Dynamics of Mechanical Systems 2009 Part 4 docx

If B is not at rest but has planar motion, then at most one point of B, in a given plane of motion, is a center of zero velocity.. To summarize, we see that if a body has planar motion,

where ωQP and ααααQP are the angular velocity and angular acceleration, respectively, of the

connecting rod QP Because QP has planar motion, we see from Figure 5.3.4 that ωQP andααα

αQP may be expressed as:

(5.3.17)

where nz (= nx× ny) is perpendicular to the plane of motion Also, the position vector QP

may be expressed as:

(5.3.18)

By carrying out the indicated operations of Eqs (5.3.15) and (5.3.16) and by using Eqs

(5.3.13) and (5.3.14), vP and aP become:

(5.3.19)and

(5.3.20)

Observe from Figure 5.3.4 that P moves in translation in the n x direction Therefore, the

velocity and acceleration of P may be expressed simply as:

ΩΩ

(5.3.24)and

(5.3.25)

Observe further from Figure 5.3.4 that x may be expressed as:

(5.3.26)and that from the law of sines we have:

(5.3.27)

By differentiating in Eqs (5.3.26) and (5.3.27), we immediately obtain Eqs (5.3.22) and(5.3.23) Finally, observe that by differentiating in Eqs (5.3.22) and (5.3.23) we obtain Eqs.(5.3.24) and (5.3.25)

If a point O of a body B with planar motion has zero velocity, then O is called a center of zero velocity If O has zero velocity throughout the motion of B, it is called a permanent center of zero velocity If O has zero velocity during only a part of the motion of B, or even for only an instant during the motion of B, then O is called an instant center of zero velocity For example, if a body B undergoes pure rotation, then points on the axis of rotation are permanent centers of zero velocity If, however, B is in translation, then there are no points of B with zero velocity If B has general plane motion, there may or may not be points of B with zero velocity However, as we will see, if B has no centers of zero velocity within itself, it is always possible to mathematically extend B to include such points In

this latter context, bodies in translation are seen to have centers of zero velocity at infinity(that is, infinitely far away)

The advantage, or utility, of knowing the location of a center of zero velocity can beseen from Eq (5.3.10):

(5.4.1)

where P and Q are points of a body B and where r locates P relative to Q If Q is a center

of zero velocity, then vQ is zero and vP is simply:

r

vP=vQ+ ×ωω r

vP= ×ωω r

This means that during the time that Q is a center of zero velocity, P moves in a circle about Q Indeed, during the time that Q is a center of zero velocity, B may be considered

to be in pure rotation about Q Finally, observe in Eq (5.4.2) that ω is normal to the plane

of motion Hence, we have:

(5.4.3)

where the notation is defined by inspection

If a body B is at rest, then every point of B is a center of zero velocity If B is not at rest but has planar motion, then at most one point of B, in a given plane of motion, is a center

of zero velocity

To prove this last assertion, suppose B has two particles, say O and Q, with zero velocity.

Then, from Eq (5.3.10), we have:

(5.4.4)

where r locates O relative to Q If both O and Q have zero velocity then,

(5.4.5)

If O and Q are distinct particles, then r is not zero Because ωω is perpendicular to r, Eq.

(5.4.5) is then satisfied only if ωω is zero The body is then in translation and all points have

the same velocity Therefore, because both O and Q are to have zero velocity, all points

have zero velocity, and the body is at rest — a contradiction to the assumption of a movingbody That is, the only way that a body can have more than one center of zero velocity,

in a plane of motion, is when the body is at rest

We can demonstrate these concepts by graphical construction That is, we can show that

if a body has planar motion, then there exists a particle of the body (or of the mathematicalextension of the body) with zero velocity We can also develop a graphical procedure forfinding the point

To this end, consider the body B in general plane motion as depicted in Figure 5.4.1 Let P and Q be two typical distinct particles of B Then, if B has a center O of zero velocity,

P and Q may be considered to be moving in a circle about O Suppose the velocities of P and Q are represented by vectors, or line segments, as in Figure 5.4.2 Then, if P and Q rotate about a center O of zero velocity, the velocity vectors of P and Q will be perpen- dicular to lines through O and P and Q, as in Figure 5.4.3.

Observe that unless the velocities of P and Q are parallel, the lines through P and Q

perpendicular to these velocities will always intersect Hence, with non-parallel velocities,

the center O of zero velocity always exists.

If the velocities of P and Q are parallel with equal magnitudes, and the same sense, then

they are equal That is,

(5.4.6)

where the last equality follows from Eq (5.3.10) with r locating P relative to Q If P and

Q are distinct, r is not zero; hence, ωω is zero B is then in translation Lines through P and

Q perpendicular to v P and vQ will then be parallel to each other and thus not intersect(except at infinity) That is, the center of zero velocity is infinitely far away (see Figure5.4.4)

If the velocities of P and Q are parallel with non-equal magnitudes, then the center of zero velocity will occur on the line connecting P and Q To see this, first observe that the relative velocities of P and Q will have zero projection along the line connecting P and Q:

That is, from Eq (5.3.10), we have:

(5.4.7)

Hence, vP/Q must be perpendicular to r (This simply means that P and Q cannot approach

or depart from each other; otherwise, the rigidity of B would be violated.) Next, observe

that if vP and vQ are parallel, their directions may be defined by a common unit vector n.

That is,

(5.4.8)

where vP, vQ, and vP/Q are appropriate scalars By comparing Eqs (5.4.7) and (5.4.8) we see

that n must be perpendicular to r Hence, when vP and vQ are parallel but with non-equal

magnitudes, their directions must be perpendicular to the line connecting P and Q

There-fore, lines through P and Q and perpendicular to v P and vQ will coincide with each other

and with the line connecting P and Q (see Figure 5.4.5).

Next, observe from Eqs (5.3.10) and (5.4.3) that, if O is the center of zero velocity, then

the magnitude of vP is proportional to the distance between O and P Similarly, the

magnitude of vQ is proportional to the distance between O and Q These observations enable us to locate O on the line connecting P and Q Specifically, from Eq (5.4.3), the distance from P to O is simply vP/ω

From a graphical perspective, O can be located as in Figure 5.4.6 Similar triangles are

formed by O, P, Q and the “arrow ends” of v P and vQ

FIGURE 5.4.3

Location of center O of zero velocity by the

inter-section of lines perpendicular to velocity vectors.

To summarize, we see that if a body has planar motion, there exists a unique point O

of the body (or the body extended) that has zero velocity O may be located at the

intersection of lines through two points that are perpendicular to the velocity vectors of

the points Alternatively, O may be located on a line perpendicular to the velocity vector

of a single point P of the body at a distance vP /ω from P (see Figure 5.4.7) Finally, when the zero velocity center O is located, the body may be considered to be rotating about O Then, the velocity of any point P of the body is proportional to the distance from

O to P and is directed parallel to the plane of motion of B and perpendicular to the line segment OP.

Consider the planar linkage shown in Figure 5.5.1 It consists of three links, or bars (B1,

B2, and B3), and four joints (O, P, Q, and R) Joints O and R are fixed while joints P and Q

may move in the plane of the linkage The ends of each bar are connected to a joint; thus,

the bars may be identified (or labeled) by their joint ends That is, B1 is OP, B2 is PQ, B3 is

QR In this context, we may also imagine a fourth bar B4 connecting the fixed joints O and

R The system then has four bars and is thus referred to as a four-bar linkage.

The four-bar linkage may be used to model many physical systems employed in anisms and machines, particularly cranks and connecting rods The four-bar linkage isthus an excellent practical example for illustrating the concepts of the foregoing sections

mech-In this context, observe that bars OP(B1) and RQ(B3) undergo pure rotation, while bar

PQ(B2) undergoes general plane motion, and bar OR(B4) is fixed, or at rest

FIGURE 5.4.5

A body B with particles P and Q having parallel

velocities with non-equal magnitudes.

FIGURE 5.4.6

Location of the center of zero velocity for a body having distinct particles with parallel but unequal velocities.

FIGURE 5.4.7

Location of the center for zero velocity knowing

the velocity of one particle and the angular

The system of Figure 5.5.1 has one degree of freedom: The rotations of bars OP(B1) and

RQ(B3) each require two coordinates, and the general motion of bar PQ(B2) requires an

additional three coordinates for a total of five coordinates Nevertheless, requiring joint P

to be connected to both B1 and B2 and joint Q to be connected to both B2 and B3 produces

four position (or coordinate) constraints Thus, there is but one degree of freedom.

A task encountered in the kinematic analyses of linkages is that of describing theorientation of the individual bars One method is to define the orientations of the bars in

terms of angles that the bars make with the horizontal (or X-axis) such as θ1, θ2, and θ3 as

in Figure 5.5.2 Another method is to define the orientation in terms of angles that the

bars make with the vertical (or Y-axis) such as φ1, φ2, and φ3 as in Figure 5.5.2 A thirdmethod is to define the orientations in terms of angles that the bars make with each other,

as in Figure 5.5.3 The latter angles are generally called relative orientation angles whereas the former are called absolute orientation angles.

Relative orientation angles are usually more meaningful in describing the configuration

of a physical system Absolute orientation angles are usually easier to work with in theanalysis of the problem In our example, we will use the first set of absolute angles θ1, θ2,and θ3

Because the system has only one degree of freedom, the orientation angles are notindependent They may be related to each other by constraint equations obtained byconsidering the linkage of four bars as a closed loop: Specifically, consider the positionvector equation:

(5.5.1)

This equation locates O relative to itself through position vectors taken around the loop

of the mechanism It is called the loop closure equation.

Let 1, 2, 3, and 4 be the lengths of bars B1, B2, B3, and B4 Then, Eq (5.5.1) may bewritten as:

(5.5.2)

where λλλλ1, λλλλ2, λλλλ3, and λλλλ4 are unit vectors parallel to the rods as shown in Figure 5.5.4 These

vectors may be expressed in terms of horizontal and vertical unit vectors nx and ny as:

Hence, by substituting into Eq (5.5.2) we have:

(5.5.4)

This immediately leads to two scalar constraint equations relating θ1, θ2, and θ3:

(5.5.5)and

(5.5.6)

The objective in a kinematic analysis of a four-bar linkage is to determine the velocityand acceleration of the various points of the linkage and to determine the angular velocitiesand angular accelerations of the bars of the linkage In such an analysis, the motion of

one of the three moving bars, say B1, is generally given The objective is then to determine

the motion of bars B1 and B2 In this case, B1 is the driver, and B2 and B3 are followers.

The procedures of Section 5.4 may be used to meet these objectives To illustrate thedetails, consider the specific linkage shown in Figure 5.5.5 The bar lengths and orientationsare given in the figure Also given in Figure 5.5.5 are the angular velocity and angular

acceleration of OP(B1) B1 is thus a driver bar and PQ(B2) and QR(B3) are follower bars

Our objective, then, is to find the angular velocities and angular accelerations of B2 and

B3 and the velocity and acceleration of P and Q.

To begin the analysis, first observe that, by comparing Figures 5.5.4 and 5.5.5, the anglesand lengths of Figure 5.5.5 satisfy Eqs (5.5.5) and (5.5.6) To see this, observe that 1, 2,

3, 4, θ1, θ2, θ3, and θ4 have the values:

x y

l1cosθ1+l2cosθ2+l3cosθ3=l4

l1sinθ1+l2sinθ2+l3sinθ3=0

(5.5.9)

Next, recall that B1 and B3 have pure rotation about points O and R, respectively, and that B2 has general plane motion

Third, let us introduce unit vectors λλλλi and ννννi (i = 1, 2, 3) parallel and perpendicular to

the bars as in Figure 5.5.6 Then, in the configuration shown, the λλλλi and ννννi may be expressed

in terms of horizontal and vertical unit vectors nx and ny as:

Because B2 has general plane motion, the velocity of Q may be expressed as:

(5.5.15)

where ω2 is the angular speed of B2 Note that Q is fixed in both B2 and B3

Because B3 has pure rotation with center R, Q moves in a circle about R Hence, v Q may

be expressed as:

(5.5.16)

where ω3 is the angular speed of B3

Comparing Eqs (5.5.15) and (5.5.16) we have the scalar equations:

(5.5.17)and

(5.5.18)Solving for ω2 and ω3 we obtain:

(5.5.19)

Hence, vQ becomes:

(5.5.20)

Observe that in calculating the angular speeds of B2 and B3 we could also use an analysis

of the instant centers as discussed in Section 5.4 Because the velocities of P and Q are perpendicular to, respectively, B1(OP) and B3(QR), we can construct the diagram shown

in Figure 5.5.7 to obtain ω2, ω3, and vQ By extending OP and RQ until they intersect, we obtain the instant center of zero velocity of B2 Then, IP and IQ are perpendicular to,

respectively, vP and vQ Triangle IOR forms a 45° right triangle Hence, the distance between

I and P is (6.098 – 2.0) m, or 4.098 m Because vP is 10 m/sec, ω2 is:

Similarly, the distance IQ is 3.67 m, and v Q is, then,

(5.5.22)Then ω3 becomes:

Because Q also moves in a circle about R, we have:

(5.5.26)

Comparing Eqs (5.5.25) and (5.5.26), we have:

(5.5.27)and

(5.5.28)Solving for α2 and α3 we obtain:

(5.5.29)

Hence, the acceleration of Q is:

(5.5.30)Observe how much more effort is required to obtain accelerations than velocities

Consider next a chain of identical pin-connected bars moving in a vertical plane and

supported at one end as represented in Figure 5.6.1 Let the chain have N bars, and let their orientations be measured by N angles θi (i = 1,…, N) that the bars make with the vertical Z-axis as in Figure 5.6.1 Because N angles are required to define the configuration and positioning of the system, the system has N degrees of freedom A chain may be considered to be a finite-segment model of a cable; hence, an analysis of the system of

Figure 5.6.1 can provide insight into the behavior of cable and tether systems

A kinematical analysis of a chain generally involves determining the velocities andaccelerations of the connecting joints and the centers of the bars and also the angularvelocities and angular accelerations of the bars To determine these quantities, it is easier

to use the absolute orientation angles of Figure 5.6.1 than the relative orientation angles

shown in Figure 5.6.2 The relative angles have the advantage of being more intuitive intheir description of the inclination of the bars.

In our discussion we will use absolute angles because of their simplicity in analysis To

begin, consider a typical pair of adjoining bars such as B j and B k as in Figure 5.6.3 Let theconnecting joints of the bars be Oj, Ok, and O as shown, and let Gj and Gk be the centers

of the bars Let nj3, nk3 and nj1, nk1 be unit vectors parallel and perpendicular, respectively,

to the bars in the plane of motion

By using this notation, the system may be numbered and labeled serially from the

support pin O as in Figure 5.6.4 Let N x, Ny, and Nz be unit vectors parallel to the X-, Y-, and Z-axes, as shown.

Because the X–Z plane is the plane of motion, the angular velocity and angular eration vectors will be perpendicular to the X–Z plane and, thus, parallel to the Y-axis.

accel-Specifically, the angular velocities and angular accelerations may be expressed as:

(5.6.1)

Next, the velocity and acceleration of G1, the center of B1, may be readily obtained by

noting that G1 moves on a circle Thus, we have:

θ θ

θ

N N-1

Observe how much simpler the expressions are when the local (as opposed to global) unit

vectors are used

Consider next the velocity and acceleration of the center G2 and the distal joint O3 of B2.From the relative velocity and acceleration formulas, we have (see Eqs (3.4.6) and (3.4.7)):

=l ˙˙ cos[ (θ θ θ−˙ sinθ) + −( θ˙˙ sinθ θ−˙ cosθ ) ]

vG2=vO2+vG O2/ 2 and aG2 =aO2+aG O2/ 2

/ =ωω ×( )l =( )l θ˙

2 2

(5.6.18)

Observe that using the local unit vectors again leads to simpler expressions (compareEqs (5.6.11) and (5.6.12) with Eqs (5.6.13) and (5.6.14)) Nevertheless, with the use of thelocal unit vectors we have mixed sets in the individual equations For example, in Eq.(5.6.11), the unit vectors are neither parallel nor perpendicular; hence, the components arenot readily added Therefore, for computational purposes, the use of the global unit vectors

(5.6.20)

where nj1, nj3, and θj are associated with the bar B j , immediately preceding B k In terms of

NX and NZ, these expressions become:

2

2 2

˙˙ cos ˙ sin ˙˙ cos ˙ sin

˙˙ sin ˙ cos ˙˙ sin ˙ cos

2 23

2 3

2 3

2 2

2 2

5.7 Instant Center, Analytical Considerations

In Section 5.4, we developed an intuitive and geometrical description of centers of zerovelocity Here, we examine these concepts again, but this time from a more analytical

perspective Consider again a body B moving in planar motion as represented in Figure 5.7.1 Let the X–Y plane be a plane of motion of B Let P be a typical point of B, and let

C be a center of zero velocity of B (That is, C is that point of B [or B extended] that has zero velocity.) Finally, let (x P , y P ) and (x C , y C ) be the X–Y coordinates of P and C, and let

P and C be located relative to the origin O, and relative to each other, by the vectors p P,

where r is the magnitude of r and θ is the inclination of r relative to the X-axis.

From Eq (4.9.4), the velocities of P and C are related by the expression:

(5.7.3)where ωω is the angular velocity of B Because C is a center of zero velocity, we have:

(5.7.4)

FIGURE 5.7.1

A body in plane motion with center

for zero velocity C.

Let nz be a unit vector normal to the X–Y plane generated by n x× ny Then, ω may beexpressed as:

(5.7.5)

where ω is positive when B rotates counterclockwise, as viewed in Figure 5.7.1.

Using Eqs (5.7.1) to (5.7.5), vP may be expressed as:

(5.7.9)

Equation (5.7.9) shows that if we know the location of a typical point P of B, the velocity

of P, and the angular speed of B, we can locate the center C of zero velocity of B.

Let Q be a second typical point of B (distinct from P) Then, from Eq (5.7.9) we have:

(5.7.10)

By comparing the terms of Eqs (5.7.9) and (5.7.10), we have:

(5.7.11)Solving for ω we obtain:

Equation (5.7.12) shows that if we know the velocities of two points of B we can determine the angular velocity of B Then, from Eq (5.7.9), the coordinates (x C , y C) of thecenter of zero velocity can be determined That is,

(5.7.13)

We can verify these expressions using the geometric procedures of Section 5.4 Consider,

for example, a body B moving in the X–Y plane with center of zero velocity C as in Figure 5.7.2 Let P and Q be two points of B whose positions and velocities are known Then, the magnitude of their velocities designated by v P and v Q are related to the angular speed ω

A body B with zero velocity center C

and typical points P and Q.

5.8 Instant Center of Zero Acceleration

We can extend and generalize these procedures to obtain a center of zero acceleration —that is, a point of a body (or the body extended) that has zero acceleration To this end,

consider again a body B moving in planar motion as depicted in Figure 5.8.1 As before, let P and Q be typical points of B, and let C be the sought-after center of zero acceleration Let (x P , y P ), (x Q , y Q ), and (x C , y C ) be the X–Y coordinates of P, Q, and C Let r locate C relative to P Let r have magnitude r and inclination θ relative to the X-axis as shown in

the figure Finally, let ω and α represent the angular speed and angular acceleration of B

Because P and C are fixed in B, their accelerations are related by the expression (see

Eq (4.9.6)):

(5.8.1)

Therefore, if the acceleration of C is zero, then the acceleration of P is:

(5.8.2)

If nz is a unit vector normal to the X–Y plane, then the angular velocity and angular

acceleration vectors may be expressed as (see Eq (5.7.5)):

(5.8.3)

Also, from Figure 5.8.1, the position vector r may be written as:

(5.8.4)Then terms α × r and ω × (ω × r) in Eq (5.8.2) are:

(5.8.5)and

Hence, Eq (5.8.2) becomes:

2

aQ= × + ×αα q ωω ωω( ×q)

Alternatively, Eqs (5.8.11) and (5.8.12) may be used to obtain the angular speed ω andthe angular acceleration α of B if the acceleration of typical points P and Q are known.

To see this, observe first that for point Q expressions analogous to Eqs (5.8.11) and (5.8.12)

are:

(5.8.14)and

(5.8.15)Next, by subtracting these expressions from Eqs (5.8.11) and (5.8.12) we have:

(5.8.20)Solving for ξ and η we obtain:

Finally, from Eq (5.8.18) we have:

(5.8.24)

Hence, α and ω2 are:

(5.8.25)

To illustrate the application of these ideas, consider a circular disk D rolling to the left

in a straight line on a surface S as in Figure 5.8.2 Let Q be the center of D, let O be the contact point (instant center of zero velocity) of D with S, and let P be a point on the periphery or rim of D Finally, let D have radius r, angular speed ω, and angular acceler-

Because O and P are also fixed on D, their velocities and acceleration may be obtained

from the expressions:

(5.8.28)and

(5.8.29)

By substituting from Eqs (5.8.26) and (5.8.27), by recognizing that ωω and αααα are ωnz and

αnz, and by carrying out the indicated operations, we obtain:

(5.8.32)

(5.8.33)

Observe that the velocity of O is zero, as expected, but the acceleration of O is not zero.

To find the point C with zero acceleration, we can use Eqs (5.8.11) and (5.8.12): specifically, for a Cartesian (X–Y) axes system with origin at O, we find the coordinates of C to be:

(5.8.34)

and

(5.8.35)

For positive values of ω and α, the position of C is depicted in Figure 5.8.3.

To verify the results, consider calculating the acceleration of C using the expression:

(5.8.40)where, from Eq (5.8.23), ∆ is:

Location of center C of zero

acceler-ation for rolling disk of Figure 5.8.2.

0

2

,,

Section 5.2 Coordinates, Constraints, Degrees of Freedom

P5.2.1: Consider a pair of eyeglasses to be composed of a frame containing the lenses and

two rods hinged to the frame for fitting over the ears How many degrees of freedom dothe eyeglasses have?

P5.2.2: Let a simple model of the human arm consist of three bodies representing the

upper arm, the lower arm, and the hands Let the upper arm have a spherical socket) connection with the chest, let the elbow be represented as a pin (or hinge), andlet the hand movement be governed by a twist of the lower arm and vertical and horizontalrotations How many degrees of freedom does the model have?

(ball-and-P5.2.3: How many degrees of freedom does a vice, as commonly found in a workshop,

have? (Include the axial rotation of the adjustment handle about its long axis and thepotential rotation of the vice itself about a vertical axis.)

P5.2.4: See Figure P5.2.4 A wheel W, having planar motion, rolls without slipping in a straight line Let C be the contact point between W and the rolling surface S How many degrees of freedom does W have? What are the constraint equations?

P5.2.5: See Problem P5.2.4 Suppose W is allowed to slip along S How many degrees of

freedom does W then have?

P5.2.6: How many degrees of freedom are there in a child’s tricycle whose wheels rollwithout slipping on a flat horizontal surface? (Neglect the rotation of the pedals abouttheir individual axes.)

Section 5.3 Planar Motion of a Rigid Body

P5.3.1: Classify the movement of the following bodies as being (1) translation, (2) rotation,

and/or (3) general plane motion

a Eraser on a chalk board

b Table-saw blade

c Radial-arm-saw blade

d Bicycle wheel of a bicycle moving in a straight line

e Seat of a bicycle moving in a straight line

f Foot pedal of a bicyclist moving in a straight line

FIGURE P5.2.4

A wheel rolling in a straight line.

W