Linear and Angular Relationships

Rotational motion and linear motion are closely connected. When an object rotates, points on the object move along circular paths. Their motion can therefore be described using both angular quantities and linear quantities.

The most important relationships are:

s = rθ

v = rω

a_t = rα

a_c = rω²

where:

• s = distance travelled along the circular path
• r = radius
• θ = angular displacement in radians
• v = tangential or linear speed
• ω = angular velocity
• a_t = tangential acceleration
• α = angular acceleration
• a_c = centripetal acceleration

These equations allow us to move between rotational and linear descriptions of the same motion.

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Angular Displacement and Linear Distance

Consider a point on a rotating wheel.

As the wheel rotates through an angular displacement θ, the point travels along a circular arc.

The length of this arc is:

s = rθ

This equation connects angular displacement to linear distance.

It applies when:

θ is measured in radians.

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Understanding s = rθ

Suppose a wheel has radius:

r = 0.50 m

and rotates through:

θ = 2 rad

Then:

s = rθ

s = (0.50)(2)

s = 1.0 m

A point on the edge of the wheel travels 1.0 m along its circular path.

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Why Radius Matters

Consider two points on the same rotating disk.

Point A is:

0.10 m

from the centre.

Point B is:

0.40 m

from the centre.

The disk rotates through:

θ = 3 rad

For Point A:

s_A = rθ

s_A = (0.10)(3)

s_A = 0.30 m

For Point B:

s_B = (0.40)(3)

s_B = 1.20 m

Both points experience the same angular displacement, but Point B travels four times farther.

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Same Angle, Different Distance

Every point on a rigid rotating object rotates through the same angular displacement.

However:

s = rθ

Therefore, the farther a point is from the axis:

the greater its linear distance travelled.

For the same θ:

s ∝ r

A point twice as far from the centre travels twice the distance.

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Finding Angular Displacement

The relationship can be rearranged:

θ = s/r

Suppose a point on the edge of a wheel travels:

s = 4.5 m

and the wheel has radius:

r = 0.30 m

Then:

θ = 4.5/0.30

θ = 15 rad

Number of revolutions:

N = θ/(2π)

N = 15/(2π)

N ≈ 2.39 revolutions

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Degrees Must Be Converted to Radians

The equation:

s = rθ

requires θ to be measured in radians.

Suppose a wheel of radius 0.40 m rotates through:

90°

First convert:

90° = π/2 rad

Then:

s = (0.40)(π/2)

s ≈ 0.628 m

Using 90 directly in the equation would give an incorrect answer.

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Angular Velocity and Tangential Velocity

Angular velocity describes how quickly an object's angular position changes.

Tangential velocity describes how quickly a point moves along its circular path.

They are related by:

v = rω

This relationship comes directly from:

s = rθ

Divide both sides by time:

s/t = r(θ/t)

Since:

v = s/t

and:

ω = θ/t

then:

v = rω

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Direction of Tangential Velocity

The instantaneous linear velocity of a point moving in a circle is directed:

tangent to the circular path

It is not directed toward the centre.

At every moment:

• the radius points from the centre to the object
• tangential velocity is perpendicular to the radius

The direction of the velocity continually changes as the object rotates.

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Same Angular Velocity, Different Linear Velocity

Every point on a rigid rotating object has the same angular velocity.

However:

v = rω

so linear speed increases with radius.

Suppose a disk rotates at:

ω = 8 rad/s

Point A:

r_A = 0.20 m

Point B:

r_B = 0.50 m

For Point A:

v_A = (0.20)(8)

v_A = 1.6 m/s

For Point B:

v_B = (0.50)(8)

v_B = 4.0 m/s

The outer point moves faster.

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What Happens at the Axis?

At the axis:

r = 0

Therefore:

v = rω

v = 0

A point exactly on the rotation axis has:

zero tangential speed

even though the object itself may have a large angular velocity.

This illustrates the difference between angular and linear motion.

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Angular Velocity from Linear Velocity

The equation can be rearranged:

ω = v/r

Suppose a point moves around a circular path at:

v = 12 m/s

with radius:

r = 0.60 m

Then:

ω = 12/0.60

ω = 20 rad/s

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Frequency and Linear Speed

Angular velocity can also be related to frequency:

ω = 2πf

Combining this with:

v = rω

gives:

v = 2πrf

This makes sense because:

2πr

is the circumference of the circle.

If an object completes f revolutions each second, it travels f circumferences each second.

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Worked Example: Rotating Platform

A platform has radius:

r = 2.0 m

and rotates at:

f = 0.50 Hz

Angular velocity:

ω = 2πf

ω = 2π(0.50)

ω = π rad/s

A person at the outer edge has tangential speed:

v = rω

v = (2.0)(π)

v ≈ 6.28 m/s

A person halfway from the centre at r = 1.0 m has:

v = (1.0)(π)

v ≈ 3.14 m/s

Both people have the same angular velocity but different tangential speeds.

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Period and Linear Speed

Since:

ω = 2π/T

and:

v = rω

we can write:

v = 2πr/T

This is another way to calculate the speed of an object undergoing uniform circular motion.

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Worked Example

A point travels in a circle of radius:

r = 3.0 m

and completes one revolution every:

T = 4.0 s

Then:

v = 2πr/T

v = 2π(3.0)/4.0

v ≈ 4.71 m/s

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Angular Acceleration and Tangential Acceleration

If angular velocity changes, the tangential velocity of points on the rotating object also changes.

The relationship is:

a_t = rα

This is the acceleration equivalent of:

v = rω

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Worked Example

A wheel has radius:

r = 0.40 m

and angular acceleration:

α = 5 rad/s²

The tangential acceleration at its edge is:

a_t = rα

a_t = (0.40)(5)

a_t = 2.0 m/s²

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Same Angular Acceleration, Different Tangential Acceleration

All points on a rigid rotating body have the same angular acceleration.

However:

a_t = rα

Therefore, points farther from the axis experience greater tangential acceleration.

For example, if:

α = 4 rad/s²

then:

Tangential acceleration increases directly with radius.

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Centripetal Acceleration

Even if angular velocity is constant, a point moving in a circle is accelerating.

Why?

Because its velocity direction is continuously changing.

This inward acceleration is called centripetal acceleration.

The familiar linear equation is:

a_c = v²/r

Centripetal acceleration always points:

toward the centre of the circular path.

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Centripetal Acceleration Using Angular Velocity

Since:

v = rω

substitute into:

a_c = v²/r

Then:

a_c = (rω)²/r

a_c = r²ω²/r

Therefore:

a_c = rω²

This is one of the most important connections between angular and linear motion.

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Worked Example

A point is:

r = 0.50 m

from the centre of a disk rotating at:

ω = 6 rad/s

Then:

a_c = rω²

a_c = (0.50)(6²)

a_c = (0.50)(36)

a_c = 18 m/s²

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The Effect of Angular Velocity

Centripetal acceleration depends on:

ω²

Therefore:

a_c ∝ ω²

If angular velocity doubles:

a_c becomes four times larger.

If angular velocity triples:

a_c becomes nine times larger.

This makes high-speed rotation capable of producing very large accelerations.

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Example: Doubling Angular Velocity

Suppose:

r = 0.40 m

ω = 5 rad/s

Then:

a_c = (0.40)(5²)

a_c = 10 m/s²

Now double the angular velocity:

ω = 10 rad/s

a_c = (0.40)(10²)

a_c = 40 m/s²

Doubling ω increased centripetal acceleration by a factor of four.

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The Effect of Radius

For the same angular velocity:

a_c = rω²

Therefore:

a_c ∝ r

A point twice as far from the axis experiences twice the centripetal acceleration if both points share the same angular velocity.

This may initially seem different from:

a_c = v²/r

but remember that for a rigid rotating object:

v increases with radius.

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Tangential and Centripetal Acceleration

When a rotating object's angular speed changes, a point on it can have two acceleration components.

Tangential acceleration:

a_t = rα

This changes the magnitude of the velocity.

Centripetal acceleration:

a_c = rω²

This changes the direction of the velocity.

These accelerations are perpendicular to each other.

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Total Acceleration

If both tangential and centripetal acceleration are present:

a_total = √(a_t² + a_c²)

because the two acceleration components are perpendicular.

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Worked Example

A point is:

r = 0.30 m

from the axis of rotation.

At one instant:

ω = 8 rad/s

and:

α = 5 rad/s²

Tangential acceleration:

a_t = rα

a_t = (0.30)(5)

a_t = 1.5 m/s²

Centripetal acceleration:

a_c = rω²

a_c = (0.30)(8²)

a_c = 19.2 m/s²

Total acceleration:

a_total = √(1.5² + 19.2²)

a_total ≈ 19.3 m/s²

Notice that the centripetal acceleration is much larger in this example.

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Constant Angular Velocity

If:

ω = constant

then:

α = 0

Therefore:

a_t = rα = 0

However:

a_c = rω²

is still present.

So an object moving in a circle at constant speed still accelerates because the direction of its velocity changes.

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Rotational Motion Producing Linear Motion

Many machines convert rotational motion into linear motion.

Examples include:

• wheels moving vehicles
• pulleys lifting loads
• gears driving machinery
• winches pulling cables
• conveyor-belt rollers
• rack-and-pinion steering
• screws and threaded mechanisms

The relationships:

s = rθ

v = rω

a = rα

allow us to calculate the resulting linear motion.

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Rolling Without Slipping

A wheel rolling along a surface combines:

• rotational motion around its centre
• linear motion of its centre

If the wheel rolls without slipping:

s = rθ

v = rω

a = rα

where v and a describe the linear motion of the wheel's centre.

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Worked Example: Bicycle Wheel

A bicycle wheel has radius:

r = 0.35 m

and rotates at:

ω = 20 rad/s

The bicycle's speed is:

v = rω

v = (0.35)(20)

v = 7.0 m/s

Convert to km/h:

7.0 × 3.6

v = 25.2 km/h

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Distance from Wheel Rotation

Suppose the same bicycle wheel rotates through:

θ = 100 rad

Then:

s = rθ

s = (0.35)(100)

s = 35 m

If there is no slipping, the bicycle moves forward 35 m.

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Number of Wheel Revolutions

Since:

1 revolution = 2π rad

the circumference of a wheel is:

2πr

After one revolution:

θ = 2π

so:

s = r(2π)

s = 2πr

This shows why the familiar circumference formula is consistent with:

s = rθ

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Worked Vehicle Problem

A car wheel has radius:

0.32 m

The car travels:

100 m

without slipping.

Find the angular displacement of the wheel.

θ = s/r

θ = 100/0.32

θ = 312.5 rad

Number of revolutions:

N = 312.5/(2π)

N ≈ 49.7 revolutions

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Vehicle Acceleration

Suppose a car accelerates at:

3.0 m/s²

and its wheel radius is:

0.30 m

For rolling without slipping:

a = rα

Therefore:

α = a/r

α = 3.0/0.30

α = 10 rad/s²

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A Complete Wheel Problem

A bicycle wheel has radius:

r = 0.40 m

It starts from rest and accelerates at:

α = 3 rad/s²

for:

t = 5 s

Final Angular Velocity

ω_f = ω_i + αt

ω_f = 0 + (3)(5)

ω_f = 15 rad/s

Final Linear Speed

v = rω

v = (0.40)(15)

v = 6.0 m/s

Angular Displacement

Δθ = 1/2 αt²

Δθ = 1/2(3)(5²)

Δθ = 37.5 rad

Linear Distance

s = rθ

s = (0.40)(37.5)

s = 15 m

Tangential Acceleration

a_t = rα

a_t = (0.40)(3)

a_t = 1.2 m/s²

Final Centripetal Acceleration

a_c = rω²

a_c = (0.40)(15²)

a_c = 90 m/s²

This single example connects rotational kinematics to linear motion.

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Pulleys

A rotating pulley can produce linear motion in a rope or cable.

If the rope does not slip:

s = rθ

v = rω

a = rα

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Worked Pulley Example

A pulley has radius:

r = 0.20 m

and rotates at:

ω = 15 rad/s

The rope moves at:

v = rω

v = (0.20)(15)

v = 3.0 m/s

If:

α = 4 rad/s²

then the rope's acceleration is:

a = rα

a = (0.20)(4)

a = 0.80 m/s²

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Winches

A winch uses a rotating drum to wind or unwind a cable.

Suppose a winch drum has radius:

0.25 m

and rotates through:

20 rad

The cable moves:

s = rθ

s = (0.25)(20)

s = 5.0 m

If the drum rotates at:

8 rad/s

the cable speed is:

v = rω

v = (0.25)(8)

v = 2.0 m/s

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Gears

Meshing gears connect rotational motion between objects.

At their contact point, the tangential speeds must be equal if there is no slipping.

Therefore:

v₁ = v₂

Since:

v = rω

then:

r₁ω₁ = r₂ω₂

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Worked Gear Example

Gear A has radius:

r_A = 5 cm

and rotates at:

ω_A = 24 rad/s

Gear B has radius:

r_B = 15 cm

Then:

r_A ω_A = r_B |ω_B|

(5)(24) = (15)|ω_B|

|ω_B| = 8 rad/s

The gears rotate in opposite directions.

If:

ω_A = +24 rad/s

then:

ω_B = −8 rad/s

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Gear Ratios

From:

r₁ω₁ = r₂ω₂

we obtain:

ω₂/ω₁ = r₁/r₂

Therefore:

• smaller gears rotate faster
• larger gears rotate slower

A small driving gear connected to a larger gear reduces angular speed.

A large driving gear connected to a smaller gear increases angular speed.

This principle is used in:

• bicycles
• cars
• clocks
• industrial machines
• mechanical transmissions

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Belt-Driven Systems

Two pulleys connected by a belt behave similarly.

If the belt does not slip:

r₁ω₁ = r₂ω₂

because both pulley edges move at the same linear speed as the belt.

Unlike directly meshing gears, an uncrossed belt usually causes both pulleys to rotate in the same direction.

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Conveyor Belts

A rotating roller can drive a conveyor belt.

Suppose a roller has:

r = 0.15 m

ω = 10 rad/s

Then:

v = rω

v = (0.15)(10)

v = 1.5 m/s

The conveyor belt moves at approximately 1.5 m/s if there is no slipping.

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Rotational Motion in Motors

Electric motors produce rotational motion.

That rotation can be converted into linear motion through:

• wheels
• gears
• pulleys
• belts
• screws
• rack-and-pinion systems

For example, a motor rotating a wheel can propel a robot forward.

The motor's angular velocity determines the wheel's linear speed through:

v = rω

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Robot Wheel Example

A robot has wheels of radius:

0.08 m

Each wheel rotates at:

25 rad/s

Then:

v = rω

v = (0.08)(25)

v = 2.0 m/s

If the motor increases angular velocity at:

α = 10 rad/s²

then:

a = rα

a = (0.08)(10)

a = 0.80 m/s²

assuming rolling without slipping.

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Ferris Wheel Example

A Ferris wheel has radius:

r = 15 m

and rotates with angular velocity:

ω = 0.10 rad/s

Every passenger has the same angular velocity.

However, because passengers are all approximately the same distance from the axis, they also have approximately the same tangential speed:

v = rω

v = (15)(0.10)

v = 1.5 m/s

Their centripetal acceleration is:

a_c = rω²

a_c = (15)(0.10²)

a_c = 0.15 m/s²

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Rotating Platform Example

A rotating platform has angular velocity:

ω = 2 rad/s

Person A stands:

1 m

from the centre.

Person B stands:

3 m

from the centre.

Angular Velocity

Both have:

ω = 2 rad/s

Tangential Speed

Person A:

v_A = (1)(2)

v_A = 2 m/s

Person B:

v_B = (3)(2)

v_B = 6 m/s

Centripetal Acceleration

Person A:

a_c = (1)(2²)

a_c = 4 m/s²

Person B:

a_c = (3)(2²)

a_c = 12 m/s²

The person farther from the centre experiences greater tangential speed and greater centripetal acceleration.

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Centripetal Force

Centripetal acceleration requires a net inward force.

Newton's Second Law gives:

F_c = ma_c

Using:

a_c = rω²

gives:

F_c = mrω²

This force is not a new type of force.

"Centripetal force" means the net inward force provided by forces such as:

• friction
• tension
• gravity
• normal force

For example, friction between tires and a road can provide the centripetal force needed for a car to turn.

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Comparing Two Ways to Calculate Centripetal Acceleration

Centripetal acceleration can be calculated using:

a_c = v²/r

or:

a_c = rω²

Use whichever equation matches the information provided.

If v is known:

a_c = v²/r

If ω is known:

a_c = rω²

Both describe exactly the same physical acceleration.

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Worked Example Using Both Methods

A point moves in a circle of radius:

r = 2.0 m

with:

ω = 3.0 rad/s

First find linear speed:

v = rω

v = (2.0)(3.0)

v = 6.0 m/s

Using angular quantities:

a_c = rω²

a_c = (2.0)(3.0²)

a_c = 18 m/s²

Using linear quantities:

a_c = v²/r

a_c = 6.0²/2.0

a_c = 18 m/s²

Both methods give the same answer.

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Rotational and Linear Quantity Pairs

The major relationships can be summarized as:

The radius connects the rotational description to the linear description.

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The Importance of Radius

Radius plays a central role because points farther from the axis must travel farther during the same rotation.

For a fixed θ:

s ∝ r

For a fixed ω:

v ∝ r

For a fixed α:

a_t ∝ r

For a fixed ω:

a_c ∝ r

Therefore, even though every point on a rigid object shares the same angular motion, its linear motion depends strongly on its distance from the axis.

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A Complete Rotating-System Problem

A rotating disk has radius:

r = 0.60 m

It starts from rest and accelerates uniformly at:

α = 4 rad/s²

for:

t = 3 s

Angular Velocity

ω = αt

ω = (4)(3)

ω = 12 rad/s

Angular Displacement

θ = 1/2 αt²

θ = 1/2(4)(3²)

θ = 18 rad

Distance Travelled at the Edge

s = rθ

s = (0.60)(18)

s = 10.8 m

Tangential Speed

v = rω

v = (0.60)(12)

v = 7.2 m/s

Tangential Acceleration

a_t = rα

a_t = (0.60)(4)

a_t = 2.4 m/s²

Centripetal Acceleration

a_c = rω²

a_c = (0.60)(12²)

a_c = 86.4 m/s²

Total Acceleration

a_total = √(a_t² + a_c²)

a_total = √(2.4² + 86.4²)

a_total ≈ 86.4 m/s²

At high angular speeds, centripetal acceleration can become much larger than tangential acceleration.

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A Strategy for Rotating-System Problems

First identify the rotational quantities:

• θ
• ω
• α
• t

Then identify the radius:

r

If necessary, use rotational kinematics to find the missing angular quantity.

Then convert to the required linear quantity.

Use:

s = rθ

for distance.

Use:

v = rω

for tangential speed.

Use:

a_t = rα

for tangential acceleration.

Use:

a_c = rω²

for centripetal acceleration.

Finally, check that the units and physical interpretation make sense.

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Common Mistakes

Confusing angular and linear quantities

Angular velocity is measured in rad/s, while linear velocity is measured in m/s.

Thinking all points on a rotating object have the same linear speed

They have the same angular velocity, but their linear speeds depend on radius.

Using degrees in s = rθ

θ must be in radians.

Forgetting radius

The radius is what connects angular quantities to linear quantities.

Using a_t and a_c as though they were the same acceleration

Tangential acceleration changes speed. Centripetal acceleration changes direction.

Thinking constant angular velocity means zero acceleration

Tangential acceleration is zero, but centripetal acceleration is still present.

Thinking centripetal acceleration points along the motion

It points toward the centre.

Thinking centripetal force is a separate type of force

It is the name given to the net inward force responsible for circular motion.

Using v = rω when slipping occurs

The rolling relationship between wheel rotation and vehicle motion assumes rolling without slipping.

Forgetting that ω is squared in a_c = rω²

Doubling angular velocity makes centripetal acceleration four times larger.

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Key Terms

Linear displacement — change in position measured along a linear path.

Angular displacement (θ) — amount through which an object rotates.

Arc length (s) — distance measured along a circular path.

Radius (r) — distance from the axis of rotation to a point.

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

Tangential velocity (v) — instantaneous linear velocity of a point moving along a circular path.

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

Tangential acceleration (a_t) — acceleration that changes the magnitude of tangential velocity.

Centripetal acceleration (a_c) — inward acceleration that changes the direction of velocity.

Centripetal force — net inward force responsible for centripetal acceleration.

Rolling without slipping — motion in which wheel rotation and forward translation are directly related.

Rigid body — object in which points maintain fixed positions relative to one another.

Axis of rotation — line around which an object rotates.

Frequency (f) — number of complete rotations per second.

Period (T) — time required for one complete rotation.

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Key Takeaways

• Linear and angular motion describe different aspects of the same rotating system.
• Angular displacement and linear distance are related by s = rθ.
• Angular displacement must be measured in radians when using s = rθ.
• Angular velocity and tangential velocity are related by v = rω.
• Angular acceleration and tangential acceleration are related by a_t = rα.
• Centripetal acceleration can be calculated using a_c = rω².
• Centripetal acceleration can also be calculated using a_c = v²/r.
• All points on a rigid rotating object have the same angular displacement.
• All points on a rigid rotating object have the same angular velocity.
• All points on a rigid rotating object have the same angular acceleration.
• Linear distance, tangential speed and tangential acceleration depend on radius.
• Points farther from the rotation axis travel farther and move faster for the same angular motion.
• Centripetal acceleration points toward the centre of rotation.
• Tangential velocity points tangent to the circular path.
• Tangential acceleration and centripetal acceleration are perpendicular.
• For rolling without slipping, s = rθ, v = rω, and a = rα connect wheel rotation with vehicle motion.
• Rotational motion can produce linear motion in wheels, pulleys, winches and conveyor systems.
• Meshing gears have equal tangential speeds at their contact point.
• Smaller gears rotate faster than larger gears when their edges move at the same speed.
• Radius is the key quantity connecting angular and linear motion.

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Check Your Understanding

1. State the relationship between angular displacement θ and arc length s.
2. A wheel of radius 0.40 m rotates through 5 rad. Calculate the distance travelled by a point on its edge.
3. A point on the edge of a 0.25 m radius disk travels 3.0 m along its circular path. Calculate the angular displacement.
4. Explain why two points on the same rotating disk can have the same angular velocity but different linear speeds.
5. A disk rotates at 8 rad/s. Calculate the tangential speed of a point 0.30 m from the centre.
6. Two points are located 0.20 m and 0.80 m from the centre of a disk rotating at 5 rad/s. Calculate the tangential speed of each point.
7. A wheel has radius 0.50 m and angular acceleration 6 rad/s². Calculate the tangential acceleration at its edge.
8. A point 0.40 m from the axis rotates at 10 rad/s. Calculate its centripetal acceleration.
9. If the angular velocity in Question 8 doubles while the radius remains unchanged, calculate the new centripetal acceleration and explain why it increases by that amount.
10. A bicycle wheel has radius 0.35 m and rotates at 18 rad/s. Calculate the bicycle's speed assuming no slipping.
11. A car travels at 24 m/s on wheels of radius 0.30 m. Calculate the angular velocity of the wheels.
12. A pulley of radius 0.20 m rotates at 12 rad/s with angular acceleration 5 rad/s². Calculate the linear speed and acceleration of a rope that does not slip on the pulley.
13. Gear A has radius 6 cm and rotates at 30 rad/s. It drives Gear B with radius 18 cm. Calculate Gear B's angular velocity and state its direction relative to Gear A.
14. A rotating platform has angular velocity 3 rad/s. Two students stand 1.0 m and 2.5 m from the centre. Compare their angular velocities, tangential speeds and centripetal accelerations.
15. A wheel of radius 0.40 m starts from rest and accelerates uniformly at 5 rad/s² for 4 s. Calculate its final angular velocity, angular displacement, linear distance travelled, final tangential speed, tangential acceleration and final centripetal acceleration.