Greenwood D.T. Advanced Dynamics

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99 Dissipative and gyroscopic forces

We see that φ is an ignorable coordinate since it does not appear in the Lagrangian function.

The corresponding generalized momentum is

∂ L

∂

= mr

φ sin

θ = β

(2.195)

where β

is a constant that is usually evaluated from initial conditions. Solving for

φ from

(2.195), we obtain

φ =

sin

(2.196)

This expression is used in obtaining the Routhian function.

R = L −β

φ =−

2mr

sin

− mgr cos θ (2.197)

The θ equation of motion is obtained from



∂ R

∂



−

∂ R

∂θ

= 0 (2.198)

It is

θ −

cos θ

sin

− mgr sin θ = 0 (2.199)

Notice that the ignorable coordinate φ and its derivatives are missing from this θ equation.

Thus, it can be integrated to give θ as a function of time if the initial conditions are known. On

the other hand, the Lagrangian procedure will result in two coupled second-order differential

equations in φ and θ and their derivatives. Of course, either procedure will yield the same

results when the equations of motion are completely integrated.

2.4 Dissipative and gyroscopic forces

Gyroscopic forces and some dissipative forces are represented in the differential equations

of motion by terms which are linear in the

qs. Suppose, for example, that the differential

equations for a system have the form



j=1

(q, t)



j=1

(q, t)

+ h

(q,

q, t) = 0(i = 1,...,n) (2.200)

The n × n coefﬁcient matrix f can be expressed as the sum of a symmetric matrix c and a

skew-symmetric matrix g where

(q, t) = c

(q, t) + g

(q, t)(i, j = 1,...,n) (2.201)

and

= c

(i, j = 1,...,n) (2.202)

=−g

(i, j = 1,...,n) (2.203)

100 Lagrange’s and Hamilton’s equations

Assuming that the c matrix is positive deﬁnite or positive semideﬁnite, the terms in the

equations of motion of the form c

are dissipative in nature while the terms g

are

gyroscopic and nondissipative.

Rayleigh’s dissipation function

Dissipative terms are not present in equations of motion obtained by using the standard

holonomic or nonholonomic forms of Lagrange’s equation because these terms are not

derivable from a potential energy function V (q, t). Rather, dissipative terms are often in-

troduced through an applied generalized force Q



where



=−



j=1

(q, t)

(i = 1,...,n) (2.204)

The cs are called damping coefﬁcients and form a symmetric matrix which is positive

deﬁnite or positive semideﬁnite for systems with passive linear dampers.

Now let us deﬁne Rayleigh’s dissipation function F(q,

q, t) using the equation

F =



i=1



j=1

(2.205)

Then we see that



=−

∂ F

∂

(i = 1,...,n) (2.206)

We can obtain the equations of motion from



∂ L

∂



−

∂ L

∂q

∂ F

∂

= 0(i = 1,...,n) (2.207)

where we assume that the damping forces are the only applied generalized forces which are

not derived from a potential energy function.

The rate at which these linear friction forces dissipate energy is

−



i=1





i=1



j=1

= 2F (2.208)

Thus, Rayleigh’s dissipation function is equal to half the instantaneous rate of dissipation

of the total mechanical energy. For passive dampers, this dissipation rate must be positive

or zero at all times.

Example 2.7 Given the system shown in Fig. 2.6 which, we note, is a linear system with

damping. First, let us obtain the differential equations of motion. The kinetic and potential

energies are

T +





(2.209)

V =



+ x



(2.210)

101 Dissipative and gyroscopic forces

Figure 2.6.

The dissipation function is equal to half the energy dissipation rate, or

F =

−

)

, c > 0 (2.211)

Let us use (2.207) in the form



∂T

∂



−

∂T

∂q

∂V

∂q

∂ F

∂

= 0 (2.212)

We obtain the following equations of motion:

+ c(

−

) +kx

= 0 (2.213)

+ c(

−

) +kx

= 0 (2.214)

Second method As an alternative approach let us introduce the generalized coordinates

+ x

)

− x

) (2.215)

The corresponding transformation equations are

= q

+ q

= q

− q

(2.216)

We see that pure q

motion with q

= 1, q

= 0impliesthatx

= x

= 1. On the other

hand, pure q

motion with q

= 1, q

= 0 means that x

=−x

= 1.

The expressions for T and V in terms of generalized coordinates are

T = m





(2.217)

V = k



+ q



(2.218)

The dissipation function involves relative velocities only and is given by half the dissipation

rate, or

F = 2c

(2.219)

Then, using (2.212), the equations of motion can be written in the form

+ kq

= 0 (2.220)

+ 2c

+ kq

= 0 (2.221)

102 Lagrange’s and Hamilton’s equations

We notice that the q

and q

motions are uncoupled, in contrast to the x motions, and

that damping is applied to the q

motion only. If, for example, we assume the initial

conditions

(0) = A, x

(0) = A,

(0) =

(0) = 0 (2.222)

the solution of (2.220) is

= A cos



t (2.223)

This sinusoidal oscillation continues indeﬁnitely, but any damped q

motion disappears as

time approaches inﬁnity.

Gyroscopic forces

Gyroscopic terms occur in the differential equations of motion and are of the form g

where the coefﬁcients g

(q, t) are skew-symmetric, that is, g

=−g

.Agyroscopic system

has equations of motion containing gyroscopic terms. These gyroscopic terms arise from T

terms in the kinetic energy (or comparable Routhian R

terms) when Lagrange’s equations

are applied.

As an example, recall that T

has the form



i=1

(q, t)

(2.224)

Then



∂T

∂





j=1

∂a

∂q

∂a

∂t

(2.225)

and

∂T

∂q



j=1

∂a

∂q

(2.226)

Hence



∂T

∂



−

∂T

∂q



j=1



∂a

∂q

−

∂a

∂q



∂a

∂t

(2.227)

and we ﬁnd that the equations of motion contain the gyroscopic terms



j=1

≡



j=1



∂a

∂q

−

∂a

∂q



(2.228)

that is, the gyroscopic coefﬁcients are

=−g

∂a

∂q

−

∂a

∂q

(2.229)

103 Dissipative and gyroscopic forces

Gyroscopic terms, when shifted to the right-hand side of the equation can be considered

to be the gyroscopic forces

=−



j=1

(i = 1,...,n) (2.230)

Note that G

is not an applied force, but is inertial in nature and always involves coupling

between two or more nonignored degrees of freedom. If the gyroscopic terms arise from

, the system must be rheonomic. On the other hand, if the Routhian procedure is used,

terms can appear in a scleronomic system.

The rate of doing work by the gyroscopic forces is



i=1

=−



i=1



j=1

= 0 (2.231)

the zero result being due to the skew-symmetry of g

.Inn-dimensional conﬁguration space,

the gyroscopic force G and the velocity

q are orthogonal.

Example 2.8 The Cartesian xyframe rotates at a constant rate  relative to the inertial XY

frame (Fig. 2.7). A particle of mass m moves in the xy-plane under the action of arbitrary

force components F

(t) and F

(t). We wish to ﬁnd the differential equations for the motion

of the particle relative to the xy frame.

We can take V = 0, so Lagrange’s equation has the form



∂T

∂



−

∂T

∂q

= Q

(2.232)

The kinetic energy is

T =

m[(

x −y)

+ (

y + x)

]

) +m(x

y − y

x) +

m

+ y

) (2.233)

(x, y)

(t)

Ω

Figure 2.7.

104 Lagrange’s and Hamilton’s equations

We ﬁnd that



∂T

∂



= m

x −m

y (2.234)

∂T

∂x

= m

y + m

x (2.235)

The x equation is

x −2m

y − m

x = F

(2.236)

Similarly,



∂T

∂



= m

y + m

x (2.237)

∂T

∂y

=−m

x +m

y (2.238)

and the y equation is

y + 2m

x −m

y = F

(2.239)

The gyroscopic terms are −2m

y in the x equation and 2m

x in the y equation. Note that

they arise from the T

(middle) term of the kinetic energy expression, they represent coupling

terms, and the coefﬁcients satisfy skew-symmetry. We frequently associate gyroscopic

effects with rotating and precessing axially symmetric bodies. In this example, however, the

gyroscopic terms for a single particle are due to Coriolis accelerations. Nevertheless, if we

should consider the detailed motions of particles in a precessing gyroscope, we would ﬁnd

that it is the Coriolis accelerations of these particles that produce the gyroscopic moment.

Coulomb friction

The nature of Coulomb friction was introduced in Chapter 1. Brieﬂy, the friction force F

between two blocks in relative sliding motion is

=−µN sgn(v

) (2.240)

where the constant µ is the coefﬁcient of sliding friction, N is the normal force between

the blocks, and the direction of F

directly opposes the relative velocity v

. During sliding

the vector sum of the normal force N and the friction force µN lies on a cone of friction

whose axis is normal to the sliding surfaces and whose semivertex angle φ is given by

tan φ = µ (2.241)

If there is no sliding, the total force lies within the cone of friction.

When the Lagrangian approach is used in the analysis of systems with Coulomb friction,

one can write



∂ L

∂



−

∂ L

∂q

= Q



(i = 1,...,n) (2.242)

105 Dissipative and gyroscopic forces

and include the generalized Coulomb friction forces in the nonpotential Q



. With this

approach, Q



might actually be a frictional moment, for example.

Another approach which is particularly applicable to systems with rotating sliding sur-

faces is to consider the effect of tangential frictional stresses at every point on the surface.

The frictional stress σ

resembles a shear stress and is equal in magnitude to µσ

where σ

is the normal stress or pressure acting at the given point on the sliding surface. The total

frictional force or moment is found by integrating the frictional stress over the entire sliding

contact area.

Example 2.9 A vertical shaft, with a hemispherical joint of radius R rotates at

φ rad/s

(Fig. 2.8). The shaft exerts a downward force F, resulting in a normal stress on the surface

of the joint given by

= σ

cos θ (2.243)

where σ

is the maximum normal stress. Assuming a coefﬁcient of sliding friction µ,we

wish to solve for the frictional moment Q



Figure 2.8.

First, let us establish a relation between σ

and the applied force F. Consider the vertical

component of the force due to the normal stress σ

acting on a circular strip of circumference

2π R sin θ and width Rdθ. This leads to the integral

F =



π/2

2π R

sin θ cos

θ dθ =

π R

(2.244)

and we ﬁnd that the maximum normal stress is

2π R

(2.245)

The frictional stress is

= µσ

3µF cos θ

2π R

(2.246)

106 Lagrange’s and Hamilton’s equations

and acts circumferentially on each circular strip with a moment arm equal to R sin θ. Thus,

we obtain the integral



=−



π/2

2πσ

sin

θ dθ

=−



π/2

3µFRsin

θ cos θ dθ =−µFR (2.247)

The negative sign indicates that the frictional moment Q



opposes the rotation φ.

Example 2.10 A particle of mass m can slide inside a straight tube which is inclined at

an angle θ from the vertical and is rigidly attached to a vertical shaft which rotates at a

constant rate  (Fig. 2.9). There is a Coulomb friction coefﬁcient µ between the particle

and the tube. We wish to ﬁnd the differential equation of motion.

Let us use Lagrange’s equation in the general form



∂T

∂



−

∂T

∂q

∂V

∂q



j=1

+ Q



(2.248)

Choose the spherical coordinates (r,θ,φ) as generalized coordinates with the constraints

θ = 0,

φ −  = 0 (2.249)

⍀

Figure 2.9.

107 Dissipative and gyroscopic forces

This approach enables one to obtain the normal constraint forces from the λs. The uncon-

strained kinetic energy is

T =

(

sin

θ) (2.250)

and the potential energy is

V = mgr cos θ (2.251)

The friction force is entirely in the radial (e

) direction so we ﬁnd that



= Q



= 0, Q



=−µN sgn(

r) (2.252)

where N is the normal force of the tube acting on the particle.

Lagrange’s equation results in the following r equation:

r − mr

sin

θ + mg cos θ = Q



r − mr

sin

θ + mg cos θ =−µN sgn(

r) (2.253)

The normal tube force is

N = N

+ N

(2.254)

and has a magnitude

N =



+ N

≥ 0 (2.255)

In order to ﬁnd expressions for N

and N

we need to write the differential equations for θ

and φ. The θ equation, obtained by using (2.248), is

θ + 2mr

θ − mr

sin θ cos θ − mgr sin θ = λ

−mr



sin θ cos θ − mgr sin θ = λ

= rN

(2.256)

Since θ is an angle, λ

must be a moment which is equal to the virtual work per unit δθ.

Similarly, the φ equation is

φ sin

θ + 2mr

φ sin

θ + 2mr

φ sin θ cos θ = λ

2mr

r sin

θ = λ

= r sin θ N

(2.257)

From (2.256) and (2.257) we obtain

=−mr

sin θ cos θ − mg sin θ (2.258)

= 2m

r sin θ (2.259)

108 Lagrange’s and Hamilton’s equations

Then, from (2.255) we obtain

N = m sin θ [4

+ (r

cos θ + g)

]

1/2

(2.260)

The required differential equation of motion is furnished by the r equation of (2.253) with

the expression for N from (2.260) substituted in its right-hand side.

Now suppose that 0 <θ <

and there are initial conditions r (0) = r

r(0) = 0. We

desire to ﬁnd the limits on 

for incipient motion either outward or inward. The differential

equation for incipient outward motion is

r = r



sin

θ − µr



sin θ cos θ − µg sin θ − g cos θ>0 (2.261)





cos θ + µ sin θ

sin θ(sin θ − µ cos θ )



, 0 ≤ µ<tan θ (2.262)

The differential equation for incipient inward motion is

r = r



sin

θ + µr



sin θ cos θ + µg sin θ − g cos θ<0 (2.263)





cos θ − µ sin θ

sin θ(sin θ + µ cos θ )



, 0 ≤ µ<cot θ (2.264)

The limits on µ can be visualized in terms of the cone of friction. For incipient motion

outward, the total reaction force of the tube acting on the particle cannot have a downward

component since it could not be counteracted by either gravity or centrifugal force effects.

Similarly, for incipient inward motion, the reaction force of the tube on the particle cannot

have an outward component since there is no counteracting force available. In both cases,

the symmetry axis of the cone of friction is normal to the tube and passes through the axis

of rotation.

Example 2.11 A particle of mass m is embedded at a distance

r from the center of a

massless disk of radius r (Fig. 2.10). The disk can roll down a plane inclined at an angle

α = 30

◦

with the horizontal. First, we wish to ﬁnd the differential equation for the rotation

angle θ, assuming no slipping. Then, for the initial conditions θ(0) = 0,

θ(0) = 0, and

assuming a Coulomb friction coefﬁcient µ =

, we wish to ﬁnd the values of the angle θ

at which slipping ﬁrst begins and then ends.

Let us obtain the differential equation of motion by using Lagrange’s equation in the

form



∂T

∂



−

∂T

∂θ

∂V

∂θ

= 0 (2.265)

The velocity of the particle as it rotates about the contact point C is

v =







+ 2r





cos θ



θ (2.266)