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  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:
  • State the forces that act on a simple pendulum
  • Determine the angular frequency, frequency, and period of a simple pendulum in terms of the length of the pendulum and the acceleration due to gravity
  • Define the period for a physical pendulum
  • Define the period for a torsional pendulum

Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.

The Simple Pendulum

A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass (Figure 15.20). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.

In the figure, a horizontal bar is shown. A string of length L extends from the bar at an angle theta counterclockwise from the vertical. The vertical direction is indicated by a dashed line extending down from where the string is attached to the bar. A circular bob of mass m is attached to the lower end of the string. The arc from the mass to the vertical is indicated by another dashed line and is a length s. A red arrow showing the time T of the oscillation of the mob is shown along the string line toward the bar. A coordinate system is shown near the bob with the positive y direction aligned with the string and pointing toward the pivot point and the positive x direction pointing tangent to the arc and away from the equilibrium position. An blue arrow from the bob toward the pivot, along the string, is labeled F sub T. A red arrow from the bob pointing down is labeled w = m g. A red arrow pointing tangent to the arc and toward equilibrium, in the minus x direction, is labeled minus m g sine theta. A red arrow at an angle theta counterclockwise from w is labeled minus m g cosine theta.
Figure 15.20 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of mgsinθmgsinθ toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

τ=L(mgsinθ);Iα=L(mgsinθ);Id2θdt2=L(mgsinθ);mL2d2θdt2=L(mgsinθ);d2θdt2=gLsinθ.τ=L(mgsinθ);Iα=L(mgsinθ);Id2θdt2=L(mgsinθ);mL2d2θdt2=L(mgsinθ);d2θdt2=gLsinθ.

The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for small angles (less than 15 degrees), sinθsinθ and θθ differ by less than 1%, so we can use the small angle approximation sinθθ.sinθθ. The angle θθ describes the position of the pendulum. Using the small angle approximation gives an approximate solution for small angles,

d2θdt2=gLθ.d2θdt2=gLθ.
(15.17)

Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is

ω=gLω=gL
(15.18)

and the period is

T=2πLg.T=2πLg.
(15.19)

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if θθ is less than about 15°.15°. Even simple pendulum clocks can be finely adjusted and remain accurate.

Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

Example 15.3

Measuring Acceleration due to Gravity by the Period of a Pendulum What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

Strategy We are asked to find g given the period T and the length L of a pendulum. We can solve T=2πLgT=2πLg for g, assuming only that the angle of deflection is less than 15°15°.

Solution

  1. Square T=2πLgT=2πLg and solve for g:
    g=4π2LT2.g=4π2LT2.
  2. Substitute known values into the new equation:
    g=4π20.75000m(1.7357s)2.g=4π20.75000m(1.7357s)2.
  3. Calculate to find g:
    g=9.8281m/s2.g=9.8281m/s2.

Significance This method for determining g can be very accurate, which is why length and period are given to five digits in this example. For the precision of the approximation sinθθsinθθ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5°0.5°.

Check Your Understanding 15.4

An engineer builds two simple pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced by 12°12°.

Physical Pendulum

Any object can oscillate like a pendulum. Consider a coffee mug hanging on a hook in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum until the oscillations die out. We have described a simple pendulum as a point mass and a string. A physical pendulum is any object whose oscillations are similar to those of the simple pendulum, but cannot be modeled as a point mass on a string, and the mass distribution must be included into the equation of motion.

As for the simple pendulum, the restoring force of the physical pendulum is the force of gravity. With the simple pendulum, the force of gravity acts on the center of the pendulum bob. In the case of the physical pendulum, the force of gravity acts on the center of mass (CM) of an object. The object oscillates about a point O. Consider an object of a generic shape as shown in Figure 15.21.

A drawing of a physical pendulum. In the figure, the pendulum is an irregularly shaped object. The center of mass, C M, is a distance L from the pivot point, O. The center of mass traces a circular arc, centered at O. The line from O to L makes an angle theta counterclockwise from the vertical. Three forces are depicted by red arrows at the center of mass. The force m g points down. Its components are minus m g sine theta which points tangent to the arc traced by the center of mass, and m g cosine theta which points radially outward.
Figure 15.21 A physical pendulum is any object that oscillates as a pendulum, but cannot be modeled as a point mass on a string. The force of gravity acts on the center of mass (CM) and provides the restoring force that causes the object to oscillate. The minus sign on the component of the weight that provides the restoring force is present because the force acts in the opposite direction of the increasing angle θθ.

When a physical pendulum is hanging from a point but is free to rotate, it rotates because of the torque applied at the CM, produced by the component of the object’s weight that acts tangent to the motion of the CM. Taking the counterclockwise direction to be positive, the component of the gravitational force that acts tangent to the motion is mgsinθmgsinθ. The minus sign is the result of the restoring force acting in the opposite direction of the increasing angle. Recall that the torque is equal to τ=r×Fτ=r×F. The magnitude of the torque is equal to the length of the radius arm times the tangential component of the force applied, |τ|=rFsinθ|τ|=rFsinθ. Here, the length L of the radius arm is the distance between the point of rotation and the CM. To analyze the motion, start with the net torque. Like the simple pendulum, consider only small angles so that sinθθsinθθ. Recall from Fixed-Axis Rotation on rotation that the net torque is equal to the moment of inertia I=r2dmI=r2dm times the angular acceleration α,α, where α=d2θdt2α=d2θdt2:

Iα=τnet=L(mg)sinθ.Iα=τnet=L(mg)sinθ.

Using the small angle approximation and rearranging:

Iα=L(mg)θ;Id2θdt2=L(mg)θ; d2θdt2=(mgLI)θ.Iα=L(mg)θ;Id2θdt2=L(mg)θ; d2θdt2=(mgLI)θ.

Once again, the equation says that the second time derivative of the position (in this case, the angle) equals minus a constant (mgLI)(mgLI) times the position. The solution is

θ(t)=Θcos(ωt+ϕ),θ(t)=Θcos(ωt+ϕ),

where ΘΘ is the maximum angular displacement. The angular frequency is

ω=mgLI.ω=mgLI.
(15.20)

The period is therefore

T=2πImgL.T=2πImgL.
(15.21)

Note that for a simple pendulum, the moment of inertia is I=r2dm=mL2I=r2dm=mL2 and the period reduces to T=2πLgT=2πLg.

Example 15.4

Reducing the Swaying of a Skyscraper In extreme conditions, skyscrapers can sway up to two meters with a frequency of up to 20.00 Hz due to high winds or seismic activity. Several companies have developed physical pendulums that are placed on the top of the skyscrapers. As the skyscraper sways to the right, the pendulum swings to the left, reducing the sway. Assuming the oscillations have a frequency of 0.50 Hz, design a pendulum that consists of a long beam, of constant density, with a mass of 100 metric tons and a pivot point at one end of the beam. What should be the length of the beam?

The figure depicts a tall building with a column on its roof and a long rod of length L that swings on a pivot point near the top of the column.

Strategy We are asked to find the length of the physical pendulum with a known mass. We first need to find the moment of inertia of the beam. We can then use the equation for the period of a physical pendulum to find the length.

Solution

  1. Find the moment of inertia for the CM:
  2. Use the parallel axis theorem to find the moment of inertia about the point of rotation:
    I=ICM+L42M=112ML2+14ML2=13ML2.I=ICM+L42M=112ML2+14ML2=13ML2.
  3. The period of a physical pendulum has a period of T=2πImgLT=2πImgL. Use the moment of inertia to solve for the length L:
    T=2πIMgL=2π13ML2MgL=2πL3g; L=3g(T2π)2=3(9.8ms2)(2s2π)2=2.98m.T=2πIMgL=2π13ML2MgL=2πL3g; L=3g(T2π)2=3(9.8ms2)(2s2π)2=2.98m.

Significance There are many ways to reduce the oscillations, including modifying the shape of the skyscrapers, using multiple physical pendulums, and using tuned-mass dampers.

Torsional Pendulum

A torsional pendulum consists of a rigid body suspended by a light wire or spring (Figure 15.22). When the body is twisted some small maximum angle (Θ)(Θ) and released from rest, the body oscillates between (θ=+Θ)(θ=+Θ) and (θ=Θ)(θ=Θ). The restoring torque is supplied by the shearing of the string or wire.

A torsional pendulum is illustrated in this figure. The pendulum consists of a horizontal disk that hangs by a string from the ceiling. The string attaches to the disk at its center, at point O. The disk and string can oscillate in a horizontal plane between angles plus Theta and minus Theta. The equilibrium position is between these, at theta = 0.
Figure 15.22 A torsional pendulum consists of a rigid body suspended by a string or wire. The rigid body oscillates between θ=+Θθ=+Θ and θ=Θθ=Θ.

The restoring torque can be modeled as being proportional to the angle:

τ=κθ.τ=κθ.

The variable kappa (κ)(κ) is known as the torsion constant of the wire or string. The minus sign shows that the restoring torque acts in the opposite direction to increasing angular displacement. The net torque is equal to the moment of inertia times the angular acceleration:

Id2θdt2=κθ; d2θdt2=κIθ.Id2θdt2=κθ; d2θdt2=κIθ.

This equation says that the second time derivative of the position (in this case, the angle) equals a negative constant times the position. This looks very similar to the equation of motion for the SHM d2xdt2=kmxd2xdt2=kmx, where the period was found to be T=2πmkT=2πmk. Therefore, the period of the torsional pendulum can be found using

T=2πIκ.T=2πIκ.
(15.22)

The units for the torsion constant are [κ]=N-m=(kgms2)m=kgm2s2[κ]=N-m=(kgms2)m=kgm2s2 and the units for the moment of inertial are [I]=kg-m2,[I]=kg-m2, which show that the unit for the period is the second.

Example 15.5

Measuring the Torsion Constant of a String A rod has a length of l=0.30ml=0.30m and a mass of 4.00 kg. A string is attached to the CM of the rod and the system is hung from the ceiling (Figure 15.23). The rod is displaced 10 degrees from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant κκ?

Figure a shows a horizontal rod, length 30.0 centimeters and mass 4.00 kilograms, hanging by a string from the ceiling. The string attaches to the middle of the rod. The rod rotates with the string in the horizontal plane. Figure b shows the rod with the details needed for finding its moment of inertia. The rod’s length, end to end, is L and its total mass is M. It has linear mass density lambda equals d m d x which also equals M over L. A small segment of the rod that has length d x at a distance x from the center of the rod is highlighted. The string is attached to the rod at the center of the rod.
Figure 15.23 (a) A rod suspended by a string from the ceiling. (b) Finding the rod’s moment of inertia.

Strategy We are asked to find the torsion constant of the string. We first need to find the moment of inertia.

Solution

  1. Find the moment of inertia for the CM:
    ICM=x2dm=L/2+L/2x2λdx=λ[x33]L/2+L/2=λ2L324=(ML)2L324=112ML2.ICM=x2dm=L/2+L/2x2λdx=λ[x33]L/2+L/2=λ2L324=(ML)2L324=112ML2.
  2. Calculate the torsion constant using the equation for the period:
    T=2πIκ;κ=I(2πT)2=(112ML2)(2πT)2;=(112(4.00kg)(0.30m)2)(2π0.50s)2=4.73N·m.T=2πIκ;κ=I(2πT)2=(112ML2)(2πT)2;=(112(4.00kg)(0.30m)2)(2π0.50s)2=4.73N·m.

Significance Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period.

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