2.1. Review of methods of momentum - ux.uis.nohirpa/DMS6021/2017/2.1. Review of methods of... · • Small oscillations, matrix formulation, Eigen value problem and numerical solutions

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Dynamics and control of mechanical systemsDate ContentDay 1(01/08)

• Review of the basics of mechanics. • Kinematics of rigid bodies - coordinate transformation, angular velocity

vector, description of velocity and acceleration in relatively moving frames. Day 2(03/08)

§ Euler angles, Review of methods of momentum and angular momentum of system of particles, inertia tensor of rigid body.

• Dynamics of rigid bodies - Euler equations, application to motion of symmetric tops and gyroscopes and problems of system of bodies.

Day 3(05/08)

• Kinetic energy of a rigid body, virtual displacement and classification of constraints.

• D’ Alembert’s principle. Day 4(07/08)

• Introduction to generalized coordinates, derivation of Lagrange's equation from D’ Alembert’s principle.

• Small oscillations, matrix formulation, Eigen value problem and numerical solutions.

Day 5(09/08)

• Modelling mechanical systems, Introduction to MATLAB®, computer generation and solution of equations of motion.

• Introduction to complex analytic functions, Laplace and Fourier transform.Day 6(11/08)

• PID controllers, Phase lag and Phase lead compensation. • Analysis of Control systems in state space, pole placement, computer

simulation through MATLAB.

DMS6021 - Dynamics and Control of Mechanical Systems

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Contents

Focuses on4 Derivation of Euler angles

4 Review of principle of impulse and angular momentum4 Angular moments in 3D and inertia tensors

4 Examples


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Euler anglesApplication areas

RR


- Aircraft and aerospace simulation- Robot simulation

- Computer graphics- Orientation of mobile phones

ROV built by UiS students 2016

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Euler anglesIntroduction4 Euler angles are the three angles used to represent a rigid body in 3D

rotations.èUsing Euler angles any rotation can be described by 3 successive

rotations about a linearly independent anglesè Specially in multibody dynamics, Euler angles are useful to

express the motion of rotating bodiesè Euler angles relate any rotating frame (non-inertial frame) fixed to

a rigid body with the fixed inertial frame through the successiverotationsNote: The "inertial frame" is an Earth-fixed set of axes that is usedas an unmoving reference.

4 Many different types of Euler angles can be driven depending onthe sequence of rotations


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Euler anglesLet’s consider the following sequence of rotation: αà b à g, for rotation about x, Y’ and Z’’ (à 1-2-3 sequence)


Determine R = RgRbRα

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Euler anglesMultibody dyanamics for robotics and simulation software in ADAMS uses the 3-1-3 sequenceShow that the following are correct for the 3-1-3 sequence, i.e.Rotation about z by α (Rzα)à rotation about x´ by b (Rx´b) à rotation about z´´ by g (Rz´´g)


úúú

û

ù

êêê

ë

é -=

úúú

û

ù

êêê

ë

é-=

úúú

û

ù

êêê

ë

é -=

10000

00

001;

10000

''

'

gggg

bbbbaa

aa

g

ba

CSSC

R

CSSCRCS

SCR

z

xz

Resultant Eulerian rotation matrix𝐶𝛼𝐶𝛾 − 𝑆𝛼𝑆𝛾𝐶𝛽 − (𝐶𝛼𝑆𝛾 + 𝑆𝛼𝐶𝛾𝐶𝛽)𝑆𝛾𝑆𝛽𝑆𝛼𝐶𝛾 + 𝐶𝛼𝑆𝛾𝐶𝛽 − 𝑆𝛼𝑆𝛾 + 𝐶𝛼𝐶𝛾𝐶𝛽 − 𝐶𝛼𝑆𝛽𝑆𝛾𝑆𝛽𝐶𝛼𝑆𝛽𝐶𝛽

S = SinC = Cos

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Review of momentum and angular momentum 4 Principle of impulse and momentum


!"!#$!"!#$momentumLinear

v

v

impulseLinear

t

tmvmvmddt

gIntegratindtdmmLawSecondsNewotn

12

2

1

2

1

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-===

==

òò å

åvFL

vaF

4 The principle says: impulse applied to an object during a time interval ( t1 – t2) is equal to the change in the object’s linear momentum

4 Note: Impulse force is a force of large magnitude acting over a short period of time

ò åå -= 2

112

1:t

tav dtFtt

FforceAverage

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Review of momentum ….4 Principle of impulse and momentum …


The linear momentum (L) of a rigid body is defined as

L = m vc where vc is velocity of the mass center

è The linear momentum vector L has a magnitude equal to (mvc) and a direction defined by vc.

The angular momentum (Hc) of a rigid body rotating with angular velocity 𝛚 about its mass center (c) is defined as

Hc = Ic𝛚Where the direction of Hc is perpendicular to the plane of rotation, and Icis mass moment of inertia about its mass center.

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Review of momentum …4 Principle of impulse and momentum ...


When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because 𝛚 = 0

è L = m vc and Hc = 0For a rigid body motion about a fixed axis passing through point O:- Linear momentum: L = mvc- Angular momentum about C: Hc = Ic𝛚- Angular momentum about the center of rotation O.

HO = ( rc × mvc) + Ic 𝛚 = IO 𝛚

where Io is calculated about OMoment of linear momentum

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Review of momentum …4 Angular momentum


!"!#$å=== FrvrvrH xdtdmxmx

dtd

dtd

momentumangularofchangeofRate

o )(

( )! dt

dmxmxdtdmx

dtdwhere

dtdmxmxx

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v

rvr

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+=

==

=

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Angular momentumabout OHo = r x mv

à Moment of linear momentum

Rate of change of the moment of momentum about point o

( ) 12

2

1

2

1 oo

H

H o

impulseAngular

t

to ddtxx

dtd

momentumangularofchangeofRate

HHHFrFrH-==Þ= òò åå

!! "!! #$Principle of angular impulseand momentum

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Review of momentum …4 Angular momentum in matrix form


Exercise: Show that the following components of the angular momentum are correct

Where r = xi + yj + zkand v = (vx, vy, vz)

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Review of momentum ….4 Consrevation of angular momentum


The angular momentum is conserved wheni.e.If the resultant moment about a fixed point Ois zero, then angular momentum remains constant,or it is said to be conserved. Note: Angular momentum may be conserved in one coordinate(e.g., x), but not necessarily in others (e.g., y or z)

and

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Review of momentum …4 Angular momentum of a rigid body in 3D- Let’s define an arbitrary angular velocity vector

as w = (wx wy wz)- Equation of angular momentum:

H = r x mv = r x m(r x w) = m(r x (r x w)), where r = (x, y, z)

4 Using equation for double cross product: A x (B x C) = B(A.C) – C(A.B)

r x (r x w) =


( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) zyx

zyx

zyx

yxyzxz

zyxzxy

zxyxzy

www

www

www

22

22

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++×-×-

×-++×-

×-×-+=×-× ω)r(rr)ω(r ( )

( )

( ) z

I

y

I

x

I

z

z

I

y

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zymxzmxym

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H

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Review of momentum …4 Inertia Tensors of rigid bodies- Expressed in matrix form


úúú

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w

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Angular speedInertia Tensor of the body about its masscenter

Angularmomentum

Note: • The term tensor refers to a higher-order vector.

A vector is written as a column, while a tensor is written as a matrix.

• The inertial tensor has the property thatIij = Iji è it is a symmetirc tensor of 2nd order and I = IT

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3

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333231

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Alternatively: Letting Ixx = I11 , Iyy = I22 , Izz = I33 , Ixy = I12 , etc.

jijij IH w×=

Inertia Tensors transform the vector w into the vector Hc

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Review of momentum …4 Inertia Tensors of rigid bodies in 3D- Inertia tensor is calculated through a point independent of the axis of

rotation, and once that is established, the angular momentum aboutany axis through that point can be determined.à If a new system of axes is used

- a new inertia tensor is obtained- the angular momentum Hc = f(w) remain the same because it is

independent of the choice of coordinate system- If the coordinate system coincides with the principal axes (axes where

the angular momentum and the angular speed coincide), calculation ofinertia tensors is especially simple.


úúú

û

ù

êêê

ë

é

=

z

y

x

II

II

00

0000 Where Ix , Iy and Iz are principal centroidal moments

If Ix = Iy = Iz, Hc and w are collinear, otherwise (in general) theprincipal moments are different and hence Hc and w are in different directions

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Review of momentum …


4 Example: Consider that the cube of dimension a has a uniform distribution of mass with density r = m/a3.

(1) Find the inertia tensor for rotation about one of the cornersas given in the figure.

(2) Assume that the cube rotates about x-axis and find theangular momentum

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Example, Inertia Tensor for Cube4 The mass of the cube is evenly distributed è the summation can be

converted to integration For example, the upper-leftelement becomes:

where denotes the mass density.

4 Thus,

4 Symmetry condition give that Ixx = Iyy = Izz, and similarly for the off-diagonal elements.

2 2

0 0 0( ),

a a a

xxI dx dy dz y zr= +ò ò ò %

3/M ar =%

( )2 2 5 22 23 30 0 0 0 0 0

.a a a a a a

xxI dx y dy dz dx dy z dz a Mar r= + = =ò ò ò ò ò ò% %


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Example, Cont’d4 The off-diagonal elements have the form

4 By symmetry, all the off-diagonal elements have the same value. Thus, the moment of inertia tensor is

4 The angular momentum for rotation about any axis through this corner.Examples:4 Rotation about x axis (w = (w, 0, 0)): L = Iw = Ma2/12 (8w, -3w, -3w)

= Ma2w (2/3, -1/4, -1/4).

4 Rotation about diagonal through O ( ):

5 21 14 40 0 0 0 0 0

, .a a a a a a

xyI dx dy dz xy xdx ydy dz a Mar r r= - = - = - = -ò ò ò ò ò ò% % %

2 2 22 1 13 4 4 2

2 2 21 2 14 3 4

2 2 21 1 24 4 3

8 3 33 8 3 . [about a corner]

123 3 8

Ma Ma MaMaMa Ma Ma

Ma Ma Ma

é ù- - - -é ùê ú ê ú= - - = - -ê ú ê úê ú ê ú- - - -ë ûë û

I

/ 3 (1, 1, 1)w=ω

2 2 28 3 3 1 23 8 3 1 2 .

12 12 63 33 3 8 1 2

Ma Ma Maw w- -é ù é ù é ù

ê ú ê ú ê ú= = - - = =ê ú ê ú ê úê ú ê ú ê ú- -ë û ë û ë û

L Iω ω

L not in samedirection asrotation axis

L is in samedirection asrotation axis


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Example, Cont’d4 If the origin is shifted to the center of the cube, the diagonal element

integrals are just as easy, simply change the limits, e.g.

but the off-diagonals are all odd functions, so the change the limits leads them to go to zero, e.g.

4 The inertia tensor is then diagonal, i.e.

4 Note that, no matter what direction w is, L is always parallel to it:

/2 /2 /2 /2 /2 /2

/2 /2 /2 /2 /2 /2, 0.

a a a a a a

xy a a a a a aI dx dy dz xy xdx ydy dzr r

- - - - - -= - = - =ò ò ò ò ò ò% %

2 21 0 00 1 0 .

6 60 0 1

Ma Maé ùê ú= =ê úê úë û

I 1

( )/2 /2 /2 /2 /2 /22 2 2 3 22 13 6/2 /2 /2 /2 /2 /22 ( / 2) .

a a a a a a

xx a a a a a aI dx y dy dz dx dy z dz a a Mar r

- - - - - -= + = =ò ò ò ò ò ò% %

2

.6Ma

= =L Iω ω


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Summary and questions

Next: Dynamics of rigid bodies - Euler equations, application to motion of symmetric tops and gyroscopes …


In this lecture, the following are covered

4 Euler angles: Example applications and derivations

4 Review of the principle of angular impulse and moment

4 Review of angular momentum and its conservation

4 Angular moments in 3D and inertia tensors

?

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Example4 Example 1Two masses A and B of 0,4 kg each with initial velocity of 2 m/s experience a moment M = 0,6 Nm.Find the speeds of the masses when time t = 4 s.


AB

C

D

M

Sol: Applying the principle of angular impulse and momentum

∫ 𝑴𝒕𝟐𝒕𝟏 dt = Change of angular momentum of mass A and B

= 2(r.m(v2 – v1)) = 2(0,3)(0,4) (v2 – 2 m/s) = 0,24v2 – 0,48

Where ∫ 𝑴𝟒𝟎 dt = 0,6 (4-0) = 2,4

è 2,4 = 0,24v2 – 0,48è v2 = 12 m/s

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Example, Cont’d4 Example 2A rod assembly rotates around its z-axis. The mass C is 10 kg and its initial velocity is 2 m/s. A moment and force bothact as shown, where M = 8t2 + 5 Nm and F = 60 N.

Find the velocity of mass C after 2 seconds.


Sol: Applying the principle of angular impulse and momentumabout z-axis (the axis of rotation)Angular impulse: ∫ 𝑀45

46 dt + ∫ (𝑟𝑥𝐹)4546 dt

= ∫ (8𝑡5 + 55= dt) + ∫ (0,75𝑥 AB 60)

5= dt

= 85,33 nm.sChange of angular momentum: Hz2 – Hz1 = r.m(v2 – v1) = 0,75(10)(v2 – 2) = 7,5v2 + 15

By the principle of angular impulse and momentum7,5v2 + 15 = 85,33 è v2 = 13,4 m/s

Documents

2.1. Review of methods of momentum - ux.uis.nohirpa/DMS6021/2017/2.1. Review of methods of... · • Small oscillations, matrix formulation, Eigen value problem and numerical solutions