dynamics of machines - part i - ifs

7/28/2019 Dynamics of Machines - Part I - IfS

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DYNAMICS OF MACHINES II

Modelling Simulation Visualization Verification

Dr.-Ing. Ilmar Ferreira Santos, Associate Professor

Department of Mechanical Engineering

Technical University of Denmark

Building 358, Room 1592800 Lyngby

Denmark

Phone: +45 45 25 62 69

Fax: +45 45 88 14 51

E-Mail: [email protected]

September 2003


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Contents

1 Introduction to Multibody Dynamics 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 From a Physical System and to a Mechanical Model . . . . . . . . . . . . . . . . 4

1.2.1 Mass and Inertia Element . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Interaction Forces represented by Stiffness Element . . . . . . . . . . . . 41.2.3 Interaction Forces represented by Damping Element . . . . . . . . . . . . 4

1.3 Steps of the Mathematical Modelling . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Theoretical and Experimental Examples . . . . . . . . . . . . . . . . . . . . . . 7

1.4.1 Dynamics of Particle An example . . . . . . . . . . . . . . . . . . . . . 71.4.2 Dynamics of System of Particles An example . . . . . . . . . . . . . . . 151.4.3 Dynamics of Ridig Body An example . . . . . . . . . . . . . . . . . . . 32

2


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Chapter 1

Introduction to Multibody Dynamics

1.1 Introduction

In many practical situations, for example when designing machines elements which focus on

reliability, information about forces and moments are of crucial importance. Such an infor-mation allows engineers to calculate strain and stresses, select material, optimize dimensionand shape of machine components.

Machine elements can be considered and treated as a system of particles, of rigid bodies or offlexible bodies, depending on the magnitude and frequency of the applied loads. These elementsor bodies are normally attached to each other, and the interaction (forces and moments) amongthese bodies can be represented by spring and damping elements. In static cases, reactionforces and moments can be directly determined using information related to external loadingand geometric properties of the bodies. Such cases are extensively treated in many STATICS

books.

In dynamic cases however, reaction forces and moments can not be directly determinedusing only loading and geometric properties of the bodies. In the dynamic cases, the reactionforces and moments are also depending on the movements described by the machine ele-ments. Thus, in dynamic cases, reaction forces and moments have to be determined usinginformation related to the external loading, the geometric properties of the bodies and the as-sociated motions described by the bodies. It means that before the student starts with the DY-NAMICS itself, part of the Mechanics responsible for describing the causes of the movements,it is necessary to describe the geometry of the movements, and count with the KINEMATICS,part of the Mechanics responsible for describing the movement of the bodies from the viewpoint

of the geometry.

For achieving reaction forces and moments simultaneously with equation of motions onewill use the axioms and principles postulated by NEWTON and EULER.

In many other practical situations, for example stability analysis of satellites, comfort analysis invehicles, vibration analysis in machines and structures, design of control system for mechatronicsystems, only equations of motion are required to conduct such studies. In these dynamiccases, it is not necessary to calculate the reaction forces and moments among the bodies, andthe equation of motion can be directly achieved by using Jourdains principle of virtual power.

3


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1.2 From a Physical System and to a Mechanical Model 4

Another possibility is also to use the energy method and Lagranges equation. Both methodswill be introduced in this section.

1.2 From a Physical System and to a Mechanical Model

1.2.1 Mass and Inertia Element

particle

rigid body

flexible body

1.2.2 Interaction Forces represented by Stiffness Element

Springs and Flexible Couplings Elasticity Theory

Journal Bearings Fluid Dynamics

Air Bearings Fluid Dynamics

Seals Fluid Dynamics

Magnetic Bearings Electromagnetism

EXPERIMENTAL linear and nonlinear coefficients

1.2.3 Interaction Forces represented by Damping Element

Journal Bearings Fluid Dynamics

Air Bearings Fluid Dynamics

Seals Fluid Dynamics

Magnetic Bearings Electromagnetism

EXPERIMENTAL linear and nonlinear coefficients

1.3 Steps of the Mathematical ModellingA description of the steps for achieving the reaction and equation of motions are presented.With this one intents to offer a helpful guidance to students on the way to correctly use theNewton-Euler, Newton-Euler-Jourdain and Lagrange methods. The steps required to obtainequations of motion and reaction forces and moments of a mechanical system composed of Nrigid bodies are summarized below:

KINEMATICS


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1.3 Steps of the Mathematical Modelling 5

1. Defining inertial reference systems Iand moving reference systems B1, B2, ..., BN. Movingsystems are always attached to bodies 1, 2 ... N (Shabana 1989). If the mechanical systemis composed ofN bodies, N moving reference systems will be used, that is, XY Z, X1 Y1Z1, X2Y2Z2, ... , XNYNZN, with their corresponding unit vectors (i,j, k), (i1, j1, k1), ...(iN,jN, kN).

2. Defining coordinate transformation matrices among moving reference frames and inertialTi and, conversely, T

Ti , with (i = 1, 2, ...N).

3. Defining position vectors and constraint equations.

4. Calculating physical parameters, such as vectors of absolute angular velocity of movingreference systems Bii and vectors of absolute linear acceleration Bii, based on geometricrelations of body motions, according to their physical restrictions (constraint equations).It should be stressed that these vectors are conveniently represented in their own movingreference system Bi, attached to the ith body.

5. Calculating physical parameters, such as vectors of absolute angular velocity of bodiesBi and vectors of absolute linear acceleration Bi, based on geometric relations of bodymotions, according to their physical restrictions (constraint equations). It should bementioned that these vectors are conveniently represented in their own moving referencesystem Bi.

6. Calculating physical parameters, such as vectors of absolute linear velocity of the centerof mass of bodies vi and vectors of absolute linear acceleration of the center of massof bodies ai, based on geometric relations of body motions, according to their physicalrestrictions (constraint equations). It should be stressed that these vectors are usuallyrepresented in the inertial reference system I,

Ivi, Iai, or in the moving reference system

Bi, Bivi, Biai.

Bivi

=Bi

vO

+Bi

i

Bi

ri

+B2

vrel

Biai

=Bi

aO

+Bi

i

(Bi

i

Bi

ri) +

Bii

Bi

ri

+ 2 Bi

i

B2

vrel

+B2

arel

GEOMETRIC PROPERTIES OF THE BODIES

7. Defining geometrical properties of the various bodies that compose the mechanical system,such as total mass of each body mi and the inertia tensor IOi with respect to point Ocomposed of moments and products of inertia.

DYNAMICS

8. Calculating vectors of linear momentum IJi of the ith body as a function of mass andvelocity of the center of mass of the body, mi Ivi.


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1.3 Steps of the Mathematical Modelling 6

9. Calculating the vector of angular momentum BiHOi of the ith body with respect to pointO based on the body angular velocity Bii and the body inertia tensor IOi with respectto point O.

10. Representing the p1 active forces and p2 active moments of excitation acting on each ofthe rigid bodies in the vector form,

p1j=1 IFEj and

p2j=1 IMEj . (Free-Body Diagram)

11. Representing the p3 reaction forces and p4 reaction moments of excitation acting on eachof the rigid bodies in the vector form,

p3j=1 IFRj and

p4j=1 IMRj . (Free-Body Diagram)

12. Calculating differential equations of motion and dynamic reactions by Newtons method(particles) or by Newton-Eulers method (rigid bodies).

NEWTON (i = 1,...,N)

p1j=1

IFEj +

p3j=1

IFRj = +mi Iai (1.1)

EULER (i = 1,...,N)

p2j=1

BiMEOi+

p4j=1

BiMROi = IOi

d

dt

Bi

i

+Bi

i

IOi Bii

+mi Bi rOiBiaOi(1.2)

where N is the total number of bodies that compose the mechanical system.

13. If only equations of motion are desired, reaction forces and moments are eliminated andonly differential equations of motion are directly calculated, based on the Principle ofVirtual Power postulated by Jourdain (Bremer 1988, Schiehlen 1986, Pfeiffer 1989). Ja-

cobians of translation Iviq T

and rotation Biiq T

are defined as a function of q, the

vector which contains the minimal coordinates of velocity, and eq.(1.3) is used as follows

NEWTON-EULER-JOURDAIN

Ni=1

Ivi

q

T p1j=1

IFEj + mi Iai

+

Bi

i

q

T p2j=1

BiMEOi + IOi

d

dt

Bi

i

+

+Bi

i

IOi Bii

+ mi Bi rOi Bi aOi

= 0 (1.3)

LAGRANGE

t

Ekinqi

Ekin

qi

+Epot

qi

= Fi (i = 1,...,Nmin) (1.4)

where

Ekin =Ni=1

12 I

vTi mi Ivi +12

Bi

Ti ICMi Bii

is the kinetic energy,

Epot is the potential energy,

Fi =p1j=1

Ivj

q

T IFEj

+p2j=1

Ii

q

T IMEj

is the generalized forces,

Nmin is the minimal coordinates.


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1.4 Theoretical and Experimental Examples 7

NUMERICAL METHODS

14. Solving differential equations of motion numerically using Runge-Kutta, Newton-Raphson,etc.

DYNAMICAL ANALYSIS

15. Analyzing the reaction forces and moments and the motions described by the multibodysystem.

VIBRATION ANALYSIS

16. Linearizing differential equations of motion for the multibody system about an equilibrium

position, and constructing mass, stiffness, damping and gyroscopic matrices. Analyzingvibration amplitudes, natural frequencies, vibration modes, stability, etc.

1.4 Theoretical and Experimental Examples

The procedure of achieving equations of motion for a particle, a system of particles, a rigidbody, a system of rigid bodies, and a system of flexible and rigid bodies is illustrated.

1.4.1 Dynamics of Particle An example

Figure 1.4.1 illustrates a simple pendulum of length l [m] and tip mass m [Kg], mounted on theextremity of a disk of radius r [m]. The mass of the arm is negligible when compared to the tipmass. The disk is mounted on the extremity of an arm of length b [m]. Distance between thecenter of disk to the arm extremity is c [m]. The disk as well the arm rotate at constant angularvelocity of [rad/s] e [rad/s], respectively. It is important to mention that all bodies (armsand disk) are rigid.

(a) Calculate the coordinate transformation matrices, which facilitate to transform the repre-sentation of the different vectors (displacement, velocity, acceleration, force, moment etc.) fromthe inertial reference frame to the moving reference frames or vice-versa. The inertial frame I

is attached to the base, point O. The moving reference frame B1 is attached to the extremityof the arm, point B. The moving reference frame B2 is attached to the disk center, point C.The moving reference frame B3 is attached to the extremity of the second arm, point D.

(b) Calculate the position vectors between points OA, AB, BC, CD and DE, representingthem with help of the most convenient reference. Indicate how to write the position vectorbetween points OE with help of the inertial reference frame I. Such a vector will be useful fordescribing the trajectory followed by the pendulum mass E.

(c) Determine the absolute angular velocity of the reference frames B1, B2 and B3, representingsuch vectors with help of the respective reference frames, i.e.

B11,B2

2

andB3

3.


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Figure 1.1: Simple pendulum, mounted on the extremity of a rotating system Spatial move-ments of a particle performing three consecutive rotations.

(d) Determine the absolute angular acceleration of the reference frames B1, B2 and B3, rep-resenting such vectors with help of the respective reference frames, i.e.

B1

1,B2

2

andB3

3.

(e) Determine the absolute linear acceleration of the point B (arm extremity), representingsuch a vector with help of the reference frame B1, i.e.

B1aB

.

(f) Determine the absolute linear acceleration of the point C (disk center), representing sucha vector with help of the reference frame B2, i.e.

B2aC

.

(g) Determine the absolute linear acceleration of the particle E, representing such a vectorwith help of the reference frame B3, i.e

B3aE

.

(h) Draw the forces which are acting on the particule E, i.e. elaborate a free-body diagram.Afterwards, write the force vectors with help of the most convenient reference frame.

(i) Write the equations responsible for describing the dynamic equilibrium of the particle E.What is the result of the set of equations?


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SOLUTION:

(a) Coordinate transformation matrices.

First rotation around the Z-axis of inertialreference frame.

Coordinate transformation matricesbetween I and B1:

T =

cos sin 0 sin cos 00 0 1

B1s = T Is

Is = TT B1s

Second rotation around the Z1-axis of themoving reference frame B1.

Coordinate transformation matricesbetween B1 and B2:

T=

cos sin 0 sin cos 0

0 0 1

B2s = T

B1s

B1s = TT B2s

Third rotation around the X2-axis of themoving reference frame B2.

Coordinate transformation matricesbetween B2 and B3:

T = 1 0 0

0 cos sin 0 sin cos

B3s = T B2s

B2s = TT B3s


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(b) Position vectors.

position vectorIrOA

:

IrOA

=

0 0 a

T

=

00

a

position vectorB1

rAB

:

B1rAB

=

0 b 0T

position vectorB2

rBC

:

B2rBC

=

0 0 cT

position vectorB2

rCD

:

B2rCD

=

0 r hT

position vectorB3

rDE

:

B3rDE

=

0 0 lT

The position vectors were written with help of the most convenient reference frames, it meansusing the reference frame in which their representation are the most compact one. For describingthe particle trajectory is necessary to write the vector

IrOE

, i.e.

IrOE = IrOA + IrAB + IrBC + IrCD + IrDE

or

IrOE

=IrOA

+ TT B1rAB + TT TT ( B2rBC + B2rCD ) + T

T TT TT B3rDE

(c) Absolute angular velocity of the moving reference frames.

Absolute angular velocity of the moving reference frame B1:

B11 = T I = cos sin 0

sin cos 00 0 1

0

0 B1 1 =

0

0 [rad/s]


B2

2= TT I + T B1 =

=

cos( + ) sin( + ) 0

sin( + ) cos( + ) 00 0 1

00

+

cos sin 0

sin cos 00 0 1

00


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B2

2=

00

+

[rad/s]


B33 = TTT I + TT B1+ T B2

= T(B21 +B2 ) =

1 0 00 cos sin

0 sin cos

0

+

=

( + )sin

( + )cos

[rad/s]

(d) Absolute angular acceleration of the moving reference frames.

Absolute angular acceleration of the moving reference frame B1:

B1

1=

d

dt(B1

1) +

B11

B1

1

=0

=

00

B1 1 =

000

[rad/s2]


B2

2=

d

dt(B2

2) +

B22

B2

2

=0=

00

+

B2

2

=

00

0

[rad/s2]


B3

3=

d

dt(B3

3) +

B33

B3

3

=0

=

B3

3=

( + )sin + ( + ) cos

( + )cos ( + ) sin

B3 3 =

( + ) cos

( + ) sin

[rad/s

2]

It is important to point out that = = 0! It is given in the beginning of the example, whenthe problem was formulated.

(e) Absolute linear acceleration of the point B.

B1aB

=B1

aO

=0+

B11

B1

1

B1

rOB

+B1

1

=0

B1rOB

+ 2 B1

1

B1

vrel

=0+

B1arel

=0


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B1

1

i1

j1

k1

0 0 0 b a

=

i1

j1

k1

0 0 b 0 0

=

0b2

0

B1aB

= 0b20

[m/s2]

(f) Absolute linear acceleration of the point C.

B1aC

=B1

aB

B2

aC

= T B1aB

=

cos sin 0 sin cos 0

0 0 1

0b2

0

B2 aC =

b2 sin b2 cos

0

[m/s2]

(g) Absolute linear acceleration of the particle E.

B3aE

=B3

aD

(I)+

B33

B3

3

B3

rDE

(II)+

B33

B3

rDE

(II I)+ 2

B33

B3

vrel

=0+

B3arel

=0 (I)

B2aD

=B2

aC

see item (f)

+B2

2

B2

2

B2

rCD

(IV)

+B2

2

B2

rCD

=0

+2 B2

2

B2

vrel

=0

+B2

arel

=0

(IV)

B2

2

i2 j2 k20 0 ( + )0 r h

=

i2 j2 k20 0 ( + )

r( + ) 0 0

=

0r( + )2

0

B2

aD

=

b2 sin

b2 cos r( + )2

0

[rad/s2]


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B3aD

= T B2aD =

1 0 00 cos sin

0 sin cos

b2 sin

b2 cos r( + )2

0

B3aD = b2 sin

(b2 cos + r( + )2)cos (b2 cos + r( + )2)sin

[rad/s2] (II)

B3

3

i3

j3

k3

( + )sin ( + )cos 0 0 l

=

i3

j3

k3

( + )sin ( + )cos l( + )sin l 0

= l( + )cos

l( + )2 sin cos l(2 + ( + )2 sin2 )

(III)

B3

3

B3rDE

=

i3

j3

k3

( + ) cos ( + ) sin 0 0 l

=

l( + ) cos

l0

Adding the terms recently calculated, i.e. (I), (II) and (III), one achieves the absolute linearacceleration of the particle E, representing such a vector with help of the reference frame B3:

B3aE

=

b2 sin 2l( + )cos

(b2 cos + r( + )2)cos l( + )2 sin cos + l

(b2 cos + r( + )2)sin + l(2 + ( + )2 sin2 )

[rad/s2]

(h) Free-body diagram of the particle E.

Vector representation of the forces acting on the par-ticle E :

IP =

0 0 mg

T

Forca na direcao da haste:

B3T =

0 0 T

T

Forca na direcao perpendicular a haste:

B3R = R 0 0

T


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(i) Dynamic equilibrium of the particle E according to the second Newtons law:

B3

F = TTTd

dt(m

IvE

) = TTT( m

=0

IvE

+ mIaE

)

B3F = mB3aE = B3R + B3T + TTT IP

B3P =

1 0 00 cos sin

0 sin cos

cos sin 0 sin cos 0

0 0 1

cos sin 0 sin cos 0

0 0 1

00

mg

=

0mg sin mg cos

R

mg sin mg cos + T

= mb2 sin 2l( + )cos

(b2 cos + r( + )2)cos l( + )2 sin cos + l

(b2cos+ r( + )2)sin + l(2 + ( + )2 sin2 )

The final result of the application of the second Newtons law is the achievement of threeequations, i.e. two for calculating the dynamic reaction forces R and T, and one additionalequation responsible for describing the movement of the particle E as a function of the time,(t).

Dynamic reaction forces:

direction X3

: R = m(b2 sin + 2l( + )cos )

direction Z3

: T = m{g cos + [b2 cos + r( + )2]sin + l[2 + ( + )2 sin2 ]}

Equation of motion for the particle E (direction Y3)

direcao Y3

: +g

lsin

(b2 cos + r( + )2)

lcos ( + )2 sin cos = 0

It is important to emphasize that the angle (t) is defined in the moving reference frameB3. The equation of motion is a non-linear second order differential equation, which has tobe numerically solved. After solving the equations by means of a Runge-Kutta integrator,computer animations are created, aiming at facilitating the visualization of the movement fordifferent initial conditions of displacement (0) and velocity (0). Figure 1.4.1 illustrates oneof frozen picture of the trajectory described by the particle E, when the arm and the disk rotatewith a constant angular velocity of

4[rad/s], starting from the initial positions = = 0 [rad].

The initial condition for the particle E are: = 4

[rad] e = 0 [rad/s].


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Figure 1.2: Trajectory described by the particle E when the arm and the disk rotate with aconstant angular velocity of

4[rad/s], starting from the initial positions = = 0 [rad],

= 4

[rad] e = 0 [rad/s].

1.4.2 Dynamics of System of Particles An example

The aim of this example is to illustrate the procedure to achieving the equation of motionand the dynamic reaction forces in a system of particles. In the case, the system is composedof 2 particles of masses m1 e m2. The dynamic reaction forces can only be predicted whenthe equations of motion are solved. Such equations are non-linear second order differentialequations and are responsible for describing the dynamic behavior of the particle in the timedomain. Such equations are numerically solved for some conditions and the results presented ingraphics and computer animations. Afterwards some of the movements simulated and animatedcan be qualitatively compared those described by a prototype of double pendulum. Two waysof achieving the equations of motion will be illustrated: firstly using only one moving referenceframe and afterwards using two moving reference frame. Such an example will clearly show

the compactness and small size of the mathematical expressions when using several movingreference frames and representing acceleration and force vectors with help of the appropriateframes. The results can easily applied to optimize the utilization of softwares, like MAPLE orMATHEMATICA, for achieving equations of motion.

SOLUTION (USING ONLY ONE MOVING REFERENCE FRAME):

The double pendulum shown in Figure 1.4.2 is built by two masses m1 [Kg] and m2 [Kg].Observing the absolute movement of particle 2, one gets an impression that such movementis extremely complicated. Nevertheless, using the moving reference frame attached to mass 1,one can see that the movement of mass 2 is a composition of two planar circular movements,


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(a) (b)

Figure 1.3: Double pendulum (a) physical system with a moving reference frame attached to

mass m1; (b) mechanical model composed of two particles of masses m1 and m2.

i.e. mass 1 rotates around point O (origin of the inertial frame) and mass 2 rotates aroundmass 1. Such an example clarifies, why moving reference frames are so useful: they facilitatethe representation of complicated movements accomplished by particles and bodies.

(1) Aiming at describing the movements of the particles, two coordinate reference frames areused, as it can be seen in Figure 1.4.2:

Inertial reference frame I, represented by the unit vectors i , j , k; Moving reference frame B1, represented by the unit vectors i1 , j1 , k1.

The moving reference frame B1 is attached to the mass m1, rotating at the angular velocity of1 [ras/s] around the Z-axis. The position of the mass m1 can be defined as function of thethe angle 1(t) and the length l1. It is taken into consideration that the arm l1 is rigid andmassless.

(2) Once one works with two reference frames, it is useful to calculate the coordinate trans-formation matrix, which makes possible to transform the representation of a vector from theinertial reference frame to the moving one:

Positive rotation of the moving ref-erence frame B1 around the Z-axis.

i1 = cos 1 i + sin 1 j + 0 k

j1 = sin 1 i + cos 1 j + 0 k

k1 = 0 i + 0 j + 1 k

i1j1k1

=

cos 1 sin 1 0 sin 1 cos 1 0

0 0 1

ijk

B1s = T

1.Is

Is = TT

1B1

s


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(3) After defining the reference frames and transformation matrix, the vectors can be easilywritten. Firstly, the position vectors l1 and l2 are written.

The vectorB1

l1 can be easily written with help of the moving reference frame B1:

B1l1 = 0

l10

Transforming its representation into the inertial frame by using the transformation matrix, onegets:

Il1 = T

T1

.B1

l1 =

cos 1 sin 1 0sin 1 cos 1 0

0 0 1

0l10

=

l1 sin 1l1 cos 1

0

The vector B1l2 can be easily written using the moving reference frame B1:

B1l2 =

l2 sin 2l2 cos 2

0

Transforming its representation into the inertial frame by using the transformation matrix, onegets:

Il2 = T

T1

.B1

l2 =

l2(sin 2 cos 1 + cos 2 sin 1)l2( sin 2 sin 1 + cos 2 cos 1)

0

=

l2 sin(1 + 2)l2 cos(1 + 2)

0

Following the absolute angular velocity and acceleration vectors of the moving reference frameB1 is defined. The absolute angular velocity of B1 can be easily written, knowing that itrotates around the Z-axis of the inertial frame:

I1

=

00

1

The absolute angular acceleration of B1 can be written as:

I1

=

00

1

The relative angular velocity of the mass m2 can be easily written with help of the referenceframe B1, knowing that it rotates around the Z1-axis of the moving reference frame. Afterwardsits representation can be transformed into the inertial frame:

B12

=

00

2

I2

= TT1

.B1

2

=


0 0 1

00

2

=

00

2


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Similarly, the absolute angular acceleration vector of the particle m2 can be described withhelp of the reference frame B1 and afterwards its representation can be transformed into theinertial frame:

B12

=

00

2

I2 = T

T1

.B1

2

=


0 0 1

00

2

=

00

2

(4) After defining the position vectors and angular velocity of the the reference frame, theabsolute linear velocity vectors of the particles m1 and m2 can be calculated. The positions ofthe masses m1 and m2 are coincident with the position of the points B and A, respectively.

IvA

=IvB

+I1

Il2 + IvRelAB

IvB

=I1

Il1 =

i j k

0 0 1l1 sin 1 l1 cos 1 0

=

1l1 cos 11l1 sin 1

0

I1

Il2 =

i j k

0 0 1l2 sin(1 + 2) l2 cos(1 + 2) 0

=

1l2 cos(1 + 2)

1l2 sin(1 + 2)0

IvRelAB = TT

1

d

dt ( B1l2) = cos 1 sin 1 0

sin 1 cos 1 00 0 1

d

dt l2 sin 2

l2 cos 20 =

=


0 0 1

2l2 cos 22l2 sin 2

0

=

2l2 cos(1 + 2)

2l2 sin(1 + 2)0

IvA

=

1l1 cos 1 1l2 cos(1 + 2) 2l2 cos(1 + 2)

1l1 sin 1 1l2 sin(1 + 2) 2l2 sin(1 + 2)0

(5) After defining the absolute linear velocity vectors,

IvA

andIvB

, the absolute linearacceleration vectors of the masses m1 and m2 can be calculated, coinciding with the points Band A, respectively.

IaA

=IaB

+I1

I1

Il2 + I1 Il2 + 2 . I1 IvRelAB + IaRelAB

Calculating each of the terms of the acceleration equation, one achieves:


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IaB

=I1

I1

Il1 + I1 Il1 =

21l1 sin 1 1l1 cos 121l1 cos 1 1l1 sin 1

0

=

aBxaBy

0

I1

I1

Il2 =

21l2 sin(1 + 2)

21l2 cos(1 + 2)

0

I1

Il2 =

1l2 cos(1 + 2)

1l2 sin(1 + 2)0

2 .I1

IvRelAB

=

212l2 sin(1 + 2)

212l2 cos(1 + 2)0

IaRelAB

= TT1d2

dt2(B1l2) =

cos 1 sin 1 0sin 1 cos 1 00 0 1

d2dt2

l2 sin 2l2 cos 20 =

=

22l2 sin(1 + 2) 2l2 cos(1 + 2)

22l2 cos(1 + 2) 2l2 sin(1 + 2)0

Adding all the terms, a very long mathematical expression is achieved for representing theabsolute linear acceleration of the particle A. Such expression is written with help of the inertialreference frame:

IaA

=

21l1 sin 1 1l1 cos 1 + 21l2 sin(1 + 2) 1l2 cos(1 + 2)

21l1 cos 1 1l1 sin 1 21l2 cos(1 + 2) 1l2 sin(1 + 2)

0

212l2 sin(1 + 2) + 22l2 sin(1 + 2) 2l2 cos(1 + 2)

212l2 cos(1 + 2) 22l2 cos(1 + 2) 2l2 sin(1 + 2)

0

=

aAxaAy

0

(6) Calculation of the reaction forces and equa-

tions of motion based on the second and thirdNewtons laws. Drawing the free-body diagramof the two particles, the forces can be representedin the vector form:

weight force vector represented with help of theinertial reference frame I:

IP

B=

0m

1g

0

IP

A=

0m

2g

0

Free-body diagram of the two particles.


20/53


Reaction force vectors acting on the pendulum arms, represented with help of the movingreference frame B1:

For mass m1:

B1T1 = 0

T10 IT1 = T

T

1 . B1T1 = T1

sin 1

T1 cos 10

For mass m2:

B1T

2=

T2

sin 2T

2cos 20

IT2 = T1 . B1T2 =

T2

sin(1 + 2)T

2cos(1 + 2)

0

Using the second Newtons law for each one of the particles, one achieves:

IFB = m1 IaB IPB + IT1 IT2 = m1 IaB

T1

sin 1 T2 sin(1 + 2)m1g T1 cos 1 + T2 cos(1 + 2)

0

= m1

aBxaBy

0

(1.5)

IFA

= m2 I

aA

IP

A+

IT

2= m

2 IaA

0

m2

g0

+

T2

sin(1 + 2)T

2cos(1 + 2)

0

=

m2

aAxm

2aAy

0

(1.6)

Solving the four equations obtained from eq.(1.5) and (1.6), one gets the analytical expressionsfor the forces acting on the arms of the pendulum T1 and T2, and additionally two equationsof motion, 1 and 2, which will be depending on the time and on the initial conditions of themovements of the masses m1 and m2.

It is important to emphasize that the solution of the problems using the inertial ref-erence frame avoids mistakes while differentiating the position vectors for obtainingthe absolute acceleration vectors. Nevertheless, such a solution results in mathemat-

ical expressions for the absolute acceleration vectors extremely long, difficult to behandled later on. In most practical situations, solving the problems with help of themoving reference frames, attached to each one of the particles, allows the achieve-ment of shorter and more compact mathematical expressions for the accelerationvectors. Such shorter expressions are much easier to be handled, interpreted andunderstood. Moreover, when working with software of symbolic manipulation, likeMAPLE and MATHEMATICA, it is very advantageous to describe such vectors withhelp of the moving reference frames. The shorter mathematical expressions for theacceleration vectors, and also the shorter mathematical expressions for the reactionforce vectors contribute to increase the rapidity of the computational calculationsof equation of motion.


21/53


SOLUTION (USING TWO MOVING REFERENCE FRAMES):

Solving the problem with help of two reference frames,one can illustrate the compactness of the mathematicalexpressions. Let us define two moving reference frames:

B1 attached to the mass 1 (point B) and B2 attachedto mass 2 (point A), as it can be seen in figure on yourright hand side.

Reference frames:

I Inertial reference frame defined by the unit vectorsi j k;

B1 Moving reference frame attached to the mass m1,defined by the unit vector i1 j1 k1 with origin in the

point B;

B2 Moving reference frame attached to the mass m2,defined by the unit vector i2 j2 k2 with origin in thepoint B;

Coordinate transformation matrices:

From the inertial reference frame I into B1 and vice-versa:

T1 =


0 0 1

B1s = T1 . Is Is = TT1 B1s

From the moving reference frame B1 into B2 and vice-versa:

T2 =


0 0 1

B2s = T

2.B1

sB1

s = TT2

B2s

Position vectors:

B1l1 =

0l10

B2l2 =

0l20

Absolute angular velocity vector of the moving reference frames B1 e B2:

angular velocity vector of B1

B11 = T1 I1 =


0 0 1

00

1

=

00

1


22/53


angular velocity vector of B2

B22 = T2T1 I1 + T2 B1 2 =


0 0 1


0 0 1

00

1

+

+


0 0 1

00

2

=

00

1 + 2

Absolute angular acceleration vectors of the moving reference frames B1 e B2:

absolute angular acceleration of B1:

B11 =

00

1

absolute angular acceleration of B2:

B22 =

00

1 + 2

Absolute linear acceleration vectors of the masses m1 and m2 (points B and A), repre-sented with help of the reference frames B1 e B2, respectively:

B1aB

=B1

a0

=0

+B1

1

B1

1

B1

l1 + B11 B1l1 + 2 . B11 B1vRel + B1aRel =0

B1aB

=B1

1

B1

1

B1

l1 + B11 B1l1 =

l11l1

21

0

B2aA

=B2

aB

+B2

2

B2

2

B2

l2 + B22 B2l2 + 2 . B22 B2vRel + B2aRel

=0B2

aA

= T2 B1aB +B2 2 B22 B2l2 + B22 B2l2

B2aA

=

l1(1 cos 2 + 12

sin 2) l2(1 + 2)

l1(1 sin 2 12

cos 2) l2(1 + 2)2

0


23/53


Reaction forces acting on the masses m1 and m2 (points B and A), represented with helpof the reference frames B1 and B2, respectively:

weight forces

B1PB = T1 0

m1 g0

= m

1g sin 1

m1 g cos 10

B2P

A= T2T1

0m

2g

0

=

m2

g sin(1 + 2)m

2g cos(1 + 2)

0

reaction forces acting on the arm of length l1, represented with help of the movingreference frame B1:

B1T

1=

0T

1

0

reaction forces acting on the arm of length l2, represented with help of the movingreference frame B2:

B2T

2=

0T

2

0

Dynamic equilibrium between the two particles:

For mass m1 (particle B)

B1

FB

= m1 B1

aB

B1

PB

+B1

T1

B1

T2

= m1 B1

aB

m1 g sin 1 T2 sin 2

m1 g cos 1 T1 + T2 cos 2

0

= m1

l11l1

21

0

(1.7)

For mass m2 (particle A)

B2

FA

= m2 B2

aA

B2

PA

+B2

T2

= m2 B2

aA

m2 g sin(1 + 2)m2 g cos(1 + 2) T2

0

= m2

l1(1 cos 2 + 12

sin 2) l2(1 + 2)

l1(1 sin 2 12

cos 2) l2(1 + 2)2

0

(1.8)


24/53


Rewriting eq.(1.7) and (1.8) in a matrix form, one achieves:

0 sin 2 m1l1 01 cos 2 0 00 0 m2(l1 cos 2 + l2) m2l2

0

1

m2l1 sin 2 0

T1T2

12

=

m1g sin 1

m1l121 m1g cos 1

m2[g sin(1 + 2) + l121 sin 2]

m2[g cos(1 + 2) + l121 cos 2 + l2(1 + 2)2]

Solving the system by Cramer, one obtains the dynamic reaction forces acting on the arms ofthe pendulum, T1 and T2, and additionally the non-linear second order differential equations ofmotion, 1 and 2, which describe the movements of the particles in the time domain:

REACTION FORCE T1

T1

=

m1g sin 1 sin 2 m1l1 0

m1l121 m1g cos 1 cos 2 0 0

m2[g sin(1 + 2) + l121 sin 2] 0 m2(l1 cos 2 + l2) m2l2m2[g cos(1 + 2) + l121 cos 2 + l2(1 + 2)

2] 1 m2l1 sin 2 0

0 sin 2 m1l1 01 cos 2 0 00 0 m2(l1 cos 2 + l2) m2l20 1 m2l1 sin 2 0

or,

T1

= (21

l21

l2m2

1m

2 gl

1l2m2

1m

2cos

1 2

1l1l22

m1m2

2cos

2 2

12l1l22

m1m2

2cos

2

22

l1l22

m1m2

2cos

2 2

1l21

l2m

1m2

2cos2

2 gl

1l2m

1m2

2cos

2cos(

1+

2)

gl1l2m

1m2

2cos

2sin

1sin

2 2

1l21

l2m

1m2

2sin2

2 gl

1l2m

1m2

2cos

1sin2

2)/

(l1l2m

1m

2 l

1l2m2

2sin2

2) (1.9)

REACTION FORCE T2

T2

=

0 m1g sin 1 m1l1 0

1 m1l121 m1g cos 1 0 0

0 m2[g sin(1 + 2) + l121 sin 2] m2(l1cos2 + l2) m2l20 m2[g cos(1 + 2) + l121 cos 2 + l2(1 + 2)

2] m2l1 sin 2 0



25/53


or,

T2

= (21

l1l22

m1m2

2 2

12l1l22

m1m2

2 2

2l1l22

m1m2

2 2

1l21

l2m

1m2

2cos

2

gl1l2m

1m2

2cos(

1+

2) gl

1l2m

1m2

2sin

1sin

2)/(l

1l2m

1m

2 l

1l2m2

2sin2

2) (1.10)

FIRST EQUATION OF MOTION 1

1

=

0 sin 2 m1g sin 1 0

1 cos 2 m1l121 m1g cos 1 0

0 0 m2[g sin(1 + 2) + l121 sin 2] m2l20 1 m2[g cos(1 + 2) + l121 cos 2 + l2(1 + 2)

2] 0


or,

1

= (gl2m

1m

2sin

1 2

1l22

m22sin

2 2

12l22

m22sin

2 2

2l22

m22sin

2

21 l1l2m22cos2sin2 gl2m22cos(1 + 2)sin2)/(l1l2m1m2 l1l2m22sin22) (1.11)

SECOND EQUATION OF MOTION 2

2 =

0 sin 2 m1l1 m1g sin 11 cos 2 0 m1l121 m1g cos 10 0 m2(l1 cos 2 + l2) m2[g sin(1 + 2) + l121 sin 2]

0 1 m2l1 sin 2 m2[g cos(1 + 2) + l121 cos 2 + l2(1 + 2)2]


or,


26/53


2

= (gl2m

1m

2sin

1 gl

1m

1m

2cos

2sin

1+ 2

1l21

m1m

2sin

2+ 2

1l22

m22sin

2+

212l22

m22sin

2+ 2

2l22

m22sin

2+ 22

1l1l2m2

2cos

2sin

2+ 2

12l1l2m2

2cos

2sin

2+

22

l1l2m2

2cos

2sin

2+ 2

1l21

m22cos2

2sin

2+ gl

2m2

2cos(

1+

2)sin

2+

gl1m22cos2cos(1 + 2)sin2 + 21 l21m22sin32 + gl1m1m2sin(1 + 2)+

gl1m2

2sin2

2sin(

1+

2))/(l

1l2m

1m

2 l

1l2m2

2sin2

2) (1.12)

Computer Program in Pascal Language An simple example

program pend2;uses crt;

{DEFINICAO DAS VARIAVEIS DO PROGRAMA}varm1, m2 : real;l1, l2 : real;t1, t2 : real;dt1, dt2 : real;ddt1, ddt2 : real;Tr1, Tr2 : real;

t, tempo, deltat, freq : real;h1x, h1y, h2x, h2y : real;arq1, arq2, arq3, arq4 : text;ni,n,fim,nj,na : integer;

{DEFINICAO DAS CONSTANTES DO PROGRAMA}constg = 9.81; {aceleracao da gravidade [m/s**2]}

{DEFINICAO DAS FUNCOES}function power(base:real;exp:integer):real;var iii:integer;pow:real;beginpow:=1;for iii:=1 to exp dobeginpow:=pow*base;end;power:=pow;end;

{BEGIN OF PROGRAM}beginclrscr;


27/53


{PARAMETERS RELATED TO THE NUMERICAL INTEGRATION}t := 0.0; {initial time [s]}tempo := 4.00; {total time [s]}n := 25000; {number of integration points}na := 1; {number of integration repetition}deltat := (tempo/na)/n; {integration step [s]}fim := 1; {criterium for closing the files}

{NAME OF THE FILES WHERE THE RESULTS WILL BE SAVED}assign(arq1,arqt1.res);assign(arq2,arqt2.res);assign(arq3,arqh1.res);assign(arq4,arqh2.res);

{PARAMETERS OF THE DOUBLE PENDULUM}m1 := 0.00046; {mass 1 [Kg]}m2 := 0.0046; {mass 2 [Kg]}l1 := 0.07; {length of the arm 1 [m]}l2 := 0.07; {length of the arm 2 [m]}

{MOVEMENT INITIAL CONDITIONS}t1 := -3.1416/4.0; {initial value of theta 1 [rad]}t2 := 3.1416/4.0; {initial value of theta 2 [rad]}dt1 := 0.0; {initial angular velocity of mass 1 [rad/s]}dt2 := 0.0; {initial angular velocity of mass 2 [rad/s]}h1x := (-l1 * sin (t1)); {initial position of mass 1 in X-direction}h1y := -(l1 * cos (t1)); {initial position of mass 1 in Y-direction}h2x := (-l1 * sin (t1) - l2 * sin (t1 + t2)); {initial position of mass 2 in X-direction}h2y := -(l1 * cos (t1) + l2 * cos (t1 + t2)); {initial postion of mass 2in Y-direction}

{OPENING FILES}

rewrite(arq1);rewrite(arq2);rewrite(arq3);rewrite(arq4);

{SAVING INITIAL CONDITIONS IN THE FILE}writeln(arq1,t, ,t1);writeln(arq2,t, ,t2);writeln(arq3,h1x, ,h1y);writeln(arq4,h2x, ,h2y);

{BEGINNING OF THE INTEGRATION IN TIME DOMAIN}

for nj:=1 to na dobeginfor ni:=2 to n dobegint := t + deltat; {time change}

{ACCELERATION OF MASS 1}ddt1:=(g*l2*m1*m2*Sin(t1) - Power(dt1,2)*Power(l2,2)*Power(m2,2)*Sin(t2) - 2*dt1*dt2*Power(l2,2)*Power(m2,2)*Sin(t2) -Power(dt2,2)*Power(l2,2)*Power(m2,2)*Sin(t2) -Power(dt1,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) -g*l2*Power(m2,2)*Cos(t1 + t2)*Sin(t2))/(-(l1*l2*m1*m2) - l1*l2*Power(m2,2)*Power(Sin(t2),2));


28/53


{ACCELERATION OF MASS 2}ddt2:=(-(g*l2*m1*m2*Sin(t1)) - g*l1*m1*m2*Cos(t2)*Sin(t1) +Power(dt1,2)*Power(l1,2)*m1*m2*Sin(t2) +Power(dt1,2)*Power(l2,2)*Power(m2,2)*Sin(t2) +2*dt1*dt2*Power(l2,2)*Power(m2,2)*Sin(t2) +Power(dt2,2)*Power(l2,2)*Power(m2,2)*Sin(t2) +2*Power(dt1,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) +

2*dt1*dt2*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) +Power(dt2,2)*l1*l2*Power(m2,2)*Cos(t2)*Sin(t2) +Power(dt1,2)*Power(l1,2)*Power(m2,2)*Power(Cos(t2),2)*Sin(t2) +g*l2*Power(m2,2)*Cos(t1 + t2)*Sin(t2) +g*l1*Power(m2,2)*Cos(t2)*Cos(t1 + t2)*Sin(t2) +Power(dt1,2)*Power(l1,2)*Power(m2,2)*Power(Sin(t2),3) +g*l1*m1*m2*Sin(t1 + t2) +g*l1*Power(m2,2)*Power(Sin(t2),2)*Sin(t1 + t2))/(-(l1*l2*m1*m2) - l1*l2*Power(m2,2)*Power(Sin(t2),2));

{DYNAMIC REACTION FORCE ACTING ON ARM 1}Tr1:=(-(Power(dt1,2)*Power(l1,2)*l2*Power(m1,2)*m2) -

g*l1*l2*Power(m1,2)*m2*Cos(t1) -Power(dt1,2)*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) -2*dt1*dt2*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) -Power(dt2,2)*l1*Power(l2,2)*m1*Power(m2,2)*Cos(t2) -Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Power(Cos(t2),2) -g*l1*l2*m1*Power(m2,2)*Cos(t2)*Cos(t1 + t2) -g*l1*l2*m1*Power(m2,2)*Cos(t2)*Sin(t1)*Sin(t2) -Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Power(Sin(t2),2) -g*l1*l2*m1*Power(m2,2)*Cos(t1)*Power(Sin(t2),2))/(-(l1*l2*m1*m2) - l1*l2*Power(m2,2)*Power(Sin(t2),2));

{DYNAMIC REACTION FORCE ACTING ON ARM 2}Tr2:=(-(Power(dt1,2)*l1*Power(l2,2)*m1*Power(m2,2)) -

2*dt1*dt2*l1*Power(l2,2)*m1*Power(m2,2) -Power(dt2,2)*l1*Power(l2,2)*m1*Power(m2,2) -Power(dt1,2)*Power(l1,2)*l2*m1*Power(m2,2)*Cos(t2) -g*l1*l2*m1*Power(m2,2)*Cos(t1 + t2) -g*l1*l2*m1*Power(m2,2)*Sin(t1)*Sin(t2))/(-(l1*l2*m1*m2) - l1*l2*Power(m2,2)*Power(Sin(t2),2));

{ANGULAR VELOCITIES OF THE MASSES 1 AND 2}dt1 := dt1 + deltat * ddt1;dt2 := dt2 + deltat * ddt2;

{ANGULAR POSITION OF THE MASSES 1 AND 2}

t1 := t1 + deltat * dt1;t2 := t2 + deltat * dt2;

{TRAJECTORY ACCOMPLISHED BY THE MASSES 1 AND 2}h1x := (-l1 * sin (t1));h1y := - (l1 * cos (t1));h2x := (-l1 * sin (t1) - l2 * sin (t1 + t2));h2y := - (l1 * cos (t1) + l2 * cos (t1 + t2));

{PRINTING RESULTS ON THE SCREEN}gotoxy(1,1);writeln(**********, ni*nj);writeln(t: ,t);


29/53


writeln;writeln(t1:5:5, ,t2:5:5);writeln(dt1:5:5, ,dt2:5:5);writeln(ddt1:5:5, ,ddt2:5:5);

{SAVING RESULTS IN THE FILES}writeln(arq1,t, ,t1);

writeln(arq2,t, ,t2);writeln(arq3,h1x, ,h1y);writeln(arq4,h2x, ,h2y);end;end;close(arq1);close(arq2);close(arq3);close(arq4);

{PROGRAM END}end.


30/53


Theoretical and Experimental Results Qualitative Comparison

Initial conditions of movement .

Following some simulation results are presented.Such results can be qualitatively compared to the pro-totype of a double pendulum designed and built.

The figure on your left hand side illustrates the ini-tial conditions of the movement for the masses A andB: 1 = 45

o = 0.785 rad, 2 = 45o = 0.785 rad,

and com 1 = 0 rad/s e 2 = 0 rad/s. The abovefigure illustrates simultaneously the behavior of theangles 1 and 2 as a function of the time e the tra-jectories accomplished by the masses A and B, dueto the initial conditions. The results were obtainedby solving the equations of motion with help of the

computational program previously presented.

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

0 0.5 1 1.5 2 2.5 3 3.5 4

arqt1.res

-0.075

-0.07

-0.065

-0.06

-0.055

-0.05

-0.045

-0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05

arqh1.res

Behavior of the angle 1 as a function of the time and trajectory accomplished by the mass B.

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

0 0.5 1 1.5 2 2.5 3 3.5 4

arqt2.res

-0.14

-0.135

-0.13

-0.125

-0.12

-0.115

-0.06 -0.04 -0.02 0 0.02 0.04 0.06

arqh2.res

Behavior of angle 2 as a function of the time and trajectory accomplished by the mass A.


31/53


Trajectory accomplished by the mass attached to the extremity of the arm, considering thefollowing initial conditions: 1 = 45

o = 0.785 rad, 2 = 45o = 0.785 rad, 1 = 0 rad/s e

2 = 0 rad/s.


o = /2 rad, 2 = 90o = /2 rad, 1 = 0 rad/s e

2 = 0 rad/s.


32/53



o = rad, 2 = 45o = 0.98 rad, 1 = 0 rad/s e

2 = 0 rad/s.

1.4.3 Dynamics of Ridig Body An example

The aim of this example is to illustrate the application of the Eulers angle for describing themovements of a top, a rigid body, shown in Figure 1.4. In the example precession, nutationand spin angles are defined with help of different moving reference frames. Afterwards it ispossible to calculate the absolute angular velocity vectors of the reference frame attached to

the top and of the top itself.

(a) (b)

Figure 1.4: Top as an example of rigid body (a) physical system; (b) mechanical model illus-trating also the reference frames.

The momentum vector is obtained, and its representation is done with help of the movingreference frame attached to the top. Such a vector is derived referred to time and equals to the


33/53


sum of external moments acting on the top, i.e. Eulers equation. The sum of the forces actingon the top is equaled to the inertia forces. In that way, the non-linear differential equations ofsecond order, responsible for describing precession, nutation and spin of the top, are achieved.Additionally, the dynamic reaction forces (contact forces between top and floor) are obtained.Finally, the equations are numerically solved, taking into consideration that the top does notslip on the floor. Some theoretical results are illustrated graphically as well as by means of

computer animations. Qualitative experimental results are also presented in the form of avideo.

Steps towards obtaining the equations of motion and dynamic reaction forces:

KINEMATICS

(1) Definition of the reference frames (inertial and moving).

(2) Calculation of the coordinate transformation matrices.

(3) Calculation the absolute angular velocity vector of the moving reference frame at-

tached to the top, .(4) Calculation of the absolute angular velocity vector of the top, .

(5) Calculation of the absolute angular acceleration vector of the top, .

(6) Calculation of the absolute linear velocity vector of the top center of mass, v.

(7) Calculation of the absolute linear acceleration vector of the top center of mass, a.

DYNAMICS

(8) Free-body diagram .

(9) Moments and products of inertia, inertia tensor I.

(10) Dynamic equilibrium [NEWTON-EULER].

NUMERICAL METHODS AND ANALYSIS

(11) Computational program for solving the equations of motion.

(12) Theoretical and qualitative experimental results.

(13) Using the program MATHEMATICA for achieving the equations of motion.

SOLUTION:

(1) Reference frames:Inertial reference frame I X, Y, Z and unity vectors i, j, k.Moving reference frame B1 X1, Y1, Z1 and unity vectors i1, j1, k1.Moving reference frame B2 X2, Y2, Z2 and unity vectors i2, j2, k2.

(2) Coordinate transformation matrices:

- rotation of the reference frame B1 around Z-axis of the inertial frame. - rotation of the reference frame B2 around X1-axis of the moving reference frame B1.


34/53


- rotation of the top around Z2-axis of the moving reference frame B2.

T

=

cos sin 0 sin cos 0

0 0 1

B1S = T

IS

T

=

1 0 00 cos sin 0 sin cos

B2

S = T B1

S

T.T =

cos sin 0 cos sin cos cos sin

sin sin sin cos cos

B2S = T

.T IS

(3) Absolute angular velocity of the moving reference B2, , in which the momentum andmoments vectors will be represented later, and in which the inertia tensor remains constant:

B2 =

B2 +

B2

I =

00

B2 = T .T I

B1 =

0

0

B2

= T B1

B2 =

00

B2 =

B2 +

B2 = T

.T I + T B1 =

sin cos

(4) Absolute angular velocity of the top, , represented with help of the moving referenceframe B2, in which the momentum and moments vectors will be represented later, and in whichthe inertia tensor remains constant:

B2 =

B2 +

B2 +

B2

B2 =

B2 +

B2 +

B2 = T

.T

I + T

B1 +

B2 =

sin

cos +


35/53


(5) Absolute angular acceleration of the top:

B2 =

d

dt(B2

) +B2

B2

d

dt(B2

) =

sin + cos

cos sin +

B2

B2 =

i2

j2

k2

sin cos

sin cos +

=

sin

0

B2 =

+ sin sin + cos

cos sin +

(6) Absolute linear velocity vector of the top center of mass:

B2v =

B2vo

= 0+

B2

B2r +

B2vrel

= 0

B2

B2r =

i2

j2

k2

sin cos 0 0 h

=

h sin

h0

B2v = h

sin

0

(7) Absolute linear acceleration vector of the top center of mass:

B2a =

B2ao

= 0

+B2

B2

B2

r +B2

B2

r + 2 vrel

= 0

+ arel

= 0

B2

B2

B2r =

h cos h2 cos sin

h2 sin2 h2


36/53


B2

B2r =

h sin + h cos

h0

B2a = h

2 cos + sin

2 cos sin

2 sin2 2

(8) Free-body diagram and forces acting on the top:

weight forceIP, represented with help of the

inertial reference frame

IP = { 0 0 mg}

T

Reaction forceIR, which is represented with

help of the inertial reference frame

IR = { RX RY RZ}

T

In case of necessity, such force vectors can be repre-sented using other reference frames.

Free-body diagram weight anddynamic reaction forces acting in thecontact point between top and floor.

(9) Tensor of inertia:

Mass moments of inertia referred to the top center of mass:

ICM

=

CMIxx 0 00 CMIyy 0

0 0CM

Izz

, where

CMIxx =

1

4mr2 +

1

12ml2

CMIyy =

1

4mr2 +

1

12ml2

CMIzz =

1

2

mr2

Mass moments of inertia referred to the contact point between top and floor:

IO

=

Ixx 0 00 Iyy 0

0 0 Izz

, where

Ixx = CMIxx + mh2

Iyy = CMIyy + mh2

Izz = CMIzz

(10) Dynamic equilibrium [NEWTON-EULER]:


37/53


NEWTON:B2

F = m= 0

B2v + m

B2a

EULER:

B2MO = B2IO ddt (B2) + B2

B2IO . B2 + B2rOCM m B2aO

= 0

Forces action on the top: R e P

In this case, the most appropriate reference frame for representing the reaction forces betweentop and floor is the inertial one. The force components in X and Y directions of R, shall notovercome the static friction forces, otherwise the top will slip on the floor.

IR = R

X

RYR

Z

B2R = T T IR = R

Xcos + R

Ysin

RX cos sin + RY cos cos + RZ sin R

Xsin sin R

Ysin cos + R

Zcos

IP =

00

mg

B2P = T T IP =

0mg sin mg cos

B2MO = B2h B2P +

00

M

=

i2 j2 k20 0 h

0 mg sin mg cos

+

00

M

=

mgh sin 0

M

IO

.d

dt(B2

) =

Ixx 0 00 Iyy 0

0 0 Izz

sin + cos

cos sin +

=

Ixx

Iyy( sin + cos )

Izz( cos sin + )

B2

IO

.B2

=B2

Ixx 0 00 Iyy 0

0 0 Izz

sin

cos +

=B2

IxxIyy sin

Izz( cos + )

=

i2 j2 k2 sin cos

Ixx Iyy sin Izz( cos + )

B2

IO

.B2

=

2(Izz Iyy)sin cos + Izz sin

(Ixx Izz)cos Izz

(Iyy Ixx)sin


38/53


B2

MO

= IO

.d

dt(B2

) +B2

IO

B2

+B2

rOCM

mB2

aO

=0

Substituting the mass moments of inertia, item (9), one achieves:

mgh sin 0

M

=

[ CM

Izz sin +2

sin2 (CM

Izz CMIyy mh2)] + (

CMIxx + mh

2)

CM

Izz + cos [CMIxx CMIzz + CMIyy + 2mh2] + sin (

CMIyy + mh

2)

sin [CM

Iyy CMIxx CMIzz ] + ( cos + )CMIzz

Based on such a system of non-linear differential equations of second order it is possible todescribe de movements of the top:

Precession

=

sin (CM

Ixx + mh2) [

CMIzz

+ cos (CM

Izz

CM

Ixx

CM

Iyy

2mh2)]

Nutation

=sin

CMIxx + mh2

{mgh + [CM

Izz

+ cos (CM

Iyy CMIzz + mh

2)]}

Spin

=

tan (CM

Iyy + mh2) [

CMIzz

+ cos (CM

Ixx

+CM

Iyy CMIzz + 2mh

2)]

+1

CMIzz

[M + sin (CM

Ixx CMIyy + CMIzz)]

In order to obtain the dynamic reaction forces between the top and floor, the equation

B2

F = mB2

a =B2

R +B2

P = TT IR + TT IP

has to be solved.

Substituting all the above calculated terms into Newtons equation, one gets:

T

T

RX

RY

RZ

=

0mg sin mg cos

+ m h

2 cos + sin

2 cos sin

2 sin2 2


39/53


40/53


(11) Computational Program in Pascal Language:

program top;uses crt;

{DEFINITION OF THE PROGRAM VARIABLES}var

psi,phi,teta : real;Dpsi,Dphi,Dteta : real;DDpsi,DDphi,DDteta : real;t,tempo,deltat,freq,r,mp,h,l,cgIxx,cgIyy: real;Ixx,Iyy,Izz,cgIzz,Mz,mom,hx,hy,hz : real;arq1,arq2,arq3,arq4,arq5,arq6,arq7,arq8 : text;i,n,fim,j,a : integer;

{DEFINITION OF THE PROGRAM CONSTANTS}constg = 9.81; {aceleracao da gravidade [m/s2]}

{BEGINNING OF THE PROGRAM}beginclrscr;

{PARAMETERS REFERRED TO THE TIME INTEGRATION}t := 0.0; {initial time [s]}tempo := 10; {total time [s]}n := 30000; {number of points}a := 1; {repetition of the integration process}deltat:= (tempo/a)/n; {time step [s]}fim := 1; {criterium to close the files}

{NAME OF THE FILES WHERE THE RESULTS WILL BE SAVED}

assign(arq1,arqpsi.res);assign(arq2,arqphi.res);assign(arq3,arqteta.res);assign(arq4,arqh.res);assign(arq5,arqdpsi.res);assign(arq6,arqdphi.res);assign(arq7,arqdteta.res);assign(arq8,arqhz.res);

{GEOMETRIC AND MASS PARAMETERS OF THE TOP}mp := 0.582; {total mass [kg]}l := 0.0215; {thickness of the disk [m]}

r := 0.05; {disk radius [m]}h := 0.0558; {length of the arm [m]}Ixx := 0.00209393; {moment of inertia Ixx around O}Iyy := 0.00209393; {moment of inertia Iyy around O}Izz := 0.00220580; {moment of inertia Izz around O}

{INITIAL CONDITIONS}psi := 0.0; {initial precession angle}teta:= (5/180)*2*3.14159; {initial nutation angle}phi := 0.0; {initial spin angle}Dpsi:= (0/180)*2*3.14159; {initial velocity of precession}Dteta:= 0.0; {initial velocity of nutation}Dphi:= 40*2*3.14159; {initial spin}


41/53


Mz := -0.001; {moment applied to the top in Z2-direction}hx := h*sin(teta)*sin(psi); {initial position of the top center of mass direction X}hy := -h*sin(teta)*cos(psi); {initial position of the top center of mass direction Y}hz := h*cos(teta); {initial position of the top center of mass direction Z}

{OPENING THE FILES}rewrite(arq1);

rewrite(arq2);rewrite(arq3);rewrite(arq4);rewrite(arq5);rewrite(arq6);rewrite(arq7);rewrite(arq8);

{SAVING THE INITIAL CONDITIONS IN THE FILES}writeln(arq1,t, ,psi);writeln(arq2,t, ,phi);writeln(arq3,t, ,teta);

writeln(arq4,hx, ,hy);writeln(arq5,t, ,Dpsi);writeln(arq6,t, ,Dphi);writeln(arq7,t, ,Dteta);writeln(arq8,t, ,hz);

{BEGINNING OF THE TIME INTEGRATION}for j:=1 to a dobeginfor i:=2 to n dobegint := t + deltat; {variacao do tempo}

{CALCULATION OF THE ANGULAR ACCELERATIONS}DDteta:= (1/Ixx)*(mp*g*h*sin(teta)+Sqr(Dpsi)*(-Izz+Iyy)*sin(teta)* cos(teta)-Izz*Dphi*Dpsi*sin(teta));DDpsi:= (1/(Iyy*sin(teta)))*(Izz*Dpsi*Dteta*cos(teta)+ Dpsi*Dteta*cos(teta)*(-Ixx-Iyy)+Izz*Dphi*Dteta);DDphi:= (1/Izz)*(-Izz*DDpsi*cos(teta)+Izz*Dpsi*Dteta*sin(teta)+ Dpsi*Dteta*(-Iyy+Ixx)*sin(teta)+Mz);

{CALCULATION OF THE ANGULAR VELOCITIES}Dpsi := Dpsi + deltat * DDpsi;Dphi := Dphi + deltat * DDphi;Dteta:= Dteta + deltat * DDteta;

{CALCULATION OF THE ANGLES}psi := psi + deltat * Dpsi;

phi := phi + deltat * Dphi;teta := teta + deltat * Dteta;

{CALCULATION OF THE TOP CENTER OF MASS}hx := h*sin(teta)*sin(psi);hy :=-h*sin(teta)*cos(psi);hz := h*cos(teta);

{PRINTING THE RESULTS ON THE SCREEN}writeln(**********, i);writeln(t: ,t, psi: ,psi);writeln(t: ,t, phi: ,phi);writeln(t: ,t, teta: ,teta);


42/53


writeln();writeln(DDpsi:5:5, ,DDteta:5:5, ,DDphi:5:5);

{SAVING THE RESULTS IN THE FILES}writeln(arq1,t, ,psi);writeln(arq2,t, ,phi);writeln(arq3,t, ,teta);

writeln(arq5,t, ,Dpsi);writeln(arq6,t, ,Dphi);writeln(arq7,t, ,Dteta);writeln(arq4,hx, ,hy);writeln(arq8,t, ,hz);

{CRITERIUM TO STOP THE INTEGRATION}if teta > = (PI/2) - arctan(r/0.0442) thenbeginwriteln(THE DISK TOUCHED THE FLOOR);close(arq1);close(arq2);close(arq3);close(arq4); close(arq5);close(arq6);close(arq7);close(arq8);j:=a;

i:=n;fim:=0;readln;end;end;end;if fim < > 0 thenbeginclose(arq1);close(arq2);close(arq3);close(arq4); close(arq5);close(arq6);close(arq7);close(arq8);end;end.

(12) Theoretical results for different initial conditions:


43/53


-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04

arqh.res

x [m]

Trajectory of the top center of mass projected intothe XY-plane as a function of the time, until thedisk touches the floor Initial conditions given inthe case (a).

y [m]

0.041

0.042

0.043

0.044

0.045

0.046

0.047

0.048

0.049

0.05

0.051

0 0.5 1 1.5 2 2.5 3

arqhz.res

t [s]

Behavior of the height of the top center of mass as

a function of the time, until the disk touches thefloor Initial conditions given in the case (a).

h [m]

CASE (a)

INITIAL CONDITIONS:

psi := 0.0;teta:= (15/180)*2*3.14159;phi := 0.0;Dpsi:= (200/180)*2*3.14159;Dteta:= 0.0;Dphi:= 7*2*3.14159;Mz := -0.025;

The above illustrations show the XY-trajectory and the height of the top center of mass as afunction of the time, until the disk touches the floor. The initial conditions are given in thecase (a). The resistance moment Mz is relative big, which forces the top spin to reduce rapidly.As consequence, the disk touches the floor after 2, 5 [s]. The behavior of the precession andnutation angles, simultaneously with the top spin in time domain are visualized in the nextfigures, for the same initial conditions, case (a).


44/53


0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

arqpsi.res

-2

0

2

4

6

8

10

12

0 0.5 1 1.5 2 2.5 3

arqdpsi.res

t [s] t [s]

[rad] [rad/s]

PRECESSION: and as a function of the time for the initial conditions given in case (a).

0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

0 0.5 1 1.5 2 2.5 3

arqteta.res

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

0 0.5 1 1.5 2 2.5 3

arqdteta.res

t [s] t [s]

[rad] [rad/s]

NUTATION: and as a function of the time for the initial conditions given in case (a).

0

10

20

30

40

50

60

70

80

0 0.5 1 1.5 2 2.5 3

arqphi.res

10

15

20

25

30

35

40

45

50

55

0 0.5 1 1.5 2 2.5 3

arqdphi.res

t [s] t [s]

[rad] [rad/s]

SPIN: and as a function of the time for the initial conditions given in case (a).


45/53


-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04

arqh.res

x [m]

y [m]

Trajectory of the top center of mass in the plane XY as a function oftime, until the disk touches the floor Initial conditions are givenin the case (b), and the top accomplishes a couple of 360-degreesprecessions until the disk touches the floor. One can clearly seethat the resistance moment Mz is small, and the disk takes longerto touch the floor.

0.041

0.042

0.043

0.044

0.045

0.046

0.047

0.048

0.049

0.05

0.051

0.052

0 1 2 3 4 5 6 7 8

arqhz.res

t [s]

h [m]

Trajectory of the top center of mass in the plane XY as a function oftime, until the disk touches the floor Initial conditions are givenin the case (b), and the top accomplishes a couple of 360-degreesprecessions until the disk touches the floor. One can clearly seethat the resistance moment Mz is small, and the disk takes longerto touch the floor, in the case 7, 5 seconds.

CASO (b)

INITIAL CONDITIONS:

psi := 0.0;teta:= (15/180)*2*3.14159;phi := 0.0;Dpsi:= (50/180)*2*3*3.14159;Dteta:= 0.0;Dphi:= 7*2*3.14159;Mz := -0.001;


46/53


-0.04

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

-0.01 -0.005 0 0.005 0.01 0.015 0.02 0.025 0.03

arqh.res

x [m]

y [m]

CASO (c)

INITIAL CONDITIONS:

psi := 0.0;teta:= (5/180)*2*3.14159;phi := 0.0;Dpsi:= 0.0;Dteta:= 0.0;Dphi:= 40*2*3.14159;Mz := -0.001;

Trajectory of the top center of mass in the plane XY as a function of time, until the disktouches the floor Initial conditions are given in the case (C), and the top accomplishes no fullprecession until the disk touches the floor.

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0.025

-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03

arqh.res

x [m]

y [m]

CASO (d)

INITIAL CONDITIONS:

psi := 0.0;teta:= (5/180)*2*3.14159;phi := 0.0;Dpsi:= 0.0;Dteta:= 0.0;Dphi:= 7*2*3.14159;Mz := -0.001;

Trajectory of the top center of mass in the plane XY as a function of time, until the disktouches the floor Initial conditions are given in the case (d), and the top accomplishes a many360-degrees precessions until the disk touches the floor.


47/53


(13) Getting equations of motions using the software MATHEMATICA.

Initializing the package which allows the cross product function


48/53


Tteta=List[{1,0,0},

{0,Cos[teta[t]],Sin[teta[t]]},

{0,-Sin[teta[t]],Cos[teta[t]]}];

MatrixForm[Tteta]

1 0 0

0 Cos[teta[t]] Sin[teta[t]]0 -Sin[teta[t]] Cos[teta[t]]

Matrizx 3 - rotation around Z and X1:

Tt=Tteta.Tpsi;

MatrixForm[Tt]

Cos[psi[t]] Sin[psi[t]] 0

-(Cos[teta[t]] Sin[psi[t]]) Cos[psi[t]] Cos[teta[t]] Sin[teta[t]]Sin[psi[t]] Sin[teta[t]] -(Cos[psi[t]] Sin[teta[t]]) Cos[teta[t]]

POSITION VECTORS

Position vector related to the top center of mass, written with help of the moving reference B2

B2rcg={0,0,h}

{0, 0, h}

VELOCITY VECTORS

Relative angular velocities (precession, nutation and spin)

Idpsi={0,0,D[psi[t],t]}

B1dteta={D[teta[t],t],0,0}

B2dphi={0,0,D[phi[t],t]}

{0, 0, psi[t]}

{teta[t], 0, 0}

{0, 0, phi[t]}

Velocity vectors written with help of the moving reference frame B2:

B2dpsi=Tt.Idpsi

B2dteta=Tteta.B1dteta

{0, Sin[teta[t]] psi[t], Cos[teta[t]] psi[t]}

{teta[t], 0, 0}


49/53


Absolute angular velocity of the reference frame B2:

B2omega=B2dpsi+B2dteta

{teta[t], Sin[teta[t]] psi[t], Cos[teta[t]] psi[t]}

Absolute angular velocity of the top

B2W=B2omega+B2dphi

{teta[t], Sin[teta[t]] psi[t], phi[t] + Cos[teta[t]] psi[t]}

Absolute linear velocity of the top center of mass written with help of B2:

B2vcg=Cross[B2omega,B2rcg]

{h Sin[teta[t]] psi[t], -(h teta[t]), 0}

ACCELERATION VECTORS

Absolute acceleration of the reference frame B2 written with help of B2:

B2Domega=D[B2omega,t]+Cross[B2omega,B2omega]

{Sin[teta[t],

Cos[teta[t]] psi[t] teta[t] + Sin[teta[t]] psi[t],

-(Sin[teta[t]] psi[t] teta[t]) + Cos[teta[t]] psi[t]}

Absolute angular acceleration of the top, written with help of the moving reference B2:

B2WW=D[B2W,t]+Cross[B2omega,B2W]

{Sin[teta[t]] phi[t] psi[t] + teta[t],

-(phi[t] teta[t]) + Cos[teta[t]] psi[t] teta[t] + Sin[teta[t]] psi[t],

-(Sin[teta[t]] psi[t] teta[t]) + phi[t] + Cos[teta[t]] psi[t]}

Absolute linear acceleration of the top center of mass, written wiht help of B2:


50/53


B2acg=Cross[B2omega,Cross[B2omega,B2rcg]]+Cross[B2Domega,B2rcg]

{2 h Cos[teta[t]] psi[t] teta[t] + h Sin[teta[t]] psi[t],

2

h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t],

2 2 2

-(h Sin[teta[t]] psi[t] ) - h teta[t] }

GEOMETRIC AND MASS PROPERTIES OF THE TOP

Geometry and mass moments of inertia referred to the top center of mass:

mp;

g;

l;

h;

r;

cgIxx;cgIyy;

cgIzz;

Inertia tensor:

Icg=List[{cgIxx,0,0},{0,cgIyy,0},{0,0,cgIzz}];

MatrixForm[Icg]

cgIxx 0 0

0 cgIyy 0

0 0 cgIzz

VECTOR REPRESENTATION OF THE FORCES AND MOMENTS

Weight force written with help of the inertial frame:

P={0,0,-mp*g}

{0, 0, -(g mp)}

Weight force written with help of the moving reference frame B2:

B2P=Tt.P

{0, -(g mp Sin[teta[t]]), -(g mp Cos[teta[t]])}

DYNAMIC EQUILIBRIUM [NEWTON-EULER]

[NEWTON] Dynamic reaction forces:


51/53


IR=-P+mp*Transpose[Tt].B2acg

2 2 2

{mp (Sin[psi[t]] Sin[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +

Cos[psi[t]] (2 h Cos[teta[t]] psi[t] teta[t] + h Sin[teta[t]] psi[t]) -

2

Cos[teta[t]] Sin[psi[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])),

2 2 2mp (-(Cos[psi[t]] Sin[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] )) +

Sin[psi[t]] (2 h Cos[teta[t]] psi[t] teta[t] + h Sin[teta[t]] psi[t]) +

2

Cos[psi[t]] Cos[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])),

2 2 2

g mp + mp (Cos[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +

2

Sin[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t]))}

[NEWTON] Reaction forces written with help of the moving reference frame B2:

B2R=Tt.IR

2 2 2

{mp Sin[psi[t]] (-(Cos[psi[t]] Sin[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] )) +


2

Cos[psi[t]] Cos[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])) +

2 2 2

mp Cos[psi[t]] (Sin[psi[t]] Sin[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +


2Cos[teta[t]] Sin[psi[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])),

mp Cos[psi[t]] Cos[teta[t]] (-(Cos[psi[t]] Sin[teta[t]]

2 2 2

(-(h Sin[teta[t]] psi[t] ) - h teta[t] )) +


2

Cos[psi[t]] Cos[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])) -

mp Cos[teta[t]] Sin[psi[t]] (Sin[psi[t]] Sin[teta[t]]

2 2 2

(-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +


2Cos[teta[t]] Sin[psi[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])) +

2 2 2

Sin[teta[t]] (g mp + mp (Cos[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +

2

Sin[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t]))),

-(mp Cos[psi[t]] Sin[teta[t]] (-(Cos[psi[t]] Sin[teta[t]]

2 2 2

(-(h Sin[teta[t]] psi[t] ) - h teta[t] )) +


2

Cos[psi[t]] Cos[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t]))) +

mp Sin[psi[t]] Sin[teta[t]] (Sin[psi[t]] Sin[teta[t]]


52/53


2 2 2

(-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +


2

Cos[teta[t]] Sin[psi[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])) +

2 2 2

Cos[teta[t]] (g mp + mp (Cos[teta[t]] (-(h Sin[teta[t]] psi[t] ) - h teta[t] ) +

2Sin[teta[t]] (h Cos[teta[t]] Sin[teta[t]] psi[t] - h teta[t])))}

[EULER] Sum of moments referred to the top center of mass:

B2SMcg=Cross[-B2rcg,B2R]+{0,0,Mz}

2 2 3 2

{g h mp Sin[teta[t]] + h mp Cos[psi[t]] Cos[teta[t]] Sin[teta[t]] psi[t] +

2 3 2 2h mp Cos[teta[t]] Sin[psi[t]] Sin[teta[t]] psi[t] +

2 2 3 2

h mp Cos[psi[t]] Cos[teta[t]] Sin[teta[t]] psi[t] +

2 2 3 2

h mp Cos[teta[t]] Sin[psi[t]] Sin[teta[t]] psi[t] -

2 2

h mp Cos[teta[t]] Sin[teta[t]] teta[t] +

2 2 2

h mp Cos[psi[t]] Cos[teta[t]] Sin[teta[t]] teta[t] +

2 2 2

h mp Cos[teta[t]] Sin[psi[t]] Sin[teta[t]] teta[t] -

2 2 2h mp Cos[psi[t]] Cos[teta[t]] teta[t] -

2 2 2 2 2

h mp Cos[teta[t]] Sin[psi[t]] teta[t] - h mp Sin[teta[t]] teta[t],

2 2

-2 h mp Cos[psi[t]] Cos[teta[t]] psi[t] teta[t] -

2 2

2 h mp Cos[teta[t]] Sin[psi[t]] psi[t] teta[t] -

2 2 2 2

h mp Cos[psi[t]] Sin[teta[t]] psi[t] - h mp Sin[psi[t]] Sin[teta[t]] psi[t], Mz}

[EULER] Variation of momentum:

H=Icg.D[B2W,t]+Cross[B2omega,(Icg.B2W)]

2

{cgIzz Sin[teta[t]] phi[t] psi[t] - cgIyy Cos[teta[t]] Sin[teta[t]] psi[t] +

2

cgIzz Cos[teta[t]] Sin[teta[t]] psi[t] + cgIxx teta[t],

-(cgIzz phi[t] teta[t]) + cgIxx Cos[teta[t]] psi[t] teta[t] -

cgIzz Cos[teta[t]] psi[t] teta[t] +

cgIyy (Cos[teta[t]] psi[t] teta[t] + Sin[teta[t]] psi[t]),

-(cgIxx Sin[teta[t]] psi[t] teta[t]) + cgIyy Sin[teta[t]] psi[t] teta[t] +

cgIzz (-(Sin[teta[t]] psi[t] teta[t]) + phi[t] + Cos[teta[t]] psi[t])}


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[EULER] Solution of the system of equation and determination of the equations of motion:

Simplify[Solve[B2SMcg==H,{teta[t],psi[t],phi[t]}]]

{{phi[t] -> (Cot[teta[t]] (-(cgIzz phi[t]) + cgIxx Cos[teta[t]] psi[t] +cgIyy Cos[teta[t]] psi[t] - cgIzz Cos[teta[t]] psi[t] +

2 2

2 h mp Cos[teta[t]] psi[t]) teta[t]) / (cgIyy + h mp) +

(Mz + cgIxx Sin[teta[t]] psi[t] teta[t] - cgIyy Sin[teta[t]] psi[t] teta[t] +

cgIzz Sin[teta[t]] psi[t] teta[t]) / cgIzz,

teta[t] -> (Sin[teta[t]] (g h mp - cgIzz phi[t] psi[t] +

2 2

cgIyy Cos[teta[t]] psi[t] - cgIzz Cos[teta[t]] psi[t] +

2 2 2

h mp Cos[teta[t]] psi[t] )) / (cgIxx + h mp),

psi[t] -> (Csc[teta[t]] (cgIzz phi[t] - cgIxx Cos[teta[t]] psi[t] -

cgIyy Cos[teta[t]] psi[t] + cgIzz Cos[teta[t]] psi[t] -2 2

2 h mp Cos[teta[t]] psi[t]) teta[t]) / (cgIyy + h mp)}}

dynamics of machines - part i - ifs

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