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199

7Inertia, Second Moment Vectors, Momentsand Products of Inertia, Inertia Dyadics

7.1 Introduction

In this chapter we review various topics and concepts about inertia. Many readers will be familiar with a majority of these topics; however, some topics, particularly thoseconcerned with three-dimensional aspects of inertia, may not be as well understood, yetthese topics will be of most use to us in our continuing discussion of mechanical systemdynamics. In a sense, we have already begun our review with our discussion of masscenters in the previous chapter. At the end of the chapter, however, we discovered thatwe need additional information to adequately describe the inertia torque of Eq. (6.9.11),

shown again here:

(7.1.1)

Indeed, the principal motivation for our review of inertia is to obtain simplified expres-sions for this torque. Our review will parallel the development in Reference 7.4 with a

basis found in References 7.1 to 7.3. We begin with a discussion about second-momentvectors a topic that will probably be unfamiliar to most readers. As we shall see, though,

second-moment vectors provide a basis for the development of the more familiar topics,particularly moments and products of inertia.

7.2 Second-Moment Vectors

Consider a particle P with mass m and an arbitrary reference point O. Consider also an

arbitrarily directed unit vector na as inFigure 7.2.1. Let p be a position vector locating Prelative to O. The second moment of P relative to O for the direction na is defined as:

(7.2.1)

T r r r r* = ( )

( )

= =

m mi i ii

N

i i i

i

N

1 1

I p n paP O D

am= ( )

2002 by CRC Press LLC

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200 Dynamics of Mechanical Systems

Observe that the second moment is somewhat more detailed than the first moment (mp)defined in Eq. (6.8.1). The second moment depends upon the square of the distance of Pfrom O and it also depends upon the direction of the unit vector na.

The form of the definition of Eq. (7.2.1) is motivated by the form of the terms of theinertia torque of Eq. (7.1.1). Indeed, for a set S of particles, representing a rigid body B(Figure 7.2.2), the second moment is defined as the sum of the second moments of theindividual particles. That is,

(7.2.2)

Then, except for the presence of na instead of or , the form of Eq. (7.2.2) is identical tothe forms of Eq. (7.7.1). Hence, by examining the properties of the second moment vector,we can obtain insight into the properties of the inertia torque. We explore these propertiesin the following subsections.

7.3 Moments and Products of Inertia

Consider again a particle P, with mass m, a reference point O, unit vector na, and a secondunit vector nb as in Figure 7.3.1. The moment and product of inertia of P relative to O forthe directions na and nb are defined as the scalar projections of the second moment vector(Eq. (7.2.1)) along na and nb. Specifically, the moment of inertia of P relative to O for the

direction na is defined as:

(7.3.1)

Similarly, the product of inertia of P relative to O for the directions na and nb is defined as:

(7.3.2)

FIGURE 7.2.1

A particle P, reference point O, and unit vector na.

FIGURE 7.2.2

A set S of particles, reference point O and unitvector na.

O

p

P(m)

n a

n

p

P (m )

P (m )

P (m )

P (m )

O

a

1 1

2 2 i i

N N

i

I I p n paS O D

aP O

i i a i

i

N

i

N

i m= = ( )==

11

I I naaP O D

aP O

a=

I I nabP O D

aP O

b=


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Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 201

Observe that by substituting for from Eq. (7.2.1) that and may be expressedin the form:

(7.3.3)

and

(7.3.4)

Observe further that (pna)2 may be identified with the square of the distance da fromP to a line passing through O and parallel to na (see Figure 7.3.2). This distance is oftenreferred to as the radius of gyration of P relative to O for the direction na.

Observe also for the product of inertia of Eq. (7.3.3) that the unit vectors na and nb maybe interchanged. That is,

(7.3.5)

Note that no restrictions are placed upon the unit vectors na and nb. If, however, na andnb are perpendicular, or, more generally, if we have three mutually perpendicular unitvectors, we can obtain additional geometric interpretations of moments and products ofinertia. Specifically, consider a particle P with mass m located in a Cartesian referenceframe R as in Figure 7.3.3. Let (x,y, z) be the coordinates of P in R. Then, from Eq. (7.2.1)the second moment vectors of P relative to origin O for the directions nx, ny, and nz are:

or

(7.3.6)

FIGURE 7.3.1A particle P, reference point O, and unit vectors

na and nb.

FIGURE 7.3.2Distance from particle P to line through O and

parallel to na.

O

p

P(m)

n a

n b

nd

P(m)

O

a

a

p

IaP O Iaa

P O IabP O

I p n p n p n p n p naaP O

a a a a am m m= ( ) = ( ) ( ) = ( )2

I p n p n p n p nabP O a b a bm m= ( ) = ( ) ( )

I p n p n p n Iab

P O

a b a ba

P Om m= ( ) ( ) = ( ) =

I p n p

n n n n n n n

x

P O

x

x y z x x y z

m

m x y z x y z

= ( )= + +( ) + +( )[ ]

I n n nxP O

x y zm y z xy xz= +( ) [ ]2 2


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and, similarly,

(7.3.7)

and

(7.3.8)

Using the definitions of Eqs. (7.3.1) and (7.3.2), we see that the various moments andproducts of inertia are then:

(7.3.9)

Observe that the moments of inertia are always nonnegative or zero, whereas the productsof inertia may be positive, negative, or zero depending upon the position of P.

It is often convenient to normalize the moments and products of inertia by dividing bythe mass m. Then, the normalized moment of inertia may be interpreted as a lengthsquared, called the radius of gyration and defined as:

(7.3.10)

(See also Eq. (7.3.3).)

Finally, observe that the moments and products of inertia of Eqs. (7.3.9) may be conve-niently listed in the matrix form:

(7.3.11)

where a and b can be x,y, or z.

FIGURE 7.3.3A particle P in a Cartesian reference frame.

P(x,y,z)

Y

Z

X

n

p

n

n

O

z

y

x

I n n nyP O

x y zm yx x z yz= + +( ) [ ]2 2

I n n nzP O

x y zm zx zy x y= + +( )[ ]2 2

I

I

I

I

I

I

I

I

I

xxP O

yxP O

zxP O

xyP O

xyP O

zyP O

xzP O

yzP O

zzP O

m x y

myx

mzx

mxy

m x y

mzy

mxz

myz

m y z

= +( )

=

=

=

= +( )

=

=

=

= +( )

2 2

2 2

2 2

,

,

,

,

,

,

k maD

aaP O= [ ]I

1 2

IabP O m

y z xy xz

yx z x yz

zx zy x y

=+( )

+( ) +( )

2 2

2 2

2 2


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7.4 Inertia Dyadics

Comparing Eqs. (7.3.6), (7.3.7), and (7.3.8) with (7.3.9) we see that the second momentvectors may be expressed as:

(7.4.1)

where the superscripts on the moments and products of inertia have been deleted. Wecan simplify these expressions further by using the index notation introduced and devel-oped in Chapter 2. Specifically, if the subscripts x,y, and z are replaced by the integers 1,2, and 3, we have:

(7.4.2)

or

(7.4.3)

where the repeated index denotes a sum over the range of the index.These expressions can be simplified even further by using the concept of a dyadic. A

dyadic is the result of a product of vectors employing the usual rules of elementary algebra,except that the pre- and post-positions of the vectors are maintained. That is, if a and bare vectors expressed as:

(7.4.4)

then the dyadic product of a and b is defined through the operations:

(7.4.5)

where the unit vector products are called dyads. The nine dyads form the basis for a generaldyadic (say, D), expressed as:

I n n n

I n n n

I n n n

xP O

xx x xy y xz z

yP O

yx x yy y yz z

zP O

zx x zy y zz z

I I I

I I I

I I I

= + +

= + +

= + +

I n n n n

I n n n n

I n n n n

1 11 1 12 2 13 3 1

2 21 1 22 2 23 3 2

3 31 1 32 2 33 3 3

P Oj j

P Oj j

P Oj j

I I I I

I I I I

I I I I

= + + =

= + + =

= + + =

I niP O

ij iI i= =( ), ,1 2 3

a n n n b n n n= + + = + +a a a b b b1 1 2 2 3 3 1 1 2 2 3 3and

ab n n n n n n

n n n n n n

n n n n n n

n n n n n n

= + +( ) + +( )

= + +

+ + +

+ + +

Da a a b b b

a b a b a b

a b a b a b

a b a b a b

1 1 2 2 3 3 1 1 2 2 3 3

1 1 1 1 1 2 1 2 1 3 1 3

2 1 2 1 2 2 2 2 2 3 2 3

3 1 3 1 3 2 2 2 3 3 3 3


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(7.4.6)

Observe that a dyadic may be thought of as being a vector whose components are vectors;hence, dyadics are sometimes called vectorvectors. Observe further that the scalar com-ponents of D (the dij of Eq. (7.4.6)) can be considered as the elements of a 3 3 matrix andas the components of a second-order tensor (see References 7.5, 7.6, and7.7).

In Eq. (7.3.12), we see that the moments and products of inertia may be assembled aselements of a matrix. In Eq. (7.4.3) these elements are seen to be Iij (i, j = 1, 2, 3). If thesematrix elements are identified with dyadic components, we obtain the inertia dyadicdefined as:

(7.4.7)

or

(7.4.8)

By comparing Eqs. (7.4.2) and (7.4.7), we see that the inertia dyadic may also be expressedin the form:

(7.4.9)

Equation (7.3.5) shows that the matrix of moments and products of inertia is symmetric(that is, Iij = Iji). Then, by rearranging the terms of Eq. (7.4.7), we see that IP/O may also be

expressed as:

(7.4.10)

The inertia dyadic may thus be interpreted as a vector whose components are second-moment vectors.

A principal advantage of using the inertia dyadic is that it can be used to generate

second-moment vectors, moments of inertia, and products of inertia. Specifically, once theinertia dyadic is known, these other quantities may be obtained simply by dot productmultiplication with unit vectors. That is,

(7.4.11)

and

(7.4.12)

D n n n n n n

n n n n n n

n n n n n n n n

= + +

+ + +

+ + + =

d d d

d d d

d d d dij i j

11 1 1 12 1 2 13 1 3

21 2 1 22 2 2 23 2 3

31 3 1 22 3 2 33 3 3

I n n n n n n

n n n n n n

n n n n n n

P O D I I I

I I I

I I I

= + +

+ + +

+ + +

11 1 1 12 1 2 13 1 3

21 2 1 22 2 2 23 2 3

31 3 1 32 3 2 33 3 3

I n nP O ij i jI=

I n I n I n IP O P O P O P O= + +1 1 2 2 3 3

I I n I n I nP O P O P O P O= + +1 1 2 2 3 3

I I n n IiP O P O

i iP O i= = =( ), ,1 2 3

I n I nijP O

iP O

j i= =( ), ,1 2 3


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Finally, for systems of particles or for rigid bodies, the inertia dyadic is developed fromthe contributions of the individual particles. That is, for a system S of Nparticles we have:

(7.4.13)

7.5 Transformation Rules

Consider again the definition of the second moment vector of Eq. (7.2.1):

(7.5.1)

Observe again the direct dependency of upon na. Suppose the unit vector na isexpressed in terms of mutually perpendicular unit vectors ni (i = 1, 2, 3) as:

(7.5.2)

Then, by substitution from Eq. (7.5.1) into (7.5.2), we have:

or

(7.5.3)

Equation (7.5.3) shows that if we know the second moment vectors for each of threemutually perpendicular directions, we can obtain the second moment vector for anydirection na.

Similarly, suppose nb is a second unit vector expressed as:

(7.5.4)

Then, by forming the projection of onto nb we obtain the product of inertia , whichin view of Eq. (7.5.3) can be expressed as:

or

(7.5.5)

I IS O P O

i

N

i==

1

I p n paP O

am= ( )

IaP O

n n n n na i ia a a a= + + =1 1 2 2 3 3

I p n p p n paP O

i i i im a a m= ( ) = ( )

I I I I IaP O

i iP O P O P O P Oa a a a= = + +1 1 2 2 3 3

n n n n nb j jb b b b= + + =1 1 2 2 3 3

IaP O

IabP O

I n I n IabP O

b aP O

i j j iP Oa b= =

I IabP O

i j ijP Oa b=


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Observe that the scalar components ai and bi of na and nb of Eqs. (7.5.2) and (7.5.3) maybe identified with transformation matrix components as in Eq. (2.11.3). Specifically, let(j = 1, 2, 3) be a set of mutually perpendicular unit vectors, distinct and noncollinear with

the ni (i = 1, 2, 3). Let the transformation matrix components be defined as:

(7.5.6)

Then, in terms of the Sij, Eq. (7.5.5) takes the form:

(7.5.7)

7.6 Parallel Axis Theorems

Consider once more the definition of the second moment vector of Eq. (7.2.1):

(7.6.1)

Observe that just as is directly dependent upon the direction of na, it is also dependentupon the choice of the reference point O. The transformation rules discussed above enableus to evaluate the second-moment vector and other inertia functions as na changes. Theparallel axis theorems discussed in this section will enable us to evaluate the second-moment vector and other inertia functions as the reference point O changes.

To see this, consider a set S of particles Pi (i = 1,, N) with masses mi as in Figure 7.6.1.Let G be the mass center of S and let O be a reference point. Let pG locate G relative toO, let pi locate Pi relative to O, and let ri locate Pi relative to G. Finally, let na be an arbitraryunit vector. Then, from Eq. (7.2.2), the second moment of S for O for the direction of na is:

(7.6.2)

FIGURE 7.6.1A set of particles with mass center G.

nj

Sij i j= n n

IkP O

ik j ijP OS S I

1 1=

I p n paP O

am= ( )

IaP O

I p n paS O

i i a i

i

N

m= ( )=

1

P (m )

P (m )

P (m )

P (m )

O

G

r

pp

nS

1 1

G

i

i

i i

a

2

N N

2


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From Figure 7.6.1, we see that pi, ri, and pG are related by:

(7.6.3)

Then, by substituting into Eq. (7.6.2) we have:

or

(7.6.4)

where is defined as:

(7.6.5)

whereM is the total mass,

,

of the particles of S; and is the second moment of a particle located at G with a massequal to the total massM of S.

Eq. (7.6.4) is often called a parallel axis theorem. The reason for this name can be seenfrom the analogous equation for the moments of inertia: that is, by examining the projec-tion of the terms of Eq. (7.6.4) along na, we have:

or

(7.6.6)

p p ri G i= +

I p r n p r

p n p p n r

r n p r n r

aP O

i G i a G i

i

N

i

i

N

G a G i G a i

i

N

i i a G

i

N

i i a i

i

N

i

i

N

m

m m

m m

m

= +( ) +( )[ ]{ }

= ( ) + ( )

+ ( ) + ( )

=

=

= =

= =

=

1

1 1

1 1

1

( ) +

+

( ) + ( )

=

= =

p n p p n r

r n p r n r

G a G G a i i

i

N

i i

i

N

a G i i

i

N

a i

m

m m

1

1 1

O

O

I I IaS O

aG O

aS G= +

IaG O

I p n paG O D

G a GM= ( )

mii

N

=

1

IaG O

I n I n I naS O

a aG O

a aS G

a = +

I I IaaS O

aaG O

aaS G= +


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The first term on the right may be developed as:

(7.6.7)

where d is pGna. Then, d is seen to be the distance between parallel lines passingthrough O and G and parallel to na (see Figure 7.6.2). Equation (7.6.6) may then be written:

(7.6.8)

By taking the projections of the terms of Eq. (7.6.4) along nb, a unit vector with a directiondifferent than na, we have for the products of inertia:

(7.6.9)

Finally, by using the transformation rules of Eqs. (7.5.5) and (7.5.7) and by successivelycombining terms of Eqs. (7.6.6) and (7.6.8), we obtain the analogous equation for inertiadyadics:

(7.6.10)

Equations (7.6.4), (7.6.6), (7.6.9), and (7.6.10) are versions of the parallel axis theorem forthe second-moment vector, for the moments of inertia, for the products of inertia, and forthe inertia dyadics. They show that if an inertia quantity is known relative to the masscenter, then that quantity can readily be found relative to any other point.

Finally, inertia quantities computed relative to the mass center are called central inertiaproperties. Observe, then, in Eq. (7.6.8) that the central moment of inertia is the minimum

moment of inertia for a given direction.

7.7 Principal Axes, Principal Moments of Inertia: Concepts

The foregoing paragraphs show that the second-moment vector may be used to generatethe moments of inertia, the products of inertia, and the inertia dyadic. Recall that the

FIGURE 7.6.2Parallel axes through O and G.

d

GO

P

n

S

a

a

G

Gd = | P N |

IaaG O

aG O

a G a G a G aM M Md= == ( )[ ] = ( ) =I n p n p n p n2 2

I I MdaaS O

aaS G= + 2

I I IabS O

abG O

abS G= +

I I IS O G O S G= +


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second-moment vector is dependent upon the direction of the unit vector na, used in thedefinition. That is, for a system of particles we have:

(7.7.1)

Observe that for an arbitrary unit vector, it is unlikely that will be parallel to na. If ithappens, however, that is parallel to na, then na is said to be a principal unit vector oran eigen unit vector.

Observe that if na is a principal unit vector and if nb is perpendicular to na, then theproduct of inertia is zero. That is,

(7.7.2)

Observe further that if na is a principal unit vector, the moment of inertia is themagnitude of . Then, we have:

(7.7.3)

The direction of a principal unit vector is called a principal direction. The correspondingmoment of inertia is called aprincipal moment of inertia or eigenvalue of inertia.

Consider again the inertia dyadic defined in Section 7.4. From Eq. (7.4.11) we have:

(7.7.4)

Hence, if na is a principal unit vector, we have:

(7.7.5)

In view of these concepts and equations, the obvious questions that arise are do principalunit vectors exist and, if they exist, how can they be found? Equation (7.7.5) is often used

as point of departure for an analysis to answer these questions and to discuss principalunit vectors in general.

To begin the analysis, let Eq. (7.7.5) be rewritten without the superscripts:

(7.7.6)

Next, let I and na be expressed in terms of mutually perpendicular unit vectors ni (i = 1,2, 3) as:

(7.7.7)

Then, Eq. (7.7.6) becomes:

I p n paS O D

i i a i

i

N

m= ( )=

1

IaS O

IaS O

Iab

S O

I I nabS O aS O b= = 0

IaaS O

IaS O

I I n I I I naaS O

aS O

a aS O

aS O

aaS O

a= = =and

I I naS O S O

a=

I n I nS O a aaS O

a =

I n I n =a aa a

I n n n n= =I aij i j a k kand

I n n n n n n n

n n

n n

= =

= =

= =

a ij i j k k ij k i j k

ij k i jk ij j i

aa a aa k k

I a I a

I a I a

I I a


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or

(7.7.8)

Equation (7.7.8) is equivalent to the three scalar equations:

(7.7.9)

where for simplicity in notation Iaa is replaced by . When the index sums of these equationsare written explicitly, they become:

(7.7.10)

Equations (7.7.10) form a set of three linear equations for the three scalar componentsai of na. If we can determine the ai we have in effect determined and found na. However,Eqs. (7.7.10) are homogeneous that is, each term on the left sides contains one and onlyone of the ai, and the right sides are zero. This means that there is no solution (except the

trivial solution ai = 0), unless the determinant of the coefficients of the ai is zero. That is,a nontrivial solution for the ai exists only if:

(7.7.11)

By expanding the determinant, we obtain:

(7.7.12)

where the coefficients II, I

II, and I

IIIare:

(7.7.13)

(7.7.14)

(7.7.15)

These coefficients are seen to be directly related to the elements Iij of the inertia matrix.Indeed, they may be identified as:

II = sum of diagonal elements of Iij

III = sum of diagonal elements of the cofactor matrix of Iij

IIII = determinant of Iij

I a I aij j i aa i in n=

I a I a I a iij j aa i ij ij i= ( ) = =( ), ,or 0 1 2 3

I a I a I a

I a I a I a

I a I a I a

11 1 12 2 13 3

21 1 22 2 23 3

31 1 32 2 33 3

0

0

0

( ) + + =+ ( ) + =

+ + ( ) =

I I I

I I I

I I I

11 12 13

21 22 23

31 32 33

0

( )( )

( )=

3 2 0 + =I I II II III

I I I I I = + +11 22 33

I I I I I I I I I I I I I II = + + 22 33 32 23 11 33 31 13 11 22 12 21

I I I I I I I I I I I I I I I I I I I III = + + 11 22 33 11 32 23 12 31 23 12 21 33 21 32 13 31 13 22


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Interestingly, these coefficients may be shown (see Reference 7.4) to be independent of thechoice of unit vectors ni in which I is expressed. As a consequence, II, III, and IIII aresometimes called invariants of I.

Equation (7.7.12) is a cubic equation for which in general has three distinct roots: 1,2, and 3. This implies that there are at least three solutions for each at the a1, a2, and a3of Eq. (7.7.10). To obtain these solutions, we can select one of the roots (say, 1) and thensubstitute 1 for in Eq. (7.7.10). This produces three equations that are dependent because1 is a solution of Eq. (7.7.11). Hence, only two of the three equations are independent.However, we can obtain a third independent equation by recalling that na is a unit vectorand thus a1, a2, and a3 also satisfy the equation:

(7.7.16)

Then, by selecting any two equations from Eq. (7.7.10) together with Eq. (7.7.16), we cansolve the resulting system of three equations and determine the values of a1, a2, and a3.

Next, by letting be 2, we can repeat the process and find a second set of components(a1, a2, and a3) of a principal unit vector na. Similarly, for = 3, we obtain a third principalunit vector.

By following this procedure we obtain the components of three principal unit vectorsdefining three principal directions of inertia of the system. The values 1, 2, and 3 arethen the corresponding principal moments of inertia. This procedure and these conceptsare illustrated in the example in the following section.

7.8 Principal Axes, Principal Moments of Inertia: Example

Consider a system whose central inertia matrix relative to mutually perpendicular unitvectors n1, n2, and n3 is:

(7.8.1)

To determine the principal values, the principal unit vectors, and the principal directions,we need to form Eq. (7.7.10). That is,

(7.8.2)

Because these equations are linear and homogeneous (right sides are zero), there is anontrivial solution for the a1, a2, and a3 only if the determinant of the coefficients is zero(Eq. (7.7.11)). Hence,

a a a12

22

32 1+ + =

I

I I I

I I I

I I Iij[ ] =

=

11 12 13

21 22 23

31 32 33

9 2 3 4 3 3 4

3 4 39 8 3 8

3 3 4 3 8 37 8

9 2 3 4 3 3 4 0

3 4 39 8 3 8 0

3 3 4 3 8 37 8 0

1 2 3

1 2 3

1 2 3

( ) + ( ) + =

( ) + ( ) + ( ) =

( ) + ( ) + ( ) =

a a a

a a a

a a a


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(7.8.3)

Expanding the determinant, we obtain:

(7.8.4)

The coefficients of Eq. (7.8.4) are seen to be the parameters II, III, and IIII of Eqs. (7.7.13),(7.7.14), and (7.7.15).

By solving Eq. (7.8.4) we obtain:

(7.8.5)

These are the principal moments of inertia. By substituting the first of these into Eq. (7.8.2),we obtain:

(7.8.6)

By eliminating fractions we have:

(7.8.7)

These equations are seen to be dependent (multiplying the first by 12 and adding theresult to the second produces the third); hence, only two of the equations are independent.A third independent equation may be obtained from Eq. (7.7.16) by requiring that na bea unit vector:

(7.8.8)

Thus, by selecting any two of Eq. (7.8.7) and appending them to Eq. (7.8.8) we candetermine a1, a2, and a3. The result is:

(7.8.9)

9 2 3 4 3 3 4

3 4 39 8 3 8

3 3 4 3 8 37 8

0

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

=

3 214 63 90 0 + =

= = = = = =1 2 33 5 6, ,

9 2 3 3 4 3 3 4 0

3 4 39 8 3 3 8 0

3 3 4 3 8 37 8 3 0

1 2 3

1 2 3

1 2 3

( ) + ( ) + ( ) =

( ) + ( ) + ( ) =

( ) + ( ) + ( ) =

a a a

a a a

a a a

2 3 0

6 15 3 018 3 13 3 0

1 2 3

1 2 3

1 2 3

a a a

a a aa a a

+ =

+ + =+ + =

a a a1

2

2

2

3

2 1+ + =

a a a1 2 32 2 2 4 6 4= = = , ,


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Hence, the principal unit vector na corresponding to the principal value = 3 is:

(7.8.10)

Similarly, by taking the second principal value = 5, substituting into Eq. (7.8.2), andusing Eq. (7.8.8), we obtain the principal unit vector:

(7.8.11)

Finally, for the third principal value = 6, the principal unit vector is:

(7.8.12)

Observe that the principal unit vectors na, nb, and nc are mutually perpendicular. Thisis important because then the products of inertia relative to the directions of na, nb, andnc are zero. To explore this further, recall from Eq. (7.7.6) that because na, nb, and nc areprincipal unit vectors, we have:

(7.8.13)

Hence, we have:

(7.8.14)

Equation (7.8.14) shows that the components of the dyadic I in the mixed directionsare zero; therefore, if we express I in terms of na, nb, and nc we have:

(7.8.15)

In view of Eqs. (7.8.1) and (7.8.15), we see that the matrices of I referred to n1, n2, andn3 and to na, nb, and nc are vastly different. That is,

(7.8.16)

and

(7.8.17)

n n n na = +

2 2 2 4 6 41 2 3

n n nb

= +3 2 1 22 3

n n n nc = +2 2 2 4 6 41 2 3

I n n I n n I n n = = =a a b b c c 1 2 3, , and

n I n n I n n I nb a c b a c ba cb acI I I = = = = = = 0

I n n n n n n

n n n n n n

= + +

= + +

1 2 3

3 5 6

a a b b c c

a a b b c c

I

I I I

I I I

I I Iij =

=

11 12 13

21 22 23

31 32 33

9 2 3 4 3 3 4

3 4 39 8 3 8

3 3 4 3 8 37 8

I

I I I

I I I

I I I

aa ab ac

ba bb bc

ca cb cc

=

=

3 0 0

0 5 0

0 0 6


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where the Greek subscripts and refer to the indices a, b, and c. We can obtain a directrelationship between these matrices by using the transformation matrices introduced inChapter 2. Specifically, from Eq. (2.11.3) the elements of the transformation matrix between

the ni (i = 1, 2, 3) and the n ( = 1, b, c) are:

(7.8.18)

Let I be expressed in the forms:

(7.8.19)

Then, the Iij and the I are related by the expressions:

(7.8.20)

From Eqs. (7.8.10), (7.8.11), (7.8.12), and (7.8.18), we see that for our example the Si are:

(7.8.21)

Then, in matrix form, Eq. (7.8.20) becomes:

(7.8.22)

and

(7.8.23)

Finally, observe that the columns of the transformation matrix [Si] are the componentsof the principal unit vectors na, nb, and nc.

S Si i i i = =n n n nand

I n n n n= =I Iij i j

I S S I I S S I ij i j i j ij= = and

Si =

2 2 0 2 2

2 4 3 2 2 4

6 4 1 2 6 4

9 2 3 4 3 3 4

3 4 39 8 3 8

3 3 4 3 8 37 8

2 2 0 2 2

2 4 3 2 2 4

6 4 1 2 6 4

3 0 0

0 5 0

0 0 6

2 2 2 4 2 40 3 2 1 2

2 2 2 4 6 4

=

3 0 0

0 5 0

0 0 6

2 2 2 4 6 4

0 3 2 1 2

2 2 2 4 6 4

9 2 3 4 3 3 4

3 4 39 8 3 8

3 3 4 3 8 37 8

2 2 0 2 2

2 4 3 2 2 4

6 4 1 2 6 4

=


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7.9 Principal Axes, Principal Moments of Inertia: Discussion

In view of these results, several questions arise:

1. Are principal unit vectors always mutually perpendicular?

2. What if the roots of the cubic equation Eq. (7.7.12) are not all distinct?

3. What if the roots are not real?

4. Do we always need to solve a cubic equation to obtain the principal momentsof inertia?

5. Will the procedures of the example always work?

To answer these questions consider again the definition of a principal unit vector, havingthe property expressed in Eq. (7.7.5). That is, if na is a principal unit vector, then:

(7.9.1)

Similarly if nb is also a principal unit vector:

(7.9.2)

If we take the scalar product of the terms of Eqs. (7.9.1) and (7.9.2) with nb and na andsubtract the results, we obtain:

(7.9.3)

Because the inertia dyadic I is symmetric, naInb and nbIna are equal, thus the leftside of Eq. (7.9.3) is zero. Then, if the principal moments of inertia Iaa and Ibb are not equal,we have:

(7.9.4)

This shows that principal unit vectors of distinct principal moments of inertia are mutuallyperpendicular.

Next, suppose that the principal moments of inertia are not all distinct. Suppose, forexample, that Iaa and Ibb are equal. Are the associated principal unit vectors still perpen-

dicular? To address this question, consider again Eq. (7.7.12):

(7.9.5)

Recall from elementary algebra that if Iaa, Ibb, and Icc are the roots of this equation then theequation may be written in the equivalent form:

(7.9.6)

I n n = ( )a aa aI ano sum on

I n n =b bb bI

n I n n I n b n n n

n n

b a a b aa b a aa a b

aa bb a b

I I

I I

=

= ( )

n na b = 0

3 2 0 + =I I II II III

( ) ( ) ( ) =I I Iaa bb cc 0


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By expanding Eq. (7.9.6) and by comparing the coefficients with those of Eq. (7.9.5), wediscover that the roots are related to the coefficients (the invariants) by the expressions:

(7.9.7)

Now, if Icc is distinct from Iaa (= Ibb), then by the reasoning of the foregoing paragraph thetwo corresponding principal unit vectors na and nc will be distinct and perpendicular.That is,

(7.9.8)

Let nb be a unit vector perpendicular to na and nc. Then, Inb will be some vector, say. However, because na, nb, and nc are mutually perpendicular, I can be expressed in theform:

(7.9.9)

Then, in view of Eq. (7.9.8) and if Inb is , we have:

(7.9.10)

Because I is also symmetric, we have:

(7.9.11)

Therefore, must be parallel to nb. That is,

(7.9.12)

where is the magnitude of . Hence, I takes the form:

(7.9.13)

Then, because II is the sum of the diagonal elements of the matrix of I (independently of

the unit vectors in which I is expressed) and in view of the first of Eq. (7.9.7), we have:

(7.9.14)

or

(7.9.15)

I I I I

I I I I I I I

I I I I

I aa bb cc

II aa bb bb cc cc aa

III aa bb cc

= + +

= + +

=

I n n I n n n n = = =a aa a c cc c a cI I, , and 0

I I n n I n n I n n= ( ) + ( ) + ( )a a b b c c

I n n n n n= + +I Iaa a a b cc c c

n nb b

=

= nb

I n n n n n n= + +I Iaa a a b b cc c c

I I I I I I I aa cc aa bb cc= + + = + +

= Ibb


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Hence, by assuming that Ibb and Iaa are equal, Eq. (7.9.13) produces an expression for I inthe form:

(7.9.16)

Then, if is any unit vector parallel to the plane of na and nb (that is, perpendicular tonc), we have:

(7.9.17)

Therefore, is a principal unit vector (see Eq. (7.7.5)). This means that all unit vectorsparallel to the plane of na and nb are principal unit vectors. Thus, if two of the principalmoments of inertia are equal, an infinite number of principal unit vectors are parallel tothe plane that is normal to the principal unit vector associated with the distinct root ofEq. (7.7.12).

We can consider the case when all three of the roots of Eq. (7.7.12) are equal by similarreasoning. In this case, only one principal unit vector is found by the procedure of the

previous sections. Let this vector be nc and let na and nb be unit vectors perpendicular tonc such that na, nb, and nc form a mutually perpendicular set. Then, I has the form:

(7.9.18)

where and are vectors to be determined and Iis the triple root of Eq. (7.7.12). BecauseI is symmetric, we have:

(7.9.19)

or, equivalently, and must be parallel to a plane normal to nc. Thus, and have theforms:

(7.9.20)

Hence, I becomes:

(7.9.21)

The matrix of I relative to na, nb, and nc is:

(7.9.22)

I n n n n n n

n n n n n n

= + +

= +( ) +

I I I

I I

aa a a aa b b cc c c

aa a a b b cc c c

I n n n n n n = +( ) +

= +

I I

I

aa a a b b cc c c

aa 0

I n n n n= + + a b c cI

n n n na b a b+ = +

= + = + a a b b a a b bn n n nand

I n n n n n n n n n n= + + + + a a a a a b b b a b b b c cI

a a

b b

0

0

0 0 1


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Because I is symmetric we have:

(7.9.23)

From Eq. (7.9.7), the invariants II, III, and IIII are:

(7.9.24)

where the right sides of these expressions are obtained from the description of II, III, andIIII in Eqs. (7.7.13), (7.7.14), and (7.7.15) and from the matrix of I in Eq. (7.9.22). By solvingfor a, b, and b, we obtain:

(7.9.25)

Hence, and are:

(7.9.26)

Finally, I becomes (see Eq. (7.9.21)):

(7.9.27)

Then, if is any vector:

(7.9.28)

Hence, when there is a triple root of Eq. (7.7.12) all unit vectors are principal unit vectors.Regarding the question of real roots of Eq. (7.7.12), suppose a root is not real. That is,

let a root have the complex form:

(7.9.29)

where and are real and i is the imaginary Let n be a principal unit vectorassociated with the complex root . Then, from Eq. (7.7.5), we have:

(7.9.30)

By following the procedures of the previous sections we can obtain n by knowing .Because is complex, the components of n will be complex. Then, n could be expressedin the form:

(7.9.31)

b

a=

I I I

I I I I

I I I

I a b

II b a a b b

III a b b

= = + +

= = + +

= = ( )

3

3 2 2

3 2

a b a bI I= = = =, ,0

a a b

I I= =n nand

I n n n n n n= + +I I Ia a b b c c

I = I

= + i

1.

I n n =

n n n= + i


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where n and n are real vectors. Hence, by substituting from Eqs. (7.9.29) and (7.9.31)into Eq. (7.9.30) (and recalling that I is real), we have:

(7.9.32)

By expanding this equation and by matching real and imaginary terms, we obtain:

(7.9.33)

By multiplying the first of these by n and the second by n and subtracting, we have:

(7.9.34)

Because I is symmetric, the left side is zero. Hence, we have the conclusion:

(7.9.35)

Therefore, because I is real and symmetric, the principal moments of inertia (the roots ofEq. (7.7.12)) are real.

Regarding the question as to whether we always need to solve a cubic equation (Eq.(7.7.12)), consider that if we know one of the roots (say, ) of an equation, we can depressthe equation and obtain a quadratic equation by dividing by 1. We can obtain a root1 if we know a principal unit vector (say, 1), as then from Eq. (7.7.5) 1 is simply In1.The question then becomes: can a principal unit vector be found without solving Eq. (7.7.5)and the associated linear system of Eqs. (7.7.10)? The answer is yes, on occasion. Specifi-cally, if is a plane of symmetry of a set S of particles, then the unit vector normal to is a principal unit vector for all reference points in .

To see this, recall that a plane of symmetry is such that for every particle on one sideof the plane there is a corresponding particle on the other side having the same mass and

at the same distance from the plane. Consider, for example, the plane depicted in Figure7.9.1 with particles P1 and P2 of S equidistant from and on opposite sides of . Let theparticles each have mass m and let their distances from be h. Let O be any referencepoint in , and let Qbe the point of intersection of with the line connecting P1 and P2.Finally, let n be a unit vector normal to .

FIGURE 7.9.1Particles on opposite sides of a planeof symmetry.

I n n n n +( ) = +( ) +( ) i i i

I n n n I n n n = = + and

n n n n n n n n = = I I 2

= 0

n

h

h

Q

O

P (m)

P (m)

1

2


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Let P1 and P2 be a typical pair of particles taken from a set S of particles for which isa plane of symmetry. Then, from Eq. (7.7.2) we see that the contributions of P1 and P2 tothe second moment of S relative to O for the direction of n are represented by:

(7.9.36)

where the last equality follows from expanding the triple products of the previous equalityand where the terms not written in Eq. (7.9.36) represent the contributions to from theother pairs of particles of S. However, by analyses identical to those shown in Eq. (7.9.36),these contributions to are also parallel to n. Therefore, is parallel to n, and thusn is a principal unit vector of S for O.

A specialization of these ideas occurs if the system of particles all lie in the same plane,as with a planar body. A unit vector normal to the plane is then a principal unit vector.

There are other ways of determining principal unit vectors without solving Eqs. (7.7.10)and (7.7.12). For homogeneous bodies occupying common geometric shapes, we cansimply refer to a table of results as in Appendix II. We can also make use of a number of

theorems for principal unit vectors stated here, but without proof (see References 7.1 and7.2 for additional information). If a line is parallel to a principal unit vector for a givenreference point, then that line is called a principal axis for that reference point. A principalaxis for the mass center as a reference point is called a central principal axis. Then, it canbe shown that:

1. If a principal axis for a reference point other than the mass center also passesthrough the mass center, then it is also a central principal axis.

2. Mutually perpendicular principal axes that include a central principal axis for

any point on the central principal axis are parallel to mutually perpendicularcentral principal axes.

3. A central principal axis is a principal axis for each of its points.

Finally, regarding the question as to whether the procedures of the example of theprevious section will always produce principal unit vectors and principal moments ofinertia, the answer is yes, but as seen above there may exist simpler procedures for anygiven problem. That is, a principal unit vector may often be identified by inspection for example, as being normal to a plane of symmetry. In this case, the task of solving acubic equation may be reduced to that of solving a quadratic equation.

To illustrate the procedure, suppose that a principal unit vector found by inspection iscalled n3. Then, let n1 and n2 be unit vectors perpendicular to n3 and also perpendicularto each other so that n1, n2, and n3 form a mutually perpendicular set. Then, because n3is a principal unit vector, the inertia dyadic I expressed in terms of n1, n2, and n3 has amatrix of components in the form:

(7.9.37)

I p n p p n p

OP n n OQ n

OQ n n OQ n

OQ n

nS O m m

m h h

m h h

m

= ( ) + ( ) +

= +( ) +( )[ ]

+ ( ) ( )[ ] +

= +

1 1 2 2

22

K

K

K

IaS O

IaS O In

S O

I

I I

I I

Iij =

11 12

21 22

33

0

0

0 0


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From Eq. (7.7.5), if na is a principal unit vector, we have:

(7.9.38)

The associated scalar equations, Eq. (7.7.10), then become:

(7.9.39)

where, as before, a1, a2, and a3 are the components of na relative to n1, n2, and n3. Byinspection, a solution to these equations is:

(7.9.40)

If, however, Iaa is not equal to I33, then a3 is zero and the equations for a1, a2, and Iaa reduce to:

(7.9.41)

Because the first two of these equations are linear and homogeneous, the third equationwill be violated unless the determinant of the coefficients of the first two equations is zero.That is,

(7.9.42)

Expanding the determinant we obtain:

(7.9.43)

Solving for Iaa we find:

(7.9.44)

When Iaa has the values as in Eq. (7.9.44), the first two equations of Eq. (7.9.41) becomedependent. Hence, by taking the first and third of Eq. (7.9.41) we have a1 and a2 to be:

(7.9.45)

I n n =a aa aI ano sum on

I I a I a

I a I I a

I I a

aa

aa

aa

11 1 12 2

21 1 22 2

33 3

0

0

0

( ) + =

+ ( ) =

( ) =

I Iaa a a a= = = =33 3 1 21 0, ,

I I a I a

I a I I a

a a

aa

aa

11 1 12 2

21 1 22 2

12

22

0

0

1

( ) + =

+ ( ) =

+ =

I I II I I

aa

aa

11 12

21 22

0( ) ( ) =

I I I I I I I aa aa2

11 22 11 22 122 0 +( ) + =

II I I I

Iaa =+

+

11 22 11 22

2

122

1 2

2 2

a I I I I aa1 12 122

11

2 1 2

= + ( )[ ]


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and

(7.9.46)

When a plane of symmetry is identified or when a system or body is planar, the analysisis two dimensional, as above. In this case, the transformation between different unit vectorsets is also simplified. Recall from Eqs. (7.5.5) and (7.5.7) that the moments and productsof inertia referred to unit vectors and ni are related by the expression:

(7.9.47)

where the transformation matrix components are defined by Eq. (7.5.6) as:

(7.9.48)

Suppose the unit vectors na, nb, and nc are principal unit vectors, with nc being normalto a plane of symmetry. Let n1, n2, and n3 be mutually perpendicular unit vectors, withn3 being parallel to nc. Let be the angle between n1 and na as in Figure 7.9.2. Then, thetransformation matrix components between the ni (i = 1, 2, 3) and the n ( = a, b, c) are:

(7.9.49)

Using Eq. (7.9.47), the moments and products of inertia Iij, relative to ni and nj, then become:

(7.9.50)

(7.9.51)

(7.9.52)

(7.9.53)

(7.9.54)

Equations (7.9.50), (7.9.51), and (7.9.52) may be expressed in an even simpler form byusing the trigonometric identities:

(7.9.55)

a I I I I I aa aa2 11 122

11

2 1 2

= ( ) + ( )[ ]

nk

I S S I k ik j ijl l

=

Sik i k= n n

n ni =

cos sin

sin cos

0

0

0 0 1

I I Iaa bb112 2= +cos sin

I I I I aa bb12 21= = ( )sin cos

I I Iaa bb222 2= +sin cos

I I I I 13 31 23 32 0= = = =

I Icc33 =

sin cos

cos cos

sin cos sin

2

2

1

2

1

22

1

2

1

22

1

22

+


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Then, I11, I12, and I22 become:

(7.9.56)

(7.9.57)

(7.9.58)

Equations (7.9.56), (7.9.57), and (7.9.58) may be interpreted geometrically using a Mohrcircle diagram as in Figure 7.9.3. In this diagram, the values of the moment of inertia I11and I22 are found along the abscissa, while the value of the product of inertia I12 is foundalong the ordinate as shown.

7.10 Maximum and Minimum Moments and Products of Inertia

We will see presently and in the following chapters that the inertia dyadic and, specifically,the moments and products of inertia are the primary parameters of inertia torques. There-fore, the questions that arise are for which reference points and for which directions do

FIGURE 7.9.2Unit vectors parallel to a plane ofsymmetry.

FIGURE 7.9.3A Mohr circle diagram for moments

and products of inertia.

nn

n

n

2b

1

a

I

2

I I

I

I2bb12 22 aa

11

bbaaI + I

II I I I aa bb aa bb

11 2 22=

+

+

cos

II Iaa bb

12 22=

sin

II I I I aa bb aa bb

22 2 22=

+

cos


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the moments and products of inertia have maximum and minimum values? To addressthese questions, consider first the parallel axis theorem as expressed in Eq. (7.6.10):

(7.10.1)

From this expression we see that if O is sufficiently far away from the mass center G, thecomponents of IG/O become arbitrarily large; thus, there are no absolute maximummoments and products of inertia. However, if the reference point O is at the mass centerG,the moments and products of inertia may have minimum values. Thus, we have thefollowing question: if the mass center G is the reference point, for which directions do themoments and products of inertia have maximum and minimum values? To answer thisquestion, consider the two-dimensional case and the Mohr circle analysis discussed at the

end of the previous section. From Figure 7.9.3, we see that the maximum and minimummoments of inertia are the principal moments of inertia Iaa and Ibb. Also, the maximumabsolute value of the product of inertia I12 occurs at 45 to the principal moment of inertiadirections with value (IaaIbb)/2. The minimum absolute value of the product of inertiais zero. This occurs in directions corresponding to the directions of the principal momentsof inertia. These results can also be seen by inspection of Eqs. (7.9.56), (7.9.57), and (7.9.58).

Do similar results hold, in general, in the three-dimensional case? The answer is yes, but the vertification is not as simple. Nevertheless, the results are readily obtained byusing the Lagrange multiplier method. For readers not familiar with this method, it issufficient to know that the method is used to find maximum and minimum values of

functions of several variables which in turn are related to each other by constraint equa-tions. For example, if the maximum or minimum value of a functionf(x,y, z) is to be foundsubject to a constraint:

(7.10.2)

then a function h(x,y, z) is defined as:

(7.10.3)

where is a parameter, called a Lagrange multiplier, to be determined along with x,y, andz such that h has maximum or minimum values. Specifically, if h is to be a maximum orminimum, the following equations need to be satisfied simultaneously:

(7.10.4)

The last of these equations is identical with Eq. (7.10.2). Thus, Eq. (7.10.4) provides the

necessary conditions for f(x, y, z) to have maximum or minimum values. The solution ofthese equations then determines the values of x, y, z, and , producing the maximum orminimum values off.

The advantage of the Lagrange multiplier method is that the constraint equation neednot be solved independently in the analysis. The disadvantage is that an additionalparameter is to be determined; however, the parameter may often be identified withan important physical or geometrical parameter in a given problem.

To apply this method in obtaining maximum and minimum central (mass center refer-ence point) moments of inertia, let I be a central inertia dyadic with components Iij relativeto an arbitrary set of mutually perpendicular unit vectors ni. Then, if na is an arbitrarily

I I IS O S G G O= +

g x y z, ,( ) = 0

h x y z f x y z g x y z

D

, , , , , , , ( ) = ( ) + ( )

= = = =h x h y h z h0 0 0 0, , ,


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directed unit vector, the moment of inertia relative to the na direction may be expressedas (see Eq. (7.5.5)):

(7.10.5)

where the ai are the ni components of na and where we are again employing the summationconvention. Because na is a unit vector, the ai must satisfy the relation:

(7.10.6)

This latter expression is a constraint equation analogous to Eq. (7.10.2). Thus, from Eq.

(7.10.3), if we wish to find the direction na such that Iaa is a maximum or minimum, thenwe can form the function h(ai,) as:

(7.10.7)

By setting the derivation of h with respect to the ak equal to zero, as in Eq. (7.10.4), we have:

or

(7.10.8)

where we have used the symmetry of the inertia dyadic components. By setting thederivative of h with respect to equal to zero, we obtain Eq. (7.10.6):

(7.10.9)

Equations (7.10.8) and (7.10.9) are identical to Eqs. (7.7.9) and (7.7.16), which determinethe principal moments of inertia and their directions. Therefore, the directions producingthe maximum and minimum central moments of inertia are the principal directions, and

the maximum and minimum moments of inertia are to be found among the principalmoments of inertia. Finally, the Lagrange multiplier is seen to be a principal momentof inertia.

We can use the same procedure to obtain the direction for the maximum products ofinertia. From Eq. (7.5.5), the product of inertia relative to the directions of unit vectors naand nb is:

(7.10.10)

I a a I aa i j ij=

a a a ai i i i= =1 1 0or

h a a a I a aiD

i j ij i i, ( ) = + ( )1

= ( ) + ( )

= +

= =

h a a a a I a a a I

a a a

a I a I a

I a a

k i k j ij i j k ij

i i k

ik i ij i jk ij i ik

kj j k

2

2

2 2 0

I a a kkj j k

= =( ) , ,1 2 3

a a Ik k

=

I a b I ab i j ij

=


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where, as before, the ai and bi are the ni components of na and nb. Then, because na andnb are unit vectors, the ai and bi must satisfy the equations:

(7.10.11)

If na and nb are also perpendicular to each other, we have:

(7.10.12)

Equations (7.10.11) and (7.10.12) form three constraint equations to be satisfied whilemaximum values of Iab of Eq. (7.10.10) are sought. Then, by generalizing the foregoingprocedure, we form a function h(ai, bi, , ,) defined as:

(7.10.13)

where , , and are Lagrange multipliers. By setting the derivative of h with respect toak and bk equal to zero, we obtain:

(7.10.14)

and

(7.10.15)

If we multiply these equations by ak and bk, respectively (and sum over k), we obtain:

(7.10.16)

(7.10.17)

(7.10.18)

(7.10.19)

where we have used Eqs. (7.10.11) and (7.10.12). By comparing Eqs. (7.10.16) and (7.10.19)and by recalling that Iik is symmetric, we have:

(7.10.20)

Then, Eqs. (7.10.14) and (7.10.15) may be rewritten in the forms:

(7.10.21)

and

(7.10.22)

1 0 1 0 = =a a b bi i i iand

a bi i = 0

h a b a b I a a b b a bi i i i ij i i i i i i, , , , ( ) = + ( ) + ( ) +1 1

= + =h a I b a bk kj j k k

2 0

= + =h b I a b ak ik i k k2 0

a I bk kj j =2 0

b I bk kj j

+ = 0

a I ak ik i + = 0

b I ak ih i

=2 0

=

I b a bkj j k k

= 2

I a b akj j k k= 2


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By adding and subtracting these equations we have:

(7.10.23)

and

(7.10.24)

These equations in turn may be expressed as:

(7.10.25)

and

(7.10.26)

Thus, (na + nb) and (nanb) are principal vectors of I corresponding to the principal values(2 ) and (2 + ). From our previous discussions, however, we know that for thegeneral three-dimensional case with three distinct principal values, the principal values

contain the maximum, the minimum, and intermediate values of the moments of inertia.For our discussion here, let these principal moments of inertia be designated as I, I,and I, with principal unit vectors n, n, and n. That is,

(7.10.27)

Then, comparing Eqs. (7.10.25) and (7.10.26) with Eq. (7.10.27) we see that (2) and(2 + ) are to be identified with I, I, and I and (na + nb) and (na nb) are in thedirection of n, n, and n. Specifically, if (2) and (2+) are identified with I andI, we have:

(7.10.28)

Then, by adding these equations we have:

(7.10.29)

However, from Eq. (7.10.16) we see that:

(7.10.30)

Therefore, the maximum absolute values of the products of inertia are the half differencesof the principal moments of inertia, and the directions where they occur are at 45 to theprincipal directions of the moments of inertia.

I a b a bkj j j k k+( ) = ( ) +( )2

I a b a bkj j j k k( ) = +( ) ( )2

I n n n n +( ) = ( ) +( )a b a b2

I n n n n ( ) = +( ) ( )a b a b2

I I I = = = ( )n I n n I n n I n , , no sum

2 2 = + = I Iand

2 2 = ( )I I

2 = = =a I b I k kj j a b ab

n I n


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7.11 Inertia Ellipsoid

Geometrical interpretations are sometimes used to obtain insight into the nature of inertiaquantities. Of these, one of the most extensively used is the inertia ellipsoid, which maybe developed from Eq. (7.5.5). Specifically, for a given reference point O the moment ofinertia of a system for the direction of a unit vector na is:

(7.11.1)

where the ai and the Iij are components of na and the inertia dyadic I relative to mutually

perpendicular unit vectors ni (i = 1, 2, 3). If the ni are principal unit vectors, the productsof inertia, Iij (ij), are zero. Eq. (7.11.1) then takes the simplified form:

(7.11.2)

Then, by dividing by Iaa, we have:

(7.11.3)

Because the denominators of this expression are all positive, we may write it in the form:

(7.11.4)

where a, b, and c are defined by inspection in comparison with Eq. (7.11.3). Eq. (7.11.4) isimmediately seen to be the equation of an ellipsoid in the Cartesian coordinate space of(a1, a2, a3). The distance from the ellipse center (at the origin O) to a point P on the surfaceis proportional to the moment of inertia of a system for the direction of a unit vector naparallel to OP. For example, if na is directed parallel to n1 such that (a1, a2, a3) is (1, 0, 0),then Iaa is I11, a principal moment of inertia. This property has led the ellipsoid of Eq.(7.11.4) to be called the inertia ellipsoid.

7.12 Application: Inertia Torques

Consider a rigid body B moving in an inertial reference frame R (see Section 6.9) asdepicted in Figure 7.12.1. Let B be considered to be composed of a set of Nparticles Pihaving masses mi (i = 1,, N). Let Gbe the mass center of B. Then, from Eqs. (6.9.9) and(6.9.11), we recall that the system of inertia forces acting on B (through particles Pi) isequivalent to a single force F* passing through G together with a couple having torque T*

where F* and T* are:

(7.12.1)

I a a I aa i j ij=

I a I a I a I aa = + +12

11 22

22 32

33

1 12

11

22

22

32

33

=( )

+( )

+( )

a

I I

a

I I

a

I Iaa aa aa

1 12

222

232

2= + +a

a

a

b

a

c

F a* = M G


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and

(7.12.2)

where aG is the acceleration of G in R; and are the angular velocity and angularacceleration, respectively, of B in R; and ri is a position vector locating Pi relative to G.

We are now in a position to further develop the terms of Eq. (7.12.2). Specifically, wecan conveniently express them in terms of the inertia dyadic of Section 7.4. Recall firstfrom Eq. (7.2.2) that the second moment of B for G, for a direction of a unit vector na, is:

(7.12.3)

(Observe the similarity of the sum of Eq. (7.12.3) with those of Eq. (7.12.2).) Then, recallfrom Eq. (7.4.11) that the second moment may be expressed in terms of the inertia dyadicof B for G as:

(7.12.4)

Consider now the terms of Eq. (7.12.2). Let and be written in the forms:

(7.12.5)

where n and n are unit vectors in the directions of and at any instant, with and then being three magnitudes of and . Then, the first term of Eq. (7.12.2) may beexpressed as:

(7.12.6)

FIGURE 7.12.1A body B moving in an inertial frame R.

R

G

B

T r r r ri i* = ( )

( )

= = m mi ii

N

i i

i

N

1 1

I m raB G

i a i

i

N

= ( )=

r ni1

IaB G

aB G= n I

= =n n and

m m

m

i ii

N

i ii

N

i i

i

N

B G

B G

B G

r r r n r

r n r

I

n I

I

i i

i

( ) = ( )

= ( )

=

=

=

= =

=

1 1

1


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where we have made use of Eqs. (7.12.4) and (7.12.5). Similarly, the second term of Eq.(7.12.2) may be expressed as:

(7.12.7)

Hence, T* becomes:

(7.12.8)

The scalar components of T* are sometimes referred to as Euler torques. We will explore

the significance of these terms in the next several chapters.

References

7.1. Kane, T. R.,Analytical Elements of Mechanics, Vol. 2, Academic Press, New York, 1962.7.2. Kane, T. R., Dynamics, Holt, Rinehart & Winston, New York, 1968.7.3. Kane, T. R., and Levinson, D. A., Dynamics: Theory and Applications, McGraw-Hill, New York,

1985.

7.4. Huston, R. L.,Multibody Dynamics, Butterworth-Heinemann, Stoneham, MA, 1990.7.5. Hinchley, F. A., Vectors and Tensors for Engineers and Scientists, Wiley, New York, 1976.7.6. Hsu, H. P., Vector Analysis, Simon & Schuster Technical Outlines, New York, 1969.7.7. Haskell, R. E., Introduction to Vectors and Cartesian Tensors, Prentice Hall, Englewood Cliffs, NJ,

1972.

Problems

Section 7.2 Second Moment Vectors

P7.2.1: A particle P with mass 3 slug has coordinates (2, 1, 3), measured in feet, in aCartesian coordinate system as represented in Figure P7.2.1. Determine the secondmoment of P relative to the origin O for the directions represented by the unit vectors nx,ny, nz, na, and nb, where na and nb are parallel to the XY plane, as shown in Figure P7.2.1.That is, find and .

FIGURE P7.2.1A particle P in a Cartesian referenceframe.

( )

= ( )=

mi ii

NB Gr r Ii

1

T I I* = ( )B G B G

I I I IxP O

yP O

zP O

aP O, , , ,

IbP O

O

Z

Y

X

Q(-1,2,4)

P(2,-1,3)

n

n

n

nn

30

45

z

b

y

ax


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P7.2.2: See Problem P7.2.1. Let point Q have coordinates (1, 2, 4). Repeat Problem P7.2.1with Q, instead of O, being the reference point. That is, find and .

P7.2.3: Consider the system S of two particles P1 and P2 located on the X-axis of a Cartesiancoordinate system as in Figure P7.2.3. Let the masses of P1 and P2 be m1 and m2, and letthe distances of P1 and P2 from the origin O be x1 and x2. Find the second moment of Prelative to O for the direction of ny. (Observe that is parallel to ny.)

P7.2.4: See Problems P7.2.1 and P7.2.3. Find a unit vector n such that is parallel to n.

P7.2.5: Let a set S of three particles P1, P2, and P3be located at the vertices of a triangle asshown in Figure P7.2.5. Let the particles have masses 2, 3, and 4 kg, respectively. Find

where as before O is the origin of the x,y, z coordinate system of Figure

P7.2.5, and nx, ny, and nz are the unit vectors shown.

P7.2.6: See Problem P7.2.5. Let na and nb be unit vectors with coordinates relative to nx,ny, and nz as:

and

Determine

P7.2.7: See Problems P7.2.5 and P7.2.6. Show that:

FIGURE P7.2.3Particles P1 and P2 on the X-axis of aCartesian reference frame.

FIGURE P7.2.5Particles at the vertices of a triangle.

I I I IxP Q

yP Q

zP Q

aP Q, , , , Ib

P Q

IyS O

Y

XO

xP (m ) P (m )

n

n

y

212

x 2 1 1

x

InP O

I I IxS O

xS O

zS O, , and

O

Z

Y

X

n

n

n

z

y

x

P (0,5,2)

P (1,1,1)

2

1

3

P (2,2,4)

n n n na x y z= +0 75 0 5 0 433. . .

n n nb y z= +0 655 0 756. .

I IaS O

bS Oand .

I I I IaS O

xS O

yS O

zS O= +0 75 0 50 0 433. . .


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and

P7.2.8: See Problem P7.2.5. Find the x, y, z coordinates of the mass center G of S. Find

P7.2.9: See Problem P7.2.8. Let G have an associated mass of 9 kg (equal to the sum of themasses of P1, P2, and P3). Find

P7.2.10: See Problems P7.2.5, P7.2.6, and P7.2.9. Show that:

and

P7.2.11: See Problem P7.2.5. Find a unit vector n perpendicular to the plane of P1, P2, andP3. Find also Show that is parallel to n.

Section 7.3 Moments and Products of Inertia

P7.3.1: See Problem P7.2.1. A particle P with mass of 3 slug has coordinates (2, 1, 3),measured in feet, in a Cartesian coordinate system as represented in Figure P7.3.1. Deter-mine the following moments and products of inertia:

P7.3.2: See Problems P7.2.2and P7.3.1. Let Q have coordinates (1, 2, 4). Repeat ProblemP7.3.1 with Q, instead of O, being the reference point. That is, determine

P7.3.3: See Problem P7.2.5. Let a set S of three particles P1, P2, and P3 be located at thevertices of a triangle as shown in Figure P7.3.3. Let the particles have masses 2, 3, and

FIGURE P7.3.1A particle P and a point Q.

I I IbS O

yS O

zS O= +0 655 0 756. .

I I I I IxS G

yS G

zS G

aS G

bS G, , , , and .

I I I I IxG O

yG O

zG O

aG O

bG O, , , , and .

I I I

I I I

I I I

I I I

xS O

xS G

xG O

yS O

yS G

yG O

zS O

zS G

zG O

aS O

aS G

aG O

= +

= +

= +

= +

I I IbS O

bS G

bG O= +

InS O. In

S O

I I I I I I IxxP O

xyP O

xzP O

yyP O

yzP O

zzP O

aaP O, , , , , , ,

I I IbbP O

abP O

baP O

, , and .

I I IxxP Q

xyP Q

xzP Q, , ,

I I I I I I IyyP Q

yzP Q

zzP Q

aaP Q

bbP Q

abP Q

baP Q, , , , , , and .


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4 kg, respectively. Find the following moments and products of inertia of S relative to theorigin O of the X-, Y-, Z-axis system of Figure P7.3.3:

P7.3.4: See Problems P7.2.5, P7.2.8, P7.2.9, and P7.3.3. For the system S shown in Figure

P7.3.3, find the following moments and products of inertia:

where G is the mass center of S, as determined in Problem P7.2.8. (Compare the

magnitudes of these results with those of Problem P7.3.3.)

P7.3.5: See Problems P7.2.9 and P7.3.3. For the system S shown in Figure P7.3.3, find the

following moments and products of inertia:

P7.3.6: See Problems P7.2.10, P7.3.3, P7.3.4, and P7.3.5. Show that:

P7.3.7: See Problems P7.2.5, P7.2.6, P7.2.7, and P7.3.3. As in Problem P7.2.6 let na and nbbe the unit vectors:

and

Find

Section 7.4 Inertia Dyadics

P7.4.1: Let vectors a, b, and c be expressed as:


I I I I I IxxS O

xyS O

xzS O

yyS O

yzS O

zzS O, , , , , and .

I I I I IxxS G

xyS G

xzS G

yyS G

yzS G, , , , ,

and IzzS G

I I I I I IxxG O

xyG O

xzG O

yyG O

yzG O

zzG O, , , , , and .

I I IijS O

ijS G

ijG O i j x y z= + =( ), , ,

n n n na x y z= +0 75 0 5 0 433. . .

n n nb y z= +0 655 0 756. .

I I IaaS O

abS O

bbS O, , .and

a n n n

b n n n

c n n n

= +

= +

= +

6 3 4

5 4 7

3 9

1 2 3

1 2 3

1 2 3


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where n1, n2, and n3 are mutually perpendicular unit vectors. Compute the followingdyadic products: (a) ab, (b) ba, (c) ca + cb, (d) c(a + b), (e) (a + b)c, and (f) ac + bc.

P7.4.2: See Problem 7.2.1. A particle P with mass 3 slug has coordinates (2, 1, 3), measuredin feet, in a Cartesian coordinate system as represented in Figure P7.4.2. Determine theinertia dyadic of P relative to the origin O, IP/O. Express the results in terms of the unitvectors nx, ny, and nz.

P7.4.3: See Problem P7.2.2. Let Q have coordinates (1, 2, 4). Repeat Problem P7.4.2 withQ instead of Obeing the reference point. That is, find IP/Q.

P7.4.4: See Problems P7.2.5 and P7.3.3. Let S be the set of three particles P1, P2, and P3located at the vertices of a triangle as shown in Figure P7.4.4. Let the particles have masses:2, 3, and 4 kg, respectively. Find the inertia dyadic of S relative to O, IS/O. Express theresults in terms of the unit vectors nx, ny, and nz.

P7.4.5: See Problems P7.2.8,P7.2.9, P7.3.4, and P7.4.4. Let Gbe the mass center of S. Findthe inertia dyadic of S relative to G, IS/G. Express the results in terms of the unit vectors

nx, ny , and nz.

P7.4.6: See Problems P7.4.4 and P7.4.5. Let G have an associated mass of 9 kg. Find theinertia dyadic of G relative to the origin O, IG/O. Express the result in terms of the unitvectors nx, ny, and nz.

P7.4.7: See Problems P7.4.5 and P7.4.6. Show that:

FIGURE P7.4.2A particle P in a Cartesian referenceframe.


O

Z

Y

X

Q(-1,2,4)

P(2,-1,3)

n

n

n

z

y

x

O

Z

Y

X

n

n

n

z

y

x

P (0,5,2)

P (1,1,1)

(units in meters)

2

3

1

P (2,2,4)

I I IS O S G G O= +


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P7.4.7: SeeProblems P7.2.5 and P7.4.4. Find the second moments of S relative to O for thedirections of nx, ny, and nz.

P7.4.8: See Problems P7.3.3 and P7.4.4. Find the following moments and products of inertiaof S for O:

P7.4.9: See Problems P7.2.6 and P7.4.4. Let na and nb be unit vectors with coordinatesrelative to nx, ny, and nz as:

Find the second moment vectors

P7.4.10: See Problems P7.2.5, P7.2.6, P7.3.7, P7.4.4, and P7.4.9. Let na and nb be the unitvectors of Problem P7.4.9. Find the following moments and products of inertia of S relativeto O:

Section 7.5 Transformation Rules

P7.5.1: Let S be a set of eight particles Pi (i = 1,, 8) located at the vertices of a cube asin Figure P7.5.1. Let the masses mi of the Pi be as listed in the figure. Determine the second-

moment vectors for the directions of the unit vectors n1, n2, and n3 shownin Figure P7.5.1.

P7.5.2: See Problem P7.5.1. Let na, nb, and nc be unit vectors with components relative ton1, n2, and n3 as:

FIGURE P7.5.1

Particles at the vertices of a cube.

I I I I IxxS O

xzS O

yyS O

yzS O

zzS O, , , , and .

n n n n

n n n

a x y z

b y z

= +

= +

0 75 0 5 0 433

0 655 0 756

. . .

. .

I IaS O

bS Oand .

I I IaaS O

abS O

bbS O, , .and

I I I1 2 3S O S O S O

, , and

O

Z

X

Y

2m

2m

2m

P

n

n

n

P

P

P

P

P

P

P

34

8

5

6

2

7

1

1

2

3

m = 2 kg,1

m = 3 kg,2

m = 1 kg,3

m = 5 kg,4

m = 4 kg5

m = 6 kg6

m = 3 kg7

m = 2 kg8

n n n

n n n n

n n n n

a

b

c

= +

= + +

= +

0 5 0 866

0 433 0 25 0 866

0 75 0 433 0 5

1 2

1 2 3

1 2 3

. .

. . .

. . .


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Determine the second moment vectors

P7.5.3: See Problem P7.5.1. Determine the moments and products of inertia (i,j = 1, 2, 3).

P7.5.4: See Problem P7.5.2. Let the transformation matrix between na, nb, nc and n1, n2, n3have elements Sj (j = 1, 2, 3; = a, b, c) defined as:

Find the Sj.

P7.5.5: See Problems P7.5.1 to P7.5.4. Find the moments and products of inertia(, = a, b, c). Also verify that:

and

P7.5.6: See Problem P7.5.3. Find the inertia dyadic IS/O. Express the results in terms of theunit vectors n1, n2, and n3 of Figure P7.5.1.

P7.5.7: See Problems P7.5.5 and P7.5.6. Verify that (, = a, b, c) is given by:

P7.5.8: See Problems P7.5.3 and P7.5.5. Verify that:

P7.5.9: A 3-ft bar B weighs 18 pounds. Let the bar be homogeneous and uniform so thatits mass center G is at the geometric center. Let the bar be placed on an XY plane so thatit is inclined at 30 to the X-axis as shown in Figure P7.5.9. It is known that the momentof inertia of a homogeneous, uniform bar relative to its center is zero for directions parallelto the bar and m2/12 for directions perpendicular to the bar where m is the bar massand is its length (see Appendix II). It is also known that the products of inertia for a barfor directions parallel and perpendicular to the bar are zero. Determine the moments andproducts of inertia:

FIGURE P7.5.9A homogeneous bar in the XY planewith center at the origin.

I I IaS O

bS O

cS O, , .and

IijS O

Sj j = n n

IS O

I I S O

i j ijS OS S=

I IijS O

i jS OS S=

IS O

I n I n S O S O=

I I I I I I11 22 33S O S O S O

aaS O

bbS O

ccS O+ + = + +

I I I I I IxxB G

yyB G

zzB G

xyB G

xzB G

yzB G, , , , , .and

Y

X

B

30O

G

= 3 ft

weight = 18 lb


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P7.5.10: A thin uniform circular disk D with mass m and radius r is mounted on a shaftS with a small misalignment, measured by the angle as represented in Figure P7.5.10.Knowing that the moments of inertia of D for its center for directions parallel to and

perpendicular to its axis are mr2/2 and mr2/4, respectively, and that the correspondingproducts of inertia of D for its axis and diameter directions are zero (see Appendix II),find the moment of inertia of D for its center G for the shaft axis direction x :

Section 7.6 Parallel Axis Theorems

P7.6.1: Consider the homogeneous rectangular parallepiped (block) B shown in FigureP7.6.1. From Appendix II, we see that the moments of inertia of B for the mass center Gfor the X, Y, and Z directions are:

where m is the mass of B and a, b and c are the dimensions as shown in Figure P7.6.1. LetB have the following properties:

Determine the moments of inertia of B relative to G for the directions of X, Y, and Z.

P7.6.2: Repeat Problem P7.6.1 with B having the following properties:

P7.6.3: See Problems P7.6.1 and P7.6.2. For the properties of Problems P7.6.1 and P7.6.2,find the moments of inertia of B for Q for the direction X, Y, and Z where Q is a vertexof B with coordinates (a, b, c) as shown in Figure P7.6.1.

FIGURE P7.5.10A misaligned circular disk on a shaft S.

FIGURE P7.6.1A homogeneous rectangular block.

IxxD G.

X

D

G S

I m b c I m a c I m a bxxB G

yyB G

zzB G= +( ) = +( ) = +( )

1

12

1

12

1

122 2 2 2 2 2, ,

m a b c= = = =12 2 4 3kg, m m m, ,

O

Z

Y

X

ba

cG

Q

m a b c= = = =15 2 5 5 3lb, ft ft ft. , ,


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P7.6.4: A body B has mass center G with coordinates (1, 3, 2), in meters, in a Cartesianreference frame as represented in Figure P7.6.4. Let the mass of B be 0.5 kg. Let the inertiadyadic of B for the origin O have the matrix given by:

where n1, n2, and n3 are parallel to the X-, Y-, and Z-axes. Determine the components ofthe inertia dyadic of B for point Q, where the coordinates of Q are (2, 6, 3), in meters.

P7.6.5: A thin, rectangular plate P weighs 15 lb. The dimensions of the plate are 20 in. by10 in. See Figure P7.6.5, and determine the moments of inertia of P relative to corner Afor the X, Y, and Z directions (see Appendix II).

P7.6.6: Repeat Problem P7.6.5 for a plate with a 5-in.-diameter circular hole centered inthe left half of the plate as represented in Figure P7.6.6.

FIGURE P7.6.4A body B in a Cartesian referenceframe.

FIGURE P7.6.5A rectangular plate in a Cartesianreference frame.

FIGURE 7.6.6A rectangular plate with an offsetcircular hole.

IijB O

I n nB O ijB O

i j ijB OI I= =

,

.

.

9 2 634 1

2 634 4 3

1 3 7

kg m2

IijB Q

G(1,3,2)

Q(2,6,3)B

Z

Y

X

O

n

n

n

3

2

1

Y

A

XO

P

G10 in.

20 in.

Y

A

XO

G10 in.

20 in.

5 in


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Sections 7.7, 7.8, 7.9 Principal Moments of Inertia

P7.7.1: Review again the example of Section 7.8. Repeat the example for an inertia matrix

given by:

P7.7.2: A 2 4-ft rectangular plate OABC is bonded to a 2-ft-square plate CDEF, forminga composite body S as in Figure P7.7.2. Let the rectangular plate weigh 40 lb and thesquare plate 20 lb.

a. Determine the x,y, z components of the mass center G of S.

b. Find the inertia dyadic of S for G. Express the results in terms of the unit vectorsnx, ny, and nz shown in Figure P7.7.2.

c. Find the principal moments of inertia of S for G.

d. Find the principal unit vectors of S for G. Express the results in terms of nx, ny,and nz.

P7.7.3: Repeat Problem P7.7.2 if the square plate CDEF weighs 30 lb.

P7.7.4: Repeat Problem P7.7.3 if the square plate CDEF weighs 10 lb.

FIGURE P7.7.2A composite plate body.

Iij =

32 2 6

2 32 9

6 9 16

slugft2

X

Y

E

D

F 2 ft

2 ft

C

B

A

O

S

2 ft

Z2 ft

4 ft

9781420041927%2Ech7

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