicle's translational kinetic energy, total rotational kinetic energy of the wheels, and total kinetic energy.
14. A spacecraft has moment of inertia 1200 kg·m². An internal reaction wheel gains 72 kg·m²/s of angular momentum. Assuming negligible external torque, calculate the change in the spacecraft body's angular velocity and explain the direction of the change.
15. An electric motor produces 120 N·m at 350 rad/s. It drives a vehicle through a 5:1 gearbox that is 90% efficient. The wheel radius is 0.32 m. Calculate the motor power, gearbox output angular velocity, output power, output torque, approximate driving force at the wheels, and vehicle speed. Then identify at least three assumptions made in this model and explain how real-world effects could change the results.