Rotational Kinetic Energy

A rotating object possesses rotational kinetic energy because its particles are moving around an axis of rotation.

Rotational kinetic energy is calculated using:

E_rot = 1/2 Iω²

where:

• E_rot = rotational kinetic energy, measured in joules (J)
• I = moment of inertia, measured in kg·m²
• ω = angular velocity, measured in rad/s

This equation is the rotational equivalent of the familiar translational kinetic-energy equation:

E_k = 1/2 mv²

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Where Rotational Kinetic Energy Comes From

Every part of a rotating object is moving.

Consider a spinning disk. Points near the centre move relatively slowly, while points near the edge move much faster.

For each small piece of the disk:

E_k = 1/2 mv²

Since:

v = rω

the kinetic energy of each piece depends on:

• its mass
• its distance from the axis
• the angular velocity

Adding the kinetic energies of all the pieces produces:

E_rot = 1/2 Iω²

The moment of inertia accounts for how all the object's mass is distributed around the rotation axis.

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Rotational and Translational Kinetic Energy

The two equations have the same mathematical structure.

In translational motion, mass measures resistance to changes in linear motion.

In rotational motion, moment of inertia measures resistance to changes in rotational motion.

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Calculating Rotational Kinetic Energy

Suppose a wheel has:

I = 4.0 kg·m²

and:

ω = 6.0 rad/s

Then:

E_rot = 1/2 Iω²

E_rot = 1/2(4.0)(6.0²)

E_rot = 2.0(36)

E_rot = 72 J

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The Effect of Moment of Inertia

For constant angular velocity:

E_rot ∝ I

If moment of inertia doubles while ω remains constant:

rotational kinetic energy doubles

For example:

Object A:

I = 2 kg·m²

ω = 5 rad/s

E_rot = 1/2(2)(5²)

E_rot = 25 J

Object B:

I = 4 kg·m²

ω = 5 rad/s

E_rot = 1/2(4)(5²)

E_rot = 50 J

At the same angular velocity, the object with the larger moment of inertia has more rotational kinetic energy.

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The Effect of Angular Velocity

Angular velocity has an even stronger effect because it is squared:

E_rot ∝ ω²

If ω doubles:

E_rot becomes four times larger

If ω triples:

E_rot becomes nine times larger

If ω is halved:

E_rot becomes one-quarter as large

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Worked Example: Doubling Angular Velocity

A flywheel has:

I = 5 kg·m²

Initially:

ω = 10 rad/s

Its rotational kinetic energy is:

E_rot = 1/2(5)(10²)

E_rot = 250 J

If its angular velocity doubles:

ω = 20 rad/s

Then:

E_rot = 1/2(5)(20²)

E_rot = 1000 J

The angular velocity doubled, but the energy increased by a factor of four.

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Why Angular Velocity Matters So Much

Because:

E_rot = 1/2 Iω²

a relatively small increase in angular velocity can produce a large increase in energy.

This is extremely important in the design and safety of:

• flywheels
• turbines
• motors
• centrifuges
• vehicle wheels
• industrial machinery

A rapidly rotating object can store a very large amount of energy.

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Moment of Inertia and Mass Distribution

Moment of inertia depends on how mass is distributed relative to the rotation axis.

For example:

Thin ring:

I = MR²

Solid disk:

I = 1/2 MR²

Solid sphere:

I = 2/5 MR²

At the same M, R, and ω, these objects have different rotational kinetic energies because their moments of inertia are different.

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Comparing a Ring and a Disk

Suppose a thin ring and a solid disk both have:

M = 8.0 kg

R = 0.50 m

ω = 10 rad/s

Thin Ring

I = MR²

I = (8.0)(0.50²)

I = 2.0 kg·m²

Therefore:

E_rot = 1/2(2.0)(10²)

E_rot = 100 J

Solid Disk

I = 1/2 MR²

I = 1/2(8.0)(0.50²)

I = 1.0 kg·m²

Therefore:

E_rot = 1/2(1.0)(10²)

E_rot = 50 J

At the same angular velocity, the ring has twice as much rotational kinetic energy.

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Why the Ring Has More Energy

The ring has more of its mass concentrated far from the axis.

This gives it a larger moment of inertia.

Since:

E_rot = 1/2 Iω²

a larger I gives greater rotational kinetic energy when ω is the same.

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Finding Angular Velocity from Energy

The rotational energy equation can be rearranged.

Starting with:

E_rot = 1/2 Iω²

Multiply by 2:

2E_rot = Iω²

Divide by I:

ω² = 2E_rot/I

Therefore:

ω = √(2E_rot/I)

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Worked Example

A flywheel stores:

E_rot = 1600 J

and has:

I = 8.0 kg·m²

Then:

ω = √(2E_rot/I)

ω = √[(2)(1600)/8]

ω = √400

ω = 20 rad/s

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Finding Moment of Inertia from Energy

Starting with:

E_rot = 1/2 Iω²

Rearrange:

I = 2E_rot/ω²

Suppose:

E_rot = 900 J

ω = 15 rad/s

Then:

I = (2)(900)/(15²)

I = 1800/225

I = 8.0 kg·m²

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Work and Rotational Energy

Just as work can change translational kinetic energy, rotational work can change rotational kinetic energy.

For constant torque:

W = τΔθ

where:

• W = work
• τ = torque
• Δθ = angular displacement in radians

The rotational work-energy theorem is:

W_net = ΔE_rot

or:

W_net = E_rot,f − E_rot,i

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Worked Example: Torque Doing Work

A constant net torque of:

12 N·m

turns a wheel through:

20 rad

The work done is:

W = τΔθ

W = (12)(20)

W = 240 J

If the wheel begins from rest and there are no other energy transfers:

E_rot,f = 240 J

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Finding Final Angular Velocity

Suppose the wheel in the previous example has:

I = 3.0 kg·m²

Then:

E_rot = 1/2 Iω²

240 = 1/2(3.0)ω²

240 = 1.5ω²

ω² = 160

ω ≈ 12.6 rad/s

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Work-Energy Method vs Dynamics Method

Some rotational problems can be solved using either forces and torques or energy.

Dynamics Approach

Use:

Στ = Iα

then rotational kinematics.

Energy Approach

Use:

W_net = ΔE_rot

The energy approach can be especially useful when:

• time is not given
• angular acceleration is not needed
• torque acts through a known angular displacement

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Rotational Power

Power measures the rate of energy transfer.

For rotational motion:

P = τω

where:

• P = power in watts
• τ = torque
• ω = angular velocity

This equation is widely used for:

• motors
• engines
• turbines
• generators
• vehicle drivetrains

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Worked Power Example

A motor produces:

τ = 40 N·m

while rotating at:

ω = 50 rad/s

Power:

P = τω

P = (40)(50)

P = 2000 W

or:

P = 2.0 kW

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Flywheels

A flywheel is designed to store rotational kinetic energy.

Flywheels often have substantial moment of inertia and rotate rapidly.

Their stored energy is:

E_rot = 1/2 Iω²

Flywheels can:

• smooth changes in machinery speed
• temporarily store energy
• provide bursts of power
• recover energy in some systems

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Why Flywheel Mass Is Often Far from the Axis

Moving mass farther from the axis increases moment of inertia.

For a point mass:

I = mr²

Therefore, distributing mass farther outward can increase the energy stored at a particular angular velocity.

However, increasing radius and rotational speed also increases mechanical stresses, so engineering design must balance energy storage with safety.

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Energy in Rolling Motion

A rolling object can have two types of kinetic energy at the same time:

1. translational kinetic energy of its centre of mass
2. rotational kinetic energy about its centre

Therefore:

E_total = E_trans + E_rot

or:

E_total = 1/2 mv² + 1/2 Iω²

For rolling without slipping:

v = rω

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Worked Example: Rolling Wheel

A wheel has:

m = 4.0 kg

I = 0.50 kg·m²

r = 0.25 m

It rolls without slipping at:

v = 3.0 m/s

First find angular velocity:

v = rω

ω = v/r

ω = 3.0/0.25

ω = 12 rad/s

Translational Kinetic Energy

E_trans = 1/2 mv²

E_trans = 1/2(4.0)(3.0²)

E_trans = 18 J

Rotational Kinetic Energy

E_rot = 1/2 Iω²

E_rot = 1/2(0.50)(12²)

E_rot = 0.25(144)

E_rot = 36 J

Total Kinetic Energy

E_total = 18 + 36

E_total = 54 J

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Rolling Objects Down a Hill

Consider an object rolling from rest down a slope.

Initially, it has gravitational potential energy.

As it descends, that energy becomes:

• translational kinetic energy
• rotational kinetic energy

Ignoring energy losses:

mgh = 1/2 mv² + 1/2 Iω²

Since:

ω = v/r

we can analyze how mass distribution affects the final speed.

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Shape Affects Rolling Motion

Consider objects with the same mass and radius rolling down the same hill.

A thin ring has:

I = MR²

A solid disk has:

I = 1/2 MR²

A solid sphere has:

I = 2/5 MR²

The object with the smaller moment of inertia relative to MR² requires a smaller fraction of its available energy for rotation.

As a result, its centre of mass can gain more translational speed.

For ideal rolling from the same height:

solid sphere reaches the bottom before a solid disk

and:

solid disk reaches the bottom before a thin ring

assuming comparable conditions.

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Deriving the Speed of a Rolling Object

Start with:

mgh = 1/2 mv² + 1/2 Iω²

For rolling without slipping:

ω = v/r

Substitute:

mgh = 1/2 mv² + 1/2 I(v²/r²)

Factor out:

mgh = 1/2 v²(m + I/r²)

Therefore:

v² = 2mgh/(m + I/r²)

This equation shows directly how moment of inertia affects the speed reached by a rolling object.

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Example: Solid Sphere Rolling Down a Hill

For a solid sphere:

I = 2/5 mr²

Substitute into the energy equation:

mgh = 1/2 mv² + 1/2(2/5 mr²)(v²/r²)

mgh = 1/2 mv² + 1/5 mv²

mgh = 7/10 mv²

Therefore:

v² = 10gh/7

and:

v = √(10gh/7)

Notice that the mass cancels.

For ideal rolling, the final speed does not depend on the object's mass.

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Rotational Energy in Vehicle Wheels

A moving vehicle has translational kinetic energy, but its rotating wheels also contain rotational kinetic energy.

The total energy therefore includes:

• translational energy of the vehicle
• rotational energy of the wheels
• rotational energy of other moving components

This means an engine must provide energy not only to accelerate the vehicle forward but also to spin its wheels.

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Why Lightweight Wheels Can Matter

Reducing the moment of inertia of a wheel reduces the energy required to reach a particular angular velocity:

E_rot = 1/2 Iω²

It can also make the wheel easier to angularly accelerate.

This is one reason engineers may try to reduce rotating mass, particularly mass located far from the axis.

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Braking a Rotating Wheel

Brakes remove rotational kinetic energy.

Suppose:

I = 5.0 kg·m²

ω_i = 20 rad/s

Initial rotational energy:

E_i = 1/2(5)(20²)

E_i = 2.5(400)

E_i = 1000 J

If the wheel stops:

E_f = 0

Therefore:

1000 J

of rotational kinetic energy must be transferred elsewhere.

Usually much of it becomes thermal energy in the braking system.

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Braking Torque and Energy

Suppose a constant braking torque of:

25 N·m

acts on the previous wheel.

Using:

W = τΔθ

the angular displacement required for the brakes to remove 1000 J is:

1000 = (25)Δθ

Δθ = 40 rad

Therefore, the wheel rotates through:

40 rad

while stopping.

In revolutions:

N = 40/(2π)

N ≈ 6.37 revolutions

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Conservation of Energy in Rotating Systems

Mechanical energy can be conserved when dissipative forces are negligible.

A typical rotational energy equation might be:

E_initial = E_final

For example:

mgh = 1/2 mv² + 1/2 Iω²

or:

E_rot,i + W = E_rot,f

Energy may move between:

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

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Angular Momentum and Rotational Energy

Angular momentum is:

L = Iω

Rotational kinetic energy is:

E_rot = 1/2 Iω²

These are related but different quantities.

If L is known, then:

ω = L/I

Substitute into the energy equation:

E_rot = 1/2 I(L/I)²

Therefore:

E_rot = L²/(2I)

This relationship is particularly useful when angular momentum is conserved.

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Energy When Moment of Inertia Changes

If angular momentum remains constant:

E_rot = L²/(2I)

Therefore:

smaller I → greater rotational kinetic energy

This may seem surprising.

Consider a skater pulling their arms inward.

• I decreases
• ω increases
• L remains constant
• E_rot increases

Where does the additional energy come from?

The skater does work while pulling their arms inward.

Chemical energy in the body is transferred into rotational kinetic energy.

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Worked Skater Energy Example

Initially:

I_i = 6 kg·m²

ω_i = 2 rad/s

Initial energy:

E_i = 1/2(6)(2²)

E_i = 12 J

Angular momentum:

L = (6)(2)

L = 12 kg·m²/s

The skater pulls inward:

I_f = 2 kg·m²

Angular momentum is conserved:

12 = (2)ω_f

ω_f = 6 rad/s

Final rotational energy:

E_f = 1/2(2)(6²)

E_f = 36 J

The rotational kinetic energy increased by:

36 − 12 = 24 J

That additional energy came from work done by the skater.

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Rotational Collisions

Angular momentum can be conserved during a rotational collision while rotational kinetic energy decreases.

Suppose a rotating disk interacts with another object and they stick together.

The final moment of inertia increases.

Angular momentum may remain constant, but some kinetic energy can be transferred into:

• heat
• sound
• deformation

This is similar to an inelastic collision in linear motion.

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Worked Rotational Collision

A disk has:

I₁ = 2 kg·m²

ω₁ = 10 rad/s

A stationary disk with:

I₂ = 3 kg·m²

is dropped onto it.

They rotate together.

Angular momentum conservation gives:

I₁ω₁ = (I₁ + I₂)ω_f

(2)(10) = (5)ω_f

ω_f = 4 rad/s

Initial rotational energy:

E_i = 1/2(2)(10²)

E_i = 100 J

Final rotational energy:

E_f = 1/2(5)(4²)

E_f = 40 J

Mechanical energy transferred to other forms:

100 − 40 = 60 J

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

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Energy Stored in Rotating Machinery

Many machines contain rotating components such as:

• shafts
• gears
• wheels
• turbines
• rotors
• fans

Every rotating component contributes rotational kinetic energy.

The total rotational energy is:

E_rot,total = 1/2 I₁ω₁² + 1/2 I₂ω₂² + ...

If different components rotate at different angular velocities, each must be calculated separately.

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Gears and Rotational Energy

Connected gears generally rotate at different angular velocities.

For meshing gears:

r₁ω₁ = r₂ω₂

A small gear may rotate rapidly while a larger gear rotates more slowly.

Each gear has rotational energy:

E_rot = 1/2 Iω²

Therefore, both the moment of inertia and angular velocity of each gear must be considered when calculating the total energy of a gear system.

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Rotational Energy and Safety

Rapidly rotating systems can store enormous amounts of energy.

Because:

E_rot ∝ ω²

doubling angular velocity produces four times the energy.

This creates important safety considerations for:

• turbine blades
• flywheels
• grinding wheels
• centrifuges
• industrial rotors

A mechanical failure at high angular speed can release a large amount of stored energy very quickly.

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A Complete Rotational Energy Problem

A solid disk has:

M = 12 kg

R = 0.50 m

It starts from rest.

A constant net torque of:

6.0 N·m

acts through:

25 rad

of angular displacement.

Moment of Inertia

For a solid disk:

I = 1/2 MR²

I = 1/2(12)(0.50²)

I = 6(0.25)

I = 1.5 kg·m²

Work Done

W = τΔθ

W = (6.0)(25)

W = 150 J

Since the disk starts from rest:

E_rot,f = 150 J

Final Angular Velocity

E_rot = 1/2 Iω²

150 = 1/2(1.5)ω²

150 = 0.75ω²

ω² = 200

ω ≈ 14.1 rad/s

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Checking with Rotational Dynamics

We can solve the same problem another way.

Newton's Second Law for rotation:

Στ = Iα

α = 6.0/1.5

α = 4.0 rad/s²

Then use:

ω_f² = ω_i² + 2αΔθ

ω_f² = 0 + 2(4.0)(25)

ω_f² = 200

ω_f ≈ 14.1 rad/s

Both approaches give the same result.

This demonstrates the consistency between:

• rotational dynamics
• rotational kinematics
• energy conservation

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A Complete Rolling-Energy Problem

A solid cylinder has:

m = 6.0 kg

r = 0.20 m

It rolls without slipping at:

v = 4.0 m/s

For a solid cylinder:

I = 1/2 mr²

Moment of Inertia

I = 1/2(6.0)(0.20²)

I = 3(0.04)

I = 0.12 kg·m²

Angular Velocity

ω = v/r

ω = 4.0/0.20

ω = 20 rad/s

Translational Kinetic Energy

E_trans = 1/2 mv²

E_trans = 1/2(6)(4²)

E_trans = 48 J

Rotational Kinetic Energy

E_rot = 1/2 Iω²

E_rot = 1/2(0.12)(20²)

E_rot = 0.06(400)

E_rot = 24 J

Total Kinetic Energy

E_total = 48 + 24

E_total = 72 J

The rolling cylinder therefore has both translational and rotational kinetic energy.

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Problem-Solving Strategy

First determine what type of motion is involved:

• rotation only
• translation only
• rolling with both translation and rotation

Identify the moment of inertia.

Then identify the angular velocity.

For pure rotation, use:

E_rot = 1/2 Iω²

For rolling motion, include:

E_total = 1/2 mv² + 1/2 Iω²

and:

v = rω

If torque and angular displacement are given, consider:

W = τΔθ

and:

W_net = ΔE_rot

If gravitational potential energy is involved, use conservation of energy.

Finally, check that every energy term is measured in joules.

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Common Mistakes

Using mass instead of moment of inertia

Rotational kinetic energy uses:

E_rot = 1/2 Iω²

not 1/2 mω².

Forgetting to square angular velocity

The equation contains ω².

Using rpm directly in the equation

Convert rpm to rad/s before using E_rot = 1/2 Iω².

Ignoring mass distribution

Different shapes can have different I even if they have the same mass and radius.

Forgetting translational kinetic energy in rolling problems

A rolling object usually has both translational and rotational kinetic energy.

Counting rotational energy twice

Be careful when deciding which parts of a system are translating and rotating.

Assuming angular momentum conservation means energy conservation

Angular momentum and kinetic energy are different quantities.

Forgetting that a person can do work while changing I

A skater's rotational kinetic energy can change even while angular momentum remains constant.

Using degrees instead of radians in rotational work

For W = τΔθ, angular displacement must be expressed in radians.

Ignoring friction or other energy losses

Real systems may transfer mechanical energy into thermal energy and sound.

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Key Terms

Rotational kinetic energy — energy possessed by an object because it is rotating, calculated using E_rot = 1/2 Iω².

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

Angular velocity (ω) — rate at which angular position changes.

Translational kinetic energy — energy associated with linear motion, calculated using E_k = 1/2 mv².

Rotational work — energy transferred by a torque acting through an angular displacement.

Work-energy theorem — principle stating that net work equals the change in kinetic energy.

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

Flywheel — rotating device designed to store rotational kinetic energy.

Rolling without slipping — motion in which v = rω.

Total kinetic energy — combined kinetic energy of all forms of motion in a system.

Angular momentum — rotational quantity calculated for a rigid body using L = Iω.

Energy conservation — principle stating that total energy cannot be created or destroyed but can be transferred or transformed.

Inelastic rotational collision — rotational interaction in which angular momentum may be conserved while kinetic energy is transferred to other forms.

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Key Takeaways

• Rotating objects possess rotational kinetic energy.
• Rotational kinetic energy is calculated using E_rot = 1/2 Iω².
• Rotational kinetic energy is measured in joules.
• Moment of inertia plays the rotational role that mass plays in translational kinetic energy.
• Angular velocity plays the rotational role that linear velocity plays.
• For constant ω, rotational energy is directly proportional to I.
• For constant I, rotational energy is proportional to ω².
• Doubling angular velocity produces four times the rotational kinetic energy.
• Mass distribution affects rotational energy through moment of inertia.
• Torque can transfer energy to a rotating object.
• For constant torque, rotational work is W = τΔθ.
• Net rotational work equals the change in rotational kinetic energy.
• Rotational power is P = τω.
• A rolling object has both translational and rotational kinetic energy.
• For rolling without slipping, v = rω.
• Different shapes can roll differently because their moments of inertia differ.
• Flywheels can store substantial amounts of rotational kinetic energy.
• Angular momentum can remain constant while rotational kinetic energy changes.
• Rapidly rotating machinery can store very large amounts of energy and therefore requires careful engineering and safety controls.
• Energy methods provide a powerful alternative to torque and kinematics when solving rotational mechanics problems.

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Check Your Understanding

1. Define rotational kinetic energy and give its SI unit.
2. A wheel has I = 3.0 kg·m² and rotates at 8.0 rad/s. Calculate its rotational kinetic energy.
3. The angular velocity of the wheel in Question 2 doubles. By what factor does its rotational kinetic energy change?
4. Two objects rotate at the same angular velocity. Object A has I = 2.0 kg·m² and Object B has I = 6.0 kg·m². Compare their rotational kinetic energies.
5. A flywheel stores 2500 J of rotational kinetic energy and has I = 5.0 kg·m². Calculate its angular velocity.
6. A rotating object has 600 J of rotational kinetic energy while rotating at 10 rad/s. Calculate its moment of inertia.
7. A thin ring and solid disk have the same mass, radius, and angular velocity. Which has greater rotational kinetic energy? Explain why.
8. A constant torque of 15 N·m turns a wheel through 30 rad. Calculate the work done. If the wheel starts from rest and there are no energy losses, determine its final rotational kinetic energy.
9. The wheel in Question 8 has I = 5.0 kg·m². Calculate its final angular velocity.
10. A motor produces 60 N·m of torque while rotating at 40 rad/s. Calculate its rotational power.
11. A 5.0 kg wheel rolls without slipping at 4.0 m/s. Its moment of inertia is 0.40 kg·m² and its radius is 0.20 m. Calculate its angular velocity, translational kinetic energy, rotational kinetic energy, and total kinetic energy.
12. Explain why a solid sphere generally accelerates down an incline faster than a thin ring of the same mass and radius when both roll without slipping.
13. A skater has I_i = 8.0 kg·m² and ω_i = 2.0 rad/s. The skater pulls inward until I_f = 2.0 kg·m². Assuming angular momentum is conserved, calculate the final angular velocity and compare the initial and final rotational kinetic energies.
14. Two disks collide and stick together. The first has I₁ = 3.0 kg·m² and rotates at 8.0 rad/s. The second has I₂ = 5.0 kg·m² and is stationary. Calculate their final angular velocity, initial rotational kinetic energy, final rotational kinetic energy, and energy transferred to other forms.
15. A solid cylinder of mass 8.0 kg and radius 0.25 m rolls without slipping down a hill from a vertical height of 5.0 m. Assuming no energy losses and using I = 1/2 mr², determine its speed and angular velocity at the bottom of the hill.