1 Particle space trajectory Frame of reference v i m a O • F k j p Newton’s second law for a particle , m = F a n i i1 = = ∑ F F KAP 5. Equations of Motion
Slide 1,m=F a n
2
O •
m m =
System of external forces:
Contact force
Body forces:
3
O
OPp
Euler’s laws – Master Laws for Dynamics Postulate I: Euler’s first
and second law of motion: There exists a frame of reference
relative to which e
P Pdm= ∫F a
(Eu2; 5.5)
for all bodies where Pdm is the mass element at particle P∈ .
4 Inertial frame of reference
The internal force
Postulate II: There is a system of internal forces { } ( ) i
Pd P= ∈Fi iF = F
where
P P P Pd dm d= −F a F
5
e i P P P P Pd d d dm= + =F F F aThe motive force (Newton). We will
here
use the term accelerating force.
The internal force
6
7
O •
i
j
i
ip
jp
ij ji= −F F
(Lem)e i P P P P Pd d d dm= + =F F F a
The local equation of motion (Lem)
”Newton’s second law”
Local implies global
, (Lem)P P P OP OP OP P Pd dm d dm= × = × ⇒∫ ∫ ∫ ∫F a p F p a
(Lem) , P P P OP P OP P Pd dm d dm⇒ = × = ×∫ ∫ ∫ ∫F a p F p a
The implication requires assumtions on continuity!
Global does not imply local!
9
(Lem)e i P P P P Pd d d dm= + =F F F a
Box 5.1: Equations of motion e
P Pdm= ∫F a
e e
i Pd =∫ F 0
Summary
10
= ∫p p
12
= ∫p p
eF
Balance of linear momentum. Motion of the center of mass.
e cm= =F l a
13
the equation of motion for the centre of mass
P P
l
14
e OP P P O
d dm dt
constant vectore O o o= ⇒ = ⇒ −M 0 L 0 L
e O O=M L
Balance of moment of momentum
When we calculate the moment we prefer fix point or the center of
mass.
O fixed in frame of reference:
O moving: e O O O= + ×M L v l
O = c: e c c=M L
15
e rel O O Oc Om= + ×M L p a
rel c c=L L
rel O O Oc Om= + ×L L p v
OP P O= −p v v
17
It is related to (‘the absolute’) moment of momentum by:
and, in particular, if O c=
By differentiating with respect to time one obtain rel O O Oc Om= +
×L L p v
A concept which will be useful in the discussion of the dynamics of
the rigid body is the relative moment of momentum defined by:
18
cose e e P P P PdP d d θ= ⋅ = ⋅v F v F
e P P
d dP d 0
Power and energy
Power of external forces:
e i kP P E+ =
e i kW P E P= − = −
: 2 k P P
iP 0=
Power theorem for rigid bodies:
A fixed in the body:
21
22
Gravity
( ) ( )
dP dm dm m m V dt
= ⋅ = ⋅ = ⋅ = ⋅ = −∫ ∫v g v g v g p g
k gE E V= +
Potential energy:
e ex g ex P P P P Pd d d d dm= + = +F F F F g e ex gP P P= +
g OcV m= − ⋅p g
Mechanical energy:
Summary
24
( ( )) c
a
F a a α p ω ω p
( ) ( )Ac A AP AP P AP AP Pm dm dm× + × × + × × × =∫ ∫p a p α p ω p
ω p
e A AP P Pdm= × =∫M p a
Ac A A Am× + + ×p a I α ω I ω
A fixed in the body
e A Ac A A A
Gyroscopic term
25
Eu 1:
Eu 2:
We will now apply the master laws to the case of a rigid
body.
- inertia tensor
The moment equation for a rigid body
e sys O Oc O O Om= × + + ×M p a I α ω I ω
e A A A= + ×M I α ω I ω
e c c c= + ×M I α ω I ω
e A Ac A A Am= × + + ×M p a I α ω I ω
Version 1: A fixed in the body:
Version 2: A fixed in the body and in space:
Version 3: A = c:
Version 4: O arbitrary:
( ) ( ) , A AP AP Pdm= × × ∈∫I a p a p a
V
( ; ) ( ; ) ( ; )A A A∨ = +I a I a I a
, , ( ) , ,A A A ∈+ = + ∀ ∈I a b I a I b a bα β α β α β V
=∧ For separate bodies:
27
We now introduce one of the fundamental concepts in rigid body
dynamics, the inertia tensor
The inertia tensor is ‘additive’ with respect to separate
bodies.
The inertia tensor may be written:
28
( ) ( ) , A AP AP Pdm= × × ∈∫I a p a p a
V
, A 0⋅ > ≠a I a a 0
, 2 2
A n A AP P P PI dm d dm 0= ⋅ = × = >∫ ∫n I n n p
6.1 Properties of the inertia tensor
, , ( ) , ,A A A ∈+ = + ∀ ∈I a b I a I b a bα β α β α β V
Linearity (tensor property):
( , ), A 1⋅ =n n n
29
( ),1 2 3=b b b b [ ]O O= b
I b b I
O O ij O 21 O 22 O 23
O 31 O 32 O 33
I I I I I I I
I I I
b I
, , ,( ) ( )O ij i O j O ij O jiI I t I t= ⋅ = =b I b
( , ), , ,iO i 1 2 3=b
Arbitrary RON-basis:
Inertia matrix:
Matrix components: Moments of inertia with respect to the
coordinate axes
1b
tB
Matrix representation of the inertia tensor
( ),o o o o 1 2 3=e e e e ( ) ( ( ) ( ) ( )),1 2 3t t t t=e e e e
o=e Re
[ ] ,A A= e
A ij 21 22 23
31 32 33
I I I
= =
e I ( )ij i A j ijI I t= ⋅ =e I e
31
Inertia tensor in the reference placement Matrix
representation
I e e I ( )0
0 0 0 11 12 13
0 0 0 0 0 A ij 21 22 23
0 0 0 31 32 33
I I I I I I I
I I I
e I
0 0 0 0 ij i A jI = ⋅e I e
time-independent
32
( )2 A AP AP AP Pdm= − ⊗∫I p 1 p p
time-dependent
[ ]A A= e
A ij 21 22 23
31 32 33
I I I
( )ij i A j ijI I t= ⋅ =e I e
33
0 T A A=I RI R
0 T T 0 T o 0 o 0 ij i A j i A j i A j i A j ijI I= ⋅ = ⋅ = ⋅ = ⋅
=e I e e RI R e R e I R e e I e
[ ] 0
I I
Thus is time-independent! [ ]A e I 34
AP AP=p Rr
( )
2 T 0 T AP AP AP P Adm= − ⊗ =∫R r 1 r r R RI R
The inertia matrices are identical, i.e.
Moments of inertia with respect to coordinate axes
; ; ;( ) 2 20 0 0 0 0 0 0 0
11 1 A 1 1 AP P 1 1 1 P 2 2 P 3 3 P PI dm dm= ⋅ = × = × + + =∫ ∫e I
e e r e e e e
x x x
; ; ; ;( ) ( ) 20 0 2 2
3 2 P 2 3 P P 2 P 3 P Pdm dm+ − = +∫ ∫e e
x x x x
; ; ; 0 0 0
AP 1 1 P 2 2 P 3 3 P= + +r e e ex x x
; ;( )2 2 33 1 P 2 P PI dm= +∫
x x
x x
Introduce Cartesian coordinates ( , , )1 2 3x x x
The moments of inertia with respect to the coordinate axes are then
given by:
Moments of inertia off-diagonal elements, inertia products
; ;31 13 3 P 1 P PI I dm= = −∫
x x
; ;( )2 AP 1 2 AP 1 AP 2 P 1 P 2 P Pdm dm⋅ − ⋅ ⋅ = −∫ ∫p e e p e p
e
x x
x x
11 12 13
31 32 33
I I I
( )2 2 22 3 1 0I dv= +∫
ρx x B0
ρx x B
ρx x B
Mass density:
, , 11 22 33 33 11 22 22 33 11I I I I I I I I I+ ≥ + ≥ + ≥ 37
Principal axis, principal moments of inertia
Is it possible to choose so that: ? If so, then =e i 12 23 31I I I
0= = =
[ ] 1
=
i I
, , ,A k k kI k 1 2 3= =I i i
The numbers , , 1 2 3I I I are called principal moments of inertia
and the axis ( , )kA i is called a principal axis of inertia at the
reduction point A.
Eigenvalue problem: [ ] [ ] [ ] [ ]( ) ( )A A AI I I= ⇔ − = ⇔ − = e
e
I i i I 1 i 0 I 1 i 0
( )1 2 3=i i i i RON-basis
38
[ ] [ ] [ ] [ ]( ) 11 12 13 1
31 32 33 3
I I I I i 0 I I I I I i 0
I I I I i 0
− − = ⇔ − = −
e e I 1 i 0
[ ] [ ]( ) det( ) det( ) A A Ap I I I 0= − = − =I e
I 1 I 1
6.2.3 Steiner’s theorem
( )2 A c Ac Ac Ac m= + − ⊗I I p 1 p p
Corollary 6.3 (Steiner’s theorem) Consider the two parallel axis (
, )A n and ( , )c n . Then , ,
2 A n c n AcI I d m= +
where Acd is the distance between the axis, i.e. ( )2 2 2
Ac Ac Acd = − ⋅p n p
40
What happens if we change the moment point from A to another point
B?
(‘The two parts theorem’). The inertia tensor with respect to the
moment point A is equal to the inertia tensor with respect to c
plus the inertia tensor with respect to A of a particle with mass m
located at c, i.e.
Free axis Corollary 6.4 Let A be a point on the axis ( , )c n then
A c=I n I n In particular ( , )c n is a principal axis if and only
if ( , )A n is a principal axis.
Proposition 6.14 A free axis is a principal axis at all its points.
All other principal axes are principal axes at precisely one
point.
41
A principal axis through the centre of mass is called a free
axis
Pdm Qdm
Use symmetry to identify principal axes! 42
Q Pdm dm=
: , ( )t t OP OP OQ Q PS S dm dm∗→ = = ⇒ =p p pB B
Mass symmetry
Pdm Qdm
Symmetry plane
Proposition 6.16 If the mass distribution has a symmetry plane Π ,
then the centre of mass of the body belongs to Π and every axis
orthogonal to Π is a principal axis at the point of intersection
with Π . In particular there exists a free axis orthogonal to Π
.
45
46
2i
Corollary 6.7 If the mass distribution of the body has two symmetry
planes then these will intersect and the line of intersection is a
free axis.
Symmetry axis An axis ( , )O n is said to be a symmetry axis of the
mass distribution if the transformation OP OP
∗ =p Sp defined by the rotation , 0 2θ θ π< <S n: , is a
symmetry transformation. If
2 n πθ =
and n 2= the symmetry is called diagonal, if n 3= trigonal and if n
4= it is called tetragonal etc. See figure below.
θ π= 2 3 πθ =
Diagonal symmetry Trigonal symmetry 47
Symmetry axis
Proposition 6.17 A symmetry axis for the mass distribution is a
free axis.
Proposition 6.18 Let ( , )O n be a symmetry axis with θ π≠ (i.e.
not diagonal symmetry) then every axis perpendicular to the
symmetry axis is a principal axis at the point of intersection with
the symmetry axis. The moments of inertia for any two such axes,
intersecting the symmetry axis at the same point, are equal.
, principal axisO i
O
i
48
Proposition 6.19 If one principal axis ( , )O ez is known then the
other two, in the plane perpendicular to ez , are determined by the
angle θ given by
arctan( ) if
I I I 0
I I I 0
6.4 The moment equation for a rigid body Matrix formulation
[ ] [ ] [ ] [ ] [ ] [ ] [ ]e A Ac A A Am = × + + × e e e e e e
ee
M p a I α ω I ω
e A Ac A A Am= × + + ×M p a I α ω I ω
( )1 2 3=e e e eRON-basis fixed in body:
o=e Re
50
Euler equations for rigid body motion
( ) ,1 2 3 Ai i i
e A 1 1 2 2 3 3M M M= + +M i i i 1 1 2 2 3 3ω ω ω= + +ω i i i
The principal frame:
[ ] [ ] [ ] [ ] [ ]e A A A = + × i i i i ii
M I α ω I ω
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
= + ×
c
51
The inertia tensor, in its turn, is determined by the principal
moments of inertia and a parallel coordinate system. , and 1 2 3I I
I
Euler equations for rigid body motion
( ) ( ) ( ), , ,
( )
2 2 2 1 3 3 1 i i
3 3 3 2 1 1 2
= + − = + − ⇒ = = ⇒ = + −
ω ω ω ω ω ω ω ω ω ω ω
( )1 2 3 Ai i i
, ( )e A 1 1 2 2 3 3 i iM M M M M t= + + =M i i i
1 1 2 2 3 3ω ω ω= + +ω i i i
sin sin cos ( ) sin cos sin ( ) ( ( ), ( ), ( ))
( )cos
1
2
3
+ = = − = ⇒ = ⇒ = =+ =
R R
ψ θ φ θ φ ω ψ ψ ψ θ φ θ φ ω θ θ ψ θ φ
φ φψ θ φ ω
The principal frame: , fixed in space or A c=A
prescribed
52
Professor of mathematics at Stockholms Högskola 1889-1891
No closed form solutions when applied to a body of arbitrary shape
and subjected to an arbitrary system of external forces. However,
there are three classical integrable cases with technical interest:
• The case of Euler (1758) - moment-free rotation
where it is assumed that the external moment sum is equal to
zero.
• The case of Lagrange (1815) - rotation of a body around a fixed
point in the gravitational field. In the case of Lagrange one
assumes that the body has a symmetry axis.
• The case of Kovalevskaya (1889) - rotation of a body around a
fixed point in the gravitational field. Solution for a
non-symmetric body.
Euler equations for rigid body motion
Joseph-Louis Lagrange (1736-1813)
1 2 3M M M 0= = =
( ) ( ) ( )
1 1 3 2 2 3
2 2 1 3 3 1
3 3 2 1 1 2
= + − = + − = + −
54
One example of this is a body rotating (around its centre of mass
c) in free space, neglecting all external forces, but gravitation.
Another example is body mounted in a suspension where the reaction
forces are producing zero moment with respect to the centre of mass
of the body. One important realization of this is the so-called
Gimbal - or Cardan - suspension and this design is, for instance,
used in gyroscopes.
Figure 6.22 The moment free motion, a) free body in the
gravitational field, b) body with moment free support at the centre
of mass.
55
, , ,, , 1 0 2 0 3 00 0 0ω ω ω≠ = =
Impose the initial conditions ( ) ( ) ( )
1 1 3 2 2 3
2 2 1 3 3 1
3 3 2 1 1 2
= + − = + − = + −
The solution is given by:
Body is initially spinning around the first principal axis
,( ) , ( ) , ( ) , 1 1 0 2 3t t 0 t 0 t 0ω ω ω ω= = = ≥
If we start the rotation around a principal axis then the body will
continue to rotate about this axis with a constant angular
speed.
cL
c
3h
1h
η
1
1
1 2I I=
Symmetry:
Precession
Spin
56
( ) ( )
= + − = + − =
Point A fixed in the body:
The time derivative of the kinetic energy of a rigid body is equal
to the power expended by the external forces.
- momentum (linear momentum) - moment of momentum (angular
momentum) of the body with respect to A
The kinetic energy for rigid bodies
rel A A Ac A A Ac Am m= + × = + ×L L p v I ω p v
k A A Ac A 1 1E m 2 2
= ⋅ + ⋅ + ⋅ ×v l ω I ω ω p v
k A 1E 2
( )
rel A AP AP P AP AP P Adm dm= × = × × =∫ ∫L p p p ω p I ω
If point A fixed in the body can be expressed as :
Point A = c : 2 k c c
1 1E m 2 2
= + ⋅v ω I ω
6.5.2 Stability of moment-free rotation
Theorem 6.6 The moment free rotation of a rigid body around a
principal axis is stable if and only if the corresponding moment of
inertia is the largest or the smallest.
, 1I stable
, 3I stable
, 2I unstable
59
We know that if the rotation is started around any one of the
principal axes then the body will continue to rotate around this
axis with constant angular speed.
The rotation around the middle sized principal moment of inertia
axis is unstable!
To flip a coin
60
The general conclusion is that only the rotation around the 3-axis
is stable.
1 2 3I I I I= = <
6.6 The case of Lagarange. The spinning top.
A
61
Famous and simple play toy - Its complicated motion and sometimes
peculiar behaviour may be analysed using the rigid body
concept.
The spinning top
cz
g
mg
ze
f
c
3i
A
62
The moment of the external forces with respect to A given by the
gravitational force:
e A Ac m= ×M p g
From the moment equation
( )Ac A A dm dt
× = =p g L I ω
we conclude that the component of the moment of momentum in the
vertical direction
,A z z A z AL constant= ⋅ = ⋅ =e L e I ω
z Ac m 0⋅ × =e p g
,z A A zL⋅ =e L
Since
cz
g
mg
ze
f
c
3i
A
63
Since the power expended by the reaction force R is zero and the
mechanical energy E of the body is conserved
A =v 0
k g A c 1E E V m constant 2
= + = ⋅ − ⋅ =ω I ω p g
We have thus found two constants of motion for the spinning top,
namely, the vertical component of the moment of momentum and the
mechanical energy.
The spinning top
c
1g
2g
θ
a
64
Assume that the body has a symmetry axis and that the fixed point A
is located on this axis. We introduce Euler angles and the
corresponding basis systems.
( , )3c i
The spinning top
c
( sin cos ) ( cos )
= − + +
= + − +
= +
θ θ ψ θ θ ψ ψ θ φ θ
ψ θ ψθ θ θ ψ θ φ
ψ θ φ
c
( sin cos ) ( cos )
= − + +
= + − +
= +
θ θ ψ θ θ ψ ψ θ φ θ
ψ θ ψθ θ θ ψ θ φ
ψ θ φ
Steady precession
67
We now specialize to the case of steady precession. We thus assume
that
0 constantθ θ= =
c
1g
2g
θ
a
,
,
Figure 6.34 The tippe top
Figure 6.35 Picture of Wolfgang Pauli and Niels Bohr studying a
Tippe Top. The picture is taken at the opening of the new institute
of physics
at the University of Lund on May 31 1951.
ω
ω
68
The ‘tippe top’ is a top consisting of slightly more than a
hemisphere resting on a cylindrical stem (concentric with the
rotation axes). The surprising thing about this top is that upon
spinning on the hemispherical portion, it spontaneously turns
itself upside- down and begins spinning on the stem. See, for
instance, https://demonstrations.wolfram.co m/TippeTop/
Exercise 1:14
The mechanism to control the deployment of a spacecraft solar panel
from position A to position B is to be designed. Determine the
transplacement, i.e. the translation vector and rotation tensor R ,
which can achieve the required change of placement. The side facing
the positive x-direction in position A must face the positive
z-direction in position B. Calculate the rotation vector n
corresponding to the rotation. (Meriam & Kraige 7/1).
70
72
Proposition 6.16 If the mass distribution has a symmetry plane Π ,
then the centre of mass of the body belongs to Π and every axis
orthogonal to Π is a principal axis at the point of intersection
with Π . In particular there exists a free axis orthogonal to Π
.
Proposition 6.14 A free axis is a principal axis at all its points.
All other principal axes are principal axes at precisely one
point.
Exercise 2a:17
74
( )a a a a 1 2 3=e e e e
( )x y z=e e e e
Fixed to the airplane
Fixed to the propeller
Fixed to the background ( )o o o o 1 2 3=e e e e
Free axis according to Proposition 6.17 ( , )xc
⇒
Exercise 2b:15
75
The solid circular disk of mass m and small thickness is spinning
freely on its shaft at the rate p. If the assembly is released in a
vertical position at 0θ = and 0θ =& determine the horizontal
components of the forces at A and B exerted by the respective
bearings on the horizontal shaft as the position 2 πθ = is passed.
Neglect the mass of the
two shafts compared with m and neglect all friction. Solve by using
the appropriate moment equations. (Meriam 7/127).
Exercise 2b:17
Ground: G
Det gick inte att hitta bilddelen med relations-ID rId8 i
filen.
Exercise 2b:17 Solution Definition of RON-bases
[ ] ,o o o
1 1= = e
e R f f R
o o 1 2→ = → =e R e f R f e
77
( )o o o o 1 2 3=e e e e
( )1 2 3=f f f f
( )1 2 3=e e e e
78
0 0 0 1
θ θ
e f e R R e e R R e
[ ] cos sin
0 0 1 0
[ ] cos sin sin cos2 2
0 0
Exercise 2b:17 Solution
Exercise 2b:17 Solution
T o oT o oT 1 1
0 0 ax ax 0 0 0 0 1 0
0 0=
fe 1
ω R R e e e e θ θ θ θ
θ θ θ θ θ
0 0
θ θ θ θ θ
0 0 1 ax 0 0 0
1 0 0
79
[ ]
sin cos cos sin (( ) ) ( cos sin sin cos )T T T
2 2
⋅ − − = = − − =
1
ω R R f f f f
sin cos cos sin ( cos sin sin cos )T
0 0 ax 0 0
0 0 0 0 0 1
− − − − =
f f
( )T 3 3
0 0 0
Exercise 2b:17 Solution
Exercise 2b:17 Solution
81
1 2 3= + + ⇒ω e e e ω ω ω
θ θ
1 2 3= + + − +ω e e e ωω ω
θ θ θ θ
rel= × +a ω a a
Exercise 2b:17 Solution
3 C 0⋅ =e M 2 A 0⋅ =f R
Free body diagrams:
Exercise 2b:17 Solution
( ) C C C 3 C 3 C0= + × ⇒ = ⋅ + × ⋅M I ω ω I ω e I ω e ω I ω
, C 1 1 1 2 2 2 3 3 3 C 1 1 1 2 2 1 3 3 3I I I I I I= + + = + +I ω
e e e I ω e e eω ω ω ω ω ω
⇒
85
l 0=
2
1
2
2
2
3
S
C CC S CA A CB B CC C S C S S C S
S C A B C s
m m
m m
− + × + × + × = × + + ×
− + + =
M p g p R p R p a I ω ω I ω
g R R R a
( )
s
S
S CA A CB B C C CC C C S S C S
A B C C s
m m
× + × = + × + × + + × + + = + −
p R p R I ω ω I ω p R I ω ω I ω
g R R a a g
Combining the equations by eliminating : CM
CA A CB B C C× + × = + ×p R p R I ω ω I ω
A B Cm m+ + =g R R a
,Sm 0=Neglecting inertia of the shaft: !C =I 0
Exercise 2b:17 Solution
C O S OC S S OC
= + × + × × =
+ × + × × = + −
a a ω p ω ω p
Exercise 2b:17 Solution
( )2 A B 2 B0 0⋅ + = ⇒ ⋅ =f R R f R
[ ] [ ] [ ] [ ]A B Cm m+ + = f f f f
g R R a
[ ] [ ] [ ] [ ] [ ] [ ] [ ] CA A CB B C C × + × = + × f f f f f f
ff f
p R p R I ω ω I ω
Equations of motion in matrix representation:
90
,2 A 0⋅ =f R A B Cm m+ + =g R R a
( sin cos )1 3 gθ θ= −g f f
Exercise 2b:17 Solution
4 b R R 0
R R mg lm
+ − = −
Equations of motion:
Five equations and five unknowns: , , , ,, , , ,A 1 A 3 B 1 B 3R R
R Rθ 92
(4), (2) sin ( ) 2
2 2
= + = +∫ ∫ ∫ π π πθ
