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College Physics

10.4 Rotational Kinetic Energy: Work and Energy Revisited

College Physics10.4 Rotational Kinetic Energy: Work and Energy Revisited
  1. Preface
  2. 1 Introduction: The Nature of Science and Physics
    1. Introduction to Science and the Realm of Physics, Physical Quantities, and Units
    2. 1.1 Physics: An Introduction
    3. 1.2 Physical Quantities and Units
    4. 1.3 Accuracy, Precision, and Significant Figures
    5. 1.4 Approximation
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
  3. 2 Kinematics
    1. Introduction to One-Dimensional Kinematics
    2. 2.1 Displacement
    3. 2.2 Vectors, Scalars, and Coordinate Systems
    4. 2.3 Time, Velocity, and Speed
    5. 2.4 Acceleration
    6. 2.5 Motion Equations for Constant Acceleration in One Dimension
    7. 2.6 Problem-Solving Basics for One-Dimensional Kinematics
    8. 2.7 Falling Objects
    9. 2.8 Graphical Analysis of One-Dimensional Motion
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
  4. 3 Two-Dimensional Kinematics
    1. Introduction to Two-Dimensional Kinematics
    2. 3.1 Kinematics in Two Dimensions: An Introduction
    3. 3.2 Vector Addition and Subtraction: Graphical Methods
    4. 3.3 Vector Addition and Subtraction: Analytical Methods
    5. 3.4 Projectile Motion
    6. 3.5 Addition of Velocities
    7. Glossary
    8. Section Summary
    9. Conceptual Questions
    10. Problems & Exercises
  5. 4 Dynamics: Force and Newton's Laws of Motion
    1. Introduction to Dynamics: Newton’s Laws of Motion
    2. 4.1 Development of Force Concept
    3. 4.2 Newton’s First Law of Motion: Inertia
    4. 4.3 Newton’s Second Law of Motion: Concept of a System
    5. 4.4 Newton’s Third Law of Motion: Symmetry in Forces
    6. 4.5 Normal, Tension, and Other Examples of Forces
    7. 4.6 Problem-Solving Strategies
    8. 4.7 Further Applications of Newton’s Laws of Motion
    9. 4.8 Extended Topic: The Four Basic Forces—An Introduction
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
  6. 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
    1. Introduction: Further Applications of Newton’s Laws
    2. 5.1 Friction
    3. 5.2 Drag Forces
    4. 5.3 Elasticity: Stress and Strain
    5. Glossary
    6. Section Summary
    7. Conceptual Questions
    8. Problems & Exercises
  7. 6 Uniform Circular Motion and Gravitation
    1. Introduction to Uniform Circular Motion and Gravitation
    2. 6.1 Rotation Angle and Angular Velocity
    3. 6.2 Centripetal Acceleration
    4. 6.3 Centripetal Force
    5. 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
    6. 6.5 Newton’s Universal Law of Gravitation
    7. 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  8. 7 Work, Energy, and Energy Resources
    1. Introduction to Work, Energy, and Energy Resources
    2. 7.1 Work: The Scientific Definition
    3. 7.2 Kinetic Energy and the Work-Energy Theorem
    4. 7.3 Gravitational Potential Energy
    5. 7.4 Conservative Forces and Potential Energy
    6. 7.5 Nonconservative Forces
    7. 7.6 Conservation of Energy
    8. 7.7 Power
    9. 7.8 Work, Energy, and Power in Humans
    10. 7.9 World Energy Use
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
  9. 8 Linear Momentum and Collisions
    1. Introduction to Linear Momentum and Collisions
    2. 8.1 Linear Momentum and Force
    3. 8.2 Impulse
    4. 8.3 Conservation of Momentum
    5. 8.4 Elastic Collisions in One Dimension
    6. 8.5 Inelastic Collisions in One Dimension
    7. 8.6 Collisions of Point Masses in Two Dimensions
    8. 8.7 Introduction to Rocket Propulsion
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  10. 9 Statics and Torque
    1. Introduction to Statics and Torque
    2. 9.1 The First Condition for Equilibrium
    3. 9.2 The Second Condition for Equilibrium
    4. 9.3 Stability
    5. 9.4 Applications of Statics, Including Problem-Solving Strategies
    6. 9.5 Simple Machines
    7. 9.6 Forces and Torques in Muscles and Joints
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  11. 10 Rotational Motion and Angular Momentum
    1. Introduction to Rotational Motion and Angular Momentum
    2. 10.1 Angular Acceleration
    3. 10.2 Kinematics of Rotational Motion
    4. 10.3 Dynamics of Rotational Motion: Rotational Inertia
    5. 10.4 Rotational Kinetic Energy: Work and Energy Revisited
    6. 10.5 Angular Momentum and Its Conservation
    7. 10.6 Collisions of Extended Bodies in Two Dimensions
    8. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  12. 11 Fluid Statics
    1. Introduction to Fluid Statics
    2. 11.1 What Is a Fluid?
    3. 11.2 Density
    4. 11.3 Pressure
    5. 11.4 Variation of Pressure with Depth in a Fluid
    6. 11.5 Pascal’s Principle
    7. 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
    8. 11.7 Archimedes’ Principle
    9. 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
    10. 11.9 Pressures in the Body
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
  13. 12 Fluid Dynamics and Its Biological and Medical Applications
    1. Introduction to Fluid Dynamics and Its Biological and Medical Applications
    2. 12.1 Flow Rate and Its Relation to Velocity
    3. 12.2 Bernoulli’s Equation
    4. 12.3 The Most General Applications of Bernoulli’s Equation
    5. 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
    6. 12.5 The Onset of Turbulence
    7. 12.6 Motion of an Object in a Viscous Fluid
    8. 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  14. 13 Temperature, Kinetic Theory, and the Gas Laws
    1. Introduction to Temperature, Kinetic Theory, and the Gas Laws
    2. 13.1 Temperature
    3. 13.2 Thermal Expansion of Solids and Liquids
    4. 13.3 The Ideal Gas Law
    5. 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
    6. 13.5 Phase Changes
    7. 13.6 Humidity, Evaporation, and Boiling
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  15. 14 Heat and Heat Transfer Methods
    1. Introduction to Heat and Heat Transfer Methods
    2. 14.1 Heat
    3. 14.2 Temperature Change and Heat Capacity
    4. 14.3 Phase Change and Latent Heat
    5. 14.4 Heat Transfer Methods
    6. 14.5 Conduction
    7. 14.6 Convection
    8. 14.7 Radiation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  16. 15 Thermodynamics
    1. Introduction to Thermodynamics
    2. 15.1 The First Law of Thermodynamics
    3. 15.2 The First Law of Thermodynamics and Some Simple Processes
    4. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
    5. 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
    6. 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
    7. 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
    8. 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  17. 16 Oscillatory Motion and Waves
    1. Introduction to Oscillatory Motion and Waves
    2. 16.1 Hooke’s Law: Stress and Strain Revisited
    3. 16.2 Period and Frequency in Oscillations
    4. 16.3 Simple Harmonic Motion: A Special Periodic Motion
    5. 16.4 The Simple Pendulum
    6. 16.5 Energy and the Simple Harmonic Oscillator
    7. 16.6 Uniform Circular Motion and Simple Harmonic Motion
    8. 16.7 Damped Harmonic Motion
    9. 16.8 Forced Oscillations and Resonance
    10. 16.9 Waves
    11. 16.10 Superposition and Interference
    12. 16.11 Energy in Waves: Intensity
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
  18. 17 Physics of Hearing
    1. Introduction to the Physics of Hearing
    2. 17.1 Sound
    3. 17.2 Speed of Sound, Frequency, and Wavelength
    4. 17.3 Sound Intensity and Sound Level
    5. 17.4 Doppler Effect and Sonic Booms
    6. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
    7. 17.6 Hearing
    8. 17.7 Ultrasound
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  19. 18 Electric Charge and Electric Field
    1. Introduction to Electric Charge and Electric Field
    2. 18.1 Static Electricity and Charge: Conservation of Charge
    3. 18.2 Conductors and Insulators
    4. 18.3 Coulomb’s Law
    5. 18.4 Electric Field: Concept of a Field Revisited
    6. 18.5 Electric Field Lines: Multiple Charges
    7. 18.6 Electric Forces in Biology
    8. 18.7 Conductors and Electric Fields in Static Equilibrium
    9. 18.8 Applications of Electrostatics
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
  20. 19 Electric Potential and Electric Field
    1. Introduction to Electric Potential and Electric Energy
    2. 19.1 Electric Potential Energy: Potential Difference
    3. 19.2 Electric Potential in a Uniform Electric Field
    4. 19.3 Electrical Potential Due to a Point Charge
    5. 19.4 Equipotential Lines
    6. 19.5 Capacitors and Dielectrics
    7. 19.6 Capacitors in Series and Parallel
    8. 19.7 Energy Stored in Capacitors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  21. 20 Electric Current, Resistance, and Ohm's Law
    1. Introduction to Electric Current, Resistance, and Ohm's Law
    2. 20.1 Current
    3. 20.2 Ohm’s Law: Resistance and Simple Circuits
    4. 20.3 Resistance and Resistivity
    5. 20.4 Electric Power and Energy
    6. 20.5 Alternating Current versus Direct Current
    7. 20.6 Electric Hazards and the Human Body
    8. 20.7 Nerve Conduction–Electrocardiograms
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  22. 21 Circuits and DC Instruments
    1. Introduction to Circuits and DC Instruments
    2. 21.1 Resistors in Series and Parallel
    3. 21.2 Electromotive Force: Terminal Voltage
    4. 21.3 Kirchhoff’s Rules
    5. 21.4 DC Voltmeters and Ammeters
    6. 21.5 Null Measurements
    7. 21.6 DC Circuits Containing Resistors and Capacitors
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  23. 22 Magnetism
    1. Introduction to Magnetism
    2. 22.1 Magnets
    3. 22.2 Ferromagnets and Electromagnets
    4. 22.3 Magnetic Fields and Magnetic Field Lines
    5. 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
    6. 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
    7. 22.6 The Hall Effect
    8. 22.7 Magnetic Force on a Current-Carrying Conductor
    9. 22.8 Torque on a Current Loop: Motors and Meters
    10. 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
    11. 22.10 Magnetic Force between Two Parallel Conductors
    12. 22.11 More Applications of Magnetism
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
  24. 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
    1. Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
    2. 23.1 Induced Emf and Magnetic Flux
    3. 23.2 Faraday’s Law of Induction: Lenz’s Law
    4. 23.3 Motional Emf
    5. 23.4 Eddy Currents and Magnetic Damping
    6. 23.5 Electric Generators
    7. 23.6 Back Emf
    8. 23.7 Transformers
    9. 23.8 Electrical Safety: Systems and Devices
    10. 23.9 Inductance
    11. 23.10 RL Circuits
    12. 23.11 Reactance, Inductive and Capacitive
    13. 23.12 RLC Series AC Circuits
    14. Glossary
    15. Section Summary
    16. Conceptual Questions
    17. Problems & Exercises
  25. 24 Electromagnetic Waves
    1. Introduction to Electromagnetic Waves
    2. 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
    3. 24.2 Production of Electromagnetic Waves
    4. 24.3 The Electromagnetic Spectrum
    5. 24.4 Energy in Electromagnetic Waves
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
  26. 25 Geometric Optics
    1. Introduction to Geometric Optics
    2. 25.1 The Ray Aspect of Light
    3. 25.2 The Law of Reflection
    4. 25.3 The Law of Refraction
    5. 25.4 Total Internal Reflection
    6. 25.5 Dispersion: The Rainbow and Prisms
    7. 25.6 Image Formation by Lenses
    8. 25.7 Image Formation by Mirrors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  27. 26 Vision and Optical Instruments
    1. Introduction to Vision and Optical Instruments
    2. 26.1 Physics of the Eye
    3. 26.2 Vision Correction
    4. 26.3 Color and Color Vision
    5. 26.4 Microscopes
    6. 26.5 Telescopes
    7. 26.6 Aberrations
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  28. 27 Wave Optics
    1. Introduction to Wave Optics
    2. 27.1 The Wave Aspect of Light: Interference
    3. 27.2 Huygens's Principle: Diffraction
    4. 27.3 Young’s Double Slit Experiment
    5. 27.4 Multiple Slit Diffraction
    6. 27.5 Single Slit Diffraction
    7. 27.6 Limits of Resolution: The Rayleigh Criterion
    8. 27.7 Thin Film Interference
    9. 27.8 Polarization
    10. 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
  29. 28 Special Relativity
    1. Introduction to Special Relativity
    2. 28.1 Einstein’s Postulates
    3. 28.2 Simultaneity And Time Dilation
    4. 28.3 Length Contraction
    5. 28.4 Relativistic Addition of Velocities
    6. 28.5 Relativistic Momentum
    7. 28.6 Relativistic Energy
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  30. 29 Introduction to Quantum Physics
    1. Introduction to Quantum Physics
    2. 29.1 Quantization of Energy
    3. 29.2 The Photoelectric Effect
    4. 29.3 Photon Energies and the Electromagnetic Spectrum
    5. 29.4 Photon Momentum
    6. 29.5 The Particle-Wave Duality
    7. 29.6 The Wave Nature of Matter
    8. 29.7 Probability: The Heisenberg Uncertainty Principle
    9. 29.8 The Particle-Wave Duality Reviewed
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
  31. 30 Atomic Physics
    1. Introduction to Atomic Physics
    2. 30.1 Discovery of the Atom
    3. 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
    4. 30.3 Bohr’s Theory of the Hydrogen Atom
    5. 30.4 X Rays: Atomic Origins and Applications
    6. 30.5 Applications of Atomic Excitations and De-Excitations
    7. 30.6 The Wave Nature of Matter Causes Quantization
    8. 30.7 Patterns in Spectra Reveal More Quantization
    9. 30.8 Quantum Numbers and Rules
    10. 30.9 The Pauli Exclusion Principle
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
  32. 31 Radioactivity and Nuclear Physics
    1. Introduction to Radioactivity and Nuclear Physics
    2. 31.1 Nuclear Radioactivity
    3. 31.2 Radiation Detection and Detectors
    4. 31.3 Substructure of the Nucleus
    5. 31.4 Nuclear Decay and Conservation Laws
    6. 31.5 Half-Life and Activity
    7. 31.6 Binding Energy
    8. 31.7 Tunneling
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  33. 32 Medical Applications of Nuclear Physics
    1. Introduction to Applications of Nuclear Physics
    2. 32.1 Medical Imaging and Diagnostics
    3. 32.2 Biological Effects of Ionizing Radiation
    4. 32.3 Therapeutic Uses of Ionizing Radiation
    5. 32.4 Food Irradiation
    6. 32.5 Fusion
    7. 32.6 Fission
    8. 32.7 Nuclear Weapons
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  34. 33 Particle Physics
    1. Introduction to Particle Physics
    2. 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
    3. 33.2 The Four Basic Forces
    4. 33.3 Accelerators Create Matter from Energy
    5. 33.4 Particles, Patterns, and Conservation Laws
    6. 33.5 Quarks: Is That All There Is?
    7. 33.6 GUTs: The Unification of Forces
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
  35. 34 Frontiers of Physics
    1. Introduction to Frontiers of Physics
    2. 34.1 Cosmology and Particle Physics
    3. 34.2 General Relativity and Quantum Gravity
    4. 34.3 Superstrings
    5. 34.4 Dark Matter and Closure
    6. 34.5 Complexity and Chaos
    7. 34.6 High-temperature Superconductors
    8. 34.7 Some Questions We Know to Ask
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  36. A | Atomic Masses
  37. B | Selected Radioactive Isotopes
  38. C | Useful Information
  39. D | Glossary of Key Symbols and Notation
  40. Index

In this module, we will learn about work and energy associated with rotational motion. Figure 10.14 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy.

The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.
Figure 10.14 The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell)

Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.15) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:

net W = ( net F ) Δ s . net W = ( net F ) Δ s . size 12{"net "W= left ("net "F right ) cdot Δs} {}
10.53

To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by rr size 12{r} {}, and gather terms:

net W = ( r net F ) Δs r . net W = ( r net F ) Δs r . size 12{"net"W= left (r" net "F right ) { {Δs} over {r} } } {}
10.54

We recognize that r net F= net τr net F= net τ size 12{r" net "F=" net "τ} {} and Δs/r=θΔs/r=θ size 12{Δs/r=θ} {}, so that

net W=net τθ.net W=net τθ. size 12{"net "W= left ("net "τ right )θ} {}
10.55

This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation net W=net τθnet W=net τθ size 12{"net "W= left ("net "τ right )θ} {} is valid in general, even though it was derived for a special case.

To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net τ=net τ= size 12{"net "W=Iα} {}, so that

net W=Iαθ.net W=Iαθ. size 12{"net "W=I ital "αθ"} {}
10.56
The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.
Figure 10.15 The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus net FΔsnet FΔs size 12{ left ("net "F right ) cdot Δs} {}. The net work goes into rotational kinetic energy.

Making Connections

Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation.

Now, we solve one of the rotational kinematics equations for αθαθ size 12{ ital "αθ"} {}. We start with the equation

ω 2 = ω 0 2 + 2 αθ . ω 2 = ω 0 2 + 2 αθ . size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {}
10.57

Next, we solve for αθαθ size 12{ ital "αθ"} {}:

αθ = ω 2 ω 0 2 2 . αθ = ω 2 ω 0 2 2 . size 12{ ital "αθ"= { {ω rSup { size 8{2} } - ω rSub { size 8{0} rSup { size 8{2} } } } over {2} } } {}
10.58

Substituting this into the equation for net WW size 12{W} {} and gathering terms yields

net W=12212I ω 0 2 .net W=12212I ω 0 2 . size 12{"net "W= { {1} over {2} } Iω rSup { size 8{2} } - { {1} over {2} } Iω rSub { size 8{0} rSup { size 8{2} } } } {}
10.59

This equation is the work-energy theorem for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term 122122 size 12{ left ( { {1} over {2} } right )Iω rSup { size 8{2} } } {} to be rotational kinetic energy KErotKErot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {} for an object with a moment of inertia II size 12{I} {} and an angular velocity ωω size 12{ω} {}:

KE rot = 1 2 2 . KE rot = 1 2 2 . size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
10.60

The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with II size 12{I} {} being analogous to mm size 12{m} {} and ωω size 12{ω} {} to vv size 12{v} {}. Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in Figure 10.16.

The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.
Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into KErotKErot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {}. It can also convert translational kinetic energy, when the bus stops, into KErotKErot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {}. The flywheel’s energy can then be used to accelerate, to go up another hill, or to keep the bus from going against friction.

Example 10.8 Calculating the Work and Energy for Spinning a Grindstone

Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of 1.00 rad(57.3º)1.00 rad(57.3º) size 12{1 "." "00"`"rad" \( "57" "." 3 \) rSup { size 8{ circ } } } {}? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.)

Strategy

To find the work, we can use the equation net W=net τθnet W=net τθ size 12{"net "W= left ("net "τ right )θ} {}. We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in KErot=122KErot=122 size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}.

Solution for (a)

The net work is expressed in the equation

net W=net τθ,net W=net τθ, size 12{"net "W= left ("net "τ right )θ} {}
10.61

where net ττ size 12{τ} {} is the applied force multiplied by the radius (rF)(rF) size 12{ \( ital "rF" \) } {} because there is no retarding friction, and the force is perpendicular to rr size 12{r} {}. The angle θθ size 12{θ} {} is given. Substituting the given values in the equation above yields

net W = rF θ= 0.320 m 200 N 1.00 rad = 64.0 Nm. net W = rF θ= 0.320 m 200 N 1.00 rad = 64.0 Nm.
10.62

Noting that 1 N·m=1 J1 N·m=1 J,

net W=64.0 J.net W=64.0 J. size 12{"net "W="64" "." 0" J"} {}
10.63
The figure shows a large grindstone of radius r which is being given a spin by applying a force F in a counterclockwise direction, as indicated by the arrows.
Figure 10.17 A large grindstone is given a spin by a person grasping its outer edge.

Solution for (b)

To find ωω size 12{ω} {} from the given information requires more than one step. We start with the kinematic relationship in the equation

ω2= ω 0 2 +2αθ.ω2= ω 0 2 +2αθ. size 12{ω rSup { size 8{2} } =ω rSub { size 8{0} rSup { size 8{2} } } +2 ital "αθ"} {}
10.64

Note that ω0=0ω0=0 size 12{ω rSub { size 8{0} } =0} {} because we start from rest. Taking the square root of the resulting equation gives

ω=2αθ1/2.ω=2αθ1/2. size 12{ω= left (2 ital "αθ" right ) rSup { size 8{1/2} } } {}
10.65

Now we need to find αα size 12{α} {}. One possibility is

α=net τI,α=net τI, size 12{α= { {"net "τ} over {I} } } {}
10.66

where the torque is

net τ=rF=0.320 m200 N=64.0 Nm.net τ=rF=0.320 m200 N=64.0 Nm. size 12{"net "τ= ital "rF"= left (0 "." "320"" m" right ) left ("200"" N" right )="64" "." 0" N" cdot m} {}
10.67

The formula for the moment of inertia for a disk is found in Figure 10.12:

I = 1 2 MR 2 = 0.5 85.0 kg 0.320 m 2 = 4.352 kg m 2 . I = 1 2 MR 2 = 0.5 85.0 kg 0.320 m 2 = 4.352 kg m 2 . size 12{I= { {1} over {2} } ital "MR" rSup { size 8{2} } =0 "." 5 left ("85" "." 0" kg" right ) left (0 "." "320"" m" right ) rSup { size 8{2} } =4 "." "352"" kg" cdot m rSup { size 8{2} } } {}
10.68

Substituting the values of torque and moment of inertia into the expression for αα size 12{α} {}, we obtain

α=64.0 Nm4.352 kgm2=14.7rads2.α=64.0 Nm4.352 kgm2=14.7rads2. size 12{α= { {"64" "." "0 N" cdot m} over {4 "." "352"" kg" cdot m rSup { size 8{2} } } } ="14" "." 7 { {"rad"} over {s rSup { size 8{2} } } } } {}
10.69

Now, substitute this value and the given value for θθ size 12{θ} {} into the above expression for ωω size 12{ω} {}:

ω = 2 αθ 1 / 2 = 2 14.7 rad s 2 1.00 rad 1 / 2 = 5.42 rad s . ω = 2 αθ 1 / 2 = 2 14.7 rad s 2 1.00 rad 1 / 2 = 5.42 rad s . size 12{ω= left (2 ital "αθ" right ) rSup { size 8{1/2} } = left [2 left ("14" "." 7 { {"rad"} over {s rSup { size 8{2} } } } right ) left (1 "." "00"" rad" right ) right ] rSup { size 8{1/2} } =5 "." "42" { {"rad"} over {s} } } {}
10.70

Solution for (c)

The final rotational kinetic energy is

KErot=122.KErot=122. size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
10.71

Both II size 12{I} {} and ωω size 12{ω} {} were found above. Thus,

KE rot = 0.5 4.352 kgm25.42 rad/s2=64.0 J. KE rot = 0.5 4.352 kgm25.42 rad/s2=64.0 J. size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left (4 "." "352"" kg" cdot m rSup { size 8{2} } right ) left (5 "." "42"" rad/s" right ) rSup { size 8{2} } ="64" "." 0" J"} {}
10.72

Discussion

The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.

Helicopter pilots are quite familiar with rotational kinetic energy. They know, for example, that a point of no return will be reached if they allow their blades to slow below a critical angular velocity during flight. The blades lose lift, and it is impossible to immediately get the blades spinning fast enough to regain it. Rotational kinetic energy must be supplied to the blades to get them to rotate faster, and enough energy cannot be supplied in time to avoid a crash. Because of weight limitations, helicopter engines are too small to supply both the energy needed for lift and to replenish the rotational kinetic energy of the blades once they have slowed down. The rotational kinetic energy is put into them before takeoff and must not be allowed to drop below this crucial level. One possible way to avoid a crash is to use the gravitational potential energy of the helicopter to replenish the rotational kinetic energy of the blades by losing altitude and aligning the blades so that the helicopter is spun up in the descent. Of course, if the helicopter’s altitude is too low, then there is insufficient time for the blade to regain lift before reaching the ground.

Problem-Solving Strategy for Rotational Energy

  1. Determine that energy or work is involved in the rotation.
  2. Determine the system of interest. A sketch usually helps.
  3. Analyze the situation to determine the types of work and energy involved.
  4. For closed systems, mechanical energy is conserved. That is, KEi+PEi=KEf+PEf.KEi+PEi=KEf+PEf. size 12{"KE" rSub { size 8{i} } +"PE" rSub { size 8{i} } ="KE" rSub { size 8{f} } +"PE" rSub { size 8{f} } } {} Note that KEiKEi size 12{"KE" rSub { size 8{i} } } {} and KEfKEf may each include translational and rotational contributions.
  5. For open systems, mechanical energy may not be conserved, and other forms of energy (referred to previously as OEOE size 12{ ital "OE"} {}), such as heat transfer, may enter or leave the system. Determine what they are, and calculate them as necessary.
  6. Eliminate terms wherever possible to simplify the algebra.
  7. Check the answer to see if it is reasonable.

Example 10.9 Calculating Helicopter Energies

A typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

Solution for (a)

The rotational kinetic energy is

KErot=122.KErot=122. size 12{"KE" rSub { size 8{"rot"} } = { {1} over {2} } Iω rSup { size 8{2} } } {}
10.73

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find KErotKErot. The angular velocity ωω size 12{ω} {} is

ω=300 rev1.00 min2π rad1 rev1.00 min60.0 s=31.4rads.ω=300 rev1.00 min2π rad1 rev1.00 min60.0 s=31.4rads. size 12{ω= { {"300"" rev"} over {1 "." "00 min"} } cdot { {2π" rad"} over {"1 rev"} } cdot { {1 "." "00"" min"} over {"60" "." 0" s"} } ="31" "." 4 { {"rad"} over {s} } } {}
10.74

The moment of inertia of one blade will be that of a thin rod rotated about its end, found in Figure 10.12. The total II size 12{I} {} is four times this moment of inertia, because there are four blades. Thus,

I=4Mℓ23=4×50.0 kg4.00 m23=1067 kgm2.I=4Mℓ23=4×50.0 kg4.00 m23=1067 kgm2. size 12{I=4 { {Mℓ rSup { size 8{2} } } over {3} } =4 times { { left ("50" "." 0" kg" right ) left (4 "." "00"" m" right ) rSup { size 8{2} } } over {3} } ="1067"" kg" cdot m rSup { size 8{2} } } {}
10.75

Entering ωω size 12{ω} {} and II size 12{I} {} into the expression for rotational kinetic energy gives

KE rot = 0.5 ( 1067 kg m 2 ) 31.4 rad/s 2 = 5.26 × 10 5 J KE rot = 0.5 ( 1067 kg m 2 ) 31.4 rad/s 2 = 5.26 × 10 5 J alignl { stack { size 12{"KE" rSub { size 8{"rot"} } =0 "." 5 left ("1067"" kg" cdot m rSup { size 8{2} } right ) left ("31" "." 4" rad/s" right ) rSup { size 8{2} } } {} # " "=5 "." "26" times "10" rSup { size 8{5} } " J" {} } } {}
10.76

Solution for (b)

Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given values of mass and velocity, we obtain

KEtrans=12mv2=0.51000 kg20.0 m/s2=2.00×105 J.KEtrans=12mv2=0.51000 kg20.0 m/s2=2.00×105 J. size 12{"KE" rSub { size 8{"trans"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } =0 "." 5 left ("1000"" kg" right ) left ("20" "." 0" m/s" right ) rSup { size 8{2} } =2 "." "00" times "10" rSup { size 8{5} } " J"} {}
10.77

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

2.00×105 J5.26×105 J=0.380.2.00×105 J5.26×105 J=0.380. size 12{ { {2 "." "00" times "10" rSup { size 8{5} } " J"} over {5 "." "26" times "10" rSup { size 8{5} } " J"} } =0 "." "380"} {}
10.78

Solution for (c)

At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:

KE rot = PE grav KE rot = PE grav size 12{"KE" rSub { size 8{"rot"} } ="PE" rSub { size 8{"grav"} } } {}
10.79

or

1 2 2 = mgh . 1 2 2 = mgh . size 12{ { {1} over {2} } Iω rSup { size 8{2} } = ital "mgh"} {}
10.80

We now solve for hh size 12{h} {} and substitute known values into the resulting equation

h = 12 2 mg = 5.26 × 10 5 J 1000 kg 9.80 m/s 2 = 53.7 m . h = 12 2 mg = 5.26 × 10 5 J 1000 kg 9.80 m/s 2 = 53.7 m . size 12{h= { { { size 8{1} } wideslash { size 8{2} } Iω rSup { size 8{2} } } over { ital "mg"} } = { {5 "." "26" times "10" rSup { size 8{5} } " J"} over { left ("1000"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} } ="53" "." 7" m"} {}
10.81

Discussion

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.

The given figure here shows a helicopter from the Auckland Westpac Rescue Helicopter Service over a sea. A rescue diver is shown holding the iron stand bar at the bottom of the helicopter, clutching a person. In the other image just above this, the blades of the helicopter are shown with their anti-clockwise rotation direction shown with an arrow and the length of one blade is given as four meters.
Figure 10.18 The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)

Making Connections

Conservation of energy includes rotational motion, because rotational kinetic energy is another form of KE KE size 12{"KE"} {} . Uniform Circular Motion and Gravitation has a detailed treatment of conservation of energy.

How Thick Is the Soup? Or Why Don’t All Objects Roll Downhill at the Same Rate?

One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?

The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy PEgravPEgrav size 12{ ital "PE" rSub { size 8{ ital "grav"} } } {}, which is converted entirely to KEKE, provided each rolls without slipping. KEKE, however, can take the form of KEtransKEtrans size 12{ ital "KE" rSub { size 8{ ital "trans"} } } {} or KErotKErot size 12{ ital "KE" rSub { size 8{ ital "rot"} } } {}, and total KEKE is the sum of the two. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can’s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure 10.19.

The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.
Figure 10.19 Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE.

Assuming no losses due to friction, there is only one force doing work—gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives

PEi=KEf.PEi=KEf. size 12{"PE" rSub { size 8{i} } ="KE" rSub { size 8{f} } } {}
10.82

More specifically,

PE grav = KE trans + KE rot PE grav = KE trans + KE rot size 12{"PE" rSub { size 8{"grav"} } ="KE" rSub { size 8{"trans"} } +"KE" rSub { size 8{"rot"} } } {}
10.83

or

mgh=12mv2+122.mgh=12mv2+122. size 12{ ital "mgh"= { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } Iω rSup { size 8{2} } } {}
10.84

So, the initial mghmgh size 12{ ital "mgh"} {} is divided between translational kinetic energy and rotational kinetic energy; and the greater II size 12{I} {} is, the less energy goes into translation. If the can slides down without friction, then ω=0ω=0 size 12{ω=0} {} and all the energy goes into translation; thus, the can goes faster.

Take-Home Experiment

Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand.

Example 10.10 Calculating the Speed of a Cylinder Rolling Down an Incline

Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Strategy

We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with vv as the only unknown.

Solution

Conservation of energy for this situation is written as described above:

mgh = 1 2 mv 2 + 1 2 2 . mgh = 1 2 mv 2 + 1 2 2 . size 12{ ital "mgh"= { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } Iω rSup { size 8{2} } } {}
10.85

Before we can solve for vv size 12{v} {} , we must get an expression for II size 12{I} {} from Figure 10.12. Because vv size 12{v} {} and ωω size 12{ω} {} are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship ω=v/Rω=v/R size 12{ω=v/R} {} into the expression. These substitutions yield

mgh=12mv2+1212mR2v2R2.mgh=12mv2+1212mR2v2R2. size 12{ ital "mgh"= { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } left ( { {1} over {2} } ital "mR" rSup { size 8{2} } right ) left ( { {v rSup { size 8{2} } } over {R rSup { size 8{2} } } } right )} {}
10.86

Interestingly, the cylinder’s radius RR and mass mm cancel, yielding

gh=12v2+14v2=34v2.gh=12v2+14v2=34v2. size 12{ ital "gh"= { {1} over {2} } v rSup { size 8{2} } + { {1} over {4} } v rSup { size 8{2} } = { {3} over {4} } v rSup { size 8{2} } } {}
10.87

Solving algebraically, the equation for the final velocity v v size 12{v} {} gives

v=4gh31/2.v=4gh31/2. size 12{v= left ( { {4 ital "gh"} over {3} } right ) rSup { size 8{1/2} } } {}
10.88

Substituting known values into the resulting expression yields

v=49.80 m/s22.00 m31/2=5.11 m/s.v=49.80 m/s22.00 m31/2=5.11 m/s. size 12{v= left [ { {4 left (9 "." "80"" m/s" rSup { size 8{2} } right ) left (2 "." "00"" m" right )} over {3} } right ] rSup { size 8{1/2} } =5 "." "11"" m/s"} {}
10.89

Discussion

Because mm size 12{m} {} and RR size 12{R} {} cancel, the result v=43gh1/2v=43gh1/2 size 12{v= left ( { {4} over {3} } ital "gh" right ) rSup { size 8{1/2} } } {} is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.) Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, 12mv2=mgh12mv2=mgh size 12{ \( 1/2 \) ital "mv" rSup { size 8{2} } `= ital "mgh"} {} and v=(2gh)1/2v=(2gh)1/2 size 12{v= \( 2 ital "gh" \) rSup { size 8{1/2} } } {}, which is 22% greater than (4gh/3)1/2(4gh/3)1/2 size 12{ \( 4 ital "gh"/3 \) rSup { size 8{1/2} } } {}. That is, the cylinder would go faster at the bottom.

Check Your Understanding

Analogy of Rotational and Translational Kinetic EnergyIs rotational kinetic energy completely analogous to translational kinetic energy? What, if any, are their differences? Give an example of each type of kinetic energy.

Solution

Yes, rotational and translational kinetic energy are exact analogs. They both are the energy of motion involved with the coordinated (non-random) movement of mass relative to some reference frame. The only difference between rotational and translational kinetic energy is that translational is straight line motion while rotational is not. An example of both kinetic and translational kinetic energy is found in a bike tire while being ridden down a bike path. The rotational motion of the tire means it has rotational kinetic energy while the movement of the bike along the path means the tire also has translational kinetic energy. If you were to lift the front wheel of the bike and spin it while the bike is stationary, then the wheel would have only rotational kinetic energy relative to the Earth.

PhET Explorations: My Solar System

Build your own system of heavenly bodies and watch the gravitational ballet. With this orbit simulator, you can set initial positions, velocities, and masses of 2, 3, or 4 bodies, and then see them orbit each other.

Figure 10.20
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