VECTOR ANALYSIS (review)
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INTRODUCTION
A vector is a quantity that has both magnitude and direction. For example, if a quantity has a magnitude A and direction , then it may be represented as a vector A as A=A
A-1
and graphically it can be shown as A
O
Figure 1.1 A vector quantity Vectors obey the parallelogram law of addition, i.e. they cannot be added algebraically like scalars. Quantities such as displacement, velocity, acceleration, force, momentum, and etc are examples of a vector quantity.
Components of a vector. Any two or more vectors whose sum is equal to a certain vector V are said to be the components of that vector.
R
V
T
V
Q
S
(a) V = R + Q
(b) V = T + S Figure 1.2 Vector components
In Figure 1.2(a), R and Q are said to be components of V while T and S form components of V in Figure 1.2(b). Notice that the vector can have its components oriented in many directions. However, for most practical situations, the vector is usually resolved into rectangular components i.e. components which are mutually perpendicular to each other as shown in Figure 1.3.
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y V = Vx + Vy
Vy
x
Vx Figure 1.3 Rectangular components of vector in 2-D. Unit vector: A unit vector is a vector whose magnitude is equal to 1. Introducing a unit vector i along the x axis and j along the y axis where
i j =1 we write V = Vx + Vy as V = Vx i + Vy j
A-2
A-2
RESOLUTION OF A PLANAR VECTOR
Using Cartesian coordinate a vector A can be resolved into components x and y as shown in Figure 1.4. y V V y
x
O
Vx
Figure 1.4 Resolution of a vector using Cartesian coordinate where Vx and VY are the x and y components of a vector V, respectively. Introducing unit vectors i and j along the x and y axis, these components can be expressed as Vx = Vx i Vy = Vy j where Vx and Vy are the magnitudes or scalar components in the corresponding x and y direction and they can be determined as Vx = Vcos
and
Vy = Vsin
The magnitude and direction of V can then be determined by writing
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V = V =
Vx2 V y2
and
tan =
Vy
A-4, 5
Vx
Alternatively, we may also look upon the vector component as a projection of the vector along a particular direction. For example, the x component Vx is the projection of the vector V along the x axis while Vy is the projection of the vector V along the y axis. Note that Vx and Vy are called the rectangular components of the vector V along the x and y axis, respectively.
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RESOLUTION OF A SPATIAL VECTOR
A general case of a vector is a vector in space which is a 3-dimensional or spatial vector as depicted in Figure 1.5 using rectangular coordinate. y Vy
V
y x
z
Vx x
Vz
Vxz
z Figure 1.5 Rectangular components of a vector in 3-D.
V = V = V x2 V y2 V z2 Vy Vx V , cos y = , cos z = z V V V
cos x =
A-6
A-7
cos x, cos y, and cos z are known as direction cosines along the respective axis. Introducing unit vectors i , j and k along the x– , y– , and z– axis, respectively, the rectangular components of V can be expressed as V = Vx i + Vy j + Vz k where
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i = j = k = 1.
Denoting the component of V on the xz plane as Vxz, we can determine its magnitude by writing
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Vxz = V cos(90o – y)
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This component Vxz can be thought as the projection of a vector V onto the plane xz. This concept is particularly useful when we want to determine the projection of a force onto or along a certain direction. Introducing unit vectors i , j and k along the x, y, and z axis, these components can now be expressed as Vy = Vy j
Vx = Vx i
Vz = Vz k
where Vx, Vy, and Vz are the magnitudes or scalar components in the corresponding x, y, and z direction and they can be determined as follows. Vx = Vcosx
Vy = Vcosy
Vz = Vcosz
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Alternatively, we may also look upon the vector component as a projection of the vector along a particular direction. For example, the x component Vx is the projection of the vector V along the x axis while Vy is the projection of the vector V along the y axis.
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Expressing a position vector using its unit vector
Let P(x1, y1, z1) and Q(x2, y2, z2) be any two points in the rectangular coordinate system. Then the vector R directing along P to Q, may be written as R = Rx i + Ry j + Rz k = Ru
where
R=
and
u=
Rx2 Ry2 Rz2 =
PQ PQ
=
( x2 x1 )2 ( y2 y1 )2 ( z2 z1 )2
( x2 x1 )i ( y2 y1 ) j ( z2 z1 )k ( x2 x1 ) 2 ( y2 y1 ) 2 ( z2 z1 ) 2
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(A-11)
Dynamics and Vibrations
Review of Vector Analysis
Example 1-1: Consider two vectors A (directed along OP) and B (directed along DE) as shown in Figure E1-1. Express each of these vectors in its rectangular components. y A ; A = 221
D
P
10
6
10
O
x 8
18
z
9
B ; B = 489
E
Figure E1-1
Solution: We may write vector A as a combination of its magnitude (A = 221) and direction (uOP) as A = A uOP where uOP =
OP = OP
(18 0)i (10 0) j (8 0)k (18 0) (10 0) (8 0)
A = (221)
2
2
2
=
1 (18i + 10j + 8k) 22.1
1 (18i + 10j + 8k) = 180i + 100j + 80k 22.1
[Ans]
Similarly, for vector B , we may write it as a combination of its magnitude (B = 489) and direction (uDE) as B = B uDE where uDE =
[18 (10)]i (9 0) j [8 (6)]k DE 1 = = (28i – 9j + 14k) 2 2 2 DE 32 . 6 (28) (9) (14)
B = (489)
1 (28i – 9j + 14k) = 420i – 135j + 210k 32.6
[Ans]
Comment: To get a unit vector u, it is recommended here that you identify the coordinate of the corresponding points first, for example, P(18, 10, 8), D(–10, 0, –6) and E(18, –9, 8)
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A-5
Dynamics and Vibrations
REVIEW OF BASIC VECTOR ALGEBRA AND VECTOR CALCULUS
Given A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k 1. Addition and subtraction A ± B = (Ax ± Bx) i + (Ay ± By) j + (Az ± Bz) k
For example: Let A = 3i + 2j – 5k and B = – 2i + 4j + 3k then A + B = [3 + (–2)] i + (2 + 4) j + (–5 + 3) k = i + 6j – 2k A – B = [3 – (–2)] i + (2 – 4) j + (–5 – 3) k = 5i – 2j – 8k
2. Multiplication 2.1. Scalar multiplication of two vectors or dot product By definition, the dot product is defined as A • B = AB cos It follows that
i • j = (1)(1)cos 90o = 0 , and similarly, j • k = k • i = 0. i • i = (1)(1)cos 0o = 1 , and similarly, j • j = k • k = 1.
Therefore,
A • B = Ax Bx + Ay By + Az Bz
For example: Using the same vectors A = 3i + 2j – 5k and B = – 2i + 4j + 3k , then A • B = 3(–2) + 2(4) + (–5)(3) = – 17 Can you compute the angle formed between these two vectors A and B?
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2.2. Vector multiplication of two vectors or cross product By definition, the cross product is defined as A B = AB sin n where n is a unit vector perpendicular to the plane containing both A and B. It follows that
i j = (1)(1)sin 90o k = k , and similarly, j k = i and k i = j .
i i = (1)(1)sin 0o n = 0 , and similarly, j j = k k = 0. Therefore, A B = (Ax i + Ay j + Az k ) (Bx i + By j + Bz k ) = (Ax By) k + (– Ax Bz) j + (– Ay Bx) k + (Ay Bz) i + (Az Bx) j + (– Az By) i = (Ay Bz – Az By) i + (Az Bx – Ax Bz) j + (Ax By – Ay Bx) k
n
A plane containing both A and B
V = AB
B
A
For example: Using the same vectors A = 3i + 2j – 5k and B = – 2i + 4j + 3k , then A B = (3i + 2j – 5k) (– 2i + 4j + 3k) = [2(3) – (–5)(4)] i + [(–5)(–2) – 3(3)] j + [3(4) – 2(–2)] k = 26i + j + 16k
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3. Differentiation Rules Some useful vector calculus involving differentiation is given as follows. 1.
d d d (A ± B) = (A) ± (B) = A + B dt dt dt
2.
d d d (A • B) = (A)•(B) + (A)• (B) = A •B + A• B dt dt dt
3.
d d d (A B) = (A) (B) + (A) (B) = A B + A B dt dt dt
4. If A = Au , where A is the magnitude and u is the unit vector, then
d (A) = A u + A u dt Note that A = 0 if A is constant in magnitude and u = 0 if the direction of u remains unchanged (i.e. there is no change in its direction).
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Dynamics and Vibrations
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Review of Vector Analysis
OMEGA THEOREM
This theorem is used when we have a vector of constant magnitude changes its direction i.e. it rotates with an absolute angular velocity . Consider point P on a rigid body which rotates about a fixed axis passing through some fixed point O. The position of point P from point O is defined by a vector R. The velocity of point P can be expressed as
v = R
v = R = R
P A
(1)
This illustrates the fact that for a constant vector R whose direction is changing due to the rotation of the body, its derivative with respect to time is equal to the vector product of the angular velocity with which it rotates and that vector R itself. It represents the rate at which a vector of constant length changes its direction.
R O
It is important to note that
Eq(1) represents the rate at which a vector of constant length changes its direction when it is rotated.
The origin O is on the axis of rotation.
It is one of the most important and useful equation in the study of dynamics.
For example: Consider unit vectors i, j, and k which has a rotation . It follows that the rate of change of these unit vectors using the Omega theorem are, respectively, obtained as
di i = =i dt
z
y j
k A
i
x
j =
dj dt
=j
dk k = =k dt
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Dynamics and Vibrations
REFERENCE COORDINATE SYSTEMS
In the analysis of problems in dynamics, there are two reference coordinate systems commonly used: global coordinate system (or fixed reference axis) and local coordinate system (or moving reference axis). (a) Global Coordinate System (or Fixed Reference Axis) Z
OXYZ is chosen to be fixed in space i.e. the origin O and the orientation of each axis is fixed. It forms what is called as global coordinate system.
K
J
O
Unit vectors along the X-, Y-, and Z-axis are, respectively: I, J, K.
Y
I
All quantities measured in this coordinate system are absolute quantities.
X
Figure 1.9 Global Coordinate System
(b) Local Coordinate System (or Moving Reference Axis). There are two types of local coordinate system. (i) Oxyz is moving in space i.e. the orientation of each axis is changing, but its origin O is fixed. This type of axis is called a local coordinate system.
Z z
y j
k
Y O
i x
X
Figure 1.10(a) Local Coordinate System with origin at a fixed point O
Unit vectors along the x-, y-, and z-axis are, respectively: i, j, k. All quantities measured in this coordinate system are relative quantities and they must be converted to global axis to obtain the absolute quantities.
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z
Z
(ii) Axyz is moving in space i.e. the origin A and the orientation of each axis is changing with time.
y j
k A
x
i
Unit vectors along the x-, y-, and z-axis are, respectively: i, j, k.
Y
O X
Figure 1.10(b) Local Coordinate System with origin at a moving point A
All quantities measured in this coordinate system are relative quantities with respect to point A and to obtain the absolute quantities, we must determine the motion of point A relative to point O and later add them together.
Example: Resolve each of the local unit vectors shown in Fig. 1.11 into or using the global coordinate. Note that the z and Z axis are established using the right-hand rule. Y
Using the sine and cosine rules, we write
y
x i
j
X
O
i = cos I + sin J j = – sin I + cos J Note k = K since z is parallel to Z.
Figure 1.11
On the other hand, can you resolve the following global unit vectors into or using the local coordinate? Z
Using the sine and cosine rules, we write
z K
J
O
Y
J = cos j + sin k K = – sin j + cos k Similarly, I = ik since X is parallel to x.
y Figure 1.12
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A-8
Dynamics and Vibrations
RATE OF CHANGE OF A VECTOR WITH RESPECT TO A ROTATING FRAME Consider a vector R(t) as shown in Figure 1.13. There are two reference frames used to characterized this vector and its derivatives. OXYZ is a fixed (or inertia) reference frame with I, J, K as unit vectors. Oxyz is a moving (i.e. rotating) reference frame which rotates at the absolute angular velocity with respect to OXYZ frame and has unit vectors i , j and k . Z
R
z
y k j Y O
i x
X
Figure 1.13 Local Coordinate System with origin at fixed point O We observe that as time t varies, the magnitude and direction of R also change i.e. its length and orientation vary with time. Writing the vector R in the Oxyz frame we have R = Rx i + Ry j + Rz k
(A-11)
Its derivative with respect to this frame is then given by
d (R)Oxyz = ( R )Oxyz = R x i + R y j + R z k dt
(A-12)
which represents the rate of change of R with respect to the rotating frame Oxyz. The derivative of the vector R with respect to the inertial frame OXYZ can be obtained by differentiating the vector R as follows:
d d d d j + Rz k (R)OXYZ = ( R )OXYZ = R x i + R y j + R z k + Rx i + Ry dt dt dt dt
(A-13)
which represents the rate of change of R with respect to the fixed reference frame. Using Omega Theorem, we may write
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d i =x i dt
Review of Vector Analysis
d j = j dt
d k = k dt
It follows that d d d j + Rz k = Rx i + Ry j + Rz k = R Rx i + Ry dt dt dt Eq(A-11) then becomes ( R )OXYZ = ( R )Oxyz + R
(A-14)
Note that the use of eq(A-14) simplifies the solution for the rate of change of a vector R with respect to a fixed inertial reference frame OXYZ when R is defined by its components along the axes of a rotating frame Oxyz.
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(I)
Dynamics and Vibrations
Motion with Respect To A Fixed Point
Let’s now consider a vector R moving in space with an absolute angular velocity rad/s and absolute angular acceleration = Ω rad/s2. R = Ru
(1)
z
where R is the magnitude and u is the unit vector that shows the direction of R.
P R = Ru
u
Differentiating eq(1) with respect to time yields
R = R u + R u = R u + R u = R u + R u R = R u + R
but
y
O
u = u
x
Figure 1.14 (2)
Differentiating eq(2) with respect to time yields
u + R u + Ω R + R R = R u + R u + Ω R + ( R u + R) = R u + R u + Ω R + R u + R = R
R = R u + 2 R u + Ω R + ( R) R = R u + 2 R u + R + ( R)
(3)
Note that the unit vector u is generally expressed using a local coordinate i , j and k which may be later be transformed into a global coordinate. If R represents a position vector of a moving point P, then eq.(2) and (3) give the absolute velocity and absolute acceleration of point P, respectively.
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Review of Vector Analysis
Motion with Respect To A Moving Point
Consider a particle P moving in space indicated by the local coordinate Axyz. At the same time, this space is moving in global coordinate OXYZ with angular velocity .
Z
Position of P relative to A:
P
z r
A
r = xi + yj + zk
z
(1)
y
x
Velocity of P relative to A:
y
x O
Y
r = x i + y j + z k + (xi + yj + zk) r = ( x i + y j + z k ) + r
X
(2a)
Figure 1.15 Acceleration of P relative to A:
r = x i + y j + z k ) + ( x i + y j + z k ) + = ( x i + y j + z k ) + ( x i + y j + z k ) +
r + r r + [( x i + y j + z k) + r]
r = ( x i + y j + z k ) + 2 ( x i + y j + z k ) + r + ( r)
(3a)
Observe that if the magnitude of the vector r is constant and its position in the local coordinates Axyz is fixed, then the terms x , y , z , x , y , and z are all equal to zero!!! Denoting terms as follows:
( r )Axyz = x i + y j + z k ( r )Axyz = x i + y j + z k
Eq(3) can be expressed as ( r )OXYZ = ( r )Axyz + r
(2b)
( r )OXYZ = ( r )Axyz + 2 ( r )Axyz + r + ( r)
(3b)
Note that the derivative terms without a subscript are all the absolute quantities i.e. with respect to the OXYZ coordinate system. Also note that if point P is a fixed point in the Axyz coordinate system, then the terms ( x i + y j + z k) and ( x i + y j + z k) are all zero.
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