Energy Conservation in Rotating Systems

The law of conservation of energy applies to rotating systems just as it does to objects moving in a straight line. Energy cannot be created or destroyed, but it can be transferred between different energy stores and transformed from one form to another.

For an ideal mechanical system with no energy losses:

E_initial = E_final

A rotating system may involve several forms of mechanical energy:

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

For a rotating object:

E_rot = 1/2 Iω²

For an object that is both translating and rotating:

E_k,total = 1/2 mv² + 1/2 Iω²

---

Energy Transformations Involving Rotation

Many rotating systems continuously transform energy.

A falling object attached to a pulley can transform:

gravitational potential energy → translational kinetic energy + rotational kinetic energy

A rolling ball moving down a hill transforms:

gravitational potential energy → translational kinetic energy + rotational kinetic energy

A motor transforms:

electrical energy → rotational kinetic energy + other energy transfers

A brake transforms:

rotational kinetic energy → thermal energy

A generator can transform:

rotational mechanical energy → electrical energy

The total energy remains conserved when all energy stores and transfers are included.

---

Mechanical Energy in Rotating Systems

Mechanical energy may include:

E_mech = E_p + E_trans + E_rot + E_elastic

Depending on the system, some terms may be zero.

For gravitational systems:

E_p = mgh

For translational motion:

E_trans = 1/2 mv²

For rotational motion:

E_rot = 1/2 Iω²

For a spring:

E_elastic = 1/2 kx²

A complete energy equation might therefore look like:

mgh = 1/2 mv² + 1/2 Iω²

---

Pure Rotation

An object rotating about a fixed axis may have rotational kinetic energy without having translational kinetic energy.

For example, a stationary flywheel spinning around its axle has:

E_rot = 1/2 Iω²

but its centre of mass is not translating.

---

Worked Example: Spinning Flywheel

A flywheel has:

I = 8.0 kg·m²

ω = 15 rad/s

Its rotational kinetic energy is:

E_rot = 1/2 Iω²

E_rot = 1/2(8.0)(15²)

E_rot = 4(225)

E_rot = 900 J

If friction is negligible, the flywheel could ideally transfer this 900 J to another part of the system as it slows to rest.

---

Translation and Rotation Together

Rolling objects possess both:

• translational kinetic energy
• rotational kinetic energy

Therefore:

E_total = 1/2 mv² + 1/2 Iω²

For rolling without slipping:

v = rω

or:

ω = v/r

This relationship allows rotational and translational energy to be connected.

---

Rolling Down a Hill

Consider a wheel starting from rest at height h.

Initially:

E_initial = mgh

At the bottom:

E_final = 1/2 mv² + 1/2 Iω²

If friction does not dissipate mechanical energy:

mgh = 1/2 mv² + 1/2 Iω²

Using:

ω = v/r

gives:

mgh = 1/2 mv² + 1/2 I(v²/r²)

---

Why Rotational Energy Cannot Be Ignored

Suppose a rolling ball moves down a hill.

If we incorrectly write:

mgh = 1/2 mv²

we are assuming all gravitational potential energy becomes translational kinetic energy.

But some energy goes into rotation.

The correct relationship is:

mgh = 1/2 mv² + 1/2 Iω²

Therefore, the rolling object's translational speed is generally lower than it would be if all the energy went into translation.

---

Shape Affects Energy Distribution

Different objects have different moments of inertia.

For common shapes:

These differences affect how gravitational potential energy is divided between translation and rotation.

---

Rolling Ring

For a thin ring:

I = mr²

Energy conservation gives:

mgh = 1/2 mv² + 1/2(mr²)(v²/r²)

Simplify:

mgh = 1/2 mv² + 1/2 mv²

mgh = mv²

Therefore:

v = √(gh)

---

Rolling Solid Cylinder

For a solid cylinder:

I = 1/2 mr²

Then:

mgh = 1/2 mv² + 1/2(1/2 mr²)(v²/r²)

mgh = 1/2 mv² + 1/4 mv²

mgh = 3/4 mv²

Therefore:

v = √(4gh/3)

---

Rolling Solid Sphere

For a solid sphere:

I = 2/5 mr²

Then:

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)

---

Which Object Reaches the Bottom First?

Compare:

Ring:

v = √(gh)

Cylinder:

v = √(4gh/3)

Sphere:

v = √(10gh/7)

Since:

10/7 > 4/3 > 1

the solid sphere reaches the greatest translational speed.

For ideal rolling from the same height:

solid sphere → fastest

solid cylinder → next

thin ring → slowest

The difference occurs because their moments of inertia distribute the available energy differently.

---

Worked Rolling Example

A solid cylinder starts from rest at a height:

h = 2.0 m

Use:

g = 9.8 m/s²

For a solid cylinder:

v = √(4gh/3)

v = √[(4)(9.8)(2.0)/3]

v = √26.13

v ≈ 5.11 m/s

If its radius is:

r = 0.25 m

then:

ω = v/r

ω = 5.11/0.25

ω ≈ 20.4 rad/s

---

Energy Distribution in the Cylinder

Suppose the cylinder has:

m = 6.0 kg

Its initial gravitational potential energy is:

E_p = mgh

E_p = (6)(9.8)(2)

E_p = 117.6 J

Translational kinetic energy:

E_trans = 1/2 mv²

E_trans = 1/2(6)(5.11²)

E_trans ≈ 78.3 J

Rotational kinetic energy:

E_rot = 117.6 − 78.3

E_rot ≈ 39.3 J

The total remains approximately:

117.6 J

Small differences may occur because of rounding.

---

Static Friction and Rolling

An important point is that friction does not always mean mechanical energy is lost.

For ideal rolling without slipping, the contact point between the wheel and the ground is instantaneously at rest relative to the ground.

The static friction responsible for producing rotational motion does not necessarily dissipate mechanical energy.

Therefore, we can still write:

E_initial = E_final

for ideal rolling.

---

Sliding Friction

If an object slides or skids, kinetic friction generally converts mechanical energy into thermal energy.

Then:

E_initial = E_final + E_thermal

or equivalently:

E_initial + W_non-conservative = E_final

Mechanical energy decreases, but total energy is still conserved.

---

Friction Does Not Destroy Energy

It is incorrect to say that friction "destroys" energy.

Friction transforms mechanical energy into other forms, particularly:

• thermal energy
• sound
• deformation

The total energy remains conserved.

What is reduced is the amount of energy available as useful mechanical energy.

---

Energy Losses in Rotating Machinery

Real rotating machines experience energy transfers caused by:

• bearing friction
• air resistance
• fluid resistance
• gear friction
• deformation
• vibration
• electrical resistance
• sound

These processes reduce the amount of input energy that becomes useful mechanical output.

---

Efficiency

Efficiency describes the fraction of input energy converted into useful output.

efficiency = useful output energy / total input energy × 100%

The same relationship can be written using power:

efficiency = useful output power / input power × 100%

---

Worked Efficiency Example

A motor receives:

5000 J

of electrical energy.

It produces:

4200 J

of useful rotational mechanical energy.

Efficiency:

efficiency = 4200/5000 × 100%

efficiency = 84%

Energy transferred to other forms:

5000 − 4200 = 800 J

---

Motor Energy Transformations

An electric motor transforms:

electrical energy → rotational mechanical energy

However, some energy is also transferred into:

• thermal energy
• sound
• vibration

Therefore:

electrical input = useful rotational output + energy transferred to other forms

---

Generator Energy Transformations

A generator performs approximately the reverse process:

rotational mechanical energy → electrical energy

Generators are used in:

• hydroelectric power stations
• wind turbines
• thermal power stations
• portable generators

Some energy is transferred into unwanted forms because of electrical resistance and mechanical friction.

---

Wind Turbines

A wind turbine transforms energy through several stages:

kinetic energy of moving air → rotational mechanical energy → electrical energy

Not all of the wind's kinetic energy can become electrical energy.

Energy remains in:

• moving air
• turbulence
• thermal energy
• sound
• mechanical vibration

---

Hydroelectric Turbines

In hydroelectric generation:

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

This is a large-scale example of conservation of energy involving rotation.

---

Vehicle Wheels

A moving vehicle contains both translational and rotational kinetic energy.

The vehicle body has primarily translational kinetic energy:

E_trans = 1/2 mv²

The wheels have:

• translational kinetic energy
• rotational kinetic energy

Therefore, accelerating a vehicle requires energy not only to increase its forward speed but also to increase the rotational speed of its wheels.

---

Why Wheel Mass Matters

A heavier wheel generally has a larger moment of inertia.

More energy is then required to reach the same angular velocity:

E_rot = 1/2 Iω²

Mass located near the wheel rim has a particularly strong effect because moment of inertia depends strongly on distance from the axis.

Reducing wheel mass can therefore reduce the energy required during acceleration.

---

Braking

When brakes are applied, the vehicle's kinetic energy decreases.

In conventional friction brakes:

translational + rotational kinetic energy → thermal energy

The brake discs and pads become hot because mechanical energy is being transferred into thermal energy.

---

Regenerative Braking

Electric and hybrid vehicles may use regenerative braking.

Instead of converting all mechanical energy into thermal energy, the motor operates as a generator.

The energy transformation becomes approximately:

vehicle kinetic energy → rotational mechanical energy → electrical energy → stored battery energy

Some energy is still lost through:

• electrical resistance
• friction
• battery inefficiency
• electronics

Regenerative braking therefore cannot recover 100% of the vehicle's kinetic energy.

---

Flywheels

Flywheels store energy in rotational motion:

E_rot = 1/2 Iω²

They can store energy when accelerated and release energy when slowed.

Flywheel systems may be used for:

• short-term energy storage
• smoothing power fluctuations
• regenerative braking
• backup power
• industrial machinery

---

Worked Flywheel Energy Problem

A flywheel has:

I = 12 kg·m²

Its angular velocity increases from:

ω_i = 10 rad/s

to:

ω_f = 30 rad/s

Initial energy:

E_i = 1/2(12)(10²)

E_i = 600 J

Final energy:

E_f = 1/2(12)(30²)

E_f = 5400 J

Energy added:

ΔE = 5400 − 600

ΔE = 4800 J

If the motor supplying this energy is 80% efficient:

input energy = useful energy / efficiency

input energy = 4800/0.80

input energy = 6000 J

Therefore:

6000 − 4800 = 1200 J

is transferred to other forms.

---

Falling Mass and Rotating Pulley

A classic combined-energy problem involves a falling mass connected to a rotating pulley.

As the mass falls:

gravitational potential energy → translational kinetic energy of mass + rotational kinetic energy of pulley

Ignoring losses:

mgh = 1/2 mv² + 1/2 Iω²

If the rope does not slip:

v = rω

Therefore:

ω = v/r

---

Worked Pulley Example

A:

4.0 kg

mass falls:

2.0 m

while turning a pulley.

The pulley has:

I = 0.50 kg·m²

r = 0.20 m

Assume no friction.

Initial gravitational energy:

E_p = mgh

E_p = (4)(9.8)(2)

E_p = 78.4 J

Energy conservation gives:

78.4 = 1/2(4)v² + 1/2(0.50)ω²

Since:

ω = v/0.20

ω = 5v

Substitute:

78.4 = 2v² + 0.25(25v²)

78.4 = 2v² + 6.25v²

78.4 = 8.25v²

v² ≈ 9.50

v ≈ 3.08 m/s

Then:

ω = 5(3.08)

ω ≈ 15.4 rad/s

---

Why the Falling Mass Is Slower Than Expected

If the pulley had negligible moment of inertia, almost all the gravitational potential energy would become translational kinetic energy.

But the real pulley also rotates.

Some of the available energy becomes:

rotational kinetic energy of the pulley

Therefore, the falling mass accelerates less and reaches a lower speed.

---

Atwood Machine with a Massive Pulley

Suppose two different masses are connected over a pulley.

As the heavier mass descends and the lighter mass rises:

• gravitational potential energy changes
• both masses gain translational kinetic energy
• the pulley gains rotational kinetic energy

A suitable energy equation is:

loss in gravitational potential energy = translational KE of masses + rotational KE of pulley

This method can often solve the problem without explicitly calculating the tensions in the rope.

---

Spring and Rotating Wheel

Rotational systems may also involve elastic potential energy.

For a spring:

E_elastic = 1/2 kx²

If a compressed spring causes a wheel to rotate:

1/2 kx² = 1/2 Iω²

in an ideal system.

If the wheel also moves forward:

1/2 kx² = 1/2 mv² + 1/2 Iω²

---

Worked Spring Example

A spring has:

k = 500 N/m

and is compressed:

x = 0.20 m

Stored elastic energy:

E_elastic = 1/2 kx²

E_elastic = 1/2(500)(0.20²)

E_elastic = 10 J

If all of this energy becomes rotational energy of a wheel with:

I = 0.20 kg·m²

then:

10 = 1/2(0.20)ω²

10 = 0.10ω²

ω² = 100

ω = 10 rad/s

---

Rotational Energy and Angular Momentum

Energy conservation and angular momentum conservation are different principles.

Rotational kinetic energy:

E_rot = 1/2 Iω²

Angular momentum:

L = Iω

Angular momentum is conserved when:

Στ_external = 0

Mechanical energy is conserved when there are no transfers into non-mechanical forms.

A system may conserve angular momentum while its rotational kinetic energy changes.

---

Skater Example

A skater pulls their arms inward.

Angular momentum remains approximately constant:

I_iω_i = I_fω_f

As I decreases, ω increases.

However:

E_rot = 1/2 Iω²

can increase.

The additional rotational energy comes from work done by the skater's muscles.

Therefore, there is no violation of energy conservation.

Chemical energy is transformed into rotational kinetic energy.

---

Rotational Collisions

Suppose two rotating disks come into contact and eventually rotate together.

Angular momentum may be conserved:

L_i = L_f

but rotational kinetic energy may decrease.

The missing mechanical energy becomes:

• thermal energy
• sound
• deformation

Total energy is still conserved.

---

Worked Rotational Collision

Disk A:

I₁ = 2 kg·m²

ω₁ = 12 rad/s

Disk B:

I₂ = 4 kg·m²

ω₂ = 0

They join together.

Angular momentum conservation:

(2)(12) = (2 + 4)ω_f

24 = 6ω_f

ω_f = 4 rad/s

Initial rotational energy:

E_i = 1/2(2)(12²)

E_i = 144 J

Final rotational energy:

E_f = 1/2(6)(4²)

E_f = 48 J

Mechanical energy transferred to other forms:

144 − 48 = 96 J

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

---

Torque and Energy

Torque can transfer energy to a rotating object.

For constant torque:

W = τΔθ

The work-energy theorem gives:

W_net = ΔE_rot

Therefore:

τ_net Δθ = 1/2 Iω_f² − 1/2 Iω_i²

This provides another way to solve rotational energy problems.

---

Worked Torque-Energy Problem

A wheel has:

I = 3.0 kg·m²

and starts from rest.

A net torque of:

12 N·m

acts through:

25 rad

Work done:

W = τΔθ

W = (12)(25)

W = 300 J

Therefore:

E_rot,f = 300 J

Now:

300 = 1/2(3)ω²

300 = 1.5ω²

ω² = 200

ω ≈ 14.1 rad/s

---

Power in Energy Transformations

Power measures how quickly energy is transferred.

Average power:

P = E/t

For rotation:

P = τω

A powerful motor can transfer large amounts of energy in a short time.

However, power and energy are not the same.

A motor may have high power but operate only briefly.

Another motor may have lower power but operate for a long time and transfer more total energy.

---

Efficiency and Power

Efficiency can also be expressed using power:

efficiency = P_useful/P_input × 100%

Suppose an electric motor receives:

P_input = 4000 W

and provides:

P_output = 3400 W

Then:

efficiency = 3400/4000 × 100%

efficiency = 85%

The remaining:

600 W

is transferred into less useful forms.

---

Bearings and Lubrication

Bearings are designed to reduce friction in rotating systems.

Lubrication helps reduce:

• friction
• heating
• wear

Reducing friction improves efficiency because more input energy remains available for useful mechanical output.

However, no real machine can completely eliminate energy losses.

---

Aerodynamic Drag

Rapidly rotating components can also transfer energy to the surrounding air.

Fans intentionally do this.

Other machines may experience it as an unwanted energy loss.

At high rotational speeds, air resistance can become significant.

This is one reason some high-speed rotating systems operate in low-pressure environments.

---

Heating in Rotating Systems

Friction can cause substantial heating in:

• brakes
• bearings
• gears
• shafts
• tires

This thermal energy must sometimes be removed using:

• ventilation
• cooling systems
• lubricants
• heat sinks

Energy conservation therefore has direct consequences for engineering design.

---

Energy Flow in a Real Rotating Machine

Consider an electric motor driving a mechanical load.

A simplified energy pathway is:

electrical energy input

↓

rotational mechanical energy

↓

useful work by the machine

At the same time, some energy is transferred into:

• thermal energy
• sound
• vibration

Therefore:

E_input = E_useful + E_losses

This is the most useful way to think about energy conservation in real machines.

---

A Complete Combined Energy Problem

A solid cylinder has:

m = 10 kg

r = 0.30 m

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

h = 4.0 m

For a solid cylinder:

I = 1/2 mr²

Initial Energy

E_initial = mgh

E_initial = (10)(9.8)(4)

E_initial = 392 J

Final Energy

E_final = 1/2 mv² + 1/2 Iω²

Moment of inertia:

I = 1/2(10)(0.30²)

I = 0.45 kg·m²

Since:

ω = v/r

ω = v/0.30

Substitute:

392 = 1/2(10)v² + 1/2(0.45)(v/0.30)²

392 = 5v² + 0.225(v²/0.09)

392 = 5v² + 2.5v²

392 = 7.5v²

v² = 52.27

v ≈ 7.23 m/s

Angular velocity:

ω = 7.23/0.30

ω ≈ 24.1 rad/s

---

Adding Energy Losses

Suppose the same cylinder loses:

50 J

of mechanical energy through rolling resistance and other effects.

Then:

E_final = 392 − 50

E_final = 342 J

Therefore:

342 = 7.5v²

v² = 45.6

v ≈ 6.75 m/s

Energy losses reduce the final speed.

---

A Complete Machine-Efficiency Problem

A motor receives:

12,000 J

of electrical energy.

Its efficiency is:

75%

Useful mechanical energy:

E_useful = 0.75(12,000)

E_useful = 9000 J

Suppose this energy accelerates a flywheel with:

I = 20 kg·m²

from rest.

Then:

9000 = 1/2(20)ω²

9000 = 10ω²

ω² = 900

ω = 30 rad/s

Energy transferred into less useful forms:

12,000 − 9000

= 3000 J

---

Solving Combined Rotational Energy Problems

A useful strategy is to begin by identifying all relevant energy stores.

Initial Energy

Ask what energy the system has initially:

• gravitational?
• elastic?
• translational?
• rotational?

Final Energy

Identify the final energy stores.

Energy Transfers

Determine whether energy is transferred through:

• friction
• motor work
• braking
• electrical systems
• external forces

Then write:

total initial energy + energy transferred in = total final energy + energy transferred out

This general relationship works even for complicated systems.

---

Choosing the System Boundary

Energy analysis depends on what is included in the system.

For example, if a wheel slows because of friction:

If the system is only the wheel, mechanical energy leaves the wheel.

If the system includes the wheel, brake, and surroundings, the energy remains within the larger system but changes form.

Carefully defining the system helps explain where the energy goes.

---

Energy Bar-Chart Thinking

Before calculating, it can help to imagine energy as amounts distributed among different stores.

For example, a rolling ball at the top of a hill may have:

• high gravitational potential energy
• zero translational kinetic energy
• zero rotational kinetic energy

At the bottom:

• lower gravitational potential energy
• greater translational kinetic energy
• greater rotational kinetic energy

If friction is present, some energy also appears as thermal energy.

This makes it easier to build the correct equation before substituting numbers.

---

Common Mistakes

Forgetting rotational kinetic energy

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

Assuming friction always causes energy loss

Static friction in ideal rolling can provide the torque needed for rotation without dissipating mechanical energy.

Saying friction destroys energy

Energy is transformed, usually into thermal energy.

Using the wrong moment of inertia

Different shapes have different moment-of-inertia equations.

Forgetting v = rω

This relationship is essential for rolling without slipping.

Assuming mechanical energy is always conserved

Mechanical energy decreases when energy is transferred into thermal energy, sound, or other forms.

Confusing total energy with mechanical energy

Total energy is always conserved. Mechanical energy may not be.

Ignoring rotational energy of pulleys

A massive pulley can store significant rotational kinetic energy.

Ignoring efficiency

Real machines do not convert all input energy into useful mechanical output.

Assuming angular momentum conservation means energy conservation

These are separate conservation principles with different conditions.

Ignoring energy supplied by a person or motor

If rotational kinetic energy increases while angular momentum remains constant, another energy store may be supplying the energy.

---

Key Terms

Conservation of energy — principle stating that energy cannot be created or destroyed, only transferred or transformed.

Mechanical energy — energy associated with motion and position, including kinetic and potential energy.

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

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

Gravitational potential energy — energy associated with position in a gravitational field, commonly calculated near Earth's surface using E_p = mgh.

Elastic potential energy — energy stored in a stretched or compressed elastic object.

Rolling without slipping — rolling motion satisfying v = rω.

Static friction — friction acting when surfaces do not slide relative to each other.

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

Thermal energy — internal energy associated with particle motion that often increases because of friction.

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

Flywheel — rotating device designed to store rotational kinetic energy.

Regenerative braking — process that recovers some mechanical energy during braking and converts it into another useful form.

Work-energy theorem — principle stating that net work changes kinetic energy.

System boundary — chosen division between the system being studied and its surroundings.

---

Key Takeaways

• Energy conservation applies to rotational systems.
• Total energy cannot be created or destroyed.
• Rotating objects can store energy as rotational kinetic energy.
• Rotational kinetic energy is E_rot = 1/2 Iω².
• Rolling objects possess both translational and rotational kinetic energy.
• For rolling without slipping, v = rω.
• Gravitational potential energy can transform into both translational and rotational kinetic energy.
• Moment of inertia affects how energy is distributed in rolling systems.
• Objects with different mass distributions can reach different speeds even when starting with the same gravitational potential energy per unit mass.
• A solid sphere generally rolls down an ideal incline faster than a solid cylinder or thin ring.
• Static friction in ideal rolling does not necessarily dissipate mechanical energy.
• Sliding friction generally transfers mechanical energy into thermal energy.
• Friction does not destroy energy.
• Real rotating machines transfer some energy into heat, sound, vibration, and deformation.
• Efficiency measures how much input energy becomes useful output.
• Motors transform electrical energy into rotational mechanical energy.
• Generators transform rotational mechanical energy into electrical energy.
• Braking transforms kinetic energy into other forms, especially thermal energy.
• Regenerative braking can recover part of a vehicle's kinetic energy.
• Flywheels store energy as rotational kinetic energy.
• Massive pulleys must be included in energy calculations because they can store rotational energy.
• Energy conservation and angular momentum conservation are different principles and should not be confused.
• Energy methods are especially useful for solving complex rotational systems involving several types of motion and energy transfer.

---

Check Your Understanding

1. State the law of conservation of energy and explain how it applies to a rotating system.
2. A flywheel has I = 6.0 kg·m² and rotates at 12 rad/s. Calculate its rotational kinetic energy.
3. Explain why a rolling wheel has both translational and rotational kinetic energy.
4. A 5.0 kg solid cylinder rolls without slipping at 4.0 m/s. Its radius is 0.20 m. Calculate its moment of inertia, angular velocity, translational kinetic energy, rotational kinetic energy, and total kinetic energy.
5. A solid sphere, solid cylinder, and thin ring of equal mass and radius roll from rest down the same incline. Predict their order at the bottom and explain your reasoning using energy conservation and moment of inertia.
6. A solid cylinder rolls from rest down a hill of vertical height 3.0 m. Use conservation of energy to calculate its speed at the bottom.
7. Explain why static friction in ideal rolling does not necessarily cause a loss of mechanical energy.
8. A rotating wheel initially contains 1200 J of rotational kinetic energy. Friction transfers 350 J into thermal energy. How much rotational kinetic energy remains?
9. A motor receives 8000 J of electrical energy and is 85% efficient. Calculate the useful mechanical energy produced and the energy transferred to other forms.
10. A flywheel has I = 15 kg·m² and accelerates from 10 rad/s to 25 rad/s. Calculate the increase in rotational kinetic energy.
11. A 3.0 kg mass falls 2.5 m while turning a pulley with I = 0.40 kg·m² and radius 0.20 m. Assuming no losses and no slipping, determine the speed of the mass immediately before it reaches the bottom.
12. Explain how conventional braking and regenerative braking differ in the way they transform a vehicle's kinetic energy.
13. Two rotating disks join together. Disk A has I = 3.0 kg·m² and ω = 10 rad/s. Disk B has I = 2.0 kg·m² and is stationary. Calculate their final angular velocity using conservation of angular momentum. Then compare the initial and final rotational kinetic energies and explain the difference.
14. A spring with k = 600 N/m is compressed by 0.25 m and releases all its energy into a wheel with I = 0.30 kg·m². Assuming no losses, calculate the wheel's angular velocity.
15. A 12 kg solid cylinder of radius 0.40 m rolls without slipping from rest down a hill with a vertical drop of 5.0 m. During the descent, 100 J of mechanical energy is transferred into thermal energy. Calculate the cylinder's final translational speed and angular velocity.