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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 –
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