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University Physics Volume 1

10.5 Calculating Moments of Inertia

University Physics Volume 110.5 Calculating Moments of Inertia
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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:
  • Calculate the moment of inertia for uniformly shaped, rigid bodies
  • Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known
  • Calculate the moment of inertia for compound objects

In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses).

Moment of Inertia

We defined the moment of inertia I of an object to be I=imiri2I=imiri2 for all the point masses that make up the object. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod (Figure 10.23) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.

In the case with the axis in the center of the barbell, each of the two masses m is a distance R away from the axis, giving a moment of inertia of

I1=mR2+mR2=2mR2.I1=mR2+mR2=2mR2.

In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is

I2=m(0)2+m(2R)2=4mR2.I2=m(0)2+m(2R)2=4mR2.

From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.

Figure A shows a barbell of the length 2 R with the masses m at the ends. It is rotating through its center. Figure B shows a barbell of the length 2 R with the masses m at the ends. It is rotating through one end.
Figure 10.23 (a) A barbell with an axis of rotation through its center; (b) a barbell with an axis of rotation through one end.

In this example, we had two point masses and the sum was simple to calculate. However, to deal with objects that are not point-like, we need to think carefully about each of the terms in the equation. The equation asks us to sum over each ‘piece of mass’ a certain distance from the axis of rotation. But what exactly does each ‘piece of mass’ mean? Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance r to the axis of rotation. However, this is not possible unless we take an infinitesimally small piece of mass dm, as shown in Figure 10.24.

Figure shows a point dm located on the X axis at distance r from the center.
Figure 10.24 Using an infinitesimally small piece of mass to calculate the contribution to the total moment of inertia.

The need to use an infinitesimally small piece of mass dm suggests that we can write the moment of inertia by evaluating an integral over infinitesimal masses rather than doing a discrete sum over finite masses:

I=imiri2becomesI=r2dm.I=imiri2becomesI=r2dm.
(10.19)

This, in fact, is the form we need to generalize the equation for complex shapes. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. This is the focus of most of the rest of this section.

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in Figure 10.25. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

Figure shows a thin rod that rotates about an axis through the center. Part of the rod of the length dx has a mass dm.
Figure 10.25 Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density λλ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ=mlorm=λl.λ=mlorm=λl.

If we take the differential of each side of this equation, we find

dm=d(λl)=λ(dl)dm=d(λl)=λ(dl)

since λλ is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact, dl=dxdl=dx in this situation. We can therefore write dm=λ(dx)dm=λ(dx), giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=r2dm=x2dm=x2λdx.I=r2dm=x2dm=x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=L/2x=L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2L/2x2λdx=λx33|L/2L/2=λ(13)[(L2)3(L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.I=L/2L/2x2λdx=λx33|L/2L/2=λ(13)[(L2)3(L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation.

A uniform thin rod with axis at the end

Now consider the same uniform thin rod of mass M and length L, but this time we move the axis of rotation to the end of the rod. We wish to find the moment of inertia about this new axis (Figure 10.26). The quantity dm is again defined to be a small element of mass making up the rod. Just as before, we obtain

I=r2dm=x2dm=x2λdx.I=r2dm=x2dm=x2λdx.

However, this time we have different limits of integration. The rod extends from x=0x=0 to x=Lx=L, since the axis is at the end of the rod at x=0x=0. Therefore we find

I=0Lx2λdx=λx33|0L=λ(13)[(L)3(0)3]=λ(13)L3=ML(13)L3=13ML2.I=0Lx2λdx=λx33|0L=λ(13)[(L)3(0)3]=λ(13)L3=ML(13)L3=13ML2.
Figure shows a thin rod that rotates about an axis through the end. Part of the rod of the length dx has a mass dm.
Figure 10.26 Calculation of the moment of inertia I for a uniform thin rod about an axis through the end of the rod.

Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four.

The Parallel-Axis Theorem

The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a parallel axis. There is a theorem for this, called the parallel-axis theorem, which we state here but do not derive in this text.

Parallel-Axis Theorem

Let m be the mass of an object and let d be the distance from an axis through the object’s center of mass to a new axis. Then we have

Iparallel-axis=Icenter of mass+md2.Iparallel-axis=Icenter of mass+md2.
(10.20)

Let’s apply this to the rod examples solved above:

Iend=Icenter of mass+md2=112mL2+m(L2)2=(112+14)mL2=13mL2.Iend=Icenter of mass+md2=112mL2+m(L2)2=(112+14)mL2=13mL2.

This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems.

Check Your Understanding 10.5

What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below?

Figure shows a cylinder of radius R that rotates about an axis through a point on the surface.

A uniform thin disk about an axis through the center

Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center (Figure 10.27).

Figure shows a uniform thin disk of radius r that rotates about a Z axis that passes through its center.
Figure 10.27 Calculating the moment of inertia for a thin disk about an axis through its center.

Since the disk is thin, we can take the mass as distributed entirely in the xy-plane. We again start with the relationship for the surface mass density, which is the mass per unit surface area. Since it is uniform, the surface mass density σσ is constant:

σ=mAorσA=m,sodm=σ(dA).σ=mAorσA=m,sodm=σ(dA).

Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment dm of radius r equidistanct from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring dA is therefore given by the length of each ring (2πr2πr) times the infinitesimmal width of each ring dr:

A=πr2,dA=d(πr2)=πdr2=2πrdr.A=πr2,dA=d(πr2)=πdr2=2πrdr.

The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to R. This radius range then becomes our limits of integration for dr, that is, we integrate from r=0r=0 to r=Rr=R. Putting this all together, we have

I=0Rr2σ(2πr)dr=2πσ0Rr3dr=2πσr44|0R=2πσ(R440)=2πmA(R44)=2πmπR2(R44)=12mR2.I=0Rr2σ(2πr)dr=2πσ0Rr3dr=2πσr44|0R=2πσ(R440)=2πmA(R44)=2πmπR2(R44)=12mR2.

Note that this agrees with the value given in Figure 10.20.

Calculating the moment of inertia for compound objects

Now consider a compound object such as that in Figure 10.28, which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object’s moment of inertia can be found from the sum of each part of the object:

Itotal=iIi.Itotal=iIi.
(10.21)

It is important to note that the moments of inertia of the objects in Equation 10.21 are about a common axis. In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius R rotating about an axis shifted off of the center by a distance L+RL+R, where R is the radius of the disk. Let’s define the mass of the rod to be mrmr and the mass of the disk to be md.md.

Figure shows a disk with radius R connected to a rod with length L.
Figure 10.28 Compound object consisting of a disk at the end of a rod. The axis of rotation is located at A.

The moment of inertia of the rod is simply 13mrL213mrL2, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is 12mdR212mdR2 and we apply the parallel-axis theorem Iparallel-axis=Icenter of mass+md2Iparallel-axis=Icenter of mass+md2 to find

Iparallel-axis=12mdR2+md(L+R)2.Iparallel-axis=12mdR2+md(L+R)2.

Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be

Itotal=13mrL2+12mdR2+md(L+R)2.Itotal=13mrL2+12mdR2+md(L+R)2.

Applying moment of inertia calculations to solve problems

Now let’s examine some practical applications of moment of inertia calculations.

Example 10.11

Person on a Merry-Go-Round A 25-kg child stands at a distance r=1.0mr=1.0m from the axis of a rotating merry-go-round (Figure 10.29). The merry-go-round can be approximated as a uniform solid disk with a mass of 500 kg and a radius of 2.0 m. Find the moment of inertia of this system.

Figure is a drawing of a child on a merry-go-round. Merry–go-round has a 2 meter radius. Child is standing one meter from the center.
Figure 10.29 Calculating the moment of inertia for a child on a merry-go-round.

Strategy This problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. The notation we use is mc=25kg,rc=1.0m,mm=500kg,rm=2.0mmc=25kg,rc=1.0m,mm=500kg,rm=2.0m.

Our goal is to find Itotal=iIiItotal=iIi.

Solution For the child, Ic=mcr2Ic=mcr2, and for the merry-go-round, Im=12mmr2Im=12mmr2. Therefore

Itotal=25(1)2+12(500)(2)2=25+1000=1025kg·m2.Itotal=25(1)2+12(500)(2)2=25+1000=1025kg·m2.

Significance The value should be close to the moment of inertia of the merry-go-round by itself because it has much more mass distributed away from the axis than the child does.

Example 10.12

Rod and Solid Sphere Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0.5 m and mass 2.0 kg. The radius of the sphere is 20.0 cm and has mass 1.0 kg.

Figure A shows a disk with radius R connected to a rod with length L. The point A is at the end of the rod opposite to the disk. Figure B shows a disk with radius R connected to a rod with length L. The point B is at the end of the rod connected to the disk.

Strategy Since we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia about each axis. In (a), the center of mass of the sphere is located at a distance L+RL+R from the axis of rotation. In (b), the center of mass of the sphere is located a distance R from the axis of rotation. In both cases, the moment of inertia of the rod is about an axis at one end. Refer to Table 10.4 for the moments of inertia for the individual objects.

  1. Itotal=iIi=IRod+ISphereItotal=iIi=IRod+ISphere;
    ISphere=Icenter of mass+mSphere(L+R)2=25mSphereR2+mSphere(L+R)2ISphere=Icenter of mass+mSphere(L+R)2=25mSphereR2+mSphere(L+R)2;
    Itotal=IRod+ISphere=13mRodL2+25mSphereR2+mSphere(L+R)2;Itotal=IRod+ISphere=13mRodL2+25mSphereR2+mSphere(L+R)2;
    Itotal=13(2.0kg)(0.5m)2+25(1.0kg)(0.2m)2+(1.0kg)(0.5m+0.2m)2;Itotal=13(2.0kg)(0.5m)2+25(1.0kg)(0.2m)2+(1.0kg)(0.5m+0.2m)2;
    Itotal=(0.167+0.016+0.490)kg·m2=0.673kg·m2.Itotal=(0.167+0.016+0.490)kg·m2=0.673kg·m2.
  2. ISphere=25mSphereR2+mSphereR2ISphere=25mSphereR2+mSphereR2;
    Itotal=IRod+ISphere=13mRodL2+25mSphereR2+mSphereR2Itotal=IRod+ISphere=13mRodL2+25mSphereR2+mSphereR2;
    Itotal=13(2.0kg)(0.5m)2+25(1.0kg)(0.2m)2+(1.0kg)(0.2m)2Itotal=13(2.0kg)(0.5m)2+25(1.0kg)(0.2m)2+(1.0kg)(0.2m)2;
    Itotal=(0.167+0.016+0.04)kg·m2=0.223kg·m2.Itotal=(0.167+0.016+0.04)kg·m2=0.223kg·m2.

Significance Using the parallel-axis theorem eases the computation of the moment of inertia of compound objects. We see that the moment of inertia is greater in (a) than (b). This is because the axis of rotation is closer to the center of mass of the system in (b). The simple analogy is that of a rod. The moment of inertia about one end is 13mL213mL2, but the moment of inertia through the center of mass along its length is 112mL2112mL2.

Example 10.13

Angular Velocity of a Pendulum A pendulum in the shape of a rod (Figure 10.30) is released from rest at an angle of 30°30°. It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point?

Figure shows a pendulum in the form of a rod with a mass of 300 grams and length of 30 centimeters. Pendulum is released from rest at an angle of 30 degrees.
Figure 10.30 A pendulum in the form of a rod is released from rest at an angle of 30°.30°.

Strategy Use conservation of energy to solve the problem. At the point of release, the pendulum has gravitational potential energy, which is determined from the height of the center of mass above its lowest point in the swing. At the bottom of the swing, all of the gravitational potential energy is converted into rotational kinetic energy.

Solution The change in potential energy is equal to the change in rotational kinetic energy, ΔU+ΔK=0ΔU+ΔK=0.

At the top of the swing: U=mghcm=mgL2(cosθ)U=mghcm=mgL2(cosθ). At the bottom of the swing, U=mgL2.U=mgL2.

At the top of the swing, the rotational kinetic energy is K=0K=0. At the bottom of the swing, K=12Iω2K=12Iω2. Therefore:

ΔU+ΔK=0(mgL2(1cosθ)0)+(012Iω2)=0ΔU+ΔK=0(mgL2(1cosθ)0)+(012Iω2)=0

or

12Iω2=mgL2(1cosθ).12Iω2=mgL2(1cosθ).

Solving for ωω, we have

ω=mgLI(1cosθ)=mgL1/3mL2(1cosθ)=g3L(1cosθ).ω=mgLI(1cosθ)=mgL1/3mL2(1cosθ)=g3L(1cosθ).

Inserting numerical values, we have

ω=9.8m/s230.3m(1cos30)=3.6rad/s.ω=9.8m/s230.3m(1cos30)=3.6rad/s.

Significance Note that the angular velocity of the pendulum does not depend on its mass.

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