Application of Matrix Algebra in Mechanics

Matrix notation is the most convenient for analyzing complicated problems, it is applied inmany areas in engineering and social sciences. A little bit of patience is required to conditionyourself to like the notation, but once you’ve familiarized yourself with the basic operations, youcan analyze simple or complicated problems in the same way without introducing separate theoriesfor each situation.

Matrix notation is used heavily today in advanced theory and undoubtedly will be the way ofthe future because it is

(1) compact and convenient to write,

(2) easily converted from mathematics to computer programs, and

(3) in an ideal form for concurrent processing.

In CE 325, there will be two basic type of notations used:

(a) Abstract Vector Notation – The subject, “Linear Algebra,” deals with abstract vectors andtheir transformation operators. The results of linear algebraic expressions are also abstractquantities and they are applicable in any coordinate systems, for example,

~F = m~a

is an abstract vector expression,~F and~a are abstract vectors andm is a scalar. The aboveequation, i.e., Newton’s Second Law, is valid in any coordinate system.

(b) Matrix Notation – The subject, “Matrix algebra,” deals with numbers because a matrix isnothing more than a set of numbers arranged in a rectangular manner. For example, the matrix

[A] = [aij ] =[a11 a12 a13a21 a22 a23

]

is a “two-by-three” matrix, meaning 2 rows and 3 columns. The subscript integers,i andj, ofthe elementaij are the row and column indices, respectively.

Matrix algebra is useful in many disciplines inside and outside of engineering. It is quiteeffective because only numbers are dealt with, the actual meaning of a particular element in amatrix is implied only by its position in the matrix. For example, in dynamics, an accelerationvector

~ap = 3.4ı+ 2.9− 7.2k

can be written in matrix form as

~ap =

3.42.9

−7.2

where the number 2.9 is implied to be they-component because it is the second number in the“column vector,” it is no longer necessary to identify the components with the basis vectorsı, andk.

In dynamics, there are many vectors to deal with, e.g., forces, moments, displacements, velocities,accelerations, impulse, momentum, angular impulse, angular momentum, angular velocities andangular accelerations. All these vectors can be written in the form of a column vector. The squarematrix is used for the moment of inertia matrix and various vector transformation matrices (rotationor translation).

– 1 / 1 –

Linear Algebra OperationsThe following are some of the frequently used (not complete) linear algebra operations:

~u+ ~v = ~v + ~u (a)(~u+ ~v) + ~w = ~u+ (~v + ~w) (b)

~u = −~v (c)~w = α~v (d)α = ~u · ~v (e)~w = ~u× ~v (f)

In the expressions (d) and (e),α is a scalar. There is no multiplication of vectors.

Expressions (e) and (f) illustrate two of the most frequently used operations between twovectors, the “dot” and “cross” product, respectively. The “dot” or “inner product” yields a scalarαbut the “cross” or “vector product” yields another vector~w which is geometrically perpendicularto both~u and~v.

Geometrical Interpretation of the Inner Product

The result,α, of the inner product of twovectors~u and~v is given by

α = ~u · ~v = |~u||~v| cos θ ,

whereα is a scalar,θ is the angle between thevectors~u and~v, and|~u| and|~v| are the lengthsof the vectors~u and~v, respectively.

If one of the vectors, say~v is a unit vectore, i.e., the length|e| = 1, then

α = ~u · e = |~u||e| cos θ = |~u| cos θ ,

θ

~v

~u

In this case,α would be the length of~u ’s projection onto a line along the direction ofe.

The inner product is an extremely useful tool because if~u·~v = 0, i.e.,cos θ = 0 orθ = 90, thetwo vectors are orthogonal (perpendicular). Mathematically, the advantages of orthogonal vectorsare numerous and they will be discussed later on.

Geometrical Interpretation of the Vector Product

If the vector product~w = ~u× ~v

is performed, the following is true about the result~w.

(a) The direction of~w is perpendicular to~u and~v. The positive direction is defined by theright-hand-rule going from~u to ~v as shownin the figure.

(b) The magnitude of~w is|~w| = |~u||~v| sinφ

whereφ is the angle between the vectors~uand~v. If φ = 0 or φ = 180, i.e.,~u and~vare parallel, then~u× ~v = ~0.

~w

~v

φ ~u

– 1 / 2 –

Definition of Basis VectorsIt is difficult to obtain numerical values

for abstract vector algebra unless a “basis”is established. Normally, we can add twovectors graphically using arrows (see figureto the right), but the length of the resultantmust be measured graphically as well.

The simplest case for vector addition iswhen the vectors are parallel, then the lengthof the result is just the arithmatic sum ordifference of the length of the vectors. ~u

~v~u + ~v

For general cases, a vector can be represented as a sum of several vectors, known ascomponents; therefore, it is most convenient to require all the vectors within the same problemto have components in a few prescribed directions. These prescribed directions are those of basisvectors. The number of basis vectors required is 2 for two-dimensional problems and 3 for three-dimensional problems.

Cartesian Coordinate system

Define a set of orthogonal unit vectorsı, andk such that

ı · ı=1, ı · =0, ı · k=0,

· ı=0, · =1, · k=0,

k · ı=0, k · =0, k · k=1,

then an arbitrary vector~v can be written interms ofı, andk as

~v = vx ı+ vy + vz k ,

x

y

z

~vk

ı

wherevx, vy andvz are scalars. To obtain the values of these scalars, use the inner product asfollows:

~v · ı = (vx ı+ vy + vz k) · ı = vx(ı · ı) + vy( · ı) + vz(k · ı) = vx(1) + vy(0) + vz(0) = vx

~v · = (vx ı+ vy + vz k) · = vx(ı · ) + vy( · ) + vz(k · ) = vx(0) + vy(1) + vz(0) = vy

~v · k = (vx ı+ vy + vz k) · k = vx(ı · k) + vy( · k) + vz(k · k) = vx(0) + vy(0) + vz(1) = vz

Summarizing, the numerical values of the scalars can be obtained as

vx = ~v · ı,vy = ~v · ,vz = ~v · k.

– 1 / 3 –

Cylindrical Coordinate system

Define a set of orthogonal unit vectorser, eθ andez such that

er · er=1, er · eθ=0, er · ez=0,eθ · er=0, eθ · eθ=1, eθ · ez=0,ez · er=0, ez · eθ=0, ez · ez=1.

er and eθ change direction as the angleθ isvaried whileez remains in the same direction.

y

z

θ

r

eθ

ez

erx

Normal-Tangent Coordinate system

Define a set of orthogonal unit vectorset,en andez = eθ × et such that

et · et=1, et · en=0, et · ez=0,en · et=0, en · en=1, en · ez=0,ez · et=0, ez · en=0, ez · ez=1.

et is always tangent to the path and it changesdirection as the scalar distance parameters isvaried. en is defined normal to the path andis selected to be the one pointing toward thecenter of curvature. For straight paths, thereis no need foren.

s

et

en

e n

e t

Spherical Coordinate system

Define a set of orthogonal unit vectorser, eθ andeφ such that

er · er=1, er · eθ=0, er · eφ=0,eθ · er=0, eθ · eθ=1, eθ · eφ=0,eφ · er=0, eφ · eθ=0, eφ · eφ=1.

er changes direction as anglesθ orφ is varied,it is always in therz-plane. eθ changesdirection only when angleθ is varied, it isalways in thexy-plane. eφ changes directiononly when angleφ is varied, it is always in therz-plane.

x

y

z

φ

θ ereφ

eθ

– 1 / 4 –

Application of Basis VectorsWith the basis vectors defined, it is now possible to calculate numerically the results. Let

~u = ux ı+ uy + uz k and~v = vx ı+ vy + vz k in the following examples.

Addition of Two Vectors

~u+ ~v = (ux + vx)ı+ (uy + vy)+ (uz + vz)k

because thex, y andz components of~u and~v are respectively parallel.

Subtraction of Two Vectors

~u− ~v = (ux − vx)ı+ (uy − vy)+ (uz − vz)k

Inner Product of Two Vectors

~u · ~v = (uxvx)ı · ı+ (uxvy )ı · + (uxvz )ı · k+ (uyvx) · ı+ (uyvy) · + (uyvz) · k+ (uzvx)k · ı+ (uzvy)k · + (uzvz)k · k

= uxvx + uyvy + uzvz .

The result is a scalar.

Vector Product of Two Vectors

~u× ~v = (uxvx)ı× ı+ (uxvy )ı× + (uxvz )ı× k

+ (uyvx)× ı+ (uyvy)× + (uyvz)× k

+ (uzvx)k × ı+ (uzvy)k × + (uzvz)k × k .

Since according to the right-hand-rule,

ı× ı = ~0, ı× = k, ı× k = −,× ı = −k, × = ~0, × k = ı,

k × ı = , k × = −ı, k × k = ~0.

It follows that

~u× ~v = (uyvz − uzvy )ı+ (uzvx − uxvz)+ (uxvy − uyvx)k .

The results shown above are valid for any set of orthogonal basis vectors other thanı, andk.

– 1 / 5 –

Matrix Algebra OperationsAddition of Two Matrices (Vectors)

Two matrices are said to be compatible for matrix addition if their row and column dimensionsare identical. If

[A] = [aij ] and [B] = [bij ]

then[C] = [A] + [B] = [cij ] = [aij + bij ] .

Since vectors are represented by column matrices, i.e., matrices with 1 column, the vectors~u and~v can be expressed as

~u = u =

ux

uy

uz

and ~v = v =

vx

vy

vz

,

respectively, and the sum~w=~u+~v is simply

~w = w =

wx

wy

wz

=

ux + vx

uy + vy

uz + vz

.

Subtraction of Two Matrices (Vectors)

Two matrices are said to be compatible for matrix subtraction if their row and columndimensions are identical. If

[A] = [aij ] and [B] = [bij ]

then[C] = [A] − [B] = [cij ] = [aij − bij ] .

In the particular case of the difference~w=~u-~v, the matrix form of~w is defined as

~w = w =

wx

wy

wz

=

ux − vx

uy − vy

uz − vz

.

Transposition of a Matrix

If a matrix [A] is defined as[A] = [aij ]

then its transpose[A]T is[A]T = [aji]

i.e., the rows and columns are switched. The transpose of the vector~u can be written as

~uT = ux uy uz .

– 1 / 6 –

Matrix Multiplication

Two matrices,[A] and[B], are said to be compatible for matrix multiplication[A][B] if thecolumn dimension of[A] is equal to the row dimension of[B]. If

[A] = [aij ], i = 1, . . . , L; j = 1, . . . ,M ;and

[B] = [bjk], j = 1, . . . ,M ; k = 1, . . . , N ;

then[C] = [cik], i = 1, . . . , L; k = 1, . . . , N ;

with the elementscik defined as

cik =M∑

j=1

aijbjk .

Inner Product of Two Vectors

The inner product,α, of two vectors~u and~v in matrix form is written as

α = ~u · ~v = uT v

in whichuT v is the multiplication of a row matrixuT and a column matrixv. The columndimension ofuT and the row dimension ofv are both 3 in this particular case.

The Determinant of a Matrix

See a reference book. Too lengthy to describe here.

Vector Product of Two Vectors

The vector product is quite similar to the matrix determinant. If~u and~v are defined as

~u =

ux

uy

uz

and ~v =

vx

vy

vz

,

then

~u× ~v =

∣∣∣∣uy vy

uz vz

∣∣∣∣∣∣∣∣ uz vz

ux vx

∣∣∣∣∣∣∣∣ux vx

uy vy

∣∣∣∣

=

uyvz − uzvy

uzvx − uxvz

uxvy − uyvx

.

The vector product can also be written as a matrix-vector product as

~u× ~v =

0 −uz uy

uz 0 −ux

−uy ux 0

vx

vy

vz

= −

0 −vz vy

vz 0 −vx

−vy vx 0

ux

uy

uz

.

Note, a matrix involved in a cross-product is antisymmetric, implying its diagonal is 0.

– 1 / 7 –

Matrix Notation for Basis Vectors

Recall a vector~v can be written numerically using basis vectorsı, andk as

~v = vx ı+ vy + vz k ,

wherevx = ~v · ı,vy = ~v · ,vz = ~v · k,

In matrix notation,

~v =

vx

vy

vz

=

vx

00

+

0vy

0

+

00vz

= vx

100

+ vy

010

+ vz

001

.

It is clear from the correspondance that

ı =

100

=

010

k =

001

.

The above representation is the most convenient form for basis vectors; it is clear that they areorthogonal (inner product equals to zero) and the length is 1. There are infinite number of otherorthogonal sets, for example, the basis vectors of the cylindrical coordinates consisting of

er =

cos θsin θ

0

, eθ =

− sin θcos θ

0

and ez =

001

are also “orthonormal”, meaning orthogonal and normalized to 1.

Equating Components of Vectors

When an unknown vector~v is said to be equal to a constant vector~p, e.g.,

~v =

vx

vy

vz

= ~p =

abc

,

it is easy to apply the inner product with unit vectorı to both sides of the equation, i.e.,

[ vx vy vz ]

1

00

= [ a b c ]

1

00

to yield vx = a .

Similarly, inner product of the equation~v = ~p with unit vectors and k will yield respectively,vy = b andvz = c. It is clear that it is allowable to equate two matrices element-for-element if theyare defined using the same set of basis vectors. If the scalar components of two column vectorsare defined using different sets of basis vectors, then a transformation (to be described in the nextsection) needs to be made before they can be equated element-for-element.

– 1 / 8 –

Change of Basis Vectors

Quite often, it is convenient to use two different sets of basis vectors to describe the samevector, e.g.,

~v = vx ı+ vy + vz k , (1)

and~v = vr er + vθ eθ + vz ez . (2)

In matrix notation, let

~v =

vx

vy

vz

xyz

=

vr

vθ

vz

rθz

in which the subscribesxyzandrθz indicates which coordinate system the scalars inside a bracketis referred to. Since the scalars inside the column vectors are obtained using two different sets ofbasis vectors, the inner product operation

er · vxyz = er · vrθz

would have to be done wither = cos θ , sin θ , 0Txyz for the left-hand-side of the equation and

with er = 1 , 0 , 0Trθz for the right-hand-side of the equation. Therefore,

[ cos θ sin θ 0 ]

vx

vy

vz

= [ 1 0 0 ]

vr

vθ

vz

yieldsvx cos θ + vy sin θ = vr ,

a scalar equation.

To simplify the correspondance between the two systems, it is best to express one set of theunit vectors(ı, , k) in terms of another set(er, eθ, ez) in the following manner:

ı = (ı · er) er + (ı · eθ) eθ + (ı · ez) ez =

ı · er

ı · eθ

ı · ez

rθz

, (3a)

= ( · er) er + ( · eθ) eθ + ( · ez) ez =

· er

· eθ

· ez

rθz

(3b)

and

k = (k · er) er + (k · eθ) eθ + (k · ez) ez =

k · er

k · eθ

k · ez

rθz

. (3c)

Now substitute the expessions in equation (3) into equation (1), then the vector~v can be expressedin therθz system as a linear combination of 3 orthogonal vectors as

~v =

vr

vθ

vz

rθz

= vx

ı · er

ı · eθ

ı · ez

rθz

+ vy

· er

· eθ

· ez

rθz

+ vz

k · er

k · eθ

k · ez

rθz

.

– 1 / 9 –

The column vectors used on the right-hand-side of the above equation are actually the basis vectorsı, andk converted to therθz system. The above expression, with 3 scalars multiplying 3 columnvectors, can be written more concisely as a matrix product with the matrix consisting of 3 columnvectors placed side-by-side as

vr

vθ

vz

rθz

=

ı · er · er k · er

ı · eθ · eθ k · eθ

ı · ez · ez k · ez

vx

vy

vz

xyz

= [Q]

vx

vy

vz

xyz

. (4)

The matrix[Q], defined above, is called a transformation matrix because it transforms a columnvector expressed in thexyzsystem to one expressed in therθz system.

To do the inversed transformation, it is just as easy to express the set of basis vectors(er, eθ, ez)in terms of the set of unit vectors(ı, , k) as

er = (er · ı) ı+ (er · ) + (er · k) k =

er · ıer · er · k

xyz

, (5a)

eθ = (eθ · ı) ı+ (eθ · ) + (eθ · k) k =

eθ · ıeθ · eθ · k

xyz

(5b)

and

ez = (ez · ı) ı+ (ez · ) + (ez · k) k =

ez · ıez · ez · k

xyz

. (5c)

The substitution of equation (5) into equation (2) allows the vector~v to be expressed in thexyzsystem as a linear combination of 3 orthogonal vectors as

~v =

vx

vy

vz

xyz

= vr

er · ıer · er · k

xyz

+ vθ

eθ · ıeθ · eθ · k

xyz

+ vz

ez · ıez · ez · k

xyz

,

or vx

vy

vz

xyz

=

er · ı eθ · ı ez · ıer · eθ · ez · er · k eθ · k ez · k

vr

vθ

vz

rθz

= [Q]T

vr

vθ

vz

rθz

. (6)

It is interesting to note that the matrices which transform a vector from one set of orthogonal basisvectors to another set of orthogonal basis vectors have the property that[Q]−1 = [Q]T , making itvery simple to obtain the inverse transformation.

– 1 / 10 –

Time Derivative of Vectors

An arbitrary vector~q can be written as~q = qe

whereq is the magnitude ande is the unit vector representing the direction of vector~q. Therefore,the time derivative of~q can be determined as

d~q

dt=

d

dt(qe) =

dq

dte+ q

de

dt.

Define now a scalarq = dq/dt to represent the rate of change of the magnitudeq. The changeof e with respect to time, on the other hand, represents the change in direction only because themagnitude ofe is 1.

Consider now the diagram to the right, thechange frome to e′ during the time interval∆t canbe represented by the rotation through the angle∆θ.Hence,

∆e = e′ − e

and the length of∆e can be derived geometrically as

|∆e| = (|e| sinφ)∆θ .

It follows that the magnitude ofde/dt can bedetermined by the limit

lim∆t→0

|∆e|∆t

=∣∣∣∣dedt

∣∣∣∣ =dθ

dt|e| sinφ = |~ω||e| sinφ .

Clearly, |~ω||e| sinφ is the magnitude of the crossproduct~ω × e.

~ω

∆θ ∆e

eφ

e’|e|sinφ

Now by inspection,∆e is also in the direction of~ω × e according to the right-hand-rule,therefore, it can be concluded that

de

dt= ~ω × e .

The final form for the time derivative of a vector~q can be expressed as

d~q

dt= qe+ q(~ω × e) = qe+ ~ω × (qe)

or

d~q

dt= qe+ ~ω × ~q .

If the above formula is applied to a vector of constant magnitude,~p = pe, with p = 0, then

dp

dt= ~ω × p .

– 1 / 11 –

Example – Matrix Operations

Transposition

If [A] =[

3 2 51 −1 0

]then [A]T =

3 1

2 −15 0

.

Matrix Multiplication

If [A] =[

3 2 51 −1 0

]is a2 × 3 matrix and [B] =

1 4 3

1 2 11 1 2

is a3 × 3

matrix, then the product[C] = [A][B] is a2 × 3 matrix. The product[C] = [B][A] is not possible

in this particular case.

[C] =[

3 2 51 −1 0

] 1 4 3

1 2 11 1 2

=[

(3)(1) + (2)(1) + (5)(1) (3)(4) + (2)(2) + (5)(1) (3)(3) + (2)(1) + (5)(2)(1)(1) + (−1)(1) + (0)(1) (1)(4) + (−1)(2) + (0)(1) (1)(3) + (−1)(1) + (0)(2)

]

=[

10 21 210 2 2

]

Matrix Multiplication 6 0 3

1 2 13 2 3

1 2 3

0 1 11 4 1

=

9 24 21

2 8 66 20 14

=

(6)(1) + (0)(0) + (3)(1) (6)(2) + (0)(1) + (3)(4) (6)(3) + (0)(1) + (3)(1)

(1)(1) + (2)(0) + (1)(1) (1)(2) + (2)(1) + (1)(4) (1)(3) + (2)(1) + (1)(1)(3)(1) + (2)(0) + (3)(1) (3)(2) + (2)(1) + (3)(4) (3)(3) + (2)(1) + (3)(1)

1 2 3

0 1 11 4 1

6 0 3

1 2 13 2 3

=

17 10 14

4 4 413 10 10

=

(1)(6) + (2)(1) + (3)(3) (1)(0) + (2)(2) + (3)(2) (1)(3) + (2)(1) + (3)(3)

(0)(6) + (1)(1) + (1)(3) (0)(0) + (1)(2) + (1)(2) (0)(3) + (1)(1) + (1)(3)(1)(6) + (4)(1) + (1)(3) (1)(0) + (4)(2) + (1)(2) (1)(3) + (4)(1) + (1)(3)

Therefore 6 0 3

1 2 13 2 3

1 2 3

0 1 11 4 1

6=

1 2 3

0 1 11 4 1

6 0 3

1 2 13 2 3

– 1 / 12 –

Example – Change of Bases

One abstract vector,~v, can be expressed in several

different bases. The numerical representations of this

vector are different if the basis vectors of each system

are aligned differently.

In the figure to the right, a vector of length|~v| = 6 is

to be written using two different bases: (a) the(ı, )system and (b) the(er, eθ) system.

ı

45o

30o

er

~v

30o

eθ

(a) Write~v using(ı, ) as the basis.

vx = ~v · ı = |~v||ı| cos θ = (6)(1) cos 45 = 4.243

vy = ~v · = |~v||| cos θ = (6)(1) cos 45 = 4.243

Hence

~v = 4.243ı+ 4.243 =

4.2434.243

xy

(b) Write~v using(er, eθ) as the basis.

vr = ~v · er = |~v||er| cos θ = (6)(1) cos 15 = 5.796

vθ = ~v · eθ = |~v||eθ| cos θ = (6)(1) cos 75 = 1.553

Hence

~v = 5.796er + 1.553eθ =

5.7961.553

rθ

To convert a vector from one basis to another, define a transformation matrix[Q] so that

vrθ = [Q]vxy

in which

[Q] =[er · ı er · eθ · ı eθ ·

]=

[(1)(1) cos 30 (1)(1) cos 60

(1)(1) cos 120 (1)(1) cos 30

]=

[0.866 0.5−0.5 0.866

].

Now the transformation can be accomplished by the matrix productvr

vθ

=

[0.866 0.5−0.5 0.866

] 4.2434.243

xy

=

0.866 × 4.243 + 0.5 × 4.243−0.5 × 4.243 + 0.866 × 4.243

=

5.7961.553

rθ

The inverse transformation[Q]−1 defined as

vxy = [Q]−1vrθ

is easily calculated as the transpose of[Q], i.e.,

[Q]−1 =[ı · er ı · eθ

· er · eθ

]=

[(1)(1) cos 30 (1)(1) cos 120

(1)(1) cos 60 (1)(1) cos 30

]=

[0.866 −0.50.5 0.866

]= [Q]T .

– 1 / 13 –

More Complicated Transformations

For the complicated transformations, it isperhaps too difficult to measure the anglesbetween basis vectors between two differentsystems to obtain the inner products neces-sary.

These transformations can generally bedecomposed into two transformations, eachstep requiring just one rotation about only oneaxis. These steps usually have to be performedin a strict order because matrix multiplicationdoes not commute.

ψ

x

y

z

a

b

c

θ

θ

ψ

θ

θxy

u

v

z , wStep 1

ψ

u

a

v , b

wψ

cStep 2

Step 1– rotation about thez- orw-axis:quqvqw

=

cos θ sin θ 0

− sin θ cos θ 00 0 1

qxqyqz

= [Q1]

qxqyqz

Step 2– rotation about thev- or b-axis:qaqbqc

=

cosψ 0 sinψ

0 1 0− sinψ 0 cosψ

quqvqw

= [Q2]

quqvqw

The total transformation[Q] can then be defined asqaqbqc

= [Q2]

quqvqw

= [Q2][Q1]

qxqyqz

= [Q]

qxqyqz

Hence

[Q] =

cosψ 0 sinψ

0 1 0− sinψ 0 cosψ

cos θ sin θ 0

− sin θ cos θ 00 0 1

=

cosψ cos θ cosψ sin θ sinψ

− sin θ cos θ 0− sinψ cos θ − sinψ sin θ cosψ

– 1 / 14 –

DYNAMICS

Dynamics is composed of 2 major topics:

(1) Kinematics Mathematical description of motion without regards to Newton’sSecond Law.

(2) Kinetics Basically Newton’s Second Law or Euler’s First Law.

Basic Definitions of “Particle” Kinematics

The following definitions are made in abstract algebra notation, therefore, they are valid in allcoordinate systems.

Displacement

The displacement of a particle is referred to as thelocation of a particle with respect to the origin.Usually, the “displacement” is represented by theposition vector~r(t).

Velocity

The average velocity over a time period∆t is definedas

~vavg =~r(t + ∆t)− ~r(t)

∆t.

while the instantaneous velocity at a given timet isdefined as

~v = lim∆t→0

~vavg =d~r

dt

P

~r

e2

e3

e1

Acceleration

The average acceleration over a time period∆t is defined as

~aavg =~v(t + ∆t)− ~v(t)

∆t.

while the instantaneous acceleration at a given timet is defined as

~a = lim∆t→0

~aavg =d~v

dt=

d2~r

dt2

Integral Equations of Motion

From the definitions of instantaneous velocity and acceleration, we can write the integral definitionsas ∫

d~v =∫

~a dt and∫

d~r =∫

~v dt .

Clearly, the above integrals preserve the vector properties. There is another integral form which iscommonly used for particles traveling along a path, i.e.,∫

~a · d~r =∫

~v · d~v ,

but this formulation retains only the tangential component. The nt-coordinate system is the best forthis particular form.

– 2 / 1 –

Kinematic Formulas in Frequently Used Bases

After defining a basis, the kinematic quantities can be expressed easily in matrix algebra notations.

Cartesian Coordinates

The basis vectors used areı, andk. The direction ofthese unit vectors are fixed so their time derivativesvanish because they have constant magnitudes and norotation.

In abstract algebra notation, the vectors written incomponent form can be expressed as

~r(t) = x(t)ı + y(t) + z(t)k

~v(t) = x(t)ı + y(t) + z(t)k

~a(t) = x(t)ı + y(t) + z(t)k

wherex(t), y(t), z(t), x(t), y(t), z(t), x(t), y(t) andz(t) are scalar functions of time. At each instant oftime, these scalar functions represents the magnitudeof the components.

x

y

z

P~r

ı

k

The same expressions in matrix algebra notation can be written as

~r(t) =

x(t)y(t)z(t)

, ~v(t) =

x(t)y(t)z(t)

and ~a(t) =

x(t)y(t)z(t)

Cylindrical (Polar) Coordinates

The basis vectors used areer, eθ andez. The directionof er and eθ changes with the angleθ but ez do notchange.

In abstract algebra notation, the position vector writtenin component form is

~r(t) = r(t)er + z(t)ez

wherer(t) andz(t) are scalar functions of time. Thereis no component in theeθ direction for the displacementvector but do remember thater is a function ofθ. Toobtain the velocity vector, take time derivative of thedisplacement vector~r(t) to yield

~v(t) =d~r

dt= r(t)er + z(t)ez + r(t)

der

dt+ z(t)

dez

dt,

x

y

z

P

~r

θ

ez eθ

er

in whichdez

dt= ~0

becauseez is fixed in thez-direction and the time derivative ofer can be obtained as

der

dt= ~ω × er = θez × er = θeθ

– 2 / 2 –

since the rotation of theer vector is represented by a rotation vector~ω = θez is in thez-direction.

Substituting the above expressions into~v(t) yields

~v(t) = r(t)er + r(t)θ(t)eθ + z(t)ez .

To obtain the acceleration vector~a(t), we take the time derivative of the velocity vector~v(t) as

~a(t) =d~v(t)dt

= rer + rder

dt+ rθeθ + rθeθ + rθ

deθ

dt+ zez + z

dez

dt

To simplify the above, obtain the time derivative ofeθ in the same manner as that forer, i.e.,

deθ

dt= ~ω × eθ = θez × eθ = θ(−er) = −θer

hence,~a(t) = rer + rθeθ + rθeθ + rθeθ − rθ2er + zez

or simply,~a(t) = (r − rθ2)er + (rθ + 2rθ)eθ + zez

The above expressions written in matrix algebra notation are

~r(t) =

r(t)0

z(t)

, ~v(t) =

r(t)r(t)θ(t)

z(t)

and ~a(t) =

r − rθ2

rθ + 2rθz

Spherical Coordinates

The basis vectors used areer, eθ and eφ. Thedirections ofer andeφ change with the anglesθandφ, but eθ changes only with the angleθ and itremains parallel to thexy-plane.

In abstract algebra notation, the position vectorwritten in component form is

~r(t) = r(t)er

wherer(t) is a scalar function of time. There isno component in theeθ nor theeφ direction forthe displacement vector buter is a function ofθandφ.

x

y

z

φ

θ ereφ

eθ

Although the equations in this coordinate system will be put to use immediately in someinteresting problems, their derivation will be performed later in this course using rotating coordinatesystems. The matrix representations of~r, ~v and~a are, respectively,

~r =

r00

rφθ

,

~v =

rrφ

rθ sinφ

rφθ

and

~a =

r − r(φ2 + θ2 sin2 φ)rφ + 2rφ− rθ2 sinφ cos φ

rθ sinφ + 2rθ sinφ + 2rθφ cos φ

rφθ

.

– 2 / 3 –

Planar nt (Normal-Tangent) Coordinates

The basis vectors used areet (tangential unit vector),en (normal unit vector) andeb (binormal unit vector).In most textbooks,et and en are in the same planewhile eb = et × en is normal to the plane.

As shown in the figureet is tangent to the curve andis defined positive in the direction of motion. Thevector en is normal to the curve with positive sensedefined towards the center of curvature. For the casewhen the path is straight, i.e., no center of curvature,en is not necessary.

en

et

en

et

Unlike the other coordinate systems, thent-coordinates require no position vector to define velocityand acceleration. Sometimes, a scalar parameters is used to determine the location of a particlealong its path. If a position vector is absolute necessary for a particular application, a second systemsuch as a cartesian or polar coordinate system, is generally introduced.

To derive an expression for velocity, lets take anarbitrary pointP as a temporary origin and use thestandard definition of the instantaneous velocity,

~v = lim∆t→0

~r(s + ∆s)− ~r(s)∆t

= lim∆t→0

∆~r

∆t

As ∆t → 0, |∆~r| → ∆s and the direction of∆~r istangent to the path, hence,

~v = lim∆t→0

∆s

∆tet = set = vet

where s(t) = v is the rate of change of the scalardistance functions. It is clear that the velocity isalways tangent to the path and there can never be anormal component of velocity using thent coordinatesystem.

P

∆~r

s

∆s

s+∆s

~r(s)

~r(s+∆s)

To obtain an expression for the acceleration, use the definition

~a =d~v

dt=

d

dt(set) = set + s

det

dt.

The derivative of the unit vectoret can be written as

det

dt= ~ω × et = θez × et = θen .

Hence,~a = set + sθen .

Define now a radius of curvatureρ such thats = ρθ or θ = s/ρ, then the expression of accelerationcan be written as

~a = set +s2

ρen .

The first term represents the rate of change of speed along the path while the second term representsthe centrifugal acceleration caused by the change of direction.

– 2 / 4 –

Three-Dimensional Space Curves

A space curve can be described by a three-dimensional position vector,~r(s), composed ofthree scalar functions,x(s), y(s) andz(s) as

~r(s) =

x(s)y(s)z(s)

,

in which x, y and z are cartesian coordinatesmeasured from an arbitrary origin. Thecoordinates are scalar functions of only one scalarparameter, the distance,s, measured from thebeginning of the space curve.

~r(s)

s

et

eneb

Let v = ds/dt be the speed along the space curve;v is the magnitude of the velocity vector,~v, which is defined as

~v =d~r

dt=

d~r

ds

ds

dt= v

d~r

ds= vet ,

in which,

et =d~r

ds,

is the tangential unit vector. Next, the acceleration vector,~a, can be derived as

~a =d~v

dt=

d

dt(vet) = vet + v

det

dt,

in which v is the rate of change of speed, or the tangential component of acceleration along thepath. The rate of change ofet with repsect tot is related to the curvature of the path, i.e.,

det

dt=

det

ds

ds

dt= v

det

ds= v

d2~r

ds2 .

The rate of change of the tangential vector,et, with the distance parameter,s, leads to the introductionof the normal unit vector,en, defined as

det

ds= κen ,

in whichκ is known as the curvature.1/κ = ρ is the instantaneous radius of curvature. With thesedefinitions, the acceleration of a particle traveling along a space curve can be written in the form,

~a = vet + vdet

dt= vet + v

(vdet

ds

)= vet + v

(v

ρen

),

or

~a = vet +v2

ρen .

A third unit vector,eb, the binormal vector, can be defined as,eb = et × en, to complete the basisfor a three-dimensional coordinate system. There is no velocity, nor acceleration components, inthe eb direction.

– 2 / 5 –

Transformation Between Coordinate SystemsThe nt and the Cartesian Coordinate Systems

Treating thent-coordinate system as a specialcase of the space curve, let~r = x(s), y(s)T andassume that the plane curve can be represented by afunction,y(x). The distance parameter,s, can thenbe defined as the integral of

ds =√

(dx)2 + (dy)2 = |dx|√

1 + (dy/dx)2 .

The derivative ofy(x) with respect tos can also beperformed by using the scale factor,∣∣∣∣dx

ds

∣∣∣∣ = 1√1 + (y′)2

,

in whichy′ = dy/dx.x

y

dy

dx

ds

θ

y(x)

To find the unit vector,et, perform a derivative with respect tos as follows:

et =d~r

ds=

dx/dsdy/ds

=

dx/ds(dy/dx)(dx/ds)

=

1√1 + (y′)2

1y′

.

To find the unit vector,en, and the radius of curvature,ρ = 1/κ, find the second derviative of~rwith respect tos as follows:

d2~r

ds2 =−1

2 (2y′)y′′(dx/ds)

[1 + (y′)2]32

1y′

+

(dx/ds)

[1 + (y′)2]12

0y′′

.

Simplify now the above expression,

d2~r

ds2 =(dx/ds)

[1 + (y′)2]32

0− y′y′′

[1 + (y′)2]y′′ − (y′)2y′′

=

y′′

[1 + (y′)2]32

(1√

1 + (y′)2

−y′

1

).

From the above expression, the “curvature,”1/ρ, anden can be expressed, respectively, as

1ρ

=

∣∣∣∣∣∣d2y

dx2

/[1 +

(dy

dx

)2]3/2

∣∣∣∣∣∣ and en =1√

1 + (y′)2

−y′

1

.

Depending on the curvature ofy(x), the normal unit vectorcan be eitheren or e′

n. Therefore, each situation must behandled with care. To simplify the expressions, use the angle,θ, as defined in the figures; the unit vectors for thetn-systemcan then be written as

et =

cos θsin θ

xy

and en = −e′n =

sin θ− cos θ

xy

,

and clearly, the unit vectors of the cartesian system are

ı =

10

xy

and =

01

xy

, en

en’

θ

etdy

dx

hence, the transformation matrix between the two systems can be established by simply using theinner products as

qt

qn

tn

=[

et · ı et · en · ı en ·

]qx

qy

xy

=[

cos θ sin θ∓ sin θ ± cos θ

]qx

qy

xy

.

– 2 / 6 –

The nt and the Polar Coordinate Systems

If r(θ) is the curve which describesthe path, then the unit vectorset and encan be found by utilizing the relationship

d~r = ~vdt =

rrθ

dt =

drrdθ

.

Since the unit vectoret is tangent to thepath, it can be defined as

et =d~r

|d~r|in which

|d~r| =√

(dr)2 + (rdθ)2

= dθ√

(dr/dθ)2 + r2 .

en

er

eθ

et

r(θ)

θO

Combining the expressions ofd~r and|d~r|, the tangential unit vector can be written as

et =1√

(dr/dθ)2 + r2

dr/dθ

r

rθ

.

As an orthogonal vector,en can be defined as

en =1√

(dr/dθ)2 + r2

−rdr/dθ

rθ

.

Note, there is another unit vector which is also orthogonal toet, but it would be pointing away fromthe center of curvature.

To define the transformation between vectors written inrθ andnt systems, recall

er =

10

rθ

and eθ =

01

rθ

,

therefore qt

qn

tn

=[

et · er et · eθ

en · er en · eθ

]qr

qθ

rθ

=1√

(dr/dθ)2 + r2

[dr/dθ r−r dr/dθ

]qr

qθ

rθ

.

– 2 / 7 –

The Spherical and the Cartesian Coordinate Systems

The spherical system and the cartesian systemhave a common origin, the two angles whichdefines the radial direction areθ andφ.

x

y

z

φ

θ ereφ

eθ

xy

z , c

b

aθ

θ

<xyz to abc>

eφ

φ

er

a

eθ , bφ

c<abc to rφθ>

To obtain the transformation fromxyzto rφθ, it is better to define an intermediate systemabcwhichis obtained by rotating the axes about thez-axis through an angle ofθ, therefore,

qa

qb

qc

=

cos θ sin θ 0− sin θ cos θ 0

0 0 1

qx

qy

qz

.

After the above transformation, theb-axis is in the direction ofeθ. We can now attain therθφsystem by simply rotating the axes about theb-axis through an angle ofφ, i.e.,

qr

qφ

qθ

=

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

qa

qb

qc

.

Now a one step transformation fromxyz to rφθ can be obtained by substituting the firsttransformation into the second transformation as

qr

qφ

qθ

=

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

qa

qb

qc

=

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

cos θ sin θ 0− sin θ cos θ 0

0 0 1

qx

qy

qz

=

cos θ sinφ sin θ sinφ cos φ

cos θ cos φ sin θ cos φ − sinφ− sin θ cos θ 0

qx

qy

qz

.

– 2 / 8 –

The Spherical and the Cylindrical Coordinate Systems

Both the spherical and the cylindrical coor-dinate systems have an azimuthal angleθ, sotheeθ unit vector is the same for both systems.Usually, the radial unit vector is identified aser for both systems, but for the purpose of thisparticular derivation, a capitalR is use for thecylindrical coordinate system.

eφ

φ

φeθ

ez

er

eR

From the figure, it is clear that the transformation can be established easily with a single-parameterφ as

qr

qφ

qθ

=

cos(π

2 − φ) 0 cos φcos φ 0 cos(π

2 + φ)0 1 0

qR

qθ

qz

=

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

qR

qθ

qz

The Cartesian and the Cylindrical Coordinate Systems

The transformation from the cartesiancoordinate system to the cylindrical coor-dinate system represents the simplest pos-sible “orthogonal transformation.” Sincethey share thez-axis, the rotation isbasically two-dimensional and can beestablished as

x θ

θ

eθ

y

er

z

qr

qθ

qz

=

cos θ cos(π

2 − θ) 0cos(π

2 + θ) cos θ 00 0 1

qx

qy

qz

=

cos θ sin θ 0− sin θ cos θ 0

0 0 1

qx

qy

qz

– 2 / 9 –

Relative Coordinates – Translating Axes

Let B be the origin of the translating coordinatesystem, then from the vector diagram we can conclude

~rA = ~rB + ~rA/B ;

~vA = ~vB + ~vA/B ; ~vA/B =d

dt~rA/B

~aA = ~aB + ~aA/B ; ~aA/B =d2

dt2~rA/B

in which the vectors with subscriptsA and B areabsolute vectors while vectors with the subscriptA/Bare relative vectors, implying “the motion of pointArelative to pointB.

X

Y

Z

B

A

z

x

y

~rA/B

O

~r A

~r B

– 2 / 10 –

Kinematics of Rigid Bodies

Applications: Analysis of cams, gears, shafts, linkages, connecting rods, etc.

Definition: A rigid body is a special case of a system of particles wherein the distances betweenall particles remain unchanged.

For a system of particles with general three-dimensional motion prescribed for each particle,there are up to3N possible degrees of freedom for a system ofN particles. With the restrictionon distance between two pointsA andB so that|~rA/B | =constant, the only possible change withrespect time for~rA/B is due to rotation. Since the rotational degrees of freedom can also berepresented by a vector, the general description of rigid body motion can be reduced to

(i) 3 translations and 3 rotations for three-dimensional problems and

(ii) 2 translations and 1 rotation for two-dimensional (plane) problems.

Motion of an Arbitrary Point on a Rigid Body

If the translation components of a rigid body isgiven at a reference pointB and the rotation vector isalso known for the rigid body, then the motion at anyarbitrary pointA on the rigid body can be determinedcompletely.

Consider first the relationship

~rA = ~rB + ~rA/B ,

in which ~rB is the known motion of the referencepointB while ~rA is the motion of the arbitrary pointA to be determined. The relative displacement~rA/B

has the additional restriction of|~rA/B | =constant fora rigid body.

A

B

O

~rB

~r A

~rA/B

To obtain the velocity at pointA, take derivative of the position vector~rA with respect to timeto yield

d~rA

dt=

d~rB

dt+

d~rA/B

dtor ~vA = ~vB + ~vA/B

where the relative velocity~vA/B is defined as

~vA/B =d~rA/B

dt= ~ω × ~rA/B

for a rigid body because|~rA/B | =constant and the change occurs only due to rotation. Hence, fora rigid body

~vA = ~vB + ~ω × ~rA/B .

The above equation indicates that once~vB and~ω are known, the velocity everywhere on therigid body can be calculated easily.

To obtain the acceleration vector, perform another time derivative to yield

d~vA

dt=

d~vB

dt+

d~ω

dt× ~rA/B + ~ω × d~rA/B

dt

or~aA = ~aB + ~α× ~rA/B + ~ω × (~ω × ~rA/B) .

– 6 / 1 –

Kinematics Described in a Rotating Frame

Sometimes it is necessary to describe the motion of a point using a coordinate system which rotates.The idea is to use rotating basis vectors so that the motion vector of the point of interest can bedescribed with the minimum change in its direction. Clearly, it is much simpler to calculate thetime derivative of vectors if their directions do not change with respect to time.

Define now two frames of reference: LetXYZbe an inertial frame of reference with fixed unitvectorsI, J andK and letxyzbe a moving frame with originB translating with respect to the

fixed originO and rotating with an angular velocity of~Ω and an angular acceleration of~Ω. For

consistency, the vector symbols~Ω and~Ω will be used mostly to describe the rotation of coordinatesystems while the vector symbols~ω and~α will be used primarily to represent the rotation a rigidbody.

TheXYZsystem is necessary because Newton’s Second Law is applicable only for absolutequantities while thexyzsystem is important because it allows the problem to be greatly simplified.At any instant of time, there is a transformation between the two coordinate systems.

Shown in the figure are theposition vectors~rA, ~rB and~rA/B . ~rA

and~rB are absolute position vectorsof points A and B, respectively, somathematically it is most convenientto express them using the fixed basisvectors I, J and K. The relativeposition vector~rA/B , however, isbest described using the rotating basisvectors ı, and k because it is aquantity local to the rotating system.

X

Y

Z

x

y

z~Ω

OB

A

~rB

~r A

~rA/B

From the above definitions of the coordinate systems, it is clear that time derivatives of vectorsusing the fixed system would be much simpler than those of the rotating system because

dI

dt= ~0 ,

dJ

dt= ~0 and

dK

dt= ~0

whiledı

dt= ~Ω× ı ,

d

dt= ~Ω× and

dk

dt= ~Ω× k .

Consider now the representation of an arbitrary vector~Q using the two systems:

Fixed Frame

Let

~Q = QFX I + QF

Y J + QFZK =

QX

QY

QZ

F

XYZ

in whichQF

X = ~Q · I , QFY = ~Q · J and QF

Z = ~Q · Kare scalar functions.

– 6 / 5 –

Rotating Frame

Let

~Q = QRx ı + QR

y + QRz k =

Qx

Qy

Qz

R

xyz

in whichQR

x = ~Q · ı , QRy = ~Q · and QR

z = ~Q · kare also scalar functions but defined with respect to a set of rotating basis vectors. Note

QX

QY

QZ

F

XYZ

and

Qx

Qy

Qz

R

xyz

are matrices but~Q is an abstract vector. Hence,d~Q/dt implied simply the time derivative of~Q,it does not imply which coordinate system is being used until the matrix is defined based on aparticular set of basis vectors.

Derivative of ~Q in the Fixed Frame

d~Q

dt= QF

X I + QFY J + QF

ZK =

QX

QY

QZ

F

XYZ

Derivative of ~Q in the Rotating Frame

d~Q

dt= QR

x ı + QRy + QR

z k + QRx

dı

dt+ QR

y

d

dt+ QR

z

dk

dt,

therefore

d~Q

dt=

Qx

Qy

Qz

R

xyz

+ ~Ω×

Qx

Qy

Qz

R

xyz

.

in which ~Ω is the angular velocity of the rotating frameR with respect to the fixed frameF .

A second term is needed for the time derivative with respect to the rotating frame because thescalar functionsQx = ~Q · ı, Qy = ~Q · andQz = ~Q · k are obtained using a set of time-dependentbasis vectors. Therefore, the first term of the matrix expression is like a partial time derivative“assuming” the basis vectors are fixed and the second term is a correction to account for the rotationof ı, andk.

– 6 / 6 –

Application – Velocity and Acceleration in Cylindrical Coordinates

The concept of a rotating frame has been applied frequently in early studies using the cylindricalcoordinate system. The unit vectorser andeθ follows the motion of the position vector which isdefined as

~r =

r0z

rθz

,

while the third unit vectorez remains unchanged in thez-direction. The angular velocity~Ω of thecylindrical coordinate system can be represented in matrix form as

~Ω = θez =

00θ

rθz

.

With the definitions of~r and~Ω, the velocity vector can be obtained as the “total” time derivative of~r as

~v =d~r

dt=

dc~r

dt+ ~Ω× ~r =

r0z

rθz

+

00θ

rθz

×

r0z

rθz

,

in which dc~r/dt is the component-by-component time derivative taken by assuming that the unitvectors are fixed. By combining the change in magnitude (first term) and the change in direction(second term), the velocity vector can be written in the cylindrical coordinate system as:

~v =

rrθz

rθz

The acceleration vector can be obtained by yet another derivative

~a =d~v

dt=

dc~v

dt+ ~Ω× ~v

=

rrθ + rθ

z

rθz

+

00θ

rθz

×

rrθz

rθz

=

rrθ + rθ

z

rθz

+

−rθ2

rθ0

rθz

or

~a =

r − rθ2

rθ + 2rθz

rθz

– 6 / 7 –

Application – Velocity and Acceleration in Spherical Coordinates

Let the position vector~r be defined as

~r =

r00

rφθ

in the spherical coordinate system. The representation is very simple because the unit vectorerfollows the position vector~r, therefore, the frame which specifies the spherical coordinate systemis rotating with the angular velocity~Ω defined as

~Ω = φeθ + θ(er cos φ− eφ sinφ) =

00φ

rφθ

+

θ cos φ−θ sinφ

0

rφθ

=

θ cos φ−θ sinφ

φ

rφθ

.

With these definitions, the velocity vector can be obtained as

~v =d~r

dt=

dc~r

dt+ ~Ω× ~r =

r00

rφθ

+

θ cos φ−θ sinφ

φ

rφθ

×

r00

rφθ

or

~v =

rrφ

rθ sinφ

rφθ

The acceleration vector can now be obtained by yet another derivative

~a =d~v

dt=

dc~v

dt+ ~Ω× ~v

=

rrφ + rφ

rθ sinφ + rθ sinφ + rθφ cos φ

rφθ

+

θ cos φ−θ sinφ

φ

rφθ

×

rrφ

rθ sinφ

rφθ

=

rrφ + rφ

rθ sinφ + rθ sinφ + rθφ cos φ

rφθ

+

−rθ2 sin2 φ− rφ2

rφ− rθ2 sinφ cos φrθφ cos φ + rθ sinφ

rφθ

or

~a =

r − r(φ2 + θ2 sin2 φ)rφ + 2rφ− rθ2 sinφ cos φ

rθ sinφ + 2rθ sinφ + 2rθφ cos φ

rφθ

– 6 / 8 –

The angular velocity~Ω of the rotating frame shown in the previous page is quite complicatedbecause there were two rotational degrees of freedomφ and θ. It will be shown here that with amuch simplified rotation of the frame the same results can be obtained, but the position vector~rwould have a much more difficult representation.

xθ

y

z

φ r

R

eφ

er

eR

ez

φ

φ

Apply now the cylindrical coordinate systemRθz. The angular velocity of the frame~Ω is simplyθez. The matrix representations of~r and~Ω are therefore

~r =

r sinφ0

r cos φ

Rθz

and ~Ω =

00θ

Rθz

,

respectively. To obtain the velocity~v using the new rotating frame, apply the formula

~v =d~r

dt=

dc~r

dt+ ~Ω× ~r

=

r sinφ + r cos φφ0

r cos φ− r sinφφ

+

00θ

×

r sinφ0

r cos φ

=

r sinφ + r cos φφrθ sinφ

r cos φ− r sinφφ

Rθz

.

Develop now a transformation between the cylindrical and spherical coordinate systems byconsulting the above figure to yield

qr

qφ

qθ

=

cos(π

2 − φ) 0 cos φcos φ 0 cos(π

2 + φ)0 1 0

qR

qθ

qz

=

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

qR

qθ

qz

.

Now transform the vector~v from theRθz to therφθ system by performing the matrix product

~v =

sinφ 0 cos φ

cos φ 0 − sinφ0 1 0

r sinφ + r cos φφrθ sinφ

r cos φ− r sinφφ

Rθz

=

r sin2 φ + r cos φ sinφφ + r cos2 φ− r sinφ cos φφr sinφ cos φ + r cos2 φφ− r cos φ sinφ + r sin2 φφ

rθ sinφ

rφθ

=

rrφ

rθ sinφ

rφθ

.

Similar results can be obtained for the acceleration vector, but clearly the amount of work neededto describe the vector~v and its time derivative is too tedious. It is perhaps better to use a framewhich is more suitable to the task (like the one shown in the previous page) even though the angularvelocity ~Ω may be more complicated.

– 6 / 9 –

Moment and Angular Momentum of a Rigid Body

Recall for a system of particles, the alternate form of Newton’s 2nd Law, that of moment andangular momentum, can be written about the originO as

∑~MO =

d

dt~HO , (1)

where the total moment about pointO for all external applied forces is defined as

∑~MO =

∑j

~rj × ~Fj (2)

and the total angular momentum about pointO for N mass particles is defined as

~HO =N∑

i=1

~ri ×mi~vi . (3)

For a rigid body, the additional constraint on the velocities is such that

~vi = ~vA + ~ω × ~ri/A , (4)

in whichA is a point on the rigid body where the velocity is known. Since Newton’s Second Lawfor translations utilizes the velocity and acceleration of the center of gravityG, it is convenient tosimplify equation (3) by substituting pointG as pointA in equation (4), i.e.,

~vi = ~vG + ~ω × ~ri/G = ~vG + ~ω × ~ρi , (5)

in which~ri/G = ~ri − ~RG = ~ρi .

To obtain~ri for equation (3), rewrite the above equation as

~ri = ~RG + ~ρi (6)

and then substitute equations (5) and (6) into equation (3) to yield

~HO =N∑

i=1

(~RG + ~ρi)×mi(~vG + ~ω × ~ρi)

=N∑

i=1

~RG ×mi~vG +N∑

i=1

~ρi ×mi~vG

+N∑

i=1

~RG ×mi(~ω × ~ρi) +N∑

i=1

~ρi ×mi(~ω × ~ρi)

= ~RG ×(

N∑i=1

mi

)~vG +

(N∑

i=1

mi~ρi

)× ~vG

+ ~RG × ~ω ×(

N∑i=1

mi~ρi

)+

N∑i=1

mi (~ρi × ~ω × ~ρi) .

– 7 / 2 –

Using again the definition of the center of gravity, i.e.,∑

mi~ρi = ~0, and letM be the total mass∑mi, we can write the angular momentum about the originO as

~HO = ~RG ×M~vG +N∑

i=1

mi(~ρi × ~ω × ~ρi) . (7)

The above expression is written inabstract algebraform, which implies it is applicable for anycoordinate system.

The Moment of Inertia Matrix of a Rigid Body

To make equation (7) easier to apply numerically, the second term on the right-hand-side,

N∑i=1

mi(~ρi × ~ω × ~ρi) ,

can be expressed as the matrix product[IG]~ω, thus separating completely the geometricalinformation contained in~ρ from the kinematic information contained in~ω. Otherwise, the entiresummation has to be recalculated every instant the angular velocity~ω of the body changes.

In vector algebra, there is an identity for any arbitrary vectors~A, ~B and ~C such that

( ~A× ~B)× ~C = ( ~A · ~C) ~B − ~A( ~B · ~C) . (8)

In our application, we shall let~A and ~C represent~ρi and ~B represents~ω, then

~ρi × ~ω × ~ρi = (~ρi · ~ρi)~ω − ~ρi(~ρi · ~ω) . (9)

In matrix form, equation (9) can be written as

~ρi × ~ω × ~ρi =(ρiT ρi

) ω − ρi(ρiT ω

)(10)

or by altering the order of matrix multiplication

~ρi × ~ω × ~ρi =(ρiT ρi

) ω − (ρiρiT) ω

=(ρiT ρi[1]− ρiρiT

) ω (11)

where[1] is the3× 3 identity matrix.

Using cartesian coordinates, let

ρi =

ρxi

ρyi

ρzi

and ~ω =

ωx

ωy

ωz

then

ρiT ρi = |~ρi|2 = [ ρxi ρyi ρzi ]

ρxi

ρyi

ρzi

=

(ρ2

xi + ρ2yi + ρ2

zi

),

ρiρiT =

ρxi

ρyi

ρzi

[ ρxi ρyi ρzi ] =

ρ2

xi ρxiρyi ρxiρzi

ρyiρxi ρ2yi ρyiρzi

ρziρxi ρziρyi ρ2zi

,

– 7 / 3 –

and

~ρi × ~ω × ~ρi =

(ρ2

xi + ρ2yi + ρ2

zi)

1 0 0

0 1 00 0 1

−

ρ2

xi ρxiρyi ρxiρzi

ρyiρxi ρ2yi ρyiρzi

ρziρxi ρziρyi ρ2zi

ωx

ωy

ωz

(12)

Performing now the summation overi, the moment of inertia matrix[IG] can be expressed as

[IG]~ω =N∑

i=1

mi(~ρi × ~ω × ~ρi)

=

N∑

i=1

mi

ρ2

yi + ρ2zi −ρxiρyi −ρxiρzi

−ρyiρxi ρ2xi + ρ2

zi −ρyiρzi

−ρziρxi −ρziρyi ρ2xi + ρ2

yi

~ω

(13)

Convenient Points to Formulate the Rotational Equation of Motion

As shown in equation (7), the angular momentum~HO is dependent of both the linearmomentum term,M~vG, and the angular momentum term[IG]~ω. Therefore, the linear momentumequation and the angular momentum equation is normally coupled. There are two special cases,however, where the two equations of motion are uncoupled.

Case 1: OriginO is the Center of Gravity G

If the center of gravityG is the originO, i.e.,~rG = ~rO, then the position vector

~RG = ~rG/O = ~rG − ~rO = ~0 .

It is clear that the first term of equation (7) vanishes and

~HO = ~HG = [IG]~ω .

Case 2: OriginO is a Fixed Point

If the originO is a fixed point, then~vO = ~0 and

~vG = ~vO + ~ω × ~rG/O = ~ω × ~RG . (14)

Substituting equation (14) into equation (7), the angular momentum~HO can be expressed as

~HO = M ~RG × ~ω × ~RG +N∑

i=1

mi(~ρi × ~ω × ~ρi) . (15)

Judging from the similarity of the two terms on the right-hand-side, we can write

~HO = [IO]~ω (16)

– 7 / 4 –

where[IO] = [IT ] + [IG] with [IG] defined by equation (13) and[IT ] defined by

[IT ]~ω = M(~RG × ~ω × ~RG)

= M

R2

y + R2z −RxRy −RxRz

−RyRx R2x + R2

z −RyRz

−RzRx −RzRy R2x + R2

y

~ω .

(17)

Equation (17) is the “parallel-axis theorem” in three dimensions. The valuesRx, Ry andRz arethex, y andz components of the vector~RG, respectively. Once the moment of inertia matrix[IG]is known at the center of gravityG, then it can be translated elsewhere by first calculating and thenadding the translation matrix[IT ].

If for some reason the moment of inertia matrix[IA] is known at pointA and it is desirable toobtain the moment of inertia matrix[IB ] at pointB. Then the steps which must be taken are

(1) calculate[IT1] using the position vector~rA/G,

(2) determine[IG] = [IA]− [IT1],(3) calculate[IT2] using the position vector~rB/G, and

(4) obtain[IB ] = [IG] + [IT2].It is important that one of the points used during the translation of axes be the pointG, it is incorrectto calculate[IB ] = [IA] + [IT ] in which [IT ] is determined using position vector~rB/A.

The Moment of Inertia Matrix of a System of Rigid Bodies

A complex machinery is often composed of several components which can be considered rigidbodies. Unlike particles, which are concentrated masses, each rigid-body component has its ownmoment of inertia matrix.

To calculate the total moment of inertia matrix for a system ofN rigid bodies, assume themoment of inertia matrix at the center of mass of each rigid body[IGi] is already known, then thetotal moment of inertia matrix of the system about pointO is

[IO] =N∑

i=1

([IGi] + [ITi]) ,

in which[ITi] is the translation matrix calculated using the position vector~RGi of thei-th rigid-bodycomponent.

If the components have simple geometries such as a slender rod, a thin plate, a cylinder or asphere, then their moment of inertia matrices about their centers of mass can be formed by simplyconsulting a reference table. The moment of inertia matrix for a symmetric body is always diagonal.

– 7 / 5 –

Rotational Transformation of Inertia Matrices

If the moment of inertia matrix[IO]xyz is known in thexyz coordinatesystem, it is sometimes important to haveit expressed in theabccoordinate system.

Consider first the definition of[IO]xyz,

HOxyz = [IO]xyzωxyz (1)

in which HOxyz and ωxyz are theangular momentum and the angular ve-locity vectors expressed in thexyzsystem,respectively.

x

y

z

a

b

c

Now if we define a coordinate transformation matrix[Q] such that

qabc = [Q] qxyz (2)

then for our vectors of interest,HO andω, we have

HOabc = [Q] HOxyz , (3)

andωabc = [Q] ωxyz (4)

orωxyz = [Q]T ωabc . (5)

Substitution of equations (1) and (5) into equation (3) yields

HOabc = [Q] HOxyz = [Q][IO]xyzωxyz = [Q][IO]xyz[Q]T ωabc ,

or the equationHOabc = [Q][IO]xyz[Q]T ωabc . (6)

Define now the new relationship

HOabc = [IO]abcωabc . (7)

It is clear by matching equations (7) and (6) that

[IO]abc = [Q][IO]xyz[Q]T . (8)

In equation (8), the transformation[Q] must be defined as shown in equation (2), i.e., fromxyzto abc. It is interesting to point out that two matrix multiplications are required to transform[IO]because it is a tensor of second rank.

– 7 / 6 –

Example – Calculation of the Moment of Inertia Matrix

Find the moment of inertia matrices[IO] and [IG]for the object represented by 3 concentrated masseswith massless rods. The masses and their respectiveposition vectors are:m1 = 4 kg, m2 = 5 kg, m3 =3 kg and~r1 = 4, 1,−2T , ~r2 = −3, 0, 3T , ~r3 =2, 1, 1T .

x y

z

OG

Solution:

Since the moment of inertia matrices of concentrated masses are zero, the moment of inertia of thesystem can be calculated by simply adding the translation matrices using the parallel-axis theorem.

[IO]

= 4

21[1]−

16 4 −8

4 1 −2−8 −2 4

+ 5

18[1]−

9 0 −9

0 0 0−9 0 9

+ 3

6[1]−

4 2 2

2 1 12 1 1

= 4

5 −4 8−4 20 28 2 17

+ 5

9 0 9

0 18 09 0 9

+ 3

2 −2 −2−2 5 −1−2 −1 5

=

20 −16 32−16 80 832 8 68

+

45 0 45

0 90 045 0 45

+

6 −6 −6−6 15 −3−6 −3 15

=

71 −22 71−22 185 571 5 128

To find the location of the center of mass for the system of 3 particles, use the definition as

(4 + 5 + 3)~RG = 4

41−2

+ 5

−303

+ 3

211

=

7710

=⇒ ~RG =

0.5830.5830.833

To find [IG] for the system of 3 concentrated masses, find the translation matrix[IT ] for the entireassembly of 3 particles from pointG to pointO using the position vector~RG:

[IT ] = 12

(0.5832 + 0.5832 + 0.8332)

1 0 0

0 1 00 0 1

−

0.34 0.34 0.49

0.34 0.34 0.490.49 0.49 0.69

= 12

1.034 −0.34 −0.49−0.34 1.034 −0.49−0.49 −0.49 0.684

=

12.4 −4.1 −5.9−4.1 12.4 −5.9−5.9 −5.9 8.21

.

Now, [IO] = [IG] + [IT ] implies [IG] = [IO]− [IT ] , or

[IG] =

58.6 −17.9 76.9−17.9 172.6 10.976.9 10.9 119.8

– 7 / 7 –

Example – Moment of Inertia of a System of Rigid Bodies

An out-of-this-world object is composed of (i) a solidsphere of mass 20 kg and radius 1.5 m placed on topof (ii) a 1-meter tall massless column with (iii) a thinring of 5 kg mass welded on the sphere’s surface atan angle of30 measured with respect to the equator.Find the3 × 3 moment of inertia matrix[IO] aboutpoint O. The radius of the ring is approximately 1.5m. 1 m

O

y

30o

x

1.5m

SOLUTION:

For the sphere, the moment of inertia matrix about its center of mass is

[IG1] =25mR2

1 0 0

0 1 00 0 1

=

25(20)(1.5)2

1 0 0

0 1 00 0 1

=

18 0 0

0 18 00 0 18

xyz

For the thin ring, use systemabc so that

[IG2]abc = mR2

0.5 0 0

0 1 00 0 0.5

=

5.625 0 0

0 11.25 00 0 5.625

abc

y

x

b

a

30o

30o

Define now the transformation fromabc to xyz as

qx

qy

qz

=

cos 30 cos 60 0

cos 120 cos 30 00 0 1

qa

qb

qc

=

0.866 0.5 0−0.5 0.866 0

0 0 1

qa

qb

qc

= [Q]

qa

qb

qc

[IG2]xyz = [Q]

5.625 0 0

0 11.25 00 0 5.625

abc

0.866 −0.5 0

0.5 0.866 00 0 1

=

0.866 0.5 0−0.5 0.866 0

0 0 1

4.87 −2.81 0

5.63 9.74 00 0 5.63

=

7.03 2.44 0

2.44 9.84 00 0 5.63

xyz

Since~RG1 = ~RG2 = 0 , 2.5 , 0T

[IT1] + [IT2] = (20 + 5)

6.25

1 0 0

0 1 00 0 1

−

0 0 0

0 6.25 00 0 0

=

156.25 0 0

0 0 00 0 156.25

[IO] = [IG1] + [IG2] + [IT1] + [IT2] =

181.28 2.44 0

2.44 27.84 00 0 179.88

xyz

– 7 / 8 –

Example – Computation of the Moment of Inertia Matrix

Find the moment of inertia matrices[IO] for the object represented by a slender rod (m1 = 10 kg)and 2 slanted thin rectangular plates (m2 = m3 = 5 kg). The thin plates have the dimensions of4 m×2 m and they are mounted onto the slender rod at an angle of30 below the horizontal axis.(Note: theabcanddefcoordinate systems are specified for intermediate calculations only.)

XY

Z

f

d

ea b

c

O

Z

2 mf

d4 m

30o 30o

c

b

OY

G3

G1

G2

Moment of Inertia Matrices About Respective Centroids:

[IG1] =

1

12 (10)(6)2 0 00 1

12 (10)(6)2 00 0 0

=

30 0 0

0 30 00 0 0

XYZ

[IG2] =

1

12 (5)(4)2 0 00 1

12 (5)(2)2 00 0 1

12 (5)(42 + 22)

abc

=

6.67 0 0

0 1.67 00 0 8.33

abc

[IG3] =

1

12 (5)(2)2 0 00 1

12 (5)(4)2 00 0 1

12 (5)(22 + 42)

def

=

1.67 0 0

0 6.67 00 0 8.33

def

Rotation Matrices

qX

qY

qZ

= [Q1]

qa

qb

qc

=

1 0 0

0 cos 30 cos 60

0 cos 120 cos 30

qa

qb

qc

qX

qY

qZ

= [Q2]

qd

qe

qf

=

0 1 0

cos 210 0 cos 120

cos 120 0 cos 30

qd

qe

qf

– 7 / 9 –

Rotational Transformation for the Rectangular Plates

[IG2]XYZ = [Q1][IG2]abc[Q1]T = [Q1]

6.67 0 0

0 1.67 00 0 8.33

abc

1 0 0

0 0.866 −0.50 0.5 0.866

=

1 0 0

0 0.866 0.50 −0.5 0.866

6.67 0 0

0 1.446 −0.8350 4.165 7.214

=

6.67 0 0

0 3.335 2.8840 2.884 6.665

[IG3]XYZ = [Q2][IG3]def[Q2]T = [Q2]

1.67 0 0

0 6.67 00 0 8.33

def

0 −0.866 −0.5

1 0 00 −0.5 0.866

=

0 1 0−0.866 0 −0.5−0.5 0 0.866

0 −1.446 −0.835

6.67 0 00 −4.165 7.214

=

6.67 0 0

0 3.335 −2.8840 −2.884 6.665

By combining the three inertia matrices, we have

3∑i=1

[IGi]XYZ =

43.34 0 0

0 36.67 00 0 13.33

XYZ

.

Translational Transformations for All Three Elements

~RG1 =

003

=⇒ [IT1] = (10)

9 0 0

0 9 00 0 9

−

0 0 0

0 0 00 0 9

=

90 0 0

0 90 00 0 0

~RG2 =

02 cos 30

4− 2 sin 30

=

01.732

3

=⇒

[IT2] = (5)

12 0 0

0 12 00 0 12

−

0 0 0

0 3 5.1960 5.196 9

=

60 0 0

0 45 −25.980 −25.98 15

~RG3 =

0−2 cos 30

4− 2 sin 30

=

0−1.732

3

=⇒

[IT3] = (5)

12 0 0

0 12 00 0 12

−

0 0 0

0 3 −5.1960 −5.196 9

=

60 0 0

0 45 25.980 25.98 15

By combining the three translation matrices (parallel-axis theorem), we have

3∑i=1

[ITi]XYZ =

210 0 0

0 180 00 0 30

XYZ

,

and the total inertia matrix[IO] can be computed as

[IO] =3∑

i=1

[IGi]XYZ+3∑

i=1

[ITi]XYZ =

253.34 0 0

0 216.67 00 0 43.33

XYZ

.

– 7 / 10 –

Moments of Inertia of Arbitrary Shape Objects

Two-Dimensional Solid Objects

Most two-dimensional objects canbe approximated by assembling a set oftriangular elements. The useful propertiesof a triangular element can be calculatedby first defining the three vertices ofthe triangle as~r1 = xk, ykT , ~r2 =xl, ylT and ~r3 = xm, ymT . Thecenter of gravity of the element can thenbe computed as

~Rg =13

(~r1 + ~r2 + ~r3) .

Its area, A, can be calculated as adeterminant of a matrix as

A =∫ ∫

dx dy =12

∣∣∣∣∣∣1 xk yk

1 xl yl

1 xm ym

∣∣∣∣∣∣ .

If a constant area density,γ, is used, themass of the triangle ism = γA.

Define now the position vectors withrespect to the center of mass as

~ρi = ~ri − ~Rg , i = 1, 2, 3;

y

x

g(xl,yl)

(xk,yk)O

x

y

2

3

4

5

g3

O

g4

g5

g1

1

G

g2

(xm,ym)

then the element’s moment of inertia matrix about its center of gravity is

[Ig] =m

12

3∑i=1

(~ρT

i ~ρi[1]− ~ρi~ρTi

)=

m

12

3∑i=1

[ρ2

iy −ρixρiy

−ρiyρix ρ2ix

].

Using the parallel axis theorem, the moment of inertia matrix,[IO], can be calculated as,[IO] =[IG] + [IT ], in which the translation matrix,[IT ], is defined as

[IT ] = m(

~RTg

~Rg[1]− ~Rg~RT

g

).

Assembling the Elements

The area of the object is first partitioned intoM triangular subareas and the nodes are numberedand theirx andy coordinates specified. The numbers used for the nodes are called global nodenumbers while the numbers of the three vertices, 1, 2 and 3, are called local node numbers.The way an element contributes to the entire object is established by three one-to-one pointersbetween the local and global nodes. For example, element 1 has nodes 1,2,6; element 2 hasnodes 2,3,6; and element 5 has nodes 5,1,6; etc. Using these nodal coordinates, the momentof inertia matrix of the trianglar elements,[Igi], can be calculated and translated to the origin,

– 7 / 11 –

O, using the position vectors,~Rgi. Therefore, the total moment of inertia and center of gravity forthe object are,

[IO] =M∑i=1

[Igi] + [ITi] and ~RG =

(M∑i=1

mi~Rgi

)/(M∑i=1

mi

),

respectively. Using the newly calculated global center of gravity,G, the moment of inertia matrixof the object can be determined atG by applying the parallel axis theorem as

[IG] = [IO]− [IT ] ,

in which

[IT ] =

(M∑i=1

mi

)(~RT

G~RG[1]− ~RG

~RTG

).

Three-Dimensional Solid Objects

Most three-dimensional objects canbe approximated by assembling a setof tetrahedron elements. The usefulproperties of a tetrahedron element canbe calculated by first defining the fourvertices of the tetrahedron as~r1 =xk, ykT , ~r2 = xl, ylT , ~r3 =xm, ymT and ~r4 = xn, ynT . Thecenter of gravity of the element can thenbe computed as

~Rg =14

(~r1 + ~r2 + ~r3 + ~r4) .

Its volume, V , can be calculated as adeterminant of a matrix as

V =∫

dV =16

∣∣∣∣∣∣∣1 xk yk zk

1 xl yl zl

1 xm ym zm

1 xn yn zn

∣∣∣∣∣∣∣ .(xk,yk)

x

y

z

(xm,ym)

g (xl,yl)(xn,yn)

O

If a uniform mass density,γ, is used, the mass of the tetrahedron ism = γV .

Define now the position vectors with respect to the center of mass as

~ρi = ~ri − ~Rg , i = 1, 2, 3, 4;

then the element’s moment of inertia matrix about its center of gravity is

[Ig] =m

20

4∑i=1

(~ρT

i ~ρi[1]− ~ρi~ρTi

)=

m

20

4∑i=1

ρ2

iy + ρ2iz −ρixρiy −ρixρiz

−ρiyρix ρ2ix + ρ2

iz −ρiyρiz

−ρizρix −ρizρiy ρ2ix + ρ2

iy

.

Using the parallel axis theorem, the moment of inertia matrix,[IO], can be calculated as,[IO] =[IG] + [IT ], in which the translation matrix,[IT ], is defined as

[IT ] = m(

~RTg

~Rg[1]− ~Rg~RT

g

).

– 7 / 12 –

Moments of Inertia of Thin ShellsThree-Dimensional Thin Shells

Most three-dimensional thin shellscan be approximated by assembling aset of two-dimensional triangular shellelements. Unlike the two-dimensional tri-angular solid elements, the shell elementvertices havex, y andz coordinates. Theuseful properties of a triangular elementcan be calculated by first defining thethree vertices of the triangle as~r1 =xk, yk, zkT , ~r2 = xl, yl, zlT and~r3 = xm, ym, zmT . The centerof gravity of the element, in thexyzcoordinate system, can then be computedas

~Rg =13

(~r1 + ~r2 + ~r3) .

To calculate the element’s area and it’smoments of inertia, it is necessary todefine a local coordinate system,abc, sothat the surface of the triangle lies on theab-plane with thec-axis perpendicular toit. First define the unit vector,ea, as

ea =(~r2 − ~r1)|~r2 − ~r1| ,

z

y

xa

b

c

x

y

z

1

2

3

4

5

6

7

8

(xk,yk,zk)(xm,ym,zm)

(xl,yl,zl)

then the unit vector,ec, as

ec =(~r2 − ~r1)× (~r3 − ~r1)|(~r2 − ~r1)× (~r3 − ~r1)| .

Because of the cross product,ec is perpendicular toea, therefore, the unit vector,eb can be obtainedas

eb = ec × ea .

Using these three unit vectors, a transformation matrix can be developed between the two coordinatesystems as follows:

[Q(abc← xyz)] =

eTa

eTb

eTc

.

Now transform the coordinates of the vertices from thexyz system to theabc system using thematrix products:

ak

bk

ck

= [Q]

xk

yk

zk

,

al

bl

cl

= [Q]

xl

yl

zl

,

am

bm

cm

= [Q]

xm

ym

zm

.

– 7 / 14 –

Becauseec is perpendicular to the surface, the ordinates,ck, cl andcm should be the same. Theelement’s area,A, can then be calculated as a determinant of a matrix as

A =∫ ∫

da db =12

∣∣∣∣∣∣1 ak bk

1 al bl

1 am bm

∣∣∣∣∣∣ .

If a constant area density,γ, is used, the mass of the triangle ism = γA.

Define now the position vectors with respect to the center of mass in theabc system as

~ρi = [Q](~ri − ~Rg) , i = 1, 2, 3;

then the element’s moment of inertia matrix about its center of gravity is

[Ig]abc =m

12

3∑i=1

(~ρT

i ~ρi[1]− ~ρi~ρTi

)=

m

12

3∑i=1

[ρ2

ib −ρiaρib

−ρibρia ρ2ia

].

After the transformation,[Ig]xyz = [Q]T [Ig]abc[Q] ,

the moment of inertia matrix,[IO], can be calculated as,[IO] = [Ig]xyz + [IT ], using the parallelaxis theorem. The translation matrix,[IT ], is defined as

[IT ] = m(

~RTg

~Rg[1]− ~Rg~RT

g

).

Assembling the Elements

The surface area of the thin shell is first partitioned intoM triangular subareas and the nodesare numbered and theirx, y andz coordinates specified. The numbers used for the nodes are calledglobal node numbers while the numbers of the three vertices, 1, 2 and 3, are called local nodenumbers. The way an element contributes to the entire object is established by three one-to-onepointers between the local and global nodes. For example, element 1 has nodes 1,7,8; element 2 hasnodes 1,2,7; and element 5 has nodes 3,4,6; etc. Using these nodal coordinates in thexyz system,a “custom made”abc coordinate system can be fitted to a particular triangular element so that themoment of inertia matrix,[Igi], can first be calculated in theabc system and then transformed to thexyz system. After[Igi] is obtained in thexyz system, it can be translated to the origin,O, usingthe parallel axis theorm and the position vector,~Rgi. Therefore, the total moment of inertia andcenter of gravity for the object are,

[IO] =M∑i=1

[Igi] + [ITi] and ~RG =

(M∑i=1

mi~Rgi

)/(M∑i=1

mi

),

respectively. Using the newly calculated global center of gravity,G, the moment of inertia matrixof the object can be determined atG by applying the parallel axis theorem as

[IG] = [IO]− [IT ] ,

in which

[IT ] =

(M∑i=1

mi

)(~RT

G~RG[1]− ~RG

~RTG

).

– 7 / 15 –

Example – Off-Balanced Rotating Object

Two spheres of mass 5 kg and radius 1.5 mare mounted on a massless rod as shown inthe figure. The position vectors ofm1 andm2 are respectively,~r1 = 0, 2, 3T and~r2 =0,−1,−3T . Find

(a) the moment of inertia matrix about pointO using thexyzsystem, and

(b) the moment vector that the spheres exerton the rod at pointO if the rod is rotatingwith an angular velocity ofω.

xm2

~ω

y

z

O

m1

Solution:

Part (a)

[IG] of both spheres 1 and 2 are

[IG]1 = [IG]2 =25mR2

1 0 0

0 1 00 0 1

=

25(5)(1.5)2

1 0 0

0 1 00 0 1

=

4.5 0 0

0 4.5 00 0 4.5

.

Now calculate[IT ] (parallel axis theorem) for both spheres to move origin to pointO.

[IT ] for sphere 1 using~RG1 = ~r1 = 0, 2, 3T is

[IT ]1 = 5

(02 + 22 + 32)

1 0 0

0 1 00 0 1

−

0 0 0

0 4 60 6 9

= 5

13 0 0

0 9 −60 −6 4

[IT ] for sphere 2 using~RG2 = ~r2 = 0,−1,−3T is

[IT ]2 = 5

(02 + (−1)2 + (−3)2)

1 0 0

0 1 00 0 1

−

0 0 0

0 1 30 3 9

= 5

10 0 0

0 9 −30 −3 1

The moment of inertia matrix[IO] using thexyzsystem is then

[IO] =([IG]1 + [IG]2

)+([IT ]1 + [IT ]2

)

=

9 0 0

0 9 00 0 9

+ 5

23 0 0

0 18 −90 −9 5

=

124 0 0

0 99 −450 −45 34

Part (b)

As the rod rotates with angular velocityω, it is convenient to attach systemxyzso that it rotateswith the masses, i.e., let the angular velocity of the coordinate system be~Ω = ω = ωJ .

Note: In this case~Ω is the same as the angular velocity of the body. For some other problems,however, the body may rotate differently from the coordinate system, so a distinction must be madebetween~Ω and~ω.

– 7 / 17 –

With thexyzsystem rotating with respect to systemXYZ, thetransformation matrix can be defined as

qX

qY

qZ

cos θ 0 sin θ

0 1 0− sin θ 0 cos θ

qx

qy

qz

= [Q]

qx

qy

qz

in whichθ = ωt.x

z Zθ

Xθ

Method 1 Use moment of inertia matrix[IO]XYZ:

[IO]XYZ = [Q][IO]xyz[Q]T = [Q]

124 0 0

0 99 −450 −45 34

cos θ 0 − sin θ

0 1 0sin θ 0 cos θ

=

cos θ 0 sin θ

0 1 0− sin θ 0 cos θ

124 cos θ 0 −124 sin θ−45 sin θ 99 −45 cos θ34 sin θ −45 34 cos θ

=

124 cos2 θ + 34 sin2 θ −45 sin θ −90 sin θ cos θ

−45 sin θ 99 −45 cos θ−90 sin θ cos θ −45 cos θ 124 sin2 θ + 34 cos2 θ

.

A moment of inertia matrix which is a function of timet asθ = ωt. Now, the angular momentumvector can be calculated as

HOX

HOY

HOZ

= [IO]XYZ

ωX

ωY

ωZ

= [IO]XYZ

0ω0

XYZ

=

−45ω sinωt

99ω−45ω cos ωt

XYZ

and the moment vector∑ ~MO about pointO is simply

∑

MOX

MOY

MOZ

=

d

dt

HOX

HOY

HOZ

=

−45ω2 cos ωt

045ω2 sinωt

XYZ

.

Method 2 Use moment of inertia matrix[IO]xyz. Calculate the angular momentum vector in thexyzsystem:

HOx

HOy

HOz

= [IO]xyz

ωx

ωy

ωz

=

124 0 0

0 99 −450 −45 34

0ω0

xyz

=

099ω−45ω

xyz

.

To calculate the moment vector, a time derivative must be taken with respect to a rotating basis:

MOx

MOy

MOz

=

dc

dt

HOx

HOy

HOz

+ ~Ω×

HOx

HOy

HOz

=

000

+

0ω0

×

099ω−45ω

=

−45ω2

00

xyz

and the same moment vector in theXYZsystem is

∑

MOX

MOY

MOZ

=

cos θ 0 sin θ

0 1 0− sin θ 0 cos θ

−45ω2

00

xyz

=

−45ω2 cos ωt

045ω2 sinωt

XYZ

.

– 7 / 18 –

Example – Badly Aligned Rectangular Plate

As the rod is rotating about theY -axis with an angularvelocity of~Ω = 4J rad/sec, the angleθ between thexand theX axes isθ = 4t. Thexyz coordinate systemis convenient for the calculation of the moment ofinertia matrix, i.e.,

[IO]xyz =112

(10)

22 0 0

0 22 + 12 00 0 12

=

3.33 0 0

0 4.17 00 0 0.83

while theXY Z system is needed to determine theabsolute bending moment.

34

5

y

Zz

X , x 1m2m

4 rad/secY

10 kg

The transformation matrix[Q(xyz ←− XY Z)] can be determined as a product between twosimpler transformation as:

[Q] =

cos 4t 0 sin 4t

0 1 0− sin 4t 0 cos 4t

1 0 0

0 0.8 −0.60 0.6 0.8

=

cos 4t 0.6 sin 4t 0.8 sin 4t

0 0.8 −0.6− sin 4t 0.6 cos 4t 0.8 cos 4t

Method 1 – Use fixed coordinate systemXYZ.

First we must evaluate the moment of inertia matrix[IO]XYZ as a function of time. This matrixfluctuates with time because the geometry is constantly changing.

[IO]XYZ = [Q][IO]xyz[Q]T

or

[IO]XYZ = [Q]

3.33 0 0

0 4.17 00 0 0.83

cos 4t 0 − sin 4t

0.6 sin 4t 0.8 0.6 cos 4t0.8 sin 4t −0.6 0.8 cos 4t

=

cos 4t 0.6 sin 4t 0.8 sin 4t

0 0.8 −0.6− sin 4t 0.6 cos 4t 0.8 cos 4t

3.33 cos 4t 0 −3.33 sin 4t

2.50 sin 4t 3.34 2.50 cos 4t0.67 sin 4t −0.50 0.67 cos 4t

=

3.33 cos2 4t + 2 sin2 4t 1.6 sin 4t −1.29 sin 4t cos 4t

1.6 sin 4t 3 1.6 cos 4t−1.29 sin 4t cos 4t 1.6 cos 4t 3.33 sin2 4t + 2 cos2 4t

.

– 7 / 19 –

Subsequently, the angular momentum in theXYZsystem can readily be calculated as

HOXYZ = [IO]XYZωXYZ =

−−− 1.6 sin 4t −−−−−− 3 −−−−−− 1.6 cos 4t −−−

040

=

6.4 sin 4t12

6.4 cos 4t

XYZ

.

The moment vector about pointO can now be obtained by differentiating the angular momentumvector as follows

∑MO

XYZ

=d

dtHOXYZ =

25.6 cos 4t0

−25.6 sin 4t

XYZ

.

It is clear that oscillating moments about theX andZ axes are caused by the mis-aligned platetrying to “right” itself. There is no torque in theY direction because the rod is rotating at a constantangular velocity and there is no friction considered.

Method 2 – Use rotating coordinate systemxyz.

In thexyzsystem,HOxyz = [IO]xyzωxyz

in which

ωxyz = [Q]T

040

XYZ

=

cos 4t 0 − sin 4t

0.6 sin 4t 0.8 0.6 cos 4t0.8 sin 4t −0.6 0.8 cos 4t

040

XYZ

=

03.2−2.4

xyz

.

Therefore

HOxyz =

3.33 0 0

0 4.17 00 0 0.83

03.2−2.4

xyz

=

013.33−2.0

xyz

.

Now to differentiate~HO with respect to a rotating coordinate system, use the formulation

MOxyz =dc

dt

013.33−2

+

03.2−2.4

×

013.33−2

=

000

+

25.600

=

25.600

xyz

The “righting” moment is about the rotatingx axis only. To obtain the results in theXYZsystem,use the transformation:

∑MO

= [Q]

25.600

xyz

=

cos 4t 0.6 sin 4t 0.8 sin 4t

0 0.8 −0.6− sin 4t 0.6 cos 4t 0.8 cos 4t

25.600

xyz

or ∑MO

=

25.6 cos 4t0

−25.6 sin 4t

XYZ

.

– 7 / 20 –