Dynamics of Mechanical Systems 2009 Part 3 pptx

Hence,unlike velocity, angular velocity cannot be considered as the derivative of a single quantity.Nevertheless, it is possible to define the angular velocity in terms of derivatives of

In matrix form these equations may be expressed as:

(4.3.8)

where N and n represent the columns of the Ni and ni, and S is the matrix triple product:

(4.3.9)where, as before, A, B, and C are:

Example 4.3.1: A 1–3–1 (Euler Angle) Rotation Sequence

Consider rotating B about n1, then n3, and then n1 again through angles θ1, θ2, and θ3 Inthis case, the configuration graph takes the form as shown in Figure 4.3.9 With the rotationangles being θ1, θ2, and θ3, the transformation matrices are:

(4.3.11)

and the general transformation matrix becomes:

(4.3.12)

FIGURE 4.3.7

A rigid body B with a general

orientation in a reference frame R.

0

0

00

00

00

Hence, the unit vectors are related by the expressions:

(4.3.13)

and

(4.3.14)

The angles θ1, θ2, and θ3 are Euler orientation angles and the rotation sequence is referred

to as a 1–3–1 sequence (The angles α, β, and γ are called dextral orientation angles or Bryan orientation angles and the rotation sequence is a 1–2–3 sequence.)

Of all kinematic quantities, angular velocity is the most fundamental and the most useful

in studying the motion of rigid bodies In this and the following three sections we willstudy angular velocity and its applications

We begin with a study of simple angular velocity, where a body rotates about a fixed axis

Specifically, let B be a rigid body rotating about an axis Z–Z fixed in both B and a reference

frame R as in Figure 4.4.1 Let n be a unit vector parallel to Z–Z as shown Simple angular velocity is then defined to be a vector parallel to n measuring the rotation rate of B in R.

To quantify this further, consider an end view of B and of axis Z–Z as in Figure 4.4.2 Let X and Y be axes fixed in R and let L be a line fixed in B and parallel to the X–Y plane.

Let θ be an angle measuring the inclination of L relative to the X-axis as shown Then, theangular velocity ω (simple angular velocity) of B in R is defined to be:

Observe that θ measures the rotation of B in R It is also a measure of the orientation of

B in R In this context, the angular velocity of B in R is a measure of the rate of change of orientation of B in R.

The simple angular acceleration α of B in R is then defined as the time rate of change of

the angular velocity That is,

(4.4.2)

If we express α and β in the forms:

(4.4.3)then α, ω, and θ are related by the expressions:

(4.4.4)

By integrating we have the relations:

(4.4.5)

where θo and ωo are values of θ and ω at some initial time to

From the chain rule for differentiation we have:

(4.4.6)Then by integrating we obtain:

Example 4.4.1: Revolutions Turned Through During Braking

Suppose a rotor is rotating at an angular speed of 100 rpm Suppose further that the rotor

is braked to rest with a constant angular deceleration of 5 rad/sec2 Find the number N

of revolutions turned through during braking

Solution: From Eq (4.4.8), when the rotor is braked to rest, its angular speed ω is zero.The angle θ turned through during braking is, then,

Hence, the number of revolutions turned through is:

Angular velocity may be defined intuitively as the time rate of change of orientation.Generally, however, no single quantity defines the orientation for a rigid body Hence,unlike velocity, angular velocity cannot be considered as the derivative of a single quantity.Nevertheless, it is possible to define the angular velocity in terms of derivatives of a set

of unit vectors fixed in the body These unit vector derivatives thus provide a measure ofthe rate of change of orientation of the body

Specifically, let B be a body whose orientation is changing in a reference frame R, as

depicted in Figure 4.5.1 Let n1, n2, and n3 be mutually perpendicular unit vectors as shown

Then, the angular velocity of B in R, written as RωB, is defined as:

(4.5.1)

The angular velocity vector has several properties that are useful in dynamical analyses.Perhaps the most important is the property of producing derivatives by vector multipli-

cation: specifically, let c be any vector fixed in B Then, the derivative of c in R is given

by the single expression:

d dt

To see this, let c be expressed in terms of the unit vectors n1, n2, and n3 That is, let

(4.5.3)

Because c is fixed in B, the scalar components c i (i = 1, 2, 3) are constants; hence, the

derivative of c in R is:

(4.5.4)Next, consider the product RωB× n1:

Because any vector V may be expressed as (V · n1)n1 + (V · n2)n2 + (V · n3)n3, we see that

the right side of Eq (4.5.8) is an identity for dn1/dt Thus, we have the result:

(4.5.9)Similarly, we have the companion results:

(4.5.10)Finally, by using these results in Eq (4.5.4) we have:

d dt d

dt

d dt

d dt

dt

d dt

d dt

Observe the pattern of the terms in Eq (4.5.1) They all have the same form, and theymay be developed from one another by simply permuting the indices.

Example 4.5.1: Simple Angular Velocity

We may also observe that Eq (4.5.1) is consistent with our earlier results on simple angular

velocity To see this, let B rotate in R about an axis parallel to, say, n1, as shown in Figure

4.5.2 Let X–X be fixed in both B and R Then, from Eq (4.4.1), the angular velocity of B

in R is:

(4.5.12)where, as before, θ measures the rotation angle From Eq (4.5.1), we see that RωB may beexpressed as:

(4.5.13)

Because n1 is fixed, parallel to axis X–X, its derivative is zero; hence, the third term in Eq.

(4.5.13) is zero The first two terms may be evaluated using Eq (3.5.7) Specifically,

(4.5.14)

By substituting into Eq (4.5.13), we have:

(4.5.15)which is identical to Eq (4.5.11)

Consider next the differentiation of a vector with respect to different reference frames

Specifically, let V be the vector and let R and be two distinct reference frames Let i

be mutually perpendicular unit vectors fixed in , as represented in Figure 4.6.1 Let V

be expressed in terms of the i as:

d dt

Because the i are fixed in , the derivative of V in is obtained by simply

differen-tiating the scalar components in Eq (4.6.1) That is,

(4.6.2)

Next, relative to reference frame R, the derivative of V is:

(4.6.3)

where the second equality is determined from Eq (4.6.2) Because the i are fixed in ,

the derivatives in R are:

(4.6.4)Hence, R dV/dt becomes:

or

(4.6.5)

Because there were no restrictions on V, Eq (4.6.5) may be written as:

(4.6.6)where any vector quantity may be inserted in the parentheses

R dV dt=RˆdVdt+R Rˆ×V

ωω

R d( ) dt=Rˆd( ) dt+R Rˆ×( )

ωω

Example 4.6.1: Effect of Earth Rotation

Equation (4.6.6) is useful for developing expressions for velocity and acceleration ofparticles moving relative to rotating bodies To illustrate this, consider the case of a rocket

R launched vertically from the Earth’s surface Let the vertical speed of R relative to the Earth (designated as E) be V0 Suppose we want to determine the velocity of R in an astronomical reference frame A in which E rotates for the cases when R is launched from

(a) the North Pole, and (b) the equator

Solution: Consider a representation of R, E, and A as in Figure 4.6.2a and b, where i,

j, and k are mutually perpendicular unit vectors fixed in E, with k being along the

north/south axis, the axis of rotation of E In both cases the velocity of R in A may be

expressed as:

(4.6.7)

where O is the center of the Earth which is also fixed in A.

a For launch at the North Pole, the position vector OR is simply (r + h)k, where

r is the radius of E (approximately 3960 miles) and h is the height of R above the surface of E Thus, from Eq (4.6.6), AVR is:

(4.6.8)

Observe that angular velocity of E in A is along k with a speed of one revolution

per day That is,

b For the launch at the Equator, the position vector OR is (r + h)i In this case, Eq.

Equation (4.6.6) is useful for establishing the addition theorem for angular velocity — one

of the most important equations of rigid body kinematics Consider a body B moving in

a reference frame , which in turn is moving in a reference frame R as depicted in Figure

4.7.1 Let V be an arbitrary vector fixed in B Using Eq (4.6.6), the derivative of V in R is:

Hence, by substituting into Eq (4.7.1) we have:

body B may itself be considered as a reference frame, Eq (4.7.4) may be rewritten in the

form:

(4.7.5)Equation (4.7.5) may be generalized to include reference frames That is, suppose a

reference frame R n is moving in a reference frame R0 and suppose that there are (n –1)intermediate reference frames, as depicted in Figure 4.7.2 Then, by repeated use of

Eq (4.7.5), we have:

(4.7.6)The addition theorem together with the configuration graphs of Section 4.3 are useful

for obtaining more insight into the nature of angular velocity Consider again a body B moving in a reference frame R as in Figure 4.7.3 Let the orientation of B in R be described

by dextral orientation angles α, β, and γ Let ni and Ni (i = 1, 2, 3) be unit vector sets fixed

in B and R, respectively Then, from Figure 4.3.8, the configuration graph relating n i and

Ni is as reproduced in Figure 4.7.4 Recall that the unit vector sets i and i are fixed

in intermediate reference frames R1 and R2 used in the development of the configurationgraph The horizontal lines in the graph signify axes of simple angular velocity (see Section4.4) of adjoining reference frames The respective rotation angles are written at the base

of the graph Therefore, from Eq (4.7.6) we have:

(4.7.7)From the configuration graph of Figure 4.7.4 we have (see Eq (4.4.1)):

(4.7.8)Hence, RωB is:

Example 4.7.1: A Simple Gyro

(See Reference 4.1.) A circular disk (or gyro), D, is mounted in a yoke (or gimbal), Y, as shown in Figure 4.7.5 Y is mounted on a shaft S and is free to rotate about an axis that

is perpendicular to S S in turn can rotate about its own axis in bearings fixed in a reference frame R Let D have an angular speed Ω relative to Y; let the rotation of Y in S be measured

by the angle φ; and let the rotation of S in R be measured by the angle ψ Let NYi and NRi

be configured so that when Y is vertical and when the plane of D is parallel to S, the unit

vector sets are mutually aligned Also in this configuration, let the orientation angles φand ψ be zero Determine the angular velocity of D in R by constructing a configuration

graph as in Figure 4.7.4 and by using the addition theorem of Eq (4.7.6)

Solution: Let Ω be the derivative of a rotation angle θ of D in Y; let NDi (i = 1, 2, 3) be

mutually perpendicular unit vectors fixed in D; and let the NDi be mutually aligned with

the NYi when θ is zero Similarly, let NSi (i = 1, 2, 3) be mutually perpendicular unit vectors

fixed in S which are aligned with the NRi when ψ is zero Then, a configuration graph

representing the various unit vector sets and the orientation angles can be constructed, as

shown in Figure 4.7.6 From this graph and from Eq (4.7.6), the angular velocity of D in

R may be expressed as:

(4.7.12)

The configuration graph is useful for expressing RωD solely in terms of one of the unit

vector sets For example, in terms of the yoke unit vectors NYi, RωD is:

(4.7.13)where Ω is (see also Problem P4.7.1)

The angular acceleration of a body B in a reference frame R is defined as the derivative

in R of the angular velocity of B in R Specifically,

(4.8.1)Unfortunately, there is not an addition theorem for angular acceleration analogous to thatfor angular velocity That is, in general,

To see this, consider three reference frames R0, R1, and R2 as in Eq (4.7.5):

(4.8.3)

By differentiating in R0 we obtain:

or

(4.8.4)Consider the last term: From Eq (4.6.6) we have:

In some occasions the angular acceleration of a body B in a reference frame R ( RαB) may

be computed by simply differentiating the scalar components of the angular velocity of

B in R ( RωB) This occurs if RωB is expressed either in terms of unit vectors fixed in R or in terms of unit vectors fixed in B To see this, consider first the case where the limit vectors

of RωB are fixed in R That is, let N i (i = 1, 2, 3) be mutually perpendicular unit vectors fixed in R and let RωB be expressed as:

(4.8.7)

Then, because the Ni are fixed in R, their derivatives in R are zero Hence, in this case,

RαB becomes:

(4.8.8)

Next, consider the case where the unit vectors of RωB are fixed in B That is, let ni (i =

1, 2, 3) be mutually perpendicular unit vectors fixed in B and let RωB be expressed as:

Then RαB is:

(4.8.10)

Observe that the last three terms each involve derivatives of unit vectors fixed in B From

Eq (4.5.2) these derivatives may be expressed as:

(4.8.11)Hence, the last three terms of Eq (4.8.10) may be expressed as:

(4.8.12)

Therefore, RααααB becomes:

(4.8.13)

Example 4.8.1: A Simple Gyro

See Example 4.7.1 Consider again the simple gyro of Figure 4.7.5 and as shown in Figure

4.8.1 Recall from Eq (4.7.13) that the angular velocity of the gyro D in the fixed frame R

may be expressed as:

(4.8.14)

where as before Ω (the gyro spin) is ; φ and ψ are orientation angles as shown in Figure

4.8.1; and NY1, NY2, and NY3 are unit vectors fixed in the yoke Y Determine the angular acceleration of D in R.

Solution: By differentiating in Eq (4.4.14) we obtain:

R B i

Y R Y R

Because the NYi (i = 1, 2, 3) are fixed in the yoke Y, their derivatives may be evaluated

from the expression:

(4.8.16)where from Eqs (4.7.12) and (4.7.13) RωY is:

(4.8 17)Hence, the R dN Yi /dt are:

is not the case, as seen in Problem P4.8.2 It happens that the NYi are the preferred unitvectors for simplicity

FIGURE 4.8.1

The gyro of Figure 4.7.5.

R Yi

4.9 Relative Velocity and Relative Acceleration of Two Points

on a Rigid Body

Consider a body B moving in a reference frame R as in Figure 4.9.1 Let P and Q be points fixed in B Consider the velocity and acceleration of P and Q and their relative velocity and acceleration in reference frame R Let O be a fixed point (the origin) of R Let vectors

p and q locate P and Q relative to O and let vector r locate P relative to Q Then, from

Figure 4.9.1, these vectors are related by the expression:

(4.9.4)

By differentiating in Eq (4.9.4), we obtain the following relations for accelerations:

(4.9.5)and

(4.9.6)

FIGURE 4.9.1

A body B moving in a reference

frame R with points P and Q fixed

Equations (4.9.4) and (4.9.6) are sometimes used to provide an interpretation of rigidbody motion as being a superposition of translation and rotation Specifically, at any

instant, from Eq (4.9.4) we may envision a body as translating with a velocity VQ and

rotating about Q with an angular velocity RωB

Example 4.9.1: Relative Movement of a Sports Car Operator’s Hands During

a Simple Turn

To illustrate the application of Eqs (4.9.4) and (4.9.5), suppose a sports car operatortraveling at a constant speed of 15 miles per hour, with hands on the steering wheel at 10o’clock and 2 o’clock, is making a turn to the right with a turning radius of 25 feet Supposethe steering wheel has a diameter of 12 inches and is in the vertical plane Suppose furtherthat the operator is turning the steering wheel at one revolution per second Find thevelocity and acceleration of the motorist’s left hand relative to the right hand

Solution: Let the movement of the car be represented as in Figure 4.9.2, and let the

steering wheel be represented as in Figure 4.9.3, where nx, ny, and nz are mutually

per-pendicular unit vectors fixed in the car Let nx be forward; nz be vertical, directed up; and

ny be to the left Let W represent the steering wheel, C represent the car, and S the road surface (fixed frame) Let O represent the center of W, and let L and R represent the

motorist’s left and right hands The desired relative velocity and acceleration (SVL/R and

saL/R) may be obtained directly from Eqs (4.9.4) and (4.9.5) once the angular velocity and

the angular acceleration of W are known From the addition theorem for angular velocity (Eq (4.7.6)), the angular velocity of the steering wheel W in S is:

(4.9.7)

From the data presented, the angular velocity of the car C in S is:

(4.9.8)

where V is the speed of C (15 mph or 22 ft/sec) and ρ is the turn radius Similarly, the

angular velocity of the steering wheel W in C is:

(4.9.9)Hence, SωW is:

By differentiating in Eq (4.9.10), the angular acceleration of W in S is:

Example 4.9.2: Absolute Movement of Sports Car Operator’s Hands

To further illustrate the application of Eqs (4.9.4) to (4.9.6), suppose we are interested indetermining the velocity and acceleration of the sports car operator’s left hand in the

previous example Specifically, find the velocity and acceleration of L in S.

Solution: Observe that the steering wheel hub O is fixed in both the steering wheel W and the car C Observe further that O moves on the 25-ft-radius curve (or circle) at 15 mph (or 22 ft/sec) Therefore, the velocity and acceleration of O are:

(4.9.17)and

(4.9.18)

Example 4.9.3: A Rod with Constrained End Movements

As a third example illustrating the use of Eqs (4.9.4) and (4.9.6) consider a rod whose

ends A and B are restricted to movement on horizontal and vertical lines as in Figure 4.9.4 Let the rod have length 13 m Let the velocity and acceleration of end A be:

(4.9.19)

where nx, ny, and nz are unit vectors parallel to the coordinate axes as shown Let theangular velocity and angular acceleration of the rod along the rod itself be zero That is,

(4.9.20)

where n is a unit vector along the rod Find the velocity and acceleration of end B for the

rod position shown in Figure 4.9.4

FIGURE 4.9.3

Sports car steering wheel W and

left and right hands L and R.

Solution: From Eq (4.9.4), the velocity of end B may be expressed as:

(4.9.21)

For the rod position of Figure 4.9.4, AB is:

(4.9.22)Let ωAB be expressed in the form:

(4.9.23)

Let VB be expressed as:

(4.9.24)Then, by substituting from Eqs (4.9.19), (4.9.22), (4.9.23), and (4.9.24) we have:

Similarly, from Eq (4.9.6), the acceleration of end A may be expressed as:

(4.9.30)Let αAB be expressed in the form:

αx= −2 083 rad sec2, αy= −0 641 rad sec2, αz= −0 213 rad sec2

a B= −29 33 m s

4.10 Points Moving on a Rigid Body

Consider next a point P moving on a body B which in turn is moving in a reference frame

R as depicted in Figure 4.10.1 Let Q be a point fixed in B Let p and q be position vectors locating P and Q relative to a fixed point O in R Let vector r locate P relative to Q Then,

from Figure 4.10.1, we have:

Suppose that an instant of interest P happens to coincide with Q Then, at that instant,

r is zero and RVP is:

A point P moving on a body B,

moving in reference frame R.

By differentiating in Eq (4.10.4), we can obtain a relation determining the acceleration

of P That is,

(4.10.7)

where, as before, RaP and RaQ are the acceleration of P and Q in R, and RαB is the acceleration

of B in R By using Eq (4.6.6), we can express R d BVP /dt as:

(4.10.8)Then, by using Eq (4.10.3), RaP may be written as:

or

(4.10.9)

Suppose again that at an instant of interest P happens to coincide with Q Then, r is

zero and RaP is:

(4.10.10)

Hence, in general, if P* is the point of B coinciding with P we have:

(4.10.11)

The term 2RωB × BVP is called the Coriolis acceleration, after the French mechanician de

Coriolis (1792–1843) who is credited with being the first to discover it We have alreadyseen this term in our analysis of the movement of a point in a plane in Chapter 3 (see Eq.(3.8.7)) This term is not generally intuitive, and it often gives rise to surprising andunexpected effects

Example 4.10.1: Movement of Sports Car Operator’s Hands

Equations (4.10.6) and (4.10.11) may also be used to determine the velocity and acceleration

of the sports car operator’s left hand of Example 4.9.2 Recall that the sports car is making

a turn to the right at 15 mph with a turn radius of 25 feet, and that the operator’s lefthand is at 10 o’clock on a 12-in.-diameter vertical steering wheel, as in Figures 4.10.2 and4.10.3 Recall further that the operator is turning the wheel clockwise at one revolutionper second, as in Figure 4.10.3

Solution: From Eq (4.10.6), the velocity of the left hand L may be expressed as:

where L* is that point of the car C where L is located at the instant of interest and where,

as before, S is the fixed road surface From Figure 4.10.3, CVL is:

From Figure 4.10.2, SaL* is:

to the surface but is still in contact with the surface with the contacting point (or points)having zero velocity relative to the surface

Rolling may be defined analytically as follows: Let S be a surface and let B be a body that rolls on S as depicted in Figure 4.11.1 (S could be a portion of a body upon which B rolls.) Let B and S be counterformal so that they are in contact at a single point Let C be the point of B that is in contact with S Rolling then occurs when: