Dynamics of Mechanical Systems 2009 Part 9 potx

11.11.1 and these simple examples show that if a potential energyfunction is known we can readily obtain the generalized applied or active forces.. The contact forces betweenthe smooth t

Let the inertia forces on B1 and B2 be represented by forces and passing through

G1 and G2 together with couples having torques and Then , , , and are:

(11.10.18)

From Eq (11.10.12) the partial velocities of G1 and G2 and the partial angular velocities

of B1 and B2 are (see also Eqs (11.4.23)):

* ˙˙ sin cos ˙˙ sin ˙ cos

˙˙ cos ˙ sin ˙˙ cos ˙ sin

m m

ll

and

(11.10.20)Finally, from Eq (11.9.6) the generalized inertia forces are:

(11.10.21)and

(11.10.22)

Observe how routine the computation is; the principal difficulty is the detail We willdiscuss this later

Example 11.10.4: Spring-Supported Particles in a Rotating Tube

Consider again the system of Section 11.7 consisting of a cylindrical tube T containing three spring-supported particles P1, P2, and P3 (or small spheres) as in Figure 11.10.5 As

before, T has mass M and length L and it rotates in a vertical plane with the angle of

rotation being θ as shown The particles each have mass m, and their positions within T are defined by the coordinates x1, x2, and x3 as in Figure 11.10.6, where  is the naturallength of each of the springs

1 2

2 1

2 2

1 2

2 1

2 2

The velocities and accelerations of P1, P2, and P3 in the fixed or inertial frame R are (see

As noted in Section 11.7, the system has four degrees of freedom represented by the

coordinates x1, x2, x3, and θ The corresponding partial velocities and partial angularvelocities are recorded in Eqs (11.7.5), (11.7.6), and (11.7.7) as:

x P

x

x P

x P

x

x P

x P

x

x x

1 1

2 1

3

1 2

2 2

3

1 3

2 3

3 3

G

x G

Finally, from Eq (11.9.6), the generalized inertia forces are:

Example 11.10.5: Rolling Circular Disk

As a final example consider again the rolling circular disk D with mass m and radius r of

Figure 11.10.7 (We first considered this system in Section 4.12 and later in Sections 8.13and 11.3.) As we observed in Section 11.3, this is a nonholonomic system having three

2 2

degrees of freedom Recall that if we introduce six parameters (say, x, y, z, θ, φ, and ψ) to

define the position and orientation of D we find that the conditions of rolling lead to the

constraint equations (see Eq (11.3.8), (11.3.9), and (11.3.10)):

This expression simply means that D must remain in contact with the surface S It is

therefore a geometric (or holonomic) constraint Equations (11.10.34) and (11.10.35), ever, are not integrable in terms of elementary functions These equations are kinematic(or nonholonomic) constraints They ensure that the instantaneous velocity of the contact

how-point C, relative to S, is zero.

To obtain the generalized inertia forces for this nonholonomic system we may simplyselect three of the six parameters as our independent variables Then the partial velocitiesand partial angular velocities may be determined from the coefficients of the derivatives

of these three variables in the expressions for the velocities and angular velocity That is,observe that the coordinate derivatives are linearly related in Eqs (11.10.34), (11.10.35),and (11.10.36) This means that we can readily solve for the nonselected coordinate deriv-atives in terms of the selected coordinate derivatives To illustrate this, suppose we want

to describe the movement of D in terms of the orientation angles θ, φ, and ψ We may

express the velocity of the mass center G as:

(11.10.38)

where N1, N2, and N3 are unit vectors parallel to the X-, Y-, and Z-axes as in Figure 11.10.7.

Then, from Eqs (11.10.34), (11.10.35), and (11.10.36), we may express vG as:

G r r r

−

˙ ˙ sin cos ˙ cos sin

˙ ˙ sin sin ˙ cos cos

From Figure 11.10.7, we see that N1, N2, and N3 may be expressed in terms of the unit

vectors n1, n2, and n3 as:

(11.10.40)

Hence, by substituting into Eq (11.10.39) vG becomes:

(11.10.41)Also, the angular velocity ωω of D relative to S may be expressed as (see Eq (4.12.2)):

(11.10.44)

The inertia force system on D may be represented by a single force F* passing through G

together with a couple with torque T* where F* and T* are (see Section 8.13, Eqs (8.13.7)

to (8.13.12)):

(11.10.45)and

(11.10.46)where

where aG is the acceleration of G relative to the fixed surface S (the inertia frame); where

ωi and αi (i = 1, 2, 3) are the n i components of ωω and the angular acceleration αααα of D relative

to S; and, finally, where:

* ˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙ ˙ cos

˙ sin cos ˙ ˙ sin ˙ sin ˙

* ˙˙ ˙ ˙ cos ˙ sin cos

˙˙ ˙ ˙ cos ˙ sin cos

˙˙ ˙ ˙ cos ˙ sin cos

˙˙ ˙ ˙ cos ˙ sin cos

The weight forces may be represented by a single vertical force W passing through G

given by:

(11.10.56)Then, from Eq (11.10.43), the generalized forces are:

(11.10.57)

11.11 Potential Energy

In elementary mechanics, potential energy is often defined as the “ability to do work.”While this is an intuitively satisfying concept it requires further development to be com-putationally useful To this end, we will define potential energy to be a scalar function ofthe generalized coordinates which when differentiated with respect to one of the coordi-nates produces the negative of the generalized force for that coordinate Specifically, we

define potential energy P(q r) as the function such that:

(11.11.1)

F mr mr mr mr

where as before, n is the number of degrees of freedom of the system (The minus sign is chosen so that P is positive in the usual physical applications.)

To illustrate the consistency of this definition with the intuitive concept, consider a

particle Q having a mass m in a gravitational field Let Q be at an elevation h above a fixed level surface S as in Figure 11.11.1 Then, if Q is released from rest in this position, the work w done by gravity as Q falls to S is:

(11.11.2)

From a different perspective, if h is viewed as a generalized coordinate, the velocity v

of Q and, consequently, the partial velocity v h of Q (relative to h) are:

which is consistent with the results of Eq (11.11.4)

As a second illustration, consider a linear spring as in Figure 11.11.2 Let the spring have

modulus k and natural length  Let the spring be supported at one end, O, and let its other end, Q, be subjected to a force with magnitude F producing a displacement x of Q,

as depicted in Figure 11.11.2 The movement of Q has one degree of freedom represented

by the parameter x The velocity v and partial velocity of Q are then:

The force S exerted on Q by the spring is:

(11.11.11)

which is consistent with Eq (11.11.9)

It happens that Eqs (11.11.5) and (11.11.10) are potential energy functions for gravity

and spring forces in general Consider first Eq (11.11.5) for gravity forces Suppose Q is

a particle with mass m Let Q be a part of a mechanical system S having n degrees of freedom represented by the coordinates q r (r = 1,…, n) Recall from Eq (11.6.5) that the contribution of the weight force on Q to the generalized force F r for the coordinate q r is:

(11.11.12)

where h is the elevation of Q above a reference level as in Figure 11.11.3.

From Eq (11.11.2), if a potential energy function P is given by mgh, Eq (11.11.1) gives the contribution to the generalized force of q r for the weight force as:

(11.11.13)

which is consistent with Eq (11.11.12)

Consider next Eq (11.11.10) for spring forces Suppose Q is a point at the end of a spring which is part of a mechanical system S as depicted in Figure 11.11.4 Let S have n degrees

of freedom represented by the coordinates q r (r = 1,…, n) Then, from Eq (11.6.7), we recall that the contribution of the spring force to the generalized force on S, for the coordinate q r, is:

(11.11.14)

where f(x) is the magnitude of the spring force due to a spring extension, or compression,

x For a linear spring f(x) is simply kx; hence, for a linear spring, is:

(11.11.15)

From Eq (11.11.10), if we let the potential energy function P be (1/2)kx2, we obtain from

Eq (11.11.1) as:

(11.11.16)

which is consistent with Eq (11.11.15)

Alternatively, for a nonlinear spring, we may let the potential energy function have theform:

(11.11.17)

Then, from Eq (11.11.1), becomes

(11.11.18)

which is consistent with Eq (11.11.14)

The definition of Eq (11.11.1) and these simple examples show that if a potential energyfunction is known we can readily obtain the generalized applied (or active) forces Indeed,the examples demonstrate the utility of potential energy for finding generalized forces.Moreover, for gravity and spring forces, Eqs (11.11.5), (11.11.10), and (11.11.17) provideexpressions for potential energy functions This, however, raises a question about otherforces; that is, what are the potential energy functions for forces other than gravitational

or spring forces? The answer is found by considering again Eq (11.11.1): If

r

q

x q

each of the q r Also, if the F r involve derivatives of the (such as with “dissipation forces”),

the integrals will not exist, because P is to depend only on the q r

Further, observe that if knowledge of the generalized forces is needed to obtain apotential energy function which in turn is to be used to obtain the generalized forces, littleprogress has been made Therefore, in the solution of practical problems as in machinedynamics, potential energy is useful primarily for gravity and spring forces Finally, itshould be noted that potential energy is not a unique function Indeed, from Eq (11.11.1),

we see that the addition of a constant to any valid potential energy function P also produces

a valid potential energy function

It may be helpful to consider another illustration Consider again the system of therotating tube containing three spring-supported particles as in Figures 11.11.5 and 11.11.6

We discussed this system in Section 11.7, and we will use the same notation here withoutrepeating the description (see Section 11.7 for the details)

Recall from Eqs (11.7.14) to (11.7.17) that the generalized forces for the coordinates x1,

In the development of these generalized forces we discovered that the only forcescontributing to these forces were gravity and spring forces The contact forces betweenthe smooth tube and the particles did not contribute to the generalized forces; hence, weshould be able to form a potential energy function for the system To this end, using Eqs.(11.11.5) and (11.11.10) we obtain the potential energy function:

(11.11.24)

where the reference level for the gravitational forces is taken through the support pin O.

From Eq (11.11.1) we can immediately determine the generalized forces That is,

(11.11.25)

By substituting from Eq (11.11.24) into (11.11.25), we obtain results identical to those ofEqs (11.11.20) to (11.11.23)

11.12 Use of Kinetic Energy to Obtain Generalized Inertia Forces

Just as potential energy can be used to obtain generalized applied (active) forces, kineticenergy can be used to obtain generalized inertia (passive) forces In each case, the forcesare obtained through differentiation of the energy functions In this section, we willestablish the procedures for using kinetic energy to obtain the generalized inertia forces

To begin the analysis, recall Eq (11.4.7) concerning the projection of the acceleration of

a particle on the partial velocity vectors:

(11.12.1)

This expression, which is valid only for holonomic systems (see Section 11.3), provides a

means for relating the generalized inertia forces and kinetic energy To this end, let P be

a particle with mass m of a holonomic mechanical system S having n degrees of freedom represented by the coordinates q r (r = 1,…, n) Then, the inertia force F* on P is:

(11.12.2)

where a is the acceleration of P in an inertial reference frame R Recall further that the

kinetic energy K of P is:

12

F*= −ma

K=12

2

v

By multiplying Eq (11.12.1) by the mass m of P, we have:

(11.12.4)

or by using Eqs (11.12.2) and (11.12.3) we have:

(11.12.5)

Consider next a set of particles P i (i = 1,…, N) as parts of a mechanical system S having

n degrees of freedom Then, by superposing (or adding together) equations as Eq (11.12.5)

for each of the particles, we obtain an expression identical in form to Eq (11.12.5) and validfor the set of particles Finally, if the set of particles is a rigid body, Eq (11.12.5) also holds

To illustrate the use of Eq (11.12.5), consider again the elementary examples of theforegoing section

Example 11.12.1: Simple Pendulum

Consider first the simple pendulum as in Figure 11.12.1, where we are using the samenotation as before Recall that this system has one degree of freedom represented by the

angle and that the velocity and partial velocity of the bob P are (see Eqs (11.10.1) and

(11.10.4)):

(11.12.6)

The kinetic energy of P is then:

(11.12.7)Then, by using Eq (11.12.5), the generalized inertia force is:

12

K q

K q

12

Example 11.12.2: Rod Pendulum

Consider next the rod pendulum of Figure 11.12.2, using the same notation as before.Recall that the mass center velocity vG and the angular velocity of the rod are (see Eqs.(11.10.6) and (11.10.7)):

Example 11.12.3: Double-Rod Pendulum

To extend this last example, consider again the double-rod pendulum of Figure 11.12.3.Unlike the two above examples, this system has two degrees of freedom, as represented

by the angles θ1 and θ2

In our earlier analysis of this system we found that the mass center velocities and theangular velocities of the rods were (see Eqs (11.10.12)):

(11.12.12)

where, as before, the unit vectors are as shown in Figure 11.12.4

The kinetic energy of the system is then:

(11.12.13)

Then, from Eq (11.12.5), the generalized inertia forces and are:

(11.12.14)and

(11.12.15)These results are identical with the Eqs (11.10.21) and (11.10.22)

2

2 2

2 1

1 2

2

m m

m

G

G B

G

G B

I I

ωωωω

2 2

2 2 2

* = −( )ml ˙˙ −( )ml ˙˙ cos( − )+( )ml ˙ sin( − )

Fθ*2= −( )1 3 ml2θ˙˙2−( )1 2 ml2θ˙˙ cos1 (θ2−θ1)−( )1 2 ml2θ˙ sin12 (θ2−θ1)

Example 11.12.4: Spring-Supported Particles in a Rotating Tube

For another example illustrating the use of kinetic energy to obtain generalized inertiaforces, consider again the system of spring-supported particles in a rotating tube as inFigure 11.12.5 (We considered this system in Sections 11.7 and 11.11.) Recall that this

system has four degrees of freedom represented by the coordinates x1, x2, x3, and θ, asshown in Figures 11.12.5 and 11.12.6

The velocities of the particles, the tube mass center velocity, and the angular velocity of

the tube in the inertia frame R are (see Eqs (11.10.23) through (11.10.27)):

(11.12.16)

where the notation is the same as we used in Section 11.7 and 11.11

The kinetic energy of the system is then:

θθθθ

θωω

12

12

121

2

12

1

2 22 2

2

2 23 2

Using Eq (11.12.5), the generalized inertia forces are then:

(11.12.18)

(11.12.19)

(11.12.20)

(11.12.21)

These results are the same as those of Eqs (11.10.30) to (11.10.33)

Example 11.12.5: Rolling Circular Disk

For an example where Eq (11.12.5) cannot be used to determine generalized inertia forces,consider again the nonholonomic system consisting of the rolling circular disk on the flat

horizontal surface as shown in Figure 11.12.7 (Recall that Eq (11.12.5) was developed for

holonomic systems [that is, systems with integrable or nonkinematic constraint equations]and, as such, it is not applicable for nonholonomic systems [that is, systems with kinematic

or nonintegrable constraint equations].)

Due to the rolling constraint, the disk has three degrees of freedom instead of six, as would

be the case if the disk were unrestrained Recall from Eqs (11.10.34), (11.10.35), and (11.10.36)that the condition of rolling (zero contact point velocity) produces the constraint equations:

2 2

3 2

The first two of these equations are nonintegrable in terms of elementary functions;therefore, the system is nonholonomic.

With three constraint equations, the disk has three degrees of freedom These mayconveniently be represented by the angles θ, φ, and ψ, and in terms of these angles themass center velocity and the disk angular velocity are (see Eqs (11.10.41) and (11.10.42)):

(11.12.25)and

(11.12.26)Hence, the kinetic energy of the disk is:

212

2

11 2 22

2 33

˙ ˙ cos ˙ ˙ sin cos

11.13 Closure

Our objective in this chapter has been to introduce the principles and procedures of

generalized dynamics Our intention was to obtain a working knowledge of the elementary

procedures More advanced procedures, such as those applicable with large multibodysystems and, to some extent, those concerned with nonholonomically constrained systems,have not been discussed because they are beyond our scope at this time The interestedreader may want to refer to References 11.2 and 11.4 and later chapters for a discussion

of these topics

The examples are intended to demonstrate the principal advantages of the generalizedprocedures over the procedures used in elementary mechanics, including: (1) non-workingconstraint forces do not contribute to the generalized forces so these forces may be simplyignored in the analysis; and (2) for holonomic systems (which include the vast majority

of systems of interest in machine dynamics), generalized inertia forces may be computedfrom kinetic energy functions This in turn means that vector acceleration need not becomputed, thus saving considerable analysis effort

In addition to these advantages, if a system possesses a potential energy function, thegeneralized active (or applied) forces may be obtained by a single derivative of thepotential energy function Moreover, for gravity and spring forces (which are prevalent

in machine dynamics), the potential energy functions may often be directly obtained fromEqs (11.11.5) and (11.11.10) In the next chapter, we will consider the application of theseprocedures in obtaining equations of motion

References

11.1 Kane, T R., Dynamics, Holt, Rinehart & Winston, New York, 1968, p 78.

11.2 Kane, T R., and Levinson, D A., Dynamics: Theory and Applications, McGraw-Hill, New York,

11.3 Kane, T R., Analytical Elements of Mechanics, Vol 1, Academic Press, New York, 1959, p 128 11.4 Huston, R L., and Passerello, C E., Another look at nonholonomic systems, J Appl Mech., 40,

101–104, 1973.

11.5 Huston, R L., Multibody Dynamics, Butterworth-Heinemann, Stoneham, MA, 1990.

11.6 Huston, R L., and Liu, C Q., Formulas for Dynamic Analysis, Marcel Dekker, New York, 2001.

Problems

Section 11.2 Coordinates, Constraints, and Degrees of Freedom

P11.2.1: Determine the number of degrees of freedom of the following systems:

a Pair of eyeglasses

b Pair of pliers or pair of scissors

c Child’s tricycle rolling on a flat horizontal surface (let the tricycle be modeled

by a frame, two rear wheels, and a front steering wheel)

d Human arm model consisting of three rigid bodies representing the upper arm,the lower arm, and the hand (with spherical joints at the wrist and shoulder and

a hinge joint at the elbow)

e Pencil writing on a sheet of paper

f Eraser, erasing a chalk board

P11.2.2: An insect A is crawling on the surface of a sphere B with radius r as represented

in Figure P11.2.2 Let the coordinates of A relative to a Cartesian X, Y, Z system with origin

O at the center of the sphere be (x, y, z) What are the constraints and degrees of freedom?

P11.2.3: A knife blade B is being held such that its cutting edge E remains horizontal (Figure P11.2.3) Let B be used to cut a soft medium such as a soft cheese Let the cheese restrict B so that B has no velocity perpendicular to its plane How many degrees of freedom does B have? What are the constraint equations?

Section 11.4 Vector Functions, Partial Velocity, and Partial Angular Velocity

P11.4.1: Consider again the double-rod pendulum of Figure 11.4.3 and as shown again inFigure P11.4.1 As before, let each rod be identical with length  Let the rod orientations

be defined by the “relative” orientation angles β1 and β2 as in Figure P11.4.1 Find the

velocities of mass centers G1 and G2, and the angular velocities of B1 and B2 Find the

partial velocities of G1 and G2 for β1 and β2 and the partial angular velocities of B1 and B2

for β1 and β2 Compare the results with those of Eqs (11.4.23) and (11.4.24)

P11.4.2: A rotating rod B has two collars C1 and C2 which can slide relative to B as indicated

in Figure P11.4.2 Let the orientation of B be and let x1 locate C1 relative to the pin O and let

x2 locate C2 relative to C1 Find the partial velocities of C1 and C2 with respect to θ, x1, and x2

P11.4.3: A box B moving in a reference frame R is oriented in R by dextral (Bryan)

orientation angles and defined as follows: let unit vectors nx, ny, and nz fixed in B be

aligned with unit vectors NX, NY, and NZ , respectively, fixed in R as depicted in Figure

P11.4.3 Then, let B be successively rotated in R about axes parallel to n x, ny, and nz throughangles α, β, and γ, bringing B into a general orientation in R Show that the angular velocity

of B in R may then be expressed as:

where s and c are abbreviations for sine and cosine (see Section 4.7).

P11.4.4: See Problem P11.4.3 Express RωB in terms of the unit vectors NX, NY, and NZ.

P11.4.5: See Problems 11.4.3 and P11.4.4 Find the partial angular velocities of B in R for

α, β, and γ (Express the results in terms of both nx, ny, and nz and NX, NY, and NZ

P11.4.6: See Problems P11.4.3, P11.4.4, and P11.4.5 Let O be the origin of a Cartesian axis system X, Y, Z fixed in R, and let Q be a vertex of B as in Figure P11.4.6 Let the velocity

of Q in R be expressed alternatively as:

and

a Express , , and in terms of , , and

b Express , , and in terms of , , and

P11.4.7: See Problem P11.4.6 Find the partial velocities of Q in R for the coordinates x, y,

z, X, Y, and Z Express the results in terms of both n x, ny, and nz and NX, NY, and NZ

P11.4.8: See Problem 11.4.6 Let the sides of the box B have lengths a, b, and c, as represented

in Figure P11.4.6, and let P be a vertex of B as shown Find the velocity of P in R Express

the results in terms of both nx, ny, and nz and NX, NY, and NZ Use either , , and or, , and for convenience

P11.4.9: See Problems P11.4.3 to P11.4.8 Find the partial velocities of P in R for the coordinates x, y, z, X, Y, Z, α, β, and γ Express the results in terms of both nx, ny, and nzand NX, NY, and NZ

Section 11.5 Generalized Forces: Applied (Active) Forces

P11.5.1: A simple pendulum with length  of 0.5 m and bob P mass of 2 kg is subjected

to a constant horizontal force F of 5 N as represented in Figure P11.5.1 Determine the

generalized active force Fθ on P for the orientation angle θ (Include the contributions of

F, gravity, and the cable tension.)

P11.5.2: Repeat Problem P11.5.1 for a force F given as follows:

a F = 4ny N

b F = 3nx + 4ny N

c F = 5nr N

d F = 5nθ N

P11.5.3: A double-rod pendulum consists of two identical pin-connected rods each having

mass m and length  as depicted in Figure P11.5.3 Let there be a particle P with mass M

at the end of the lower rod Assuming the pin connections are frictionless, find thegeneralized active forces for the orientation angles θ1 and θ2 shown in Figure P11.5.3

P11.5.4: Repeat Problem P11.5.3 for the “relative” orientation angles β1 and β2 shown inFigure P11.5.4

P11.5.5: See Problem P11.5.3 A triple-rod pendulum consists of three pin-connected rods

each having mass m and length  as depicted in Figure P11.5.5 Let there be a particle P with mass M at the end of the lower rod Assuming the pin connections are frictionless,

find the generalized active forces for the orientation angles θ1, θ2, and θ3 shown in FigureP11.5.3

β

P11.5.6: Repeat Problem P11.5.5 for the relative orientation angles β1, β2, and β3 shown inFigure P11.5.6.

P11.5.7: See Problem P11.5.5 Let there be linear torsion springs at the pin joints of the

triple-rod pendulum Let these springs have moduli k1, k2, and k3, and let the resultingmoments generated by the springs be parallel to the pin axes and proportional to therelative angles β1, β2, and β3 between the rods as shown in Figure P11.5.6 Find thecontribution of the spring moments to the generalized active forces for the angles θ1, θ2,and θ3

P11.5.8: Repeat Problem P11.5.7 for the relative orientation angles β1, β2, and β3

P11.5.9: An elongated box kite K, depicted in Figure P11.5.9A, is suddenly subjected to turbulent wind gusts creating forces on K as represented in Figure P11.5.9B Let the wind

forces be modeled by forces F1, F2,…, F8, whose directions are shown in Figure P11.5.9Band whose magnitudes are:

Let the orientation of K be defined by dextral orientation angles α, β, and γ, and let theangular velocity ωω of K be expressed as (see Problem P11.4.3):

where n1, n2, and n3 are unit vectors parallel to the edges of K as shown in Figure P11.5.9B.

Finally, let the velocity v of the tether attachment point O of K be:

Find the generalized forces Fα, Fβ, and Fγ due to the wind forces F1, F2,…, F8

P11.5.10: Repeat Problem P11.5.9 by first replacing F1, F2,…, F8 by a single force F passing

through O together with a couple with torque T; then, use Eq (11.5.7) to determine Fα,