vehicle dynamics ch 2 & 3

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Chapter 2

Kinematics Fundamentals

2.1 Introduction

This chapter discusses the kinematics of motion, that is, looking at the nature of the motion withoutexamining the forces that cause the motion. We will focus on the two- and three-dimensional kine-matics of particles, as well as planar kinematics of rigid bodies. Three-dimensional kinematics of rigidbodies will be discussed in Chapter 9.

We begin this chapter by looking at coordinate systems and the kinematics of particles. Themotion of particles is purely translational. Then, rotating reference frames, angular velocity andangular acceleration are discussed. Relative velocity and acceleration equations and motion observedfrom a rotating coordinate system, are developed. Instant centers are introduced, as they are crucialto the analysis of vehicles and mechanisms.

Kinematic analysis serves two purposes: First and foremost, it is a precursor to kinetic analysis,a topic that will be discussed in the next chapter. We cannot analyze the kinetics of a system withoutfirst studying its kinematics. In addition, kinematic analysis by itself is a valuable tool and is widelyused in the design of mechanisms and vehicle suspensions, as well as in motion planning.

Chapter 3 discusses applications of kinematics to rolling, ground vehicles, and mechanisms.

2.2 Position, Velocity and Acceleration

When studying the kinematics of a particle, that is, the translational motion of a point, we needto describe its position, velocity and acceleration. The description must be made with respect toa reference point or origin using a coordinate system. A coordinate system or coordinate frame ischaracterized by a set of coordinate axes, the positive directions of these axes and unit vectors alongthese axes. Over the years, several types of coordinate systems have been developed. We selectthe coordinate system of frame that will make the analysis easier and more meaningful. Selecting a

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40 CHAPTER 2. KINEMATICS FUNDAMENTALS

coordinate frame is, in essence, selecting the set of motion variables.

Figure 2.1: A particle and its path.

Figure 2.1 depicts a particle that is moving and its position at time t. The position is measuredfrom a reference point that is fixed and the position vector is denoted by r (t). Consider next theposition of the same particle at time a time increment t later, at time t + t. The position vectorat t + t is r (t + t). The velocity v (t) and acceleration a (t) of the particle are defined as

v (t) =dr

dt= lim

t0

r

t= lim

t0

r (t + t) r (t)t

a (t) =dv

dt= lim

t0

v (t + t) v (t)t

(2.2.1)

The rate of change of acceleration is of interest in several applications. Examples includevehicle dynamics and human motion analysis. The commonly used terms for the rate of change ofacceleration are shock or jerk. For example, the occupants of a vehicle get shaken and thus experiencediscomfort if the acceleration profile undergoes a sudden change, such as when accelerating or brakingrapidly and when taking sharp turns. A vehicle and its components wear out sooner when they arerepeatedly subjected to sudden acceleration and sudden changes in acceleration.

2.3 Reference Frames: Single Rotation in a Plane

Coordinate systems are used in kinematics to observe motion. We decide on which coordinate systemto use by considering the nature of the motion. This section develops relationships between differentcoordinate systems and transformations from one coordinate system to another. Only right-handedcoordinate systems are considered here. When we point our right hand towards the positive directionof one of the axes (say x, with unit vector i) and rotate our fingers towards the positive direction ofthe second axis (say y, with unit vector j), the thumb points in the positive direction of the third axisz, with unit vector k = i j.

Consider plane motion and a planar coordinate system XY , as shown in Fig. 2.2. The unitvectors along the X and Y directions are I and J, respectively. Also shown in the same figure is a


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2.3. REFERENCE FRAMES: SINGLE ROTATION IN A PLANE 41

Figure 2.2: The XY and xy coordinate systems.

coordinate system xy (with unit vectors i and j) that is obtained by rotating the XY axes by an angle

in the counterclockwise direction. The rotation is about the Z axis (not shown here) perpendicularto the plane. Without loss of generality, we can use the same reference point for both coordinatesystems. The relationship between the unit vectors of the two coordinate systems is

i = cos I + sin J j = sin I + cos J

I = cos i sin j J = sin i + cos j (2.3.1)

and, the unit vector perpendicular to the plane of motion is common to both coordinate systems, sothat k = K. The above relationships can be expressed in matrix form as

ij

= cos sin

sin cos I

J I

J

= cos sin sin cos ij (2.3.2)

Define the rotation matrix [R] as

[R] =

cos sin

sin cos

(2.3.3)

The matrix [R] is unitary; its determinant is equal to 1 and its inverse is equal to its transpose,[R]1 = [R]T. The two relationships in Eq. (2.3.2) are inverse transformations.

Next, consider a point P on the plane (Fig. 2.3) and express the coordinates of point P as(XP, YP) in the XY coordinates and (xP, yP) in the xy frame, as shown in Fig. 2.3. The vector rP,which denotes the position of point P, can then be written in terms of the two coordinate systems as

rP = XPI + YPJ = xPi + yPj (2.3.4)

It is of interest to explore the relationship between the components of the two descriptions ofthe position vector rP. To this end, introduction of Eq. (2.3.2) to the above equation results in

rP = XPI + YPJ = XP (cos i

sin j) + YP (sin i + cos j)


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Figure 2.3: Representation of vector rP in XY and xy coordinates.

= (XP cos + YP sin ) i + (YP cos XP sin )j = xPi + yPj (2.3.5)from which we conclude that

xP = XP cos + YP sin yP = YP cos XP sin (2.3.6)

At this stage, we introduce the column vector representations of the unit vectors and of thelocation of point P in the two reference frames:

{xyz rP} = xPyP

XY ZrP = XPYP (2.3.7)

and express the transformation between the two column vectors asxPyP

= [R]

XPYP

or {xyz rP} = [R]

XY ZrP

(2.3.8)

The inverse relationship between the two vectors isXY ZrP

= [R]1{xyz rP} or

XY ZrP

= [R]T{xyz rP} (2.3.9)

The above relationships can be generalized to three dimensions by redefining the column vectorrepresentations and the rotation matrix as

{xyz rP} = xPyP

zP

XY ZrP =

XPYP

ZP

[R] =

cos sin 0 sin cos 0

0 0 1

(2.3.10)

We see that the same relationship that governs the position vectors is also valid for the unitvectors. Indeed, expressing the unit vectors in column vector format [i j k]T and [I J K]T we canwrite

i

jk

= [R]

IJK

IJK

= [R]T

i

jk

(2.3.11)


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2.4. COLUMN VECTOR REPRESENTATION 43

Example 2.1

The XY Zcoordinate system is rotated about the Zaxis by 30 clockwise to obtain the xyz coordinates.

If a vector in the xyz coordinates has the value r = 3i 4j, express r in terms of the XY Z frame.Using column vector notation, r is

{xyz r} =

34

[a]

The transformation angle is = 30, so that the matrix between the two coordinate systems is

[R] =

cos sin

sin cos

=

0.8660 0.50000.5000 0.8660

[b]

The second of Eq. (2.3.9) can be used to express r in the XY Z frame, with the result

XY Zr

= [R]T{xyz r} =

0.8660 0.50000.5000 0.8660

3

4

=

0.5981

4.9641

[c]

The results can be checked by noting that the magnitude of {xyz r} and XY Zr must be thesame. Therefore,

{xyz r}T{xyz r} = 32 + 42 = 25 XY ZrT XY Zr = 0.59812 + 4.96412 = 25.00001 [d]where the difference is due to roundoff error.

2.4 Column Vector Representation

This section discusses two notations to represent vectors. Consider a coordinate system with unitvectors e1, e2, and e3, which form a mutually orthogonal set. Also consider two vectors r and qdefined as

r = r1e1 + r2e2 + r3e3 q = q1e1 + q2e2 + q3e3 (2.4.1)

A set of vectors described this way is referred as geometric vectors or spatial vectors. The dot andcross products of these vectors yield

r q = r1q1 + r2q2 + r3q3

r q = (r2q3 r3q2) e1 + (r3q1 r1q3) e2 + (r1q2 r2q1) e3 (2.4.2)The previous section demonstrated that the vectors r and q can be expressed in column vector formatas

{r} =

r1r2r3

{q} =

q1q2q3

(2.4.3)


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The column vectors are also referred to as algebraic vectors. Using this description, we canexpress the dot product of two geometric vectors in column vector format as

r q = {r}T{q} (2.4.4)The skew-symmetric matrix [r] associated with the column vector {r} is a compact way of expressinga cross product. Define it as

[r] =

0 r3 r2r3 0 r1

r2 r1 0

(2.4.5)

so that

r q = [r] {q} =

r2q3 r3q2r3q1 r1q3r1q2 r2q1

(2.4.6)

Note that because r q = q r, the relationship [r] {q} = [q] {r} also holds.

In dynamics, and particularly in kinematics, we frequently encounter the vector product r (r q). The expression is commonly shortened to r r q, with the understanding that the crossproduct between r and q is performed first. Using the notation introduced above,

r (r q) = [r] [r] {q} (2.4.7)The matrix multiplications in [r] [r] {q} can be performed in any order.

Another use of the column vector notation arises when taking derivatives of a function with

respect to a set of variables, or when taking the derivative of a scalar with respect to a vector. Considera vector {q} = [q1 q2 . . . q n]T of dimension n, where the elements q1, q2, . . . , q n are variables that areindependent of each other, and a scalar S which is a function of these variables, S = S(q1, q2, . . . , q n).The derivative of S with respect to the the vector {q} is defined as the n-dimensional row vectordS/d{q}, whose elements have the form

dS

d{q} =

dS

dq1

dS

dq2. . .

dS

dqn

(2.4.8)

The derivative of one column vector with respect to another can be obtained in a similarfashion. Consider the column vector {v} of order m, where {v} = [v1 v2 . . . vm]T, where the elementsof {v} are functions of q1, q2, . . . , q n. The derivative of {v} with respect to {q} is a matrix of orderm n having the form

d{v}d{q} =

dv1dq1

dv1dq2

. . . dv1dqndv2dq1

dv2dq2

. . . dv2dqn

. . .dvmdq1

dvmdq2

. . . dvmdqn

(2.4.9)

For the special case when the scalar S is in quadratic form and expressed as S = {q}T [D] {q},where the elements of the matrix [D] are not functions of the variables q1, q2, . . . , q n, the derivative


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2.5. COMMONLY USED COORDINATE SYSTEMS 45

of S w.r.t. {q} has the form

dS

d

{q

}

=d{q}T [D] {q}

d

{q

}

= {q}T [D] + {q}T [D]T (2.4.10)

and when the matrix [D] is symmetric we obtain

dS

d{q} =d{q}T [D] {q}

d{q} = 2{q}T [D] (2.4.11)

Example 2.2

Consider the scalar S = 3x2 + 4y2 5xy and express it in terms of a symmetric matrix [D].

Writing the variables in vector form as {q} = [x y]T and taking the derivative of S w.r.t. {q}gives

{v}T = dSd{q} = [6x 5y 8y 5x] = 2{q}

T [D] [a]

from which it follows that

{v} = 2 [D]T {q} = 2 [D] {q} [b]and

d{v}d{q} = 2 [D] =

6 5

5 8

[c]

Thus, the matrix [D] is

[D] =

3 2.5

2.5 4

[d]

2.5 Commonly Used Coordinate Systems

This section discusses four coordinate systems that are commonly used to describe motion. One ofthese, rectilinear coordinates involves unit vectors that are fixed in space, and the other three aremoving coordinate systems.

2.5.1 Rectilinear Coordinates

The axes of rectilinear coordinate system are fixed in direction. The unit vectors along the coordinateaxes are also fixed and hence their derivatives are zero. Consider a coordinate system XY Z with unitvectors I, J and K along the X, Y and Z axes, respectively. Another commonly used coordinatesystem is xyz with unit vectors i, j and k. Also consider a reference point O, and a point P, as shownin Fig. 2.4. The position vector rP, which describes the position of point P, has the form

rP = XPI + YPJ + ZPK (2.5.1)


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Figure 2.4: Rectilinear coordinates.

in which XP, YP and ZP are the coordinates of point P, that is, their distance from the origin Oalong the X, Y and Z axes. To obtain the velocity of point P, denoted by vP, the above expressionis differentiated with respect to time. Noting that the time derivative of the unit vectors is zero,

vP =drPdt

= XPI + YPJ + ZPK + XPI + YPJ + ZPK = XPI + YPJ + ZPK (2.5.2)

with the overdots denoting differentiation with respect to time. Similarly, the acceleration of point P,which is denoted by aP, has the form

aP =

dvP

dt =

XPI +YPJ +

ZPK (2.5.3)

The advantage of using a rectilinear coordinate system is its simplicity and the ease with whichcomponents of the motion in other directions are viewed. The simplicity of rectilinear coordinates,however, is also their disadvantage, as rectilinear coordinates provide no information about the natureof the path that is followed.

Rectilinear coordinates are useful when components of the motion can be separated from eachother. A common application is projectile motion.

Example 2.3

A basketball player wants to shoot the basketball into the hoop. The player is at a distance L fromthe basket and the basket is at a height h from the players chest, from where the player launches theball. The player wants the ball to travel as a projectile and reach a height of 5h/4 before it begins itsdescent towards the basket, as shown in Fig. 2.5. Calculate the initial velocity v0 and angle withwhich the player needs to launch the ball.

Neglecting the aerodynamics, the acceleration of the projectile in the x and z directions is

ax = 0 az =

g [a]


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Figure 2.5: Basketball on a trajectory.

and the initial velocities in the x and z directions are

vx0 = v0 cos vz0 = v0 sin [b]

It follows that the expressions for velocity and acceleration can be written separately in the x and zdirections as

vx = v0 cos x = v0 cos t

vz = v0 sin gt z = v0 sin t 12

gt2 [c]

Point A is the point where the peak amplitude is reached. At this point, the vertical velocityis zero, or vz = v0 sin gtA = 0. Solution for the time tA it takes to reach this point becomes

tA =v0 sin

g

[d]

and the height reached at this point is

z = v0 sin v0 sin

g 1

2g

v0 sin

g

2=

1

2

v20 sin2

g=

5

4h [e]

Next, consider the time it takes to reach the basket, that is, point B. The horizontal distancetraversed L, can be expressed as x = L = v0 cos tB . Solution of this equation for the time to reach Bgives

tB =L

v0 cos

[f]

Introduction of the above expression into the height at time t = tB gives the height at time tB as

z(tB) = h = v0 sin L

v0 cos 1

2g

L

v0 cos

2[g]

There are two unknowns, v0 and . The two equations that need to be solved are Eqs. [e] and[g]. The solution can be simplified by introducing the variables u = v0 cos and w = v0 sin . Equation[e] can be rewritten as

1

2

w2

g=

5

4h w2 = 5

2gh [h]


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Introducing this result into Eq. [g] and rearranging leads to a quadratic equation in terms of u

u2

5g

2hLu +

1

2

gL2

h= 0 [i]

which can be solved as

u =1

2

gL2

h

5

2

5

2 2

=

1

2

gL2

h

5

2

1

2

[j]

Note that there are two solutions. After calculating u and w, the next step is to solve for thelaunch angle using the relation

w

u=

v0 sin

v0 cos = tan [k]

from which the angle is obtained as = tan1(w/u). We can then introduce the value of to anyone of the expressions for u or w to find the launch speed.

This example can be used as a a parametric study to determine the best options for maximizingpossibilities of scoring a basket.

2.5.2 Normal-Tangential Coordinates

Normal-tangential coordinates take into consideration the properties of the path taken by the movingbody, which is extremely useful. On the other hand, normal-tangential coordinates are not very usefulin describing position.

Figure 2.6: Particle on a curved path.

Consider a particle moving along a curved path. The normal-tangential coordinate system isa moving coordinate system attached to the particle. Two principal directions describe the motion,normal and tangential. To obtain these directions, consider the position of the particle after it hastraveled distances s and s + s along the path, as shown in Fig. 2.6. The associated position vectors,measured from a fixed location, are denoted by r (s) and r (s + s), respectively. Define by r thedifference between r (s) and r (s + s), thus

r = r (s + s)

r (s) (2.5.4)


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As s becomes small, r and s have the same length and become parallel to each other. Further,r becomes aligned with the tangent to the curve. The tangential direction is taken as the directiontangent to the curve with the positive direction in the same direction as the velocity. The unit vectorin the tangential direction is defined as

et = lims0

r

s=

dr

ds(2.5.5)

The tangential direction is shown in Fig. 2.7. The unit vector et changes direction as the particlemoves.

The velocity is obtained by differentiating the displacement vector with respect to time. Usingthe chain rule for differentiation,

Figure 2.7: Normal and tangential directions.

v (t) =dr

dt=

dr

ds

ds

dt(2.5.6)

Using the definition of et from Eq. (2.5.5) and noting that the speed v is the rate of change of thedistance traveled along the path, v = ds/dt, the expression for velocity becomes

v (t) = vet (2.5.7)

The second principal direction is defined as normal to the curve and directed toward the centerof curvature of the path, and it is shown in Fig. 2.7. This direction is defined as the normal direction(n), and the associated unit vector is denoted by en. The center of curvature associated with a certainpoint on a path lies along a line perpendicular to the path at that point. An infinitesimal arc of thecurve in the vicinity of that point can be viewed as a circular path, with the center of curvature atthe center of the circle. The radius of the circle is called the radius of curvature and is denoted by .The two unit vectors introduced above are orthogonal, that is, et en = 0.

Differentiation of Eq. (2.5.7) with respect to time gives the acceleration of the particle a (t) as

a (t) = v (t) = vet + vet (2.5.8)


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Figure 2.8: Infinitesimal change in tangential direction.

The derivative of et is needed to calculate the acceleration. To this end, consider Fig. 2.7 and

displace the particle by an infinitesimal distance ds along the path. The unit vectors associated withthe new location are et (s + ds) and en (s + ds). The center of curvature remains the same as theparticle is moved infinitesimally, so the arc length can be expressed as ds = d in which d is theinfinitesimal angle traversed as the particle moves by a distance ds. Defining the vector connectinget (s + ds) and et (s) by det, so that det = et (s + ds) et (s). From Fig. 2.8 the angle betweenet (s + ds) and et (s) is small, so that

|det| sin d |et (s)| d = ds

(2.5.9)

or

detds = 1 (2.5.10)The radius of curvature is a measure of how much a curve bends. For motion along a straight

line, the curve does not bend and the radius of curvature has the value of infinity. For plane motion,using the coordinates x and y such that the curve is described by y = y(x), the expression for theradius of curvature can be shown to be

1

=

d2y/dx2

1 + (dy/dx)2

3/2

(2.5.11)

The absolute value sign in the above equation is necessary because the radius of curvature is defined asa positive quantity. Considering the sign convention adopted above, the derivative of the unit vectorin the tangential direction becomes

detds

=en

(2.5.12)

Using the chain rule, the time derivative of et becomes

et =detds

ds

dt=

v

en (2.5.13)


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Introduction of this relationship to Eq. (2.5.8) yields

a (t) = v (t) = vet +v2

en (2.5.14)

The first term on the right in this equation is the component of the acceleration due to a change inspeed, referred to as tangential acceleration (at). The second term is the contribution due to a changein direction, referred to as the normal acceleration (an). The acceleration expression can be writtenas

a (t) = atet + anen (2.5.15)

with at = v and an = v2/.

The normal and tangential directions define the instantaneous plane of motion, also knownas the osculating plane. The velocity and acceleration vectors lie on this plane. The orientation of

osculating plane changes direction (twists), as the particle moves.

Example 2.4 - Road Curvature Design

Roads that change direction have to be designed with a curvature. The amount of curvature dependson the maximum normal acceleration that a vehicle can have and not slide. When designing a curvingroad, two important considerations are the amount of curvature and the variation of the curvature asthe vehicle enters and leaves a curve.1

Figure 2.9: Connecting two roads by a quarter circle.

Suppose you are designing a connection between two roads. The roads are perpendicular, asshown in Fig. 2.9. Geometrically, the easiest way of designing the connection is to form a rectangle,whose sides are a and b and to fit a quarter ellipse (quarter circle of radius

2b if a = b) into the

rectangle. The disadvantage of this design is that, even for a vehicle moving with constant speed, the

1Banking of the curve is another important factor. We will discuss this issue in the next chapter, when discussingkinetics.


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lateral acceleration will be zero before entering the curve and v2/ immediately after. This substantialjump in acceleration, even if it can be sustained by the friction between the tires and the road surface,contributes to shock (or jerk, defined earlier as the time derivative of acceleration) and can causediscomfort. Also, the driver may have to conduct a rapid steer maneuver, which has a destabilizing

effect.

A wiser choice of curved road design is one where the radius of curvature gradually changes.Two examples of such a curve are shown in Fig. 2.10. The slow initial change in curvature is usuallycompensated for by a higher curvature in the middle of the curve. Regardless of understeer, which isincluded in the design of most vehicles, a driver driving into a curve usually has to turn the steeringwheel more to accommodate the smaller radius of curvature in the middle of the curve. The secondcurve in Fig. 2.10, where the vehicle first turns away from the curve (seems contrary to intuition), isused to maximize the smallest radius of curvature of the path. This countersteer action is also whata bicycle rider (or speed skater) does when taking a turn.

Figure 2.10: Improved road curvature design.

Example 2.5

The motion of a point is described in Cartesian coordinates xy as x (t) = 2t2 + 4t, y (t) = 0.1t3 +cos t,

z (t) = 3t. Find the radius of curvature and normal and tangential accelerations at t = 0.

To find the radius of curvature, we need to first calculate the normal and tangential directions,as well as the speed and the acceleration components in terms of the normal and tangential coordinates.The position, velocity and acceleration vectors are

r (t) =

2t2 + 4t

i +

0.1t3 + cos t

j + 3tk [a]

v (t) = dr (t) /dt = (4t + 4) i +

0.3t2 sin tj + 3k [b]a (t) = dv (t) /dt = 4i + (0.6t

cos t)j [c]


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The speed is

v =

v v =

(4t + 4)2 + (0.3t2 sin t)2 + 9 [d]and the unit vector in the tangential direction an be written as et = v/v. At t = 0 the velocity and

speed becomev ( 0 ) = 4i + 3k v (0) =

42 + 32 = 5 [e]

so that the unit vector in the tangential direction is

et =v

v=

4

5i +

3

5k [f]

At this point, as accelerations have yet not been taken into consideration, there is not enoughinformation to determine en. The value of the acceleration at t = 0 is

a (0) = 4i

j [g]

The tangential acceleration can be obtained from

at (0) = a (0) et = (4i j) (4i + 3k) /5 = 3.2 [h]

and the normal acceleration becomes

an = anen = a atet = (4i j) 3.2 (4i + 3k) /5 = 1.44i j 1.92k [i]

The magnitude of the normal acceleration is

an =

an an = 1.442 + 1 + 1.922 = 2.6 [j]

so the unit vector in the normal direction is

en =anan

=1.44i j 1.92k

2.6[k]

Taking the dot product with the unit vector in the tangential direction, we can confirm that the twounit vectors are orthogonal to each other.

The radius of curvature can be calculated using

=v2

an=

52

2.6= 9.615 [l]

2.5.3 Cylindrical Coordinates

Cylindrical coordinates and their two-dimensional counterpart, polar coordinates, are preferred whenmotion is along a curved path, the distance of a point from an origin is of interest and one componentof the motion can be separated from the other two. A common use of polar coordinates is in orbitalmechanics.

Consider a point P and an inertial coordinate system XY Z with point O acting as the referencepoint from which P is observed, as shown in Fig. 2.11a. The position of point P can be described by


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Figure 2.11: Cylindrical coordinates.

first taking the projection of P onto the XY plane, denoted by P. The distance ZP from points Pto P is along the vertical direction and it is one of the parameters describing the motion. Next, drawa line from point O towards P and call the direction of this line as the radial direction. The radialdirection is also denoted as the r-direction and the associated unit vector along is denoted by er. Thedistance r from points O to P is the second parameter describing the motion. The third parameter isthe angle between the radial direction and X-axis and is denoted by and measured in radians. Theunit vector along the radial direction, er, is

er = cos I + sin J (2.5.16)

The transverse direction lies on the XY plane and it is perpendicular to the radial direction.Its positive direction is along the direction of a positive rotation of . The unit vector along thetransverse direction, denoted by e, is along this direction and it obeys the rule er e = K. FromFig. 2.11b, e is

e = cos J sin I (2.5.17)

The position of point P is expressed in cylindrical coordinates as

rP = XPI + YPJ + ZPK = Rer + ZPK (2.5.18)

The unit vectors er and e change direction as point P moves. To obtain the velocity, we need todifferentiate the above equation

vP = rP = Rer + Rer + ZPK (2.5.19)

which requires the derivative of the unit vector in the radial direction. To calculate this derivative,consider the projection of the motion onto the XY plane and that the particle has moved to point Q,whose projection is Q. Consequently, the coordinate system has moved by , as shown in Fig. 2.12.

The unit vectors of the new coordinate system are denoted by er ( + ) and e ( + ) andrelated to er () and e () by

er ( + ) = er ()cos + e ()sin


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Figure 2.12: Polar coordinate system moved by .

e ( + ) = er ()sin + e ()cos (2.5.20)

Using a small angles assumption of sin , cos 1, and taking the limit as approacheszero, the derivatives of the unit vectors become

lim0

er ( + ) er ()

=derd

= e

lim0

e ( + ) e ()

=ded

= er (2.5.21)

The time derivatives of the unit vectors become

er =derd

d

dt= e e =

ded

d

dt= er (2.5.22)

which, when substituted in the expression for velocity in Eq. (2.5.19), results in

v = Rer + Re + ZPK (2.5.23)

The first term on the right side corresponds to a change in the radial distance and the second term toa change in angle.

In a similar fashion we can find the expression for acceleration. Differentiation of Eq. (2.5.23)yields

aP = vP = Rer + Rer + Re + Re + Re + ZPK (2.5.24)

Substituting in the values for the derivatives of the unit vectors and combining terms gives

aP =

R R2

er +

R + 2R

e + ZPK (2.5.25)

We can attribute a physical meaning to the acceleration terms. The first term, R, describesthe rate of change of the component of the velocity in the radial direction. The second term, R2,


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is the centripetal acceleration. This term is always in the negative radial direction, as R is alwayspositive. The term R describes the acceleration due to a change in the angle . The next term, 2R,is known as the Coriolis acceleration, named after the French military engineer Gustave G. Coriolis(1792-1843). The Coriolis acceleration is due to two sources. Both deal with a changing distance in a

rotating system, hence the coefficient 2.

Cylindrical coordinates are most suitable to use when one component of the motion (which isselected as the Z (or z)direction) is separable from the others.

Example 2.6

For the mechanism in Fig. 2.13, the crank is at an angle = 30 and is rotating at the rate of = 0.2rad/s, which is increasing by = 0.1 rad/s2. The crank causes the slotted link to rotate. Using

cylindrical coordinates, calculate r and r associated with point P on the slotted link for the specialcase when b = a.

Figure 2.13: Crank and slotted link.

As in any kinematics problem, the analysis begins with examining the position, continues onto velocity analysis, and then accelerations. Because OP B is an isosceles triangle, = /2, so that

=

2 =

2[a]

The polar coordinates for the slotted link r, , the normal-tangential coordinates for the crankt, n and the inertial coordinates X, Y are shown in Fig. 2.14. The length r can be shown to be

r = 2a cos [b]

The velocity of the tip of the crank is

vP = vet = aet = 2aet [c]

where v = a. The velocity of point P in terms of polar coordinates is

vP = rer + re [d]


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Figure 2.14: Coordinate systems for crank and slotted link.

Equating Eqs. [c] and [d] leads to the expression for r. From Fig. 2.14, the unit vector in the tangentialdirection becomes

et = sin er + cos e [e]so that

vP = 2aet = 2a sin er + 2a cos e [f]and considering Eq. [c] gives

r = 2a sin = 2a cos r

=

2[g]

Note that r can also be obtained by direct differentiation of Eq. [c].

To find the second derivatives of r and of , we can either differentiate the above equation,which yields

r = 2a sin + 2a

2

cos =

2 [h]or we can obtain the acceleration terms by equating the normal-tangential and polar components ofthe acceleration. In normal-tangential coordinates, the acceleration is

aP = vet +v2

en [i]

where v = a and = a, so thataP = aet + a

2en [j]

The acceleration components in polar coordinates are

aP = r r2 er + r + 2r e [k]

and the value for can be obtained by relating the components of the unit vectors in the two coordinatesystems.

It should be noted that when b = a, the solution becomes much more complicated from analgebraic point of view, as r, and are related by

r cos = a + b cos r sin = b sin [l]

This example shows that we can obtain solutions to kinematics problems either by selecting coordinatesystems, or by finding algebraic relationships that describe the geometry and by differentiating theseequations.


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2.5.4 Spherical Coordinates

Spherical coordinates express position in terms of one position and two angular coordinates. An

important use of spherical coordinates is describing the position of a point on Earth in terms of thepoints latitude and longitude. The configuration of spherical coordinates is shown in Fig. 2.15a.There are several different conventions used to define the principal directions associated with sphericalcoordinates. The parameters used here are the absolute distance R of the point from a reference pointO, and two angles and , referred to as the azimuthaland zenith angles, respectively. The parameterR here (total distance from reference point) is different than the R (distance from reference point toprojection onto the ZY plane) used in cylindrical coordinates. The azimuthal angle is the same as thepolar angle in cylindrical coordinates.

Figure 2.15: a) Spherical coordinate system, b) side view.

The principal directions are referred to as the radial, azimuthaland zenith. The radial directionconnects reference point O and point P, with the positive direction as outward. The correspondingunit vector is denoted by eR, so that the position vector for P has the form

rP = ReR (2.5.26)

To define the azimuthal and zenith directions, it is necessary to first select and orient an inertialXY Z coordinate system. In Earth geometry, the equatorial plane is the XY plane with the Z axistowards the north. Projection of point P onto the XY plane is denoted by P. Next, rotate the XY Z

coordinates about the Z axis by the azimuthal angle to get an xyz coordinate system, noting thatthe x axis goes through point P. In Earth geometry, the azimuth angle is the longitude. The zenithangle is defined as the angle that the z axis makes with the radial direction, as shown in Fig. 2.15b.In Earth coordinates, the zenith angle is known as the colatitude or 90 minus the latitude.

The unit vector in the radial direction can be expressed in terms of the xyz coordinates as

eR = sin i + cos k = sin i + cos K (2.5.27)

Noting that the unit vector in the x direction is

i = cos I + sin J (2.5.28)


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the unit vector in the radial direction in terms of the XY Z coordinates becomes

eR = sin cos I + sin sin J + cos K (2.5.29)

Figure 2.16: Top view of spherical coordinate system.

As shown in Fig. 2.16, the unit vector in the azimuthal direction is selected as similar to itscounterpart in cylindrical coordinates, the polar direction, so that

e = j = sin I + cos J (2.5.30)

We can show that eR e = 0, so the two unit vectors are orthogonal. The unit vector associated withthe zenith angle satisfies the relationship

e = e eR = cos cos I + cos sin J sin K = cos i sin k (2.5.31)

The unit vectors eR, e and e form a mutually orthogonal set with eR and e lying on the xz (or xZ)plane. We need to obtain the derivatives of the unit vectors associated with the spherical coordinatesin order to calculate velocities and accelerations. The procedure is tedious and only the results arestated here:

eR = sin e + e e = (sin eR + cos e) e = eR + cos e (2.5.32)

vP = rP =d

dt(ReR) = ReR + R sin e + Re (2.5.33)

aP =

R R2 R2 sin2

eR +

R sin + 2R sin + 2R cos

e

+R + 2R R2 sin cos e (2.5.34)


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Figure 2.17: Airplane tracked by radar.

Figure 2.18: Side view, with shifted axes.


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Example 2.7

An airplane is traveling in the Y Z plane and, at the instant shown, it is executing a maneuver so

that it is at the bottom of a vertical loop that has a radius of curvature of 1500 m. The speed of theairplane is constant at 550 km/hr. A radar is tracking the airplane. Using spherical coordinates anda radar at point O, find the values of R and .

Since the azimuthal angle is not of interest (but its derivative is), without loss of generality itcan be set equal to zero in Fig. 2.17, so that the xyz and XY Z coordinate systems coincide. Shiftingthe reference point to O, as shown in Fig. 2.18, OP P forms a 5-12-13 triangle, so = tan1(12/5) =67.38. The unit vectors are

eR =12

13I +

5

13K e = J e = e eR = 5

13I 12

13K [a]

and the radial distance is R =

12002 + 5002 = 1300 m.

The next step is velocity analysis. The velocity of the airplane is v = 550J km/h, and in m/sit is

550km

hr= 550

km

hr

1000 m

1 km

1 hr

3600 s= 152.77

m

s[b]

The velocity is in the Y direction. The unit vectors in the radial and zenith directions do not havecomponents in the Y direction. Comparing with Eq. (2.5.33) leads to the conclusion that

v eR = R = 0 v e = R sin = 1300 1213

= 152.77 m/s v e = R = 0 [c]

Solving for gives = 152.77/1200 = 0.1273 rad/s.

The acceleration analysis is next. The aircraft is moving with constant speed on a curved path.Considering normal-tangential coordinates, the only component of the acceleration is in the normaldirection (Z axis) and

a =v2

an =

152.772

1500K = 15.56K m/s2 [d]

The components of the acceleration in the radial, azimuthal and zenith directions are

a eR = 15.56K

12

13I +

5

13K

= 5.985 m/s2 a e = 0

a e = 15.57K 513 I 1213 K = 14.36 m/s2 [e]From Eq. (2.5.34) and considering from Eq. [c] that R = 0, = 0, the components of the

acceleration are

a eR = R R2 cos2 a e = R cos a e = R R2 sin cos [f]We solve for the second derivatives ofR, and by equating Eqs. [e] and [f]. The azimuthal componentyields = 0 and the radial and zenith directions give

R = 5.985 + R2 cos2 = 5.985 + 1300

0.12732

(

12

13

)2 = 23.94 m/s2


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=1

1300

14.37 + 1300 (0.1273)2 5

13 12

13

= 0.0053 rad/s2 [g]

2.6 Moving Reference Frames

Section 2.3 discussed transformation of a coordinate system into another by a single rotation about oneof the coordinate axes. Then, we considered coordinate systems commonly used in dynamics. Thesecoordinates are obtained by coordinate transformations from initial frames. This section extends theconcept of coordinate transformations to the general case of rotations about coordinate axes.

Rotating coordinate systems are essential to the study of dynamics. There are several caseswhere, either by choice or by necessity, we need to use a coordinate system that rotates to describe

motion. For example, when describing certain motions in the vicinity of the Earth, such as hurricaneformation and satellite launching, it is necessary to consider the rotation of the Earth in the mathe-matical model. It is impossible to use a realistic fixed reference frame to conduct such analysis. Thetreatment of rigid body motion also is facilitated by use of rotating coordinate frames.

This section explores relationships between different reference frames and associated unit vec-tors. Two notations will commonly be used to distinguish between different reference frames:

1. Denote one of the frames by XY Z and the other by xyz with unit vectors I, J, K and i, j, k,respectively. In general, XY Z denotes a fixed frame and xyz one that moves. Intermediate

frames are usually referred to as XYZ, XYZ, and so on.

2. In the second description, which is especially useful when several frames are involved, a letteris assigned to each frame: for example, frames A and B. The coordinate axes of the framesare called a1a2a3 and b1b2b3 and the unit vectors along the coordinate axes are defined asa1, a2, and a3 and b1, b2, and b3. Intermediate frames are defined in a similar fashion.

Consider a vector q as viewed in the B frame. This vector can be expressed as

q = q1b1 + q2b2 + q3b3 (2.6.1)

in which q1, q2, and q3 are the components of q. Because the unit vectors are orthogonal, we canexpress each component as qj = q bj , (j = 1, 2, 3), so that

q = (q b1) b1 + (q b2) b2 + (q b3) b3 (2.6.2)

The vector q can be resolved in the A frame, as well. However, referring to the componentsqi (i = 1, 2, 3) will not make it possible to distinguish them from their counterparts associated withthe B frame. Rather, these components can be expressed as

q = Aq1a1 +Aq2a2 +

Aq3a3 =Bq1b1 +

Bq2b2 +Bq3b3 (2.6.3)


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2.6. MOVING REFERENCE FRAMES 63

Figure 2.19: Direction angles of a vector.

so that in column vector format q becomes

Aq

=

Aq1 Aq2 Aq3T

Bq

=

Bq1 Bq2 Bq3T

(2.6.4)

This notation is similar to the notation used in Sec. 2.3 with XY Z and xyz as the coordinate axes.

It is of interest to investigate relationships between the unit vectors of two coordinate systems.Consider the vector q and the unit vector e along it, so that q = qe, where q is the magnitude of q.The angles that q makes with the axes of a coordinate system are denoted by 1, 2 and 3 and arecalled direction angles, as shown in Fig. 2.19. The direction cosines of q (and of e) are denoted byc1, c2 and c3 and they are the cosines of the angles that e makes with the coordinate axes

e = cos 1b1 + cos 2b2 + cos 3b3 (2.6.5)

The direction cosines are written as cj = ebj = cos j , (j = 1, 2, 3), so that q = qc1b1+qc2b2+qc3b3.Now, consider the unit vectors of the A and B frames. Considering the above equation, we

can write

bj = (bj a1) a1 + (bj a2) a2 + (bj a3) a3

aj = (aj b1) b1 + (aj b2) b2 + (aj b3) b3 (2.6.6)

The direction cosine between two coordinate axes aj and bk is the cosine of the angle betweenthe two axes, and is denoted by cjk = aj

bk = cos jk (j, k = 1, 2, 3) . Define the column vectors

containing the unit vectors of the two frames as {a} and {b}, where{a} = [a1 a2 a3]T {b} = [b1 b2 b3]T (2.6.7)

and the direction cosine matrix [c] c11 c12 c13c21 c22 c23

c31 c32 c33

(2.6.8)

from which we can show that

{b

}= [c]T

{a

}(2.6.9)


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Equation (2.6.9) can be inverted to yield

{a} = [c]T{b} (2.6.10)

Comparing Eq. (2.6.10) with Eq. (2.6.9) leads to the conclusion that the direction cosine matrix isunitary (also called orthonormal), that is, its inverse is equal to its transpose, or

[c]1 = [c]T [c] [c]T = [1] (2.6.11)

where [1] is the identity matrix. Note that the determinant of a general unitary matrix is 1, but forthe direction cosine matrix det [c] = 1.

Equation (2.6.9) also applies when relating the components of a vector resolved along the axesof two coordinate systems. Indeed, consider the vector q and its column vector representation in theA and B frames. We can show that

Bq

= [c]T

Aq

Aq

= [c]

Bq

(2.6.12)

The definition of direction cosine above is not universally accepted. Some texts instead definethe direction cosine by cjk = bj ak = cos kj (j, k = 1, 2, 3).

A total of nine direction cosines are defined, one for each angle between the j-th and k-th coordinate axes (i, j = 1, 2, 3). These nine direction cosines are not independent of each other.Equation (2.6.11) represents six independent equations that relate the direction cosines (six becauseof the symmetry of [c] [c]T), reducing the number of independent direction cosines to three (9 directioncosines - 6 relational equations).2 It follows that, at most, three independent parameters are necessary

to represent the transformation from any given configuration of coordinate axes to another one. Twoissues that are of interest are:

How do we select these three rotation parameters? Given the three parameters, how do we calculate the direction cosine matrix? Given the direction cosine matrix, what are the three parameters associated with that matrix?

The next section will consider two approaches for the selection of the rotation parameters.

2.7 Selection of Rotation Parameters

This section considers two approaches for selecting the rotation parameters. The first involves threerotations about independent axes. The second approach uses a single rotation about an axis whoseorientation is specified.

2We can demonstrate this by writing [c] as three column vectors [{c1}{c2}{c3}]. These vectors are orthogonal vectorsand they represent the direction angles of the axes of the transformed coordinates. It follows that Eq. (2.6.11) representsthe six possible dot products among these vectors.


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2.7. SELECTION OF ROTATION PARAMETERS 65

2.7.1 Transformation by Three Rotation Angles

In the first approach, the three parameters are selected as three independent rotations. This neces-

sitates selection of the axes about which these rotations are made. To this end, there are infinitechoices. These axes are selected in a way to simplify calculations. It should be noted that any twosuccessive rotations have to be conducted about axes that are not parallel to each other. Otherwise,the rotation angles will not be independent.

The choices of selecting axes about which rotations are made are narrowed down by carrying outthe rotations about the axes of coordinate frames. The procedure is demonstrated as follows. Denotethe initial position of the axes by frame A, with axes a1a2a3, and rotate frame A counterclockwiseby angle about one of the axes to get a rotated frame B, whose axes are b1b2b3. This rotationconvention ensures that det [c] = 1. Rotation about the a1 axis is called a 1 rotation, about the a2 axisa 2 rotation and rotation about the a3 axis a 3 rotation. These rotations are illustrated in Fig. 2.20.

Figure 2.20: Rotation types: a) a 1 rotation; b) a 2 rotation; c) a 3 rotation.

For a 1 rotation the unit vectors of the A and B frame are related by

b1 = a1 b2 = cos a2 + sin a3 b3 = sin a2 + cos a3 (2.7.1)

or, in matrix form

{b

}= [c]1

{a

}= [R]

{a

} {Bq

}= [R]

{Aq

}(2.7.2)

where [R] = [c]1 is called the rotation matrix and has the form

For a 1 rotation [R] =

1 0 00 cos sin

0 sin cos

(2.7.3)

When the rotation is performed about the a2 axis, that is, a 2 rotation, the rotated vectorsbecome

b1 = cos a1

sin a3 b2 = a2 b3 = sin a1 + cos a3 (2.7.4)


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and the rotation matrix has the form


cos 0 sin 0 1 0

sin 0 cos

(2.7.5)

For a 3 rotation, the rotated vectors are

b1 = cos a1 + sin a2 b2 = sin a1 + cos a2 b3 = a3 (2.7.6)

and the rotation vector is


cos sin 0 sin cos 0

0 0 1

(2.7.7)

For all three rotations above, the determinant of the rotation matrix is 1, or det[R] = 1. This verifiesour earlier statement that the determinant of the direction cosine matrix is unity.

Consider now the three rotation angles needed for a general rotation sequence and how we canaccomplish the three rotations. To this end, two approaches can be identified: body-fixed rotationsequence and space-fixed rotation sequence.

Body-Fixed Rotation Sequence

This rotation sequence can be visualized by considering a box and performing the rotations about aset of axes attached to the box. Begin with an initial coordinate set a1a2a3, align the box with it androtate the box about one of the axes by an angle 1. Let us call the resulting orientation of the boxthe D frame with the axes d1d2d3, which are now aligned with the box. There are three choices in theselection of the rotation axis.

The procedure is shown in Fig. 2.21a for a 3 rotation, that is, a rotation about the a3 axis.The associated rotation matrix is denoted by [R1]. Note that the subscript 1 denotes that this is thefirst rotation and not the axis about which the rotation takes place. As a result of the rotation, pointsP QR on the box move to their new positions PQR in Fig. 2.21b. These positions do not changerelative to the box.

Next, rotate the box about one of the d1d2d3 axes by an angle 2 to obtain the new orientationof the box, denoted as the H frame, with axes h1h2h3. The rotation matrix associated with thisrotation is denoted by [R2]. Two axes can be chosen for this rotation. If the first rotation is carriedout about, say, a3, then d3 = a3 and the second rotation can only be carried out about d1 or d2.Otherwise, the two rotations cannot be distinguished from each other.

The third rotation is carried out by rotating the box about one of the axes h1h2h3 by 3 toobtain the final frame B with axes b1b2b3. For the same reason described above, there are two axes torotate about. The associated rotation matrix is [R3]. It follows that there are 3 2 2 = 12 differentways to carry out a body-fixed rotation sequence. These possible ways of selection are known as Eulerangle sequences and they are denoted by the number of the axes about which the rotations are made.


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Figure 2.21: Body-fixed rotation about a3: a) initial position, b) after rotation by .

For example, if the first rotation is about the a3 axis, the second rotation is about the d1 axis andthe third rotation is about the h2 axis, the rotation sequence is called 3-1-2. Chapter 9 will quantifyrotation sequences and discuss their applications in more detail.

Consider next the combined rotation. For the three rotations above

{d} = [R1] {a} {h} = [R2] {d} {b} = [R3] {h} (2.7.8)

Combining the three rotations gives

{b} = [R3] [R2] [R1] {a} = [R] {a} (2.7.9)

and the combined rotation matrix is [R] = [R3] [R2] [R1].

Space-Fixed Rotation Sequence

Here, we consider an initial coordinate system and perform all three rotations about the axes of theinitial coordinates. Let a1a2a3 be the initial axes. The first rotation is about one of a1, a2 or a3 axes

to yield the d1d2d3 axes, with {d} = [R1]{a}. Then, the coordinate axes d1d2d3 are rotated aboutone of the a1, a2 or a3, but not about the same axis about which the first rotation was made. Notethat the axis for the second rotation is not one of the axes of the rotated frame. The third rotationis carried out about another one of a1, a2 or a3. We can show that the combined rotation matrix hasthe form

{b} = [R] {a} = [R1] [R2] [R3] {a} (2.7.10)

where the combined rotation matrix is in reverse order of the body-fixed transformations. Space-fixedrotations are not used as frequently as body-fixed rotations in dynamics. They come in handy whendealing with kinematics problems, such as robotics and mechanisms.


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Inversion

We have obtained values for the rotation matrix [R] (and hence direction cosine matrix [c]) in terms of

three rotation angles. The inverse problem, that is, given [R] or [c] to find the three rotation angles, isnot as straightforward. To solve the inverse problem one needs to be given, in addition to [ R] (or [c]),the sequence under which the individual rotation transformations are made. Given this information,the rotation matrix [R] is constructed and the unknowns are solved for. In general, we look at hegeneral form of [R] and begins with the entries of [R] that are the simplest to solve for.

2.7.2 Resolving a Rotated Vector

So far, we have developed two coordinate systems and looked at the same vector in the two coordinatesystems. That is, if the coordinates of a particular vector are

{Aq

}= [Aq1

Aq2Aq3]

T in the A frame,

and the coordinates of that same vector are {Bq} = [Bq1 Bq2 Bq3]T in the B frame, thenBq

= [R]

Aq

(2.7.11)

Figure 2.22: a) Initial position of vector, b) rotated vector.

Another problem of interest in kinematics is to take a vector, to rotate the vector and to findthe coordinates of the rotated vector in the original coordinate system. Consider the A frame andattach a box (or a rectangle) to it, as shown in Fig. 2.22a. Let us select point P on the box and select

the vector to rotate as going from O to P. The initial position of the vector is qi or {A

qi}. Next,rotate the vector qi (or the box to which the vector is attached). For illustrative purposes, Fig. 2.22bshows a rotation by about the a1 axis. The rotated vector is denoted by qf or {Aqf}, and thisvector connects points O and P, where P is the point to which P moves after the rotation. Notethat rotating the vector is equivalent to rotating the box (or rectangle) in which the vector is defined.

Consider now a second frame B, which is obtained by rotating the A frame the same way thevector qi was rotated. The initial and rotated vectors can also be expressed in terms of the B frame as{Bqi} and {Bqf}. Making use of the coordinate transformation relationships we can relate the initialand final vectors in the two coordinate systems as

Bqi = [R]Aqi Bqf = [R]Aqf (2.7.12)


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Because the vector is attached to the reference frame, the initial vector in the A frame has thesame coordinates as the rotated vector in the B frame. Hence,

Bqf = Aqi (2.7.13)Combining the above two equations, the initial and final positions of the vector in the initial A frameare expressed as

Aqf

= [R]T

Aqi

(2.7.14)

Let us compare equations (2.7.11) and (2.7.14). In Eq. (2.7.11) the vector is fixed, but isviewed from two different coordinate frames. In Eq. (2.7.14) the vector is rotated, but its initial andfinal locations are viewed from the same reference frame. In one case the transformation matrix is [R]and in the other it is [R]T, denoting inverse transformations. It follows that viewing a fixed vector

from a rotated frame is exactly the opposite of rotating that vector (in the same amount) and viewingthe rotated vector from the fixed frame.

2.7.3 Single Rotation about a Specified Axis

This rotation approach is based on Eulers theorem, which states that the most general transformationof a rigid body with one fixed point can be described as a single rotation about a certain axis goingthrough the fixed point. The axis about which the rotation is made is called the principal line and isdenoted by n, as shown in Fig. 2.23a. The unit vector along this axis is expressed as n. The rotationangle is called the principal angle and is denoted by .

Figure 2.23: a) General rotation of a vector, b) top view, c) side view.

The three direction cosines of the axis of rotation {n} = [n1 n2 n3]T are not independent andthey obey the relationship n21 + n

22 + n

23 = 1, so that given two direction cosines the third can be

ascertained to within a value. Hence, only two of the parameters are independent. The rotationangle becomes the third parameter. When the sign of the third direction cosine is changed, this


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means the principal line is pointing in the opposite direction and the same transformation can beachieved by reversing the sign of .

Given a vector qi and rotating it by about axis n, the rotated vector, denoted by qf, can

be expressed as

qf = cosqi + (1 cos )(qi n)n + sin n qi (2.7.15)or, in column vector notation

{qf} = cos {qi} + (1 cos){n}{n}T{qi} + sin [n] {qi} (2.7.16)or

{qf} =

cos [1] + (1 cos ){n}{n}T + sin [n] {qi} (2.7.17)where [n] is the skew-symmetric matrix associated with the vector {n}.

The validity of the above relationship can be demonstrated by defining a coordinate systemwith directions n, a1, a2 and rotating the coordinate system about n by . Referring to the rotatedcoordinate system as n, a1, a

2, as shown in 2.23b, the two sets of vectors are related by

a1 = cosa1 + sin a2 a

2 = sina1 + cos a2 (2.7.18)

Figure 2.23c shows the side view. The angle is not a rotation parameter; rather, it is theangle between the principal line and the vector q. Now, consider the initial and rotated vectors qiand qf and denote the magnitude of these vectors by q. Because qf is obtained by rotating qi aboutn, the two vectors have the same components along the initial and rotated axes. These vectors arewritten as

qi = qcos n qsin a2 qf = qcos n qsin a2 (2.7.19)Substitution of the value for a2 from the above equation leads to

qf = qcos n + qsin sina1 qsin cosa2 (2.7.20)The above equation can be rewritten as

qf = qcos(cos n sin a2) + q(1 cos)cos n + qsin sina1 (2.7.21)The first term in the above equation is recognized as cos qi. Noting that cos = qi n, the secondterm can be written as (1 cos )(qi n)n. Finally, evaluating

sinn qi = sinn q(cos n + sin a2) = qsinsin a1 (2.7.22)The conclusion is that Eq. (2.7.15) and Eq. (2.7.21) are equivalent.

Equations (2.7.16) - (2.7.17) give the relationship between the initial and final positions of avector in column vector format. Hence, considering an initial frame and rotate it by angle about nto get a rotated frame, the direction cosine matrix between the two frames has the form

[c] = [R]T = cos [1] + (1 cos) {n}{n}T + sin [n] (2.7.23)which is a useful way of calculating the rotation matrix between two reference frames. Given a rotationmatrix [R], calculation of the associated principal line and rotation angle will be discussed in Chapter9.


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2.7.4 Finite Rotations Do Not Commute

The preceding analysis leads to the conclusion that the order in which rotations of coordinates are

performed makes a difference in the orientation of the transformed coordinate system. This holds truewhether the rotations are performed as a body-fixed rotation sequence or as space-fixed rotations.We can verify this visually, by taking a book and rotating it about two axes in different sequences.The procedure is illustrated in Fig. 2.24 for a body-fixed rotation sequence. Begin with the XY Zframe, where the book lies on the XZ plane. Rotate the book by 90 about the X axis to get theXYZ coordinates and then rotate about the Z axis by 90 to get the xyz coordinates. Repeatingthe procedure with first rotating about the Z and then X coordinates leads to a different orientationof the book.

Figure 2.24: Finite rotations do NOT commute.

The conclusion is that sequences of finite rotations in three dimensions cannot be expressedas vectors, as the commutativity rule does not hold. It follows that there does not exist an angularposition vector to differentiate in order to obtain angular velocity, except for the special case of planemotion. The subsequent sections will provide definitions for the angular velocity of a reference frame.

Example 2.8

Given the rotation matrix

[R] =

0.1768 0.8839 0.43300.9186 0.3062 0.2500

0.3536 0.3536 0.8660

[a]

find the rotation angles if the coordinate axes are obtained by a 3-1-3 transformation.


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For a 3-1-3 transformation with angles 1, 2 and 3, the combined rotation matrix is expressedas

[R] = [R3][R2][R1] [b]

where

[R1] =

cos 1 sin 1 0 sin 1 cos 1 0

0 0 1

[R2] =

1 0 00 cos 2 sin 2

0 sin 2 cos 2

[R3] =

cos 3 sin 3 0 sin 3 cos 3 0

0 0 1

[c]

Carrying out the algebra, the combined rotation matrix has the form

[R] =

c3c1 s3c2s1 c3s1 + s3c2c1 s3s2s3c1 c3c2s1 s3s1 + c3c2c1 c3s2

s2s1 s2c1 c2

[d]

where the compact notation is used, where s denotes the sine and c to denotes the cosine functions.For example c1 = cos 1. When proceeding to identify the rotation angles, it is convenient to beginwith the simplest expression, R33

R33 = cos 2 = 0.8660 = 2 = 30 [e]where the rotation angles in the range of 180 to 180. Comparing the other simple elements in therotation matrix yields

R13R33

= tan 3 =0.433

0.250=

3 = 3 = 60 or 120 [f]

R31

R32 = tan 1 = 1 = 2 = 45

or 135

[g]

Begin with the assumption that 2 = 30. Examining R31 and R11 leads to

R31 = sin 2 sin 1 = 0.5sin 1 = 0.3536 R13 = sin 3 sin 2 = 0.5sin 3 = 0.4330 [h]It follows that the sines of both these angles are negative, and we conclude from Eqs. [f] and [g] that

1 = 45 2 = 30 3 = 120 [i]

The next step is to check the accuracy of the assumption that was made. Calculating any oneof the remaining elements of [R], say R11, gives

R11 = cos 3 cos 1 sin 3 cos 2 sin 1 = 12

22

+ 32

32

22

= 0.1768 [j]

But R11 = 0.1768, so the assumption made earlier about 2 is incorrect. It follows that 2 = 30.Following the same procedure as in Eq. [h] leads to

R31 = sin 2 sin 1 = 0.5sin 1 = 0.3536 R13 = sin 3 sin 2 = 0.5sin 3 = 0.4330 [k]and the conclusion is

1 = 45 2 = 30 3 = 120 [l]

Substituting these values into R11 or in any other element of [R], the correctness of results inEq. [l] is confirmed.


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Example 2.9

An xyz coordinate system is obtained by rotating the XY Z coordinates about a line n that makes

an angle of 30

with the X axis, 60

with the Y axis and is perpendicular to the Z axis, as shownin Fig. 2.25. Calculate the direction matrix [c] between XY Z and xyz when the rotation angle is = 66 and the coordinates of a point defined by rP = 2I after rotation in the XY Z frame.

Figure 2.25: Rotation axis n.

The direction cosines of the principal line are

nX =

3

2nY =

1

2nZ = 0 [a]

and the associated [n] matrix is

[n] =

0 nZ nYnZ 0 nX

nY nX 0

= 1

2

0 0 1

0 0 31 3 0

[b]

Noting that cos = cos (66) = 0.4067, sin = sin(66) = 0.9135, and using Eq. (2.7.23) resultsin

[c] = 0.4067

1 0 00 1 0

0 0 1

+ 0.5933 1

4

3

10

3

10

T

0.9135 1

2

0 0 1

0 0 31 3 0

= 0.8517 0.2569

0.4568

0.2569 0.5551 0.79120.4568 0.7912 0.4067

[c]

To find the location of point P after the rotation, we can use the relationship {qf} = [c]{qi},where {qi} = [2 0 0]T, with the result

{qf} = [c]{qi} =

0.8517 0.2569

0.45680.2569 0.5551 0.7912

0.4568 0.7912 0.4067

20

0

=

1.70340.5138

0.9136

[d]

Note that the rotated vector

{qf

}is expressed in terms of the XY Z frame.


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The same result can also be obtained by successive coordinate transformations. Noting thatfor body fixed transformations the rotations are conducted about one of the coordinate axes, the XY Zframe is rotated so that one of the axes of the rotated frame is the n axis. This can be accomplishedby rotating about the Z axis by 30, and calling the rotated axes XYZ. The associated rotation (a

3 rotation) matrix is

[R1] =

3/2 1/2 0

1/2 3/2 00 0 1

[e]

and X axis is the same as the n axis in Fig. 2.25. Then, rotate the XYZ frame and the vectorrP = 2I about the X

axis by 66 to get the xyz frame. The associated rotation (a 1 rotation)matrix is

[R2] =

1 0 00 0.4067 0.9135

0 0.9135 0.4067

[f]

The position of the vector rP after the rotation, in terms of the XYZ frame, is

{XYZqf} = [R2]T {XYZqi} [g]

Noting that XY

Z

= [R1]

XY

Z

[h]

so that {XYZqf} = [R1]

XY Zqf

, Eq. [g] can then be expressed in the XY Z frame as

XY Zqf

= [R1]T

[R2]T

[R1]XY Z

qi

= [R]T XY Z

qi

[i]

where the rotation matrix between the initial and final frames are

[R] = [R1]T [R2] [R1] [j]

The above result can be explained by noting that the first rotation [R1] merely shifts to adifferent coordinate frame but does not rotate the vector rP. The second rotation rotates the vectorrP. So, the two rotations are completely different in nature. Carrying out the matrix multiplications,Eq. [j] gives the same results as Eqs. [c] and [d].

2.8 Rate of Change of a Vector, Angular Velocity

The previous section demonstrated that consecutive rotations of coordinate frames by finite angles donot lend themselves to representation as vectors. Hence, we do not have a vector to differentiate inorder to represent rotation rates. This section explores ways of looking at rotation rates and definingthe angular velocity vector.

When looking at the rate of change of a quantity, we must distinguish between derivativestaken in different reference frames. For example, consider a moving vehicle and attach a reference


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2.8. RATE OF CHANGE OF A VECTOR, ANGULAR VELOCITY 75

frame to the vehicle. When something inside the vehicle moves, say a passenger throws a ball up anddown, someone inside the vehicle sees the ball as moving up and down only. An observer outside thevehicle sees the ball moving with the vehicle as well as moving with respect to the vehicle.

We can define rates of rotations and angular velocities in a variety of ways. One is to useinfinitesimal values for the angles in a body fixed (or space-fixed) rotation sequence. This approach isintuitive, but not mathematically sound. Two more rigorous approaches are presented here; one thatmakes use of column vector notation and one that is based on Eulers theorem. This section beginswith the definition of angular velocity for plane motion and then considers the three-dimensional case.

2.8.1 Angular Velocity for Plane Motion

Consider the two reference frames XY Z and xyz in Section 2.3, where the XY Z coordinates are

rotated by about the Z axis to arrive at the xyz coordinate system. Let us consider XY Z to be afixed reference frame and evaluate the time derivative of the unit vectors associated with the movingframe xyz. Differentiating Eq. (2.3.1) leads to

d

dti = ( sin I + cos J) = j d

dtj = ( cos I sin J) = i (2.8.1)

and ddt k = 0, as the z direction is the same as the Z direction, which is fixed. The above relationships

can be expressed in terms of the cross product involving the angular velocity vector = k = k andby writing

d

dti = i = k i = j d

dtj = j = i d

dtk = k k = 0 (2.8.2)

Indeed, when dealing with a single rotation about a fixed axis, the rotation can be representedas a vector, k. Taking the derivative of this vector gives the angular velocity = k. The crossproducts above are demonstrated in Fig. 2.26.

Figure 2.26: Cross products of unit vectors in xy plane.

The rate of change of a vector in the fixed and moving frames is of interest. Consider first avector that is fixed in the xyz frame. This vector can be any type of vector, such as position, velocityor acceleration, linear or angular momentum or angular velocity. To an observer attached to the xyz


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frame, this vector does not move. As the xyz coordinate system is moving, an observer in the XY Zsystem sees the vector move.

The vector of interest is q = qxi + qyj = qXI + qYJ. Because this vector is fixed in the xyz

frame, its time derivative becomes

d

dtq = q = qxi + qy j = qx i + qy j = q (2.8.3)

The rate of change of a vector that is fixed in a moving frame is obtained by the cross product betweenthe angular velocity of the moving frame and the vector itself. This relationship is valid for both two-or three-dimensional motion, as will be demonstrated soon.

Next, we consider the case when the vector q is not fixed in the moving frame xyz but insteadit is changing with respect to the moving frame. The derivatives qx, qy are no longer zero and

d

dt q = qxi + qyj + q (2.8.4)Denoting by qrel = qxi + qyj the local derivative or relative rate of change of q, or the rate of changeof the vector q in the moving frame, we can write the above equation as

d

dtq = q = qrel + q (2.8.5)

The total derivative of a vector is comprised of two components: i) the change of the vectoras viewed from the moving frame, plus ii) the change due to the rotation of the moving frame. Theabove equation is known as the transport theorem and it is valid for any vector that is being observedfrom a moving (rotating) frame. The name transport reflects the fact that the derivative is being

transported from one reference frame to another. The transport theorem is depicted in Fig. 2.27.

Figure 2.27: Derivative of a vector using transport theorem.

2.8.2 Angular Velocity for Three-Dimensional Motion

As discussed earlier, for three-dimensional rotation, the cumulative effect of rotation sequences cannotbe described as vectors and thus no rotation vector exists for us to differentiate. Instead, we defineangular velocity for three-dimensional motion by means of a column vector formulation.


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Begin with the XY Z and xyz frame representations. The frame xyz rotates with respect tothe XY Z frame. Consider a vector q that is represented in terms of the two reference frames as

q = qxi + qyj + qzk = qXI + qYJ + qZK (2.8.6)

Using the notation in Section 2.3, the vector q in column vector format has the form

XY Zq

=

qXqY

qZ

{xyz q} =

qxqy

qz

(2.8.7)

Two column vector representations are needed to indicate that the vector is expressed in two differentreference frames. The two vectors are related by

{xyz q} = [R]XY Zq or XY Zq = [R]T {xyz q} (2.8.8)where [R] is the 3 3 rotation matrix. The form for [R] for plane motion is given in Eq. (2.3.3).

Differentiation of the term on the right in Eq. (2.8.8) results inXY Zq

= [R]T {xyz q} + [R]T {xyz q} (2.8.9)

The term {xyz q} is the derivative of the vector as viewed from the moving frame. The term [R] isthe derivative or the rotation vector, and

XY Zq

is the derivative of the vector viewed from the

non-moving (inertial) frame, that is, the total derivative. Introducing the notation for the derivativeof q as v = q = ddt q, the rate of change vector v in terms of the XY Z and xyz frames becomes

XY Zv

and {xyz v}, respectively. This way, {xyz v} = [R]XY Zv.Note that while the relationship

XY Zv

=

XY Zq

(2.8.10)

is valid, the similar-looking counterpart for xyz frame is not,

{xyz v} = {xyz q} (2.8.11)This is because {xyz v} describes the rate of change vector in terms of the coordinates of the movingframe, while {xyz q} denotes the local derivative of{xyz q} as viewed from the moving (rotating) frame.Introducing Eq. (2.8.9) to Eq. (2.8.10) and left multiplying by [R] gives

[R]

XY Zv

= {xyz v} = [R][R]T {xyz q} + {xyz q} (2.8.12)

Let us examine the matrix product [R][R]T more closely. For the special case of plane motion(i.e., two-dimensional)

[R][R]T =

cos sin

sin cos

sin cos cos sin

=

0 11 0

(2.8.13)

which is recognized as the matrix representation [] of the angular velocity = k in the xyz frame.It follows that the above equation is yet another way to define the angular velocity vector, and we canwrite

{xyz v

}= []

{xyz q

}+

{xyz q

}(2.8.14)


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which is the representation of the transport theorem in column vector notation. Note that all of theterms in the above equation are expressed in terms of parameters associated with the xyz frame.

The same procedure can be carried out for the rotation matrix in three dimensions. Regardless

of the rotation sequence, we can show that [R][R]T is a skew-symmetric matrix. Consider the identity[R] [R]T = [1] and differentiate it, which yields

[R] [R]T + [R] [R]T = [0] (2.8.15)

The two terms in the preceding equation are transposes of each other. Indeed, denoting by[W] = [R] [R]T, it follows that [W]T = [R] [R]T. A matrix which, when added to its transpose, yieldsa null matrix must be skew-symmetric. The matrix [W] can be expressed as

[W] = [R] [R]T =

0 z yz 0 x

y

x0

(2.8.16)

Since [W] is skew-symmetric, it can be recognized as the matrix representation of a vector =xi + yj + zk used when expressing a cross product, [W] = []. We refer to [W] = [] as the angularvelocity matrix.

The vector is defined as the angular velocity vector of the reference frame xyz with respectto the XY Z frame and x, y and z as the instantaneous angular velocities or components of theangular velocity vector. Thus, Eq. (2.8.12) is verified for three-dimensional motion.

It must be emphasized that, for three-dimensional rotations, the angular velocity vector is

a defined quantity and that it is not the derivative of another vector. For this reason, the angularvelocity vector is referred to as nonholonomic, a term that is associated with expressions that cannotbe integrated to another expression. The way we arrive at the angular velocity vector is completelydifferent from the derivation of the expression for translational velocity, or the rate of change of anydefined vector.

When using the A and B frame notation, the angular velocity vector of the B frame withrespect to the A frame is written as AB , where the superscripts denote the frames that are related.The angular velocity of the A frame with respect to the B frame is BA and BA = AB . In addition,the derivative of a vector obtained in a certain reference frame is denoted by a left superscript, suchas B ddt q or

Bv. It follows that the transport theorem can be written as

A ddt

q = B ddt

q + AB q (2.8.17)

2.8.3 Other Definitions of Angular Velocity

The definition of angular velocity in the preceding subsection is not the only way angular velocity canbe defined. Two additional definitions are presented here. Consider the frames A and B. The angularvelocity of frame B with respect to frame A is defined by

AB = 1b1 + 2b2 + 3b3 = b2 b3b1 + b3

b1b2 + b1 b2b3 (2.8.18)


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This definition can be verified by analyzing the expressions for the rates of change of the unit vec-tors. While the above definition is more abstract than the way we arrived at Eq. (2.8.16), it can besubstituted more easily into mathematical operations that involve angular velocity.

Yet another definition of angular velocity can be obtained from Eulers theorem and from therelationship between initial and rotated vectors, as given in Eq. (2.7.15). Let us rewrite that equationby replacing qi by q and by replacing qf by q

, which gives

q = cosq + (1 cos)(q n) n + sin n q (2.8.19)

A special application of Eq. (2.8.19) is for small rotations over small time intervals. Indeed, as approaches zero,

d cos 1 sin q q dq (2.8.20)

which, when substituted into Eq. (2.8.19) yields

dq = d (n q) (2.8.21)

Because the infinitesimal rotation is taken about a single axis at that particular instant, divisionof d by dt yields the angular velocity expression in the form

d

dt= = n (2.8.22)

so that dq/dt = q.

2.8.4 Additive Properties of Angular Velocity

Now that we have defined angular velocity as a vector, we can use the additive properties of vectorsand obtain the angular velocity of a reference frame by adding up the angular velocities associatedwith the rotations that lead to that reference frame. As an illustration, consider an initial frame XY Zand rotate it by an angle 1 about the X axis to obtain a X

YZ frame. The angular velocity of theXYZ frame with respect to the initial frame XY Z is recognized as simple angular velocity. Denotethis angular velocity by 1 and express it as

1 = 1I = 1I (2.8.23)

Next, rotate the XYZ frame about the Z axis by an angle 2 to obtain the xyz frame. Theangular velocity of the xyz frame with respect to the intermediate XYZ frame is also simple angularvelocity and it can be expressed as

2 = 2K = 2k (2.8.24)

The angular velocity of the xyz frame with respect to the XY Z frame can then be expressed as thesum of the two angular velocities

= 1 + 2 = 1I + 2K (2.8.25)


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and we can use coordinate transformations to express the angular velocity in terms of the unit vectorsofXY Z, XYZ, or xyz. In general, it is more convenient to express the angular velocity of a referenceframe in terms of the unit vectors of that frame. For the example above, the angular velocity of thexyz frame is expressed as

= 1I + 2k (2.8.26)

Since the transformation from XYZ to xyz is a 3-rotation, and from Eq. (2.7.5) we write I =cos 1i sin 1j, thus the expression for angular velocity becomes

= 1 (cos 1i sin 1j) + 2k = 1 cos 1i 1 sin 1j + 2k (2.8.27)

It is clear that cannot be expressed as the derivative of another vector, even though the twocomponents of the angular velocity, 1 and 2, are differentiable when evaluated individually. Thesituation does not change if is written in terms of the fixed reference frame XY Z.

When using the A and B frame notations, if there is an additional reference frame, say D, webegin with the A frame, rotate it to obtain the D frame and then rotate the D frame to obtain the Bframe. The angular velocities are related by

AB = AD + DB (2.8.28)

As discussed earlier, we usually attach the moving reference frame to the body. The angularvelocity of the reference frame and the angular velocity of the body are then the same. There are caseswhen it is preferable to not attach the moving reference frame to the body. An important application

is rotating axisymmetric bodies. Axisymmetric bodies are analyzed at length in Chapters 9 and 11.A simple illustration is presented here.

Consider Fig. 2.28a and the elbow-shaped pendulum that swings in the xy (or XY) plane. Thez and Z axes are along the same direction and the relation between the xyz and XY Z coordinates isshown in Fig. 2.28b. A disk is rotating with angular velocity with respect to the pendulum. Weattach the xyz frame to the elbow and write the angular velocities of the disk and the reference frameas

elbow = frame = K = k disk = frame + disk/frame = k + i (2.8.29)

The angular velocity of the disk is expressed in terms of coordinates of a reference frame notattached to the disk. It is important to make the distinction between the angular velocities of thereference frame and of the body. This distinction will be discussed again in the next section, withinthe context of angular acceleration. From now on, the angular velocity of a reference frame will bedenoted by f.

In this section, we have defined angular velocity in a number of ways, discussed what it isphysically and derived expressions for derivatives of vectors in moving reference frames. What wehave not done is to come up with a general way to quantify angular velocity as a function of rotationalparameters. The quantification issue will be addressed in Chapter 9, within the context of rotationsof rigid bodies.


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Figure 2.28: Spinning disk on a rotating elbow arm.

Example 2.10

Obtain the angular velocities of the double link in Fig. 2.29 that is supported by a rotating column.

Figure 2.29: a) Double link on a rotating column, b) reference frames for first link.

The XY Z frame rotates with the column and the angular velocity of the column is K.Attaching an xyz frame to the first link, and noting that the link makes an angle of 1 with thevertical, the angular velocity of the first link becomes

1 = K + 1I = K + 1i [a]

where K = sin 1j + cos 1k, so that the angular velocity of the first link becomes

1 = 1i + sin 1j + cos 1k [b]

To obtain the angular velocity of the second link, observe that the angle 2 is also measured


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from the vertical, so that its value is independent of the orientation of link 1. The angular velocityof the second link is found similarly, by replacing the number 1 with 2 in Eq. [b], as well as the unitvectors with their primed counterparts, with the result

2 = K + 2I = 2i

+ sin 2j

+ cos 2k

[c]

2.9 Angular Acceleration and Second Derivatives

This section extends the developments of the previous section and obtains expressions for angularacceleration and for other second derivatives.

2.9.1 Angular Acceleration

An important application of the transport theorem is the calculation of the derivative of angularvelocity, known as angular acceleration. The angular acceleration of a coordinate frame, denoted by, is defined as

=d

dt (2.9.1)

Note that the time derivative here is being taken in the inertial (non-moving) reference frame. Weneed to make the distinction between the reference frame whose angular velocity is considered and thecoordinate axes used to express the angular velocity. Two scenarios are possible:

1. The angular velocity components of the moving frame are expressed in terms of the coordinatesof the moving frame. This is the most widely encountered case. The moving frame is attachedto the body, as shown in Fig. 2.30.

Figure 2.30: Reference frame attached to body.

Let us consider the moving xyz frame and express the angular velocity and angular accelerationof the xyz frame as

= xi + yj + zk = xi + yj + zk (2.9.2)


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2.9. ANGULAR ACCELERATION AND SECOND DERIVATIVES 83

Differentiation of the angular velocity results in

= xi + yj + zk + = xi + yj + zk (2.9.3)

so that the components of the angular acceleration are time derivatives of the components of theangular acceleration

x = x y = y z = z (2.9.4)

Interestingly, we get the same result when the angular velocity is expressed in terms of thecoordinates of the inertial frame. Expressing as = XI + YJ + ZK, it follows thatX = X, Y = Y, Z = Z.

2. The angular velocity components of the moving frame are not expressed in terms of the coordi-nates of the moving frame. An example is shown in Fig. 2.28. This approach is used primarilywhen dealing with rotating axisymmetric bodies and in other cases when it is convenient to doso.

Here, we need to make a distinction between the angular velocities of the two reference frames.As mentioned Sec. 2.8, the notation f denotes that angular velocity of the reference frame usedto express the vector that is differentiated. The transport theorem for the angular accelerationbecomes

= = rel + f (2.9.5)Note that in this case i = i (i = x,y ,z). As an illustration, consider Fig. 2.28, which depictsan elbow-shaped rod and a disk rotating with respect to the collar. The xyz frame is attachedto the elbow-shaped rod, whose angular velocity is elbow = k = k. The disk rotates aboutthe x-axis and the angular velocity of the disk with respect to the collar is disk/elbow = i. The

angular velocity of the disk can then be written as

disk = elbow + disk/elbow = k + i (2.9.6)

To obtain the angular acceleration, observe that the angular velocity of the reference framewhose coordinates are used to express disk is the xyz frame, so that f = elbow = k and thatthis reference frame is not attached to the disk. Differentiating the angular velocity term givesthe angular acceleration as

disk = disk = elbow + disk/elbow (2.9.7)

Evaluation of the first term results in

elbow = elbowrel + f elbow = k (2.9.8)and the cross product term vanishes because f = elbow. The second term becomes

disk/elbow = disk/elbowrel

+ f disk/elbow

= i + k (k + i) = i + j (2.9.9)so that the angular acceleration of the disk is

disk = i + j + k (2.9.10)


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For the general case involving more than one reference frame, say A, C, F, B

AB = AC + CF + FB (2.9.11)

the angular acceleration

A

B

is obtained by differentiating each term individually, as shown below.A d

dtAC = AC A

d

dtCF = CF + AC CF

A d

dtFB = FB + AF FB (2.9.12)

The above results ar

vehicle dynamics ch 2 & 3

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