Skip to Content
OpenStax Logo
University Physics Volume 1

11.1 Rolling Motion

University Physics Volume 111.1 Rolling Motion
Buy book
  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Describe the physics of rolling motion without slipping
  • Explain how linear variables are related to angular variables for the case of rolling motion without slipping
  • Find the linear and angular accelerations in rolling motion with and without slipping
  • Calculate the static friction force associated with rolling motion without slipping
  • Use energy conservation to analyze rolling motion

Rolling motion is that common combination of rotational and translational motion that we see everywhere, every day. Think about the different situations of wheels moving on a car along a highway, or wheels on a plane landing on a runway, or wheels on a robotic explorer on another planet. Understanding the forces and torques involved in rolling motion is a crucial factor in many different types of situations.

For analyzing rolling motion in this chapter, refer to Figure 10.20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. You may also find it useful in other calculations involving rotation.

Rolling Motion without Slipping

People have observed rolling motion without slipping ever since the invention of the wheel. For example, we can look at the interaction of a car’s tires and the surface of the road. If the driver depresses the accelerator to the floor, such that the tires spin without the car moving forward, there must be kinetic friction between the wheels and the surface of the road. If the driver depresses the accelerator slowly, causing the car to move forward, then the tires roll without slipping. It is surprising to most people that, in fact, the bottom of the wheel is at rest with respect to the ground, indicating there must be static friction between the tires and the road surface. In Figure 11.2, the bicycle is in motion with the rider staying upright. The tires have contact with the road surface, and, even though they are rolling, the bottoms of the tires deform slightly, do not slip, and are at rest with respect to the road surface for a measurable amount of time. There must be static friction between the tire and the road surface for this to be so.

Figure a is a photograph of a person riding a bicycle. The camera followed the bike, so the image of the bike and rider is sharp, the background is blurred due to bike’s motion. Figure b is a photograph of a bicycle wheel rolling on the ground, with the camera stationary relative to the ground. The wheel and spokes are blurred at the top but clear at the bottom.
Figure 11.2 (a) The bicycle moves forward, and its tires do not slip. The bottom of the slightly deformed tire is at rest with respect to the road surface for a measurable amount of time. (b) This image shows that the top of a rolling wheel appears blurred by its motion, but the bottom of the wheel is instantaneously at rest. (credit a: modification of work by Nelson Lourenço; credit b: modification of work by Colin Rose)

To analyze rolling without slipping, we first derive the linear variables of velocity and acceleration of the center of mass of the wheel in terms of the angular variables that describe the wheel’s motion. The situation is shown in Figure 11.3.

Figure a shows a free body diagram of a wheel, including the location where the forces act. Four forces are shown: M g is a downward force acting on the center of the wheel. N is an upward force acting on the bottom of the wheel. F is a force to the right, acting on the center of the wheel, and f sub s is a force to the left acting on the bottom of the wheel. The force f sub s is smaller or equal to mu sub s times N. Figure b is an illustration of a wheel rolling without slipping on a horizontal surface. Point P is the contact point between the bottom of the wheel and the surface. The wheel has a clockwise rotation, an acceleration to the right of a sub C M and a velocity to the right of v sub V M. The relations omega equals v sub C M over R and alpha equals a sub C M over R are given. A coordinate system with positive x to the right and positive y up is shown. Figure c shows wheel in the center of mass frame. Point P has velocity vector in the negative direction with respect to the center of mass of the wheel. That vector is shown on the diagram and labeled as minus R omega i hat. It is tangent to the wheel at the bottom, and pointing to the left. Additional vectors at various locations on the rim of the wheel are shown, all tangent to the wheel and pointing clockwise.
Figure 11.3 (a) A wheel is pulled across a horizontal surface by a force FF. The force of static friction fS,|fS|μSNfS,|fS|μSN is large enough to keep it from slipping. (b) The linear velocity and acceleration vectors of the center of mass and the relevant expressions for ωandαωandα. Point P is at rest relative to the surface. (c) Relative to the center of mass (CM) frame, point P has linear velocity Rωi^Rωi^.

From Figure 11.3(a), we see the force vectors involved in preventing the wheel from slipping. In (b), point P that touches the surface is at rest relative to the surface. Relative to the center of mass, point P has velocity Rωi^Rωi^, where R is the radius of the wheel and ωω is the wheel’s angular velocity about its axis. Since the wheel is rolling, the velocity of P with respect to the surface is its velocity with respect to the center of mass plus the velocity of the center of mass with respect to the surface:

vP=Rωi^+vCMi^.vP=Rωi^+vCMi^.

Since the velocity of P relative to the surface is zero, vP=0vP=0, this says that

vCM=Rω.vCM=Rω.
(11.1)

Thus, the velocity of the wheel’s center of mass is its radius times the angular velocity about its axis. We show the correspondence of the linear variable on the left side of the equation with the angular variable on the right side of the equation. This is done below for the linear acceleration.

If we differentiate Equation 11.1 on the left side of the equation, we obtain an expression for the linear acceleration of the center of mass. On the right side of the equation, R is a constant and since α=dωdt,α=dωdt, we have

aCM=Rα.aCM=Rα.
(11.2)

Furthermore, we can find the distance the wheel travels in terms of angular variables by referring to Figure 11.4. As the wheel rolls from point A to point B, its outer surface maps onto the ground by exactly the distance travelled, which is dCM.dCM. We see from Figure 11.4 that the length of the outer surface that maps onto the ground is the arc length RθRθ. Equating the two distances, we obtain

dCM=Rθ.dCM=Rθ.
(11.3)
A wheel, radius R, rolling on a horizontal surface and moving to the right at v sub C M is drawn in two positions. In the first position, point A on the wheel is at the bottom, in contact with the surface, and point B is at the top. The arc length from A to B along the rim of the wheel is highlighted and labeled as being R theta. In the second position, point B on the wheel is at the bottom, in contact with the surface, and point A is at the top. The horizontal distance between the wheel’s point of contact with the surface in the two illustrated positions is d sub C M. The arc length A B is now on the other side of the wheel.
Figure 11.4 As the wheel rolls on the surface, the arc length RθRθ from A to B maps onto the surface, corresponding to the distance dCMdCM that the center of mass has moved.

Example 11.1

Rolling Down an Inclined Plane A solid cylinder rolls down an inclined plane without slipping, starting from rest. It has mass m and radius r. (a) What is its acceleration? (b) What condition must the coefficient of static friction μSμS satisfy so the cylinder does not slip?

Strategy Draw a sketch and free-body diagram, and choose a coordinate system. We put x in the direction down the plane and y upward perpendicular to the plane. Identify the forces involved. These are the normal force, the force of gravity, and the force due to friction. Write down Newton’s laws in the x- and y-directions, and Newton’s law for rotation, and then solve for the acceleration and force due to friction.

Solution

  1. The free-body diagram and sketch are shown in Figure 11.5, including the normal force, components of the weight, and the static friction force. There is barely enough friction to keep the cylinder rolling without slipping. Since there is no slipping, the magnitude of the friction force is less than or equal to μSNμSN. Writing down Newton’s laws in the x- and y-directions, we have
    Fx=max; Fy=may. Fx=max; Fy=may.

    A diagram of a cylinder rolling without slipping down an inclined plane and a free body diagram of the cylinder. On the left is an illustration showing the inclined plane, which makes an angle of theta with the horizontal. The cylinder is shown to be at rest at the top, then moving along the incline when it is lower. On the right is a free body diagram. The x y coordinate system is tilted so that the positive x direction is parallel to the inclined plane and points toward its bottom, and the positive y direction is outward, perpendicular to the plane. Four forces are shown. N j hat acts at the center of the cylinder and points in the positive y direction. m g sine theta i hat acts at the center of the cylinder and points in the positive x direction. Minus m g cosine theta j hat acts at the center of the cylinder and points in the negative y direction. Minus f sub s i hat acts at the point of contact and points in the negative x direction.
    Figure 11.5 A solid cylinder rolls down an inclined plane without slipping from rest. The coordinate system has x in the direction down the inclined plane and y perpendicular to the plane. The free-body diagram is shown with the normal force, the static friction force, and the components of the weight mgmg. Friction makes the cylinder roll down the plane rather than slip.

    Substituting in from the free-body diagram,
    mgsinθfS=m(aCM)x,Nmgcosθ=0,fSμSN,mgsinθfS=m(aCM)x,Nmgcosθ=0,fSμSN,
    we can then solve for the linear acceleration of the center of mass from these equations:
    (aCM)x=g(sinθμScosθ).(aCM)x=g(sinθμScosθ).

    However, it is useful to express the linear acceleration in terms of the moment of inertia. For this, we write down Newton’s second law for rotation,
    τCM=ICMα. τCM=ICMα.

    The torques are calculated about the axis through the center of mass of the cylinder. The only nonzero torque is provided by the friction force. We have
    fSr=ICMα.fSr=ICMα.

    Finally, the linear acceleration is related to the angular acceleration by
    (aCM)x=rα.(aCM)x=rα.

    These equations can be used to solve for aCM,α,andfSaCM,α,andfS in terms of the moment of inertia, where we have dropped the x-subscript. We write aCMaCM in terms of the vertical component of gravity and the friction force, and make the following substitutions.
    aCM=gsinθfSmaCM=gsinθfSm

    fS=ICMαr=ICMaCMr2fS=ICMαr=ICMaCMr2
    From this we obtain
    aCM=gsinθICMaCMmr2,=mgsinθm+(ICM/r2).aCM=gsinθICMaCMmr2,=mgsinθm+(ICM/r2).

    Note that this result is independent of the coefficient of static friction, μSμS.
    Since we have a solid cylinder, from Figure 10.20, we have ICM=mr2/2ICM=mr2/2 and
    aCM=mgsinθm+(mr2/2r2)=23gsinθ.aCM=mgsinθm+(mr2/2r2)=23gsinθ.

    Therefore, we have
    α=aCMr=23rgsinθ.α=aCMr=23rgsinθ.
  2. Because slipping does not occur, fSμSNfSμSN. Solving for the friction force,
    fS=ICMαr=ICM(aCM)r2=ICMr2(mgsinθm+(ICM/r2))=mgICMsinθmr2+ICM.fS=ICMαr=ICM(aCM)r2=ICMr2(mgsinθm+(ICM/r2))=mgICMsinθmr2+ICM.

    Substituting this expression into the condition for no slipping, and noting that N=mgcosθN=mgcosθ, we have
    mgICMsinθmr2+ICMμSmgcosθmgICMsinθmr2+ICMμSmgcosθ

    or
    μStanθ1+(mr2/ICM).μStanθ1+(mr2/ICM).

    For the solid cylinder, this becomes
    μStanθ1+(2mr2/mr2)=13tanθ.μStanθ1+(2mr2/mr2)=13tanθ.

Significance

  1. The linear acceleration is linearly proportional to sinθ.sinθ. Thus, the greater the angle of the incline, the greater the linear acceleration, as would be expected. The angular acceleration, however, is linearly proportional to sinθsinθ and inversely proportional to the radius of the cylinder. Thus, the larger the radius, the smaller the angular acceleration.
  2. For no slipping to occur, the coefficient of static friction must be greater than or equal to (1/3)tanθ(1/3)tanθ. Thus, the greater the angle of incline, the greater the coefficient of static friction must be to prevent the cylinder from slipping.
Check Your Understanding 11.1

A hollow cylinder is on an incline at an angle of 60°.60°. The coefficient of static friction on the surface is μS=0.6μS=0.6. (a) Does the cylinder roll without slipping? (b) Will a solid cylinder roll without slipping?

It is worthwhile to repeat the equation derived in this example for the acceleration of an object rolling without slipping:

aCM=mgsinθm+(ICM/r2).aCM=mgsinθm+(ICM/r2).
(11.4)

This is a very useful equation for solving problems involving rolling without slipping. Note that the acceleration is less than that for an object sliding down a frictionless plane with no rotation. The acceleration will also be different for two rotating cylinders with different rotational inertias.

Rolling Motion with Slipping

In the case of rolling motion with slipping, we must use the coefficient of kinetic friction, which gives rise to the kinetic friction force since static friction is not present. The situation is shown in Figure 11.6. In the case of slipping, vCMRω0vCMRω0, because point P on the wheel is not at rest on the surface, and vP0vP0. Thus, ωvCMR,αaCMRωvCMR,αaCMR.

Figure a shows a free body diagram of a wheel, including the location where the forces act. Four forces are shown: M g is a downward force acting on the center of the wheel. N is an upward force acting on the bottom of the wheel. F is a force to the right, acting on the center of the wheel, and f sub k is a force to the left acting on the bottom of the wheel. The force f sub k is equal to mu sub k times N. Figure b is an illustration of the wheel rolling and slipping on a horizontal surface. The wheel has a clockwise rotation, an acceleration to the right of a sub C M and a velocity to the right of v sub V M. omega does not equal v sub C M over R and alpha does not equal a sub C M over R. A coordinate system with positive x to the right and positive y up is shown.
Figure 11.6 (a) Kinetic friction arises between the wheel and the surface because the wheel is slipping. (b) The simple relationships between the linear and angular variables are no longer valid.

Example 11.2

Rolling Down an Inclined Plane with Slipping A solid cylinder rolls down an inclined plane from rest and undergoes slipping (Figure 11.7). It has mass m and radius r. (a) What is its linear acceleration? (b) What is its angular acceleration about an axis through the center of mass?

Strategy Draw a sketch and free-body diagram showing the forces involved. The free-body diagram is similar to the no-slipping case except for the friction force, which is kinetic instead of static. Use Newton’s second law to solve for the acceleration in the x-direction. Use Newton’s second law of rotation to solve for the angular acceleration.

Solution

A diagram of a cylinder rolling and slipping down an inclined plane and a free body diagram of the cylinder. On the left is an illustration showing the inclined plane, which makes an angle of theta with the horizontal. The cylinder is shown to be at rest at the top, then moving along the incline when it is lower. On the right is a free body diagram. The x y coordinate system is tilted so that the positive x direction is parallel to the inclined plane and points toward its bottom, and the positive y direction is outward, perpendicular to the plane. Four forces are shown. N j hat acts at the center of the cylinder and points in the positive y direction. m g sine theta i hat acts at the center of the cylinder and points in the positive x direction. Minus m g cosine theta j hat acts at the center of the cylinder and points in the negative y direction. Minus f sub k i hat acts at the point of contact and points in the negative x direction.
Figure 11.7 A solid cylinder rolls down an inclined plane from rest and undergoes slipping. The coordinate system has x in the direction down the inclined plane and y upward perpendicular to the plane. The free-body diagram shows the normal force, kinetic friction force, and the components of the weight mg.mg.

The sum of the forces in the y-direction is zero, so the friction force is now fk=μkN=μkmgcosθ.fk=μkN=μkmgcosθ.

Newton’s second law in the x-direction becomes

Fx=max, Fx=max,
mgsinθμkmgcosθ=m(aCM)x,mgsinθμkmgcosθ=m(aCM)x,

or

(aCM)x=g(sinθμKcosθ).(aCM)x=g(sinθμKcosθ).

The friction force provides the only torque about the axis through the center of mass, so Newton’s second law of rotation becomes

τCM=ICMα, τCM=ICMα,
fkr=ICMα=12mr2α.fkr=ICMα=12mr2α.

Solving for αα, we have

α=2fkmr=2μkgcosθr.α=2fkmr=2μkgcosθr.

Significance We write the linear and angular accelerations in terms of the coefficient of kinetic friction. The linear acceleration is the same as that found for an object sliding down an inclined plane with kinetic friction. The angular acceleration about the axis of rotation is linearly proportional to the normal force, which depends on the cosine of the angle of inclination. As θ90°θ90°, this force goes to zero, and, thus, the angular acceleration goes to zero.

Conservation of Mechanical Energy in Rolling Motion

In the preceding chapter, we introduced rotational kinetic energy. Any rolling object carries rotational kinetic energy, as well as translational kinetic energy and potential energy if the system requires. Including the gravitational potential energy, the total mechanical energy of an object rolling is

ET=12mvCM2+12ICMω2+mgh.ET=12mvCM2+12ICMω2+mgh.

In the absence of any nonconservative forces that would take energy out of the system in the form of heat, the total energy of a rolling object without slipping is conserved and is constant throughout the motion. Examples where energy is not conserved are a rolling object that is slipping, production of heat as a result of kinetic friction, and a rolling object encountering air resistance.

You may ask why a rolling object that is not slipping conserves energy, since the static friction force is nonconservative. The answer can be found by referring back to Figure 11.3. Point P in contact with the surface is at rest with respect to the surface. Therefore, its infinitesimal displacement drdr with respect to the surface is zero, and the incremental work done by the static friction force is zero. We can apply energy conservation to our study of rolling motion to bring out some interesting results.

Example 11.3

Curiosity Rover The Curiosity rover, shown in Figure 11.8, was deployed on Mars on August 6, 2012. The wheels of the rover have a radius of 25 cm. Suppose astronauts arrive on Mars in the year 2050 and find the now-inoperative Curiosity on the side of a basin. While they are dismantling the rover, an astronaut accidentally loses a grip on one of the wheels, which rolls without slipping down into the bottom of the basin 25 meters below. If the wheel has a mass of 5 kg, what is its velocity at the bottom of the basin?

A photograph of the NASA rover Curiosity during testing at the Jet Propulsion Laboratory.
Figure 11.8 The NASA Mars Science Laboratory rover Curiosity during testing on June 3, 2011. The location is inside the Spacecraft Assembly Facility at NASA’s Jet Propulsion Laboratory in Pasadena, California. (credit: NASA/JPL-Caltech)

Strategy We use mechanical energy conservation to analyze the problem. At the top of the hill, the wheel is at rest and has only potential energy. At the bottom of the basin, the wheel has rotational and translational kinetic energy, which must be equal to the initial potential energy by energy conservation. Since the wheel is rolling without slipping, we use the relation vCM=rωvCM=rω to relate the translational variables to the rotational variables in the energy conservation equation. We then solve for the velocity. From Figure 11.8, we see that a hollow cylinder is a good approximation for the wheel, so we can use this moment of inertia to simplify the calculation.

Solution Energy at the top of the basin equals energy at the bottom:

mgh=12mvCM2+12ICMω2.mgh=12mvCM2+12ICMω2.

The known quantities are ICM=mr2,r=0.25m,andh=25.0mICM=mr2,r=0.25m,andh=25.0m.

We rewrite the energy conservation equation eliminating ωω by using ω=vCMr.ω=vCMr. We have

mgh=12mvCM2+12mr2vCM2r2mgh=12mvCM2+12mr2vCM2r2

or

gh=12vCM2+12vCM2vCM=gh.gh=12vCM2+12vCM2vCM=gh.

On Mars, the acceleration of gravity is 3.71m/s2,3.71m/s2, which gives the magnitude of the velocity at the bottom of the basin as

vCM=(3.71m/s2)25.0m=9.63m/s.vCM=(3.71m/s2)25.0m=9.63m/s.

Significance This is a fairly accurate result considering that Mars has very little atmosphere, and the loss of energy due to air resistance would be minimal. The result also assumes that the terrain is smooth, such that the wheel wouldn’t encounter rocks and bumps along the way.

Also, in this example, the kinetic energy, or energy of motion, is equally shared between linear and rotational motion. If we look at the moments of inertia in Figure 10.20, we see that the hollow cylinder has the largest moment of inertia for a given radius and mass. If the wheels of the rover were solid and approximated by solid cylinders, for example, there would be more kinetic energy in linear motion than in rotational motion. This would give the wheel a larger linear velocity than the hollow cylinder approximation. Thus, the solid cylinder would reach the bottom of the basin faster than the hollow cylinder.

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
Citation information

© Sep 19, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.