Dynamical Modeling of Movement
Anteprima
ESTRATTO DOCUMENTO
3.1. CENTER OF MASS 28
−
r = P O
define the vector which point straight the center of each infinitesimal
dV dm.
element of mass This two quantities are related by a proportional
coefficient called density, which appears to change with respect to the given
∈
P S.
point Thus we obtain: dm = ρ(r)dV
The total mass of the rigid body is:
Z Z ZZZ
M = dm = ρ(r)dV = ρ(x, y, z)dxdydz
S S S
f (M, t) S.
Remark. Let be a scalar or vector field, defined on the solid It is
S
called integral of field, with respect to the mass distribuition of the quantity:
Z ZZZ
f (t, r)dm = f (t, r)ρ(r)dxdydz
S S
It can be shown that the timederivative operation can be lead within the sign of
integral by this equality: Z Z ∂f (t, r)
d f (t, r)dm = dm
dt ∂t
S S
This remark will help us to express some of the following quantities.
3.1 Center of mass {O, },
R = e , e , e N
Definition. In a given reference frame let us consider
1 2 3 −
m r G O
particles of mass and position expressed by . The center of mass of
i i
these particles, coinciding with the center of gravity in a uniform gravitational
field, si defined by: N
P m r
i i
i=1
−
G O = g e + g e + g e =
1 1 2 2 3 3 N
P m
i
i=1
We will replace discrete distributions of mass by a continuos ones:
N Z
X −→ S
i=1 −→
m dm
i −→
r r
i 28
3.1. CENTER OF MASS 29
R
M =
Definition. Given a solid S, whose mass is dm, with relative position
S
~
−
P O = OP = r,
of each infinitesimal element expressed by the center of mass
is defined by: R rdm
S
−
G O = R dm
S
A very important property of the center of mass is that it is independent on
0
O
the frame reference. Let be another reference frame centered in the new one
we calculate the center of mass with respect to. Ad absurdum, we state that
this change of reference modify the position of the center of mass too, that we
0
G
note with point . By definition:
0 0 0
R R R
− − −
r dm (P O )dm (P O + O O )dm
0 0 S S S
−
G O = = = =
R R R
dm dm dm
S S S
0
R R
− −
(P O)dm (O O )dm 0 0
S S −
− − = G O
+ = G O + O O
= R R
dm dm
S S 0
≡
G G
G
It’s clear that the center of mass is an ideal point in which we can place
a specific frame reference, and further it will be clear why. Now we need to
express all the kinetic quantity introduced for particles. {O, }
S R = e , e , e
Definition. Given a solid and a reference frame its linear
1 2 3
R
momentum with respect to is: Z
p = v dm
S/R P /R
S
v
where express velocity for each infinitesimal element.
P /R S A,
Definition. Given a solid and a pole its angular momentum is, with
{O, }:
R = e , e , e
respect to a reference frame 1 2 3
Z − ×
σ (A) = (P A) v dm
S/R P /R
S
v
where express velocity for each infinitesimal element.
P /R S,
Observe that the pole has not to be within the solid and the definition is
R.
true wheater the pole is fixed or moving with respect to Infact a fundamental
property of the angular momentum is that it can be expressed in terms of
29
3.2. KOENIG’S THEOREMS 30
A B:
another pole. For any given points and
Z Z
− × − − ×
σ (A) = (P A) v dm = (P B + B A) v dm =
S/R P /R P /R
S S
Z Z
− × − ×
= (B A) v dm + (P B) v dm =
P /R P /R
S S
Z
− × − ×
= (B A) v dm + σ (B) = (B A) p + σ (B)
P /R S/R S/R S/R
S
Now, in order to achieve better expression of these two quantities without cal
S,
culating integrals on the solid we have to rely on Koenig’s theorem, which is
the content of next paragraph.
3.2 Koenig’s Theorems
S,
Definition. Let us consider a solid moving with respect to the usual reference
R. R S
The Koenig’s reference frame associated to the movement of with
k
{O, }
R = e , e , e
respect to is:
1 2 3 {G, }
R = k , k , k
k 1 2 3
Note that the Koenig’s reference has a drag velocity of the point G; if the solid
R, v
is translating with respect to it is said that is the drag velocity of the
G/R
S.
solid {O, }, {G, },
R = e , e , e R = k , k , k
Remark. For the reference frames 1 2 3 k 1 2 3
P S
and a given point belonging to the solid we can write composition of vectorial
v ω
quantities of linear velocity and angular velocity with respect of these two
references. Infact: − − − − −
P O = P G + G O = (G O) + (P G)
Applying timederivative: d d
d 0 0
− − −
(P O) = (O O) + (P O )
dt dt dt
that is to say that: v = v + v
P /R G/R P /Rk
S
In particular, given a solid with its Koenig’s frame and an absolute reference
R,
frame the Galileo theorem for the composition of velocity states that:
~
0
×
v = v + ω O P
P /R G/R S/R
30
3.2. KOENIG’S THEOREMS 31
Respectively for angular velocity, we know that this vectorial quantity is
independent by the application point (is a free a vector, on the contrary of
velocity. acceleration, force which are all applied ones) and its coordinates
R
dependens on the basis we want to express. Note that the axes of are not
k
S,
fixed with the movement of the solid it only has a common point G with it.
R G
As its directions remain fixed with respect to whatever the movement of
{k } {e }),
, k , k , e , e
(no rotation of axis with respect to the movement of
1 2 3 1 2 3
R R
the Koenig’s reference with respect to is a translation. First we find the
k S:
expression of the linear momentum for a solid −
Z Z d(P O)
p = v dm = dm =
S/R P /R dt
S S
R −
P Odm
− −
Z d(P O) d d(G O)
1 S
= M dm = M = M = M v
G/R
M dt dt M dt
S
Now, using the property of the precedent paragraph, we can express angular
G.
momentum with respect to another pole, let it be From the previous point
S,
we know the expression for the linear momentum of the solid so that:
− ×
σ (A) = (G A) M v + σ (G)
S/R G/R P /R
σ (G).
We have now to calculate the quantity By definition of linear momen
P /R
tum and using the folliwing remark about the Koenig’s reference:
Z Z
− × − ×
σ (G) = (P G) v dm = (P G) (v + v )dm =
P /R P /R G/R P /Rk
S S
Z
Z − ×
− × + (P G) v dm
= (P G) v P /Rk
G/Rdm S
S {z } {z }

 =0 =σ (G)
P /Rk ∈
P S
The second term is simply the angular momentum of all the points of
R
the solid, with the pole coinciding with the origin of , in which every point of
k
S R
have only rotation speed (no translation with respect to ). Let us consider
k
the first one; it can be assumed that velocity is a costant (is the drag velocity
of the solid) with respect to the infinitesimal element, so we can write:
Z
− ×
(P G) dm v
G/R
S
Furthermore: R −
(P G)dm
Z S
− −
(P G)dm = M = M (G G) = 0
R dm
S S
31
3.3. KINETICS OF ROTATIONS 32
So it’s finally: σ (G) = σ (G)
S/R S/Rk
All the calculation are the resultant of the well known theorem of Koening,
which states: S G,
Theorem. Given a solid with center of mass placed in fixed or moving with
R, R
respect to its linear momentum with respect to is equal to linear momentum
G M A
of the particle of mass and its angular momentum in a given pole with
R A G
respect to is the total of the angular momentum in of the particle of mass
M S
and the angular momentum in G of the solid in its movement with respect
R G.
to the Koenig’s reference , which is a rotation around the fixed point In
k
formulas: p = M v (3.1)
S/R G/R
σ (A) = M v + σ (G) (3.2)
S/R G/R S/Rk
3.3 Kinetics of rotations
In this section our goal is to express the specific quantities describing motion
S
and dinamics of a rigid body when it is rotating around the motionless point
∈
O S, throught which passes the rotation axis, and where is centered the fixed
R.
reference frame Within this context, is obviously that drag velocity is nils
ω
(no translation) and there would be a laying on the instantaneous rotation
n(t) O t.
axis with direction expressed by versor passing by for every istant We
will need some preliminary results to go on. Relying on what said until here,
S
now we calculate angular momentum of the solid with respect to the reference
R: Z Z
~ ~ ~
× × ×
σ (O) = OP v dm = OP ω OP dm
Σ/R
P /R
Σ/R Σ Σ
The angular velocity is independent on the integral, so the previous relation
defines a linear dependence between the angular momentum and the angular
velocity. It can be written that:
σ (O) = ) (3.3)
L(ω
Σ/R Σ/R
tensor of inertia.
where is the so called
L V 3
→ (1, 1)
Remark. Let be a linear application (or a tensor) trasforming a
R
V
∈
v w. I
given vector in another one Said its associated matrix, we can write
R:
with respect to a given reference [w] = [I v]
R R
32
3.3. KINETICS OF ROTATIONS 33
B {i,
= j, k} R,
Being the related basis with the associated matrix with regard to
B is: h i
[I] = L(i) L(j) L(k)
R
V
∈
v
For any given vector (which here take the same place of the quantity
ω in 3.3), the inertia tensor is in this way:
Z
~ ~
× ×
= OP v OP dm
L(v) Σ {O,
R = i, j, k},
With respect to the reference frame we can express vectors
using coordinates: ~
OP = x i + x j + x k
1 2 3 I
For we want to express the inertia tensor with the associated matrix with
R,
respect to following the previous remark, we aim to express its columns by
{i, j, k}.
applying on each versor Recalling the Gibb’s formula we also know
L
that:
~ ~ ~ ~ ~ ~
× × · − ·
OP v OP = (
OP OP ) v (
OP v) OP =
~ ~ ~
2
k k − ·
= OP ( OP v) OP
I,
Thus, in order to compute the first column of we must calculate the application
B. i:
in the element of the basis Consider its first element,
L Z
~ ~ ~
2
k k − ·
= OP ( OP i) OP dm =
L(i) Σ
Z
21 22 23 −
= (x + x + x )i x (x i + x j + x k) dm =
1 1 2 3
Σ Z
22 23 − −
(x + x )i x x j x x k dm
= 1 2 1 3
Σ I
The latter formula states the first three elements of the first column of are:
Z 22 23
I = x + x dm
11 Σ
Z
−
I = x x dm
21 1 2
Σ
Z
−
I = x x dm
31 1 3
Σ
Iterating this thinking on the remaing to columnas, we obtain all the coefficients
I.
of By easy calculation it’s clear that, whatever is chosen the basis, the
33
3.3. KINETICS OF ROTATIONS 34
I = I
associated matrix is symmetric, so that . Therefore we can write it in
ij ji
this manner:
I I I
11 12 13
·
I = I I
22 23
· · I 33
where the remaing unexpressed terms are given by:
Z 21 23
I = x + x dm
22 Σ
Z
−
I = x x dm
23 2 3
Σ
Z 21 22
−
I = x + x dm
33 Σ
Observe that, thought the inertia tensor it’s an invariant upon the chosen basis
(only changes elements when it’s computed its associated matrix in a given
O,
basis), it depends on the chosen pole such as the angular momentum that,
not by chance, is calculated by means of the inertia tensor as stated in 3.3.
Thus, the fully representation of the inertia matrix (using tensorial product) is:
Z
~ ~ ~
2
k k − ⊗
I = OP Id OP OP dm
3
O S
Now, in order to apply Koenig’s theorems, we aim to calculate the inertia tensor
with respect to a special point, the baricenter: in this scope, the inertia tensor
central inertia tensor.
is called Σ (O, n̂) Σ.
Definition. Let it be a solid and an axis, not necessary crossing
∈
P Σ, H
For every calling the point which represent the orthogonal projection
P n̂, Σ n̂
of on it is called moment of inertia of with respect to the quantity:
Z ~ 2
k k
I = HP dm
O,n̂ Σ
Regarding this definition is almost prooved the following result:
B {i,
= j, k} O,
Theorem. In a given basis and a pole the diagonal components
I Σ
of the inertia matrix are respectively the moments of inertia of with respect
O
(O, ī), (O, j̄) (O, k̄).
to the axes and B {i,
= j, k} O,
Definition. In a given basis and a pole the products of inertia
I
are the nondiagonal components of the matrix of inertia .
O
With these definition we have characterized the six indipendent coefficients
of the inertia matrix, distinguished in moments of inertia and products of inertia.
34
3.3. KINETICS OF ROTATIONS 35
{O,
Σ, R = i, j, k} (O, n̂)
Theorem. Given a solid let it be and an axis of
Σ
symmetry for and its mass distribution. If one of the vectors of the basis
n, I
coincides with then the product of inertia of along the correspondig row
O
and column are nils. (O, î) Σ,
Proof. Let’s assume that is an axis of symmetry for and we have to
B {i,
O = j, k}.
calculate inertia matrix with respect to the pole and the basis I
We must compute the terms outside the diagonal, and we choose for istance :
12
Z
−
I = x x dm
12 1 2
Σ B;
M (m , m , m )
Let’s consider , with coordinates with respect to regarding
1 1 2 3
M
the symmetry axis there is an opposite point which is expressed exactly like
2
−m −m
(m , , ). Σ,
This thinking is extendeble to every point of that can be
1 2 3
parted in two subsets containing opposite point with respect to the simmetry
axis: −
+ ∪
Σ = Σ Σ
It comes out that: Z Z
− −
I = x x dm = x x dm =
12 1 2 1 2
−
+ ∪Σ
Σ Σ
Z
Z
−
= x x dm + x x dm =
1 2 1 2
− +
Σ Σ
Z
Z
−
= m m dm + m (−m ) dm = 0
1 2 1 2
− +
Σ Σ
I = I
It also holds that for the inertia matrix symmetry. The same result
12 21
I
may be obtained for ; therefore in this case the inertia matrix will be:
13
I 0 0
11
I = 0 I I
22 23
0 I I
23 33
From the latter result we can diagonalize inertia matrix if there are unless
Σ.
two axis of symmetry for In general, since inertia tensor is symmetryc, it
always exists a basis which allows to express inertia matrix as a diagonal matrix.
This comes as a direct application of the spectral theorem of the linear algebra,
which states that any real symmetric linear operator is diagonalizable: in other
words, it exists always at least an orthonormal basis in which the matrix of
inertia results to be diagonal. 35
3.3. KINETICS OF ROTATIONS 36
Theorem. The inertia tensor is symmetrical, therefore it exist always a basis
L
B {
= î, ĵ, k̂} I
in which it’s associated matrix is diagonal. The axis associated
O
O
with the versors and passing throught are called principal axes of inertia.
Σ
Generally, this basis is affixed to and the resulting inertia matrix is com
G Σ,
puted with rispect to the center of mass of the solid which is also the
center of the affixed reference frame. Σ
Theorem. It is called a revolution solid if the inertia tensor has infinitive
principal axis of inertia. In this situation there will be at least one direction
{i, },
k j, k
where lies an axis of revolution; for every orthonormal basis the
r r
inertia matrix is written as:
I 0 0
11
I = 0 I 0
11
0 0 I 33
Therefore, in this special cases, only two free parameters are needed.
k Σ, z
Proof. Let’s consider the versor a revolution axis for and let be the axis
r
of an orthonormal frame made up as:
{i, }
j, k r
This is a set of principal axes of inertia, so the associated inertia matrix is
0 0 }
{i, k z
j , k another reference frame. Since is its axis this
diagonal. Let be r
r z
will be another set of principal axes of inertia. Furthermore, whilst the to
axes of the two referencies are not changing orientation, it can be said that the
θ z
second is obtain as rotation of angle of the first along is axis:
−
cos θ sin θ 0 T
10 01
1 R = R
R = sin θ cos θ 0
0
0 0 1
Whereas the rotation matrix between them expresses the change of basis matrix,
it’s known by linear algebra the relation that ties matrix of linear application
36
3.3. KINETICS OF ROTATIONS 37
in different reference frames: 01 10
[I] = R [I] R =
0 0
R R
− cos θ sin θ 0
A 0 0
cos θ sin θ 0 − =
= sin θ cos θ 0
0 B 0
sin θ cos θ 0
0 0 1
0 0 C
0 0 1 2
2 −
A cos θ + B sin θ A sin θ cos θ B cos θ sin θ 0
2
2
−
= A sin θ cos θ B cos θ sin θ A cos θ + B sin θ 0
0 0 C
Since even the new reference frame is principal inertia axes one, so the non
diagonal terms has to be nils. That means:
−
A sin θ cos θ B sin θ cos θ = 0
A = B,
The equation is solved for and that justify the structure of the inertia
matrix in this particular case.
The meaning of a revolution axis is that the mass of the solid is homoge
neously distributed with respect to this one. Few of the most famous geometrical
shapes has an axis of revolution: sphere and cylinder are same examples. With
regards to the inertia matrix, to find out a revolution axis equals to a remarkable
x
property of the diagonal coefficientes (for shapes with revolution axis, the and
y axes are symmetry ones, so inertia matrix is diagonal). With these results, it
is easy for some solids to write the inertia matrix in a suitable reference frame.
Moreover, is possible to change the pole of inertia matrix; the HugensSteiner
theorem, also known as the parallel axis theorem, allows to express the moment
I) d
of inertia (diagonal terms of referred to a parallel axis distant from the
0
O
starting one, passing from another point . S
Theorem. The moment of inertia of a solid about any its axis is equal to
the moment of inertia about a parallel axis through the center of mass plus the
mass of the solid times the squared distance between the axes. In formulas:
2
I = I + M d
0 0
O ,n G,n 0
G, O
Proof. Let assume that in the respective referencies of the parallel axes
z I
lie both on the direction. So the inertia moment is:
0 0
O ,n
Z 02 02
I = x + y dm
0 0
O ,n S
37
3.3. KINETICS OF ROTATIONS 38
Figure 3.2: Parallel axes
0 0 0 0
(x , y , z ) O
Given a point of coordinate in the reference of , in the reference
(x +
frame of the center of mass the same point is written with coordinates
d, y, z). Thus: Z Z
02 02 2 2
I = x + y dm = (x + d) + y dm =
0 0
O ,n S S
Z Z Z Z
2 2 2 2 2 2
= x + 2dx + d + y dm = x + y dm + d dm + 2d xdm =
S S S S
Z Z Z
1
2 2 2
= x + y dm +d dm +2dM xdm
M
S S S
 {z }  {z }
I M
G,n x
The latter term indicates the coordinate of the center of mass, but since the
G,
coordinate in the integral is expressed with respect to the reference frame in
x
the is nil. So:
G 2
I = I + M d
0 0
O ,n G,n
The physical interpretation of moment of inertia is the resistance of a body
to a change in the magnitude of angular velocity, so it describes the attitude to
a solid body to change its initial orientation when a force is applied upon it. On
the other hand, the products of inertia create coupling in rotation. A moment
of forces about a principal axis a principal axis produces rotation about this axis
alone. Due to the nonzero product of inertia, a moment of force about any axis
other than a principal axis produces simultaneous rotations about orthogonal
I
axes. Here is a rewiev of the main inertia moments , which mean they are
GM
z
expressed with respect to the center of mass, and a axis passing across it:
M
Observe that being equal the mass , the sphere has the lower amount of inertia
regard with rotations, as other many properties that in sphere seems to acquire
optimal values. 38
3.4. KINETIC ENERGY 39
Geometric shape Inertia moment
2
M R I = M R
Hoop or cylindrical shell, mass , radius GM 12 2
M R I = M R
Solid cylinder, mass , radius GM 1 2
M L
M L I =
Long slender rod, mass , length GM 12
25 2
M R I = M R
Solid sphere, mass , radius GM 1 2 2
M a, b I = M (a + b )
Rectangul plate, mass , sides GM 12 23 2
M R I = M R
Cave sphere, mass , radius GM 3 2
M R
M R I =
Cone, mass , basis radius GM 10
Table 3.1: Inertia moments
3.4 Kinetic energy
S R,
Definition. Given a solid moving with respect to a reference frame its
kinetic energy is: Z
1 2
kv k
T = dm
S/R P /R
2 S
As one can understand by definition, the kinetic energy is a quantity related
to the whole rigid body with respect to an observer. Indeed, it does not depend
on the chosen basis because it’s a scalar quantity. As well as linear and angular
momentum, also the kinetic energy has its decomposition, without calculating
S.
the integral of each single point of S
Theorem. The kinetic energy associated to the movement of a solid with
R
respect to a reference frame is the sum following contributes:
• R G S,
the kinetic energy with respect to of the center of mass of taken
as a particle with assigned mass pair to the solid.
• S
The kinetic energy associated to the movement of with respect to the
G.
Koenig’s reference frame, which is a rotation around the point
In formula: 1 1
2
kv k ·
T = M + σ (G) ω (3.4)
S/R G/R S/R S/R
2 2
Proof. By the principle of galilean relativity:
v = v + v
P /R Rk /R P /Rk
39
3.4. KINETIC ENERGY 40
We get: 2
kv k · · · ·
= v v = v v + v v + 2 v v =
P /R P /R P /R Rk /R Rk /R P /Rk P /Rk Rk /R P /Rk
2 2
kv k kv k ·
= + + 2 v v
P /Rk P /Rk Rk /R P /Rk
Then by definition:
Z Z
1 1
2 2 2
kv k kv k kv k ·
T = dm = + + 2 v v dm =
S/R P /R R/Rk P /Rk Rk /R P /Rk
2 2
S S Z
Z Z
1
1 2 2 ·
kv k kv k
+ + v v dm
dm dm
= Rk /R P /Rk
P /Rk Rk /R
2 2 S
S S 
 {z }  {z } {z }
=0
T M
S/Rk
Let us consider the last term: ~
Z Z Z
d d
d
GP ~ −
v dm = dm = GP dm = (G G) = 0
P /Rk dt dt dt
S S S
v v
Considering that is the same of , we obtain:
Rk /R G/R
1 2
kv k
T = M + T
S/R G/R S/Rk
2 T
that is the thesis. In particular, the explicit calculus of leads to:
S/Rk
Z Z
1
1 2
kv k ·
v
T dm = v
= dm
P /Rk P /Rk
S/Rk P /Rk
2 2
S S S
By the Galileo theorem, the linear velocity of any point of with respect to its
~
×
v = ω GP
attached frame reference is , hence exploiting a property
S/R
P /R k
k
of the mixt product:
Z Z
1
1 ~
· · ×
v v dm = v (ω GP ) dm =
S/R
P /Rk P /Rk P /Rk
2 2
S S
Z Z
1 1
~ ~ ×
· × ·
= GP v dm =
ω ( GP v ) dm = ω
S/R P /Rk S/R P /R
k
2 2
S S
1 1
· ·
= ω σ (G) = σ (G) ω
S/R S/R S/R S/R
2 2
40
3.5. TIME DERIVATIVES OF LINEAR AND ANGULAR MOMENTUM 41
3.5 Time derivatives of linear and angular mo
mentum {O,
R = i, j, k} S
Definition. Let it be the inertial reference frame a let be a
S
rigid body. Then the rate of change of the linear momentum of with respect
R
to is: Z
dp
S/R
A := = a dm
S P
dt S S R
The rate of change of the angular momentum of with respect to and a pole
A is given by: Z
dσ (A) ~
S/R ×
=
δ (A) := AP a dm
S/R P
dt S
In particular the rate of change of angular momentum has this property to
B:
change its pole. In fact, chosen a new pole
Z Z Z ~
~ ~ ~ ×
× × BP a dm =
δ (A) = (
AB + BP ) a dm = AB a dm + P
S/R P P
S S S
~ ×
= AB A + δ (B)
S/R S/R
A B:
So, changing the pole from to ~ ×
δ (A) = δ (B) + AB A (3.5)
S/R S/R S/R
G S,
Theorem (Eulero’s equations). Considering the center of mass of with
R,
respect to the usual frame reference the rate of change of linear and angular
momentums is given by: A = M a (3.6)
S G
d ×
σ (A) + v M v
δ = (3.7)
S/R A/R G/R
S/R dt
A M S.
where is the pole of the angular momentum and is the mass of the solid
Proof. For the first one: Z Z dv
P /R
A = a dm = dm =
S P dt
S S
2 2
Z
d d
~ ~
= OP dm = M OG = M a G
2 2
dt dt
S
For the second one, by definition:
Z Z
d d
~ ~
× ×
δ (A) = AP v dm = AP v dm =
S/R P /R P /R
dt dt
S S
~
Z Z d AP d
~ × ×
= AP a dm + v dm := σ (A) [1]
P /R P /R S/R
dt dt
S S 41
3.5. TIME DERIVATIVES OF LINEAR AND ANGULAR MOMENTUM 42
Observe that:
d d d d
~ ~ ~ ~ ~
− −
AP = (
AO + OP ) = OP OA = v v
P /R A/R
dt dt dt dt
d ~ × × − × −v ×
AP v = v v v v = v
P /R P /R P /R A/R P /R A/R P /R
dt
Substituing on the main one:
Z Z
~ × − ×
= AP a dm v v dm =
P /R A/R P /R
S S
Z
− × − ×
= δ (A) v v dm = δ (A) v M v
P /R A/R P /R P /R A/R G/R
S [1]:
By making equal this latter one with d
− × σ (A)
δ (A) v M v = S/R
P /R A/R G/R dt
d ×
δ (A) = σ (A) + v M v
P /R S/R A/R G/R
dt
This system of equations allowing a simplified description of motion of rigid
bodies. Consider the equation 3.7: it can be simplified in two different situations:
A R.
1. is affixed with the reference frame That means:
d
⇒
v = 0 δ (A) = σ (A)
A/R S/R S/R
dt
≡
A G:
2. d
δ (G) = σ (G)
S/R S/R
dt
To summarize, we have to methods to compute the timederivative of angular
A, A
momentum. Given a pole if it is known the angular momentum in is proper
σ (G),
to use 3.7. Otherwise, expecially if it is known it is better to switch
S/R
A G B G:
pole from to and to use 3.5, identifying with
~ ×
δ (A) = δ (G) + AG A
S/R S/R S/R
To relate human body motion to forces, we need to know the inertial properties
of body segments: mass, tensor of inertia, locations of the center of mass. This
body segment parameters.
whole set of parameters is known as Body seg
ment parameters are difficult to measure. Although studies to quantify body
segment parameters have more than a 300year hisyory, we are still far from
42
3.5. TIME DERIVATIVES OF LINEAR AND ANGULAR MOMENTUM 43
characterization of the inertial properties of the body segments in various pop
ulations which differ in sex, age, race, health, physical activity and so on. The
dynamic modelling of human movement consists in representing the body as a
kinematic multilink chain, made up by links (or segments) with known inertial
characteristics. For this reason, one has to divide the body in such a way that
the individual parts move similarly to rigid bodies. When this segmentation
is realized according to landmarks, the individual body parts are called body
segments (unfortunatly many authors used different segmentation protocols in
their research). The determination of the body segment parameters requires
experimental measurements, either directly on anatomical specimens, or indi
rectly in vivo. From these data, authors have used statistic methods to establish
regression equations aiming to estimate the body segment parameters of any
subject as a function of several characteristic like mass, height, lenght et cetera.
On anatomical specimens, the mass of each segment can be determined by direct
measurement after separation. On volunteers, the mass of each body segment
can be determined by measurement of the water volume displaced when it is
plunged in water: this requires an hypothesis about the density of the segment.
Figure 3.3: Measures of volumes
On anatomical specimen, the center of mass of each segment can be determined
by finding the equilibrium position when it is hanged by a thread (no rotation
occurs when the force exerted by the thread and the gravity are aligned).
On anatomical specimens, the moment of inertia about an axis can be deter
mined by measuring the frequency of the free oscillations of the segment, when
articulated by a hinge joint at one extremity (knowing the location of the center
of mass).
On volunteers, the inertial characteristics can be derived from a geometrical
modelling of the segment (more or less fine), assuming a constant density.
Many authors have established regression to fit the body segment parameters
to the actual subject under study. Nevertheless, the results of such regression
43
3.5. TIME DERIVATIVES OF LINEAR AND ANGULAR MOMENTUM 44
Figure 3.4: Anatomical specimens
Figure 3.5: Modelization of a segment
tables can be used only when the subject characteristic (sex, age, physical sta
tus) are close to the avarage value of the sample that had been used to build
the regression equations. 44
3.5. TIME DERIVATIVES OF LINEAR AND ANGULAR MOMENTUM 45
Figure 3.6: Body segments approximations
45
Chapter 4
Rigid multibody dynamics
Dynamics is the study of the relationship between forces and motion, in
order to describe mechanical system with causeeffect relationships. The prin
ciple of inertia, stated by Galileo Galilei and then the fundamental principle of
dynamics, stated by Newton are the basic of classical mechanics. By experi
menting with balls rolling on planes of different types, if the plan is very rough,
the ball stops quickly. On the contrary, if the plan is very smooth or covered
with a film of oil for example,nthe ball travels a much greater distance before
it stops. Galileo had then the idea of "friction forces". It is finally from this
principle that Netwon introduced, in 1687, the description of the causes of the
movement, and introduces the basic concept of force and mechanical action: the
movement of an isolated system is rectilinear and uniform, any deviation from
this law of motion is due to a force. In the absence of external force, the body
continues its trajectory and retains its speed.
4.1 Dynamic and static principles
Σ
Definition. Let it be a material system (a set of material points or solid
R
bodies) and a galilean reference frame (inertial frame). Then for every point
∈
P Σ are defined the following abstract vectors:
" #
A Σ
~
D =
Σ/R δ (P )
Σ/R
" #
v
~
O Σ/R
=
Σ/R ω (P )
Σ/R
" #
F ext
~
F =
e→Σ M
P,ext
46
4.1. DYNAMIC AND STATIC PRINCIPLES 47
Σ
Note that external mechanical actions on are the sum of all actions exerted
by an external element on one element of the system under study.
Theorem (Cardinal equations of dynamics). It exist at least one reference
R, Σ:
frame named galilean reference frame in which for any material system
~ ~
D F
=
In other terms:
A = R (Theorem of the linear momentum) (4.1)
e→Σ
Σ
δ = M (Theorem of the angular momentum) (4.2)
O,e→Σ
Σ/R Σ,
This fundamental principle, applied to a given system provides after
projection only 6 independant equations, which are differential equations of
Σ
the second order with regards to the position parameters of with respect to
R. Besides, they involve the unknow components of external joints forces. In
general, these equations are coupled and nonlinear, so that it is not possible
to obtain an analytical solution for the position parameters, expecially ad there
are often more unkwons that equations. In terms of unkwons, we use additional
empirical laws, in particular the friction laws that help to model the joint forces.
Usually, to find position parameters as a function of time, the system is solved
Σ
numerically. Now let us consider the special case where is reduced to a unique
solid or a particle. S
Definition. A solid (or a particle) is in static equilibrium with respect to an
R
inertial frame if it stays at rest: ~
O = 0
S/R
~ ~
O D
∀t,
S = 0
If is at rest, then and since depends directly on it, is
S/R Σ/R
as well zero, so the folliwing theorem is proved, and it’s a necessary condition
S
to have static equilibrium for a solid under the actions of forces.
S
Theorem. A solid is in static equilibrium with respect to an inertial frame
R if: ~
F ∀t
= 0
e→Σ ~
F
In pratice, it consists in canceling the elements of in any point, for
e→Σ
O,
example obtaining the following system:
⇒
F =0 3 scalar equations after projection
ext ⇒
M =0 3 scalar equations after projection
P,ext 47
4.1. DYNAMIC AND STATIC PRINCIPLES 48
At maximum we have 6 independent equations to describe the equilibrium of
S R.
with respect to Nevertheless, more than 6 unknows take part in the
equilibrium of a olid, which are components of joint forces. The system is the
statically indeterminate
named and another theory is required to calculate
Σ,
all the unknowns: theory of elasticity. For a material system the definition
of equilibrium is similary as the previous.
S
Σ = S S
Theorem. Let it be a material system, composed of solids or
i i
i Σ
material points. A necessary condition for static equilibrium of with respect
R S
to an inertial frame is that each part of is in static equilibrium with respect
i
R:
to ~
F = 0
e→S
i
S Σ.
Some (or all) forces external to can be forces internal to the system The
i
~
F ×
= 0 6 N
condition leads to after projection, if the system is composed
e→S
i
N S
of parts . In this perspective it is possible to formulate the actionreaction
i
principle: S ∈
Σ = S O Σ
Theorem. Let it be a material system, then at any give point
i
i
we have: −F
F =
→S →S
S S
i j j i 6
i = j
−M
M =
→Sj →Si
O,Si O,Sj
Proof. The prove is given for a particular system, and the thesis can be affirmed
Σ
by that without loss of generality. Let consider a system composed by 2 solids
S S
and , interacting between them (internal forces) and subject to external
1 2 S S S
forces by other to rigid bodies, , which contacts with the only , and ,
e1 1 e2
S
which contacts with the only . The cardinal equations of dynamics applied
2
Σ
to tells us that: ~ ~ ~ ~
D F F F
= = +
→S →S
e→Σ S S
Σ/R e1 1 e2 2
S S S
The same one applied to involves also the interation of on , considered
1 2 1
now as external: ~ ~ ~ ~
D F F F
= = +
→S →S
e→S S S
S /R 1 e1 1 2 1
1
S
Applied to :
2 ~ ~ ~ ~
D F F F
= = +
→S →S
e→S S S
S /R 2 e2 2 1 2
2
Yet: ~ ~ ~
D D D
= +
Σ/R S /R S /R
1 2
Matching together:
~ ~ ~ ~ ~ ~
F F F F F F
+ = + + +
→S →S →S →S →S →S
S S S S S S
e1 1 e2 2 e1 1 2 1 e2 2 1 2
48
4.1. DYNAMIC AND STATIC PRINCIPLES 49
Canceling equal quantities we have:
~ ~
F F
+ = 0
→S →S
S S
1 2 2 1
which implies the thesis. S
Remark. The kinetic energy for a solid with respect to a inertial frame
R
reference is the following quantity: Z
1 2
kv k
T = dm
S/R P /R
2 S T S:
We have proved that the timederivative of is the power of this
S/R
information is involved in the proof of the following result. S
Theorem. The power of the external forces exerted on a rigid body with
R
respect to a inertial frame is given by:
d
~ ~ ~
P F O
·
= T = (4.3)
e→S
e→S S/R
S/R
dt
Proof. Z
1 d
d 2
kv k
T = dm =
S/R P /R
dt 2 dt S
Z Z
1 ∂ · ·
(v v )dm = v a dm
= P /R P /R P /R P
2 ∂t
S S
A S,
Consider an affixed point to we can use the Galileo theorem to express
every point’s velocity with regards to this pole, that a priori is not fixed.
~
×
v = v + ω AP
P /R A/R S/R
Z Z ~
· × ·
(v a dm = (v + ω AP ) a dm =
P /R P A/R S/R P
S S
Z Z ~
· × ·
= v a dm + ω AP a dm =
A/R P S/R P
S S
Z Z ~
· · ×
= v a dm +ω AP a dm
A/R P S/R P
S S
 {z }  {z }
A δ
Thus: d ~ ~
O D
· · ·
T = v A + ω δ (A) =
S/R A/R S/R S/R S/R S/R S/R
dt
Since the timederivative of the kinetic energy is by definition the power, apply
49
4.2. JOINT MODELING 50
~ ~
D F
=
ing the cardinal equations ( ) we finally have:
e→S
S/R
~ ~ ~
P O F
·
=
e→S e→S
S/R
R,
Definition. For an inertial frame the work of the external forces exerted to
S t t t > t
a solid between the istances and , is defined by:
1 2 2 1
t
Z 2 P
W dt
= e→S/R
e→S/R t 1
T
Knowing that the kinetic energy is a primitive of the power, applying the
fundamental theorem of calculus we have:
W −
= T (t ) T (t )
2 1
e→S/R S/R S/R
S S
Definition. Consider to rigid bodies and and a inertial reference frame
1 2
R sharing one or more points of contact. The mutual power is the rate of energy
R:
exchanged between the two and is defined as a quantity independent on
~ ~
~ ~
P F F
O O
· ·
= =
→S →S
1↔2 S S
S /S S /S
1 2 2 1
1 2 2 1
~
O S
is the vector containing linear and angular velocity of the solid
where 2
S /S
2 1 ~
O
S ).
with respect (and viceversa with
1 S /S
1 2
∪
Σ = S S
In fact, having isolated (no external actions) but with contact,
1 2 R:
we can write with respect to a inertial system
d d d P P
T = T + T = + =
→S →S
Σ/R S /R S /R S S
1 2 2 1 1 2
dt dt dt
~ ~ ~ ~
F O F O
· ·
= +
→S →S
S S /R S S /R
2 1 2 1 2 1
~ ~
F F
−
=
Applying the actionreaction principle we know that :
→S →S
S S
1 2 2 1
d ~ ~ ~ ~
F O F O
· − ·
T = =
→S →S
Σ/R S S /R S S /R
2 1 2 2 1 1
dt
~ ~ ~ ~ ~
F O O F O
· − ·
= ( ) =
→S →S
S S /R S /R S S /S
2 1 2 1 2 1 2 1
4.2 Joint modeling
The connection between to solids involves forces which detailed study is del
icate. These forces depend on the nature on the nature of the materials in
contact, on their rugosity and on the local deformations of the concatc surfaces.
50
4.2. JOINT MODELING 51
When it comes to complex mechanical systems, a computationallyefficient tech
nique is needed to model the connections. Here comes the mathematical repre
joints.
sentation of different types of connections known as
Figure 4.1: Joints
S S I.
Let us consider two rigid bodies and , with one contact point in Ac
1 2
tually, because of this contact, a small area of deformation exist around the
I. S S
contact point Then the mechanical actions exerted by on at the point
1 2
I 3
should be modeled by a force vector and a moment vector of components
each: " #
R
~
F →S
S
1 2
=
→S
S
1 2 M
I, →S
S
1 2
P I,
We can always define a plane , tangent to both contact surface in and
R M
decompose the two vectors and in one normal component
I,
→S →S
S S
1 2 1 2
P
(along the normal axis to the plane ) and one tangential component (in the
{t,
P n, b},
plane ). With the basis the previous relation is expressed as:
" # " #
R T + N
~
F →S
S 1 2
= =
→S
S
1 2 M M + M
I, →S
S I,t I,n
1 2 N,
The normal component of the resultant force, is responsible for the non
compenetration of two contacting bodies, given at microscopic level by electro
static interaction and plays an important role in the friction laws, established
T
empirically by Coulomb. If tangential component is nil, the normal compo
51
4.2. JOINT MODELING 52
N
nent is perpendicular with the line of action, hence there is no momentum.
" " #
#
R N
~
F →S
S 1 2
= =
→S
S
1 2 0
M
I, →S
S
1 2
Let us consider a general case of a contact through a surface. The resultant force
R comes from elementary forces distributed on the surface contact, represented
I m
here by a single force applied in a point of the surface. Let consider a mass
placed on the horizontal ground, that we try to drag by exerting an horizontal
F S S
force on it. In our formalism is the ground and is the mass, and
1 2
R is so called vincular reaction, because is due to the contact that implies
→S
S
1 2
a vinculus on the possible displacements. The experiment shows that if the
F
magnitude of is too low, the solid remains motionless.
Figure 4.2: Force diagram
This experiment allows bringing out laws, named friction laws, which describe
to different situations: when the body is not sliding that means that, for this
couple of bodies, the exerted force is less than the maximum resistance friction
T
force. In this case the mass is in static equilibrium, and the friction force is
F,
equal in each istant to and it holds that:
kTk kNk
< µ
S
µ
where is the static coefficient of friction, which depends only on the nature
S
and on the surface quality of materials in contact. When the body start to
slide, the friction force became constant, depending on a coefficient of dynamic
v
friction, the normal component, and the direction of the sliding velocity :
g
v
g
kNku −
T = µ u =
k kv k
g
Let us give interpretation of the other forces devoloped by contact. Consider a
wheel in contact with ground: to make it pivoting the experiments shows one has
to create torque, which magnitude is higher than a minimal value corresponding
52
4.2. JOINT MODELING 53
Figure 4.3: Constitutive relation of friction
kM k. kM k <
to the pivoting friction torque No pivoting as long as as
I,n I,n
kNk.
λ When this condition is no more observed, there is pivoting and:
p 0
kM k kNk
= λ (Pivoting friction force)
p
I,n 0
λ λ are coefficient of pivoting friction
The coefficients of pivoting friction and
p p
in static and dynamics, consistent with a length. To roll a wheel on a horizontal
plane, one has to create a torque, which magnitude is higher than a minimal
kM k.
value corresponding to the rolling friction torque No rolling as long as
I,t
kM k kNk.
< δ When this condition is no more observed, there is rolling and:
p
I,t 0
kM k kNk
= δ (Rolling friction force)
p
I,t 0
λ λ
The coefficients of pivoting friction and are coefficient of pivoting friction
p p
in static and dynamics, consistent with a length. Very often, the pivoting and
M = 0
rolling friction are negligeable, so and the mechanical actions of
→S
I,S
1 2 R
contact between rigid bodies are simplified as a force applied to the contact
I.
point Now we are ready to describe mathematically joints, which can be
modeled as multiple point or surfaces of contact, imposing the degrees of freedom
∪
S S
of the system .
1 2 S S
Remark. Consider two solids, and , contacting each other. If:
1 2
~ ~ ~ ~
P F O F O
· ·
= = = 0
1↔2 →S →S
S S /S S S /S
1 2 1 2 2 1 2 1
then the joint is said ideal. 53
4.2. JOINT MODELING 54
~
F
Modelling joints action mean to introduce a number of variables in ,
→S
S
1 2
S S
each one blocking a degree of freedom od the solid with respect to (and
2 1
viceversa). Let us describe the following joints:
• Ball and socket joint
Figure 4.4: Ball and socket joint
The geometry of the joint leads to: " #
ω
~
O S /S
1 2
=
S /S
2 1 v A/S
1
v = 0.
where Let us calculate the mutual power:
A/S
1 ~ ~
P O F
· · ·
= = F v + M ω = 0
1↔2 →S →S →S
S /S S S A/S A,S S /S
2 1 2 1 1 2 1 1 2 1 2
v = 0, M = 0.
while the equality stays till So that a ball and
→S
A/S A,S
1 1 2
A
socket joint of center is ideal, it is enough that:
" #
F
~
F S /S
1 2
=
S /S
2 1 M
A,S /S
2 1
M = 0, F
where while is nondescript, so this joint allows three
A,S /S S /S
2 1 1 2 S
degrees of freedom to the attached bodies. To know the orientation of 2
(and viceversa) other equations are necessary, because the whole problem
involves 6 unknowns.
• Cylindrical joint
The geometry of the joint leads to:
" # " #
ω θ̇u
~
O S /S
1 2
= =
S /S
2 1 v λ̇u
A/S
1
λ
In fact the joint allows displacements only on the direction along the axis
(A, u), u.
and the only possible axis of rotation is indeed Nonetheless this
joint is not screwnut, for there is no relation between linear and angular
54
4.2. JOINT MODELING 55
Figure 4.5: Cylindrical joint
velocity (this case will be seen further). Let us calculate:
~ ~
P O F
· · ·
= = F v + M ω =
1↔2 →S →S →S
S /S S S A/S A,S S /S
2 1 2 1 1 2 1 1 2 1 2
· ·
= λ̇ F u + θ̇ M u = 0
→S →S
S A,S
1 2 1 2
λ̇ θ̇,
For every give linear and angular velocity and to be ideal is enough
that: ⊥ ⊥
F u M u
and
S /S A,S /S
1 2 2 1 F M
The condition above implies two equation that relate and ,
S /S A,S /S
1 2 2 1
so the number of unkwowns drop from 6 to 4: the cylindrical joint allows
two degrees of freedom (the other four are blocked by the four indipendent
component of the reaction force and momentum).
As a generalization, the number of mechanical actions components transmit
table by an ideal joint is equal to 6 minus the number of the joint’s degrees of
freedom. If the joint block one degree of freedom in translation then the joint
is able to transmit a force component in that direction. If the joint block one
degree of freedom in rotation around an axis, the joint is able to trasmit a mo
kinetostatic duality.
ment component about this axis. We speak of On the
other hand the number of degrees of freedom (DOF) of a rigid body tied with
costraints is given by: −
DOF = 6 n ~
F
n n = 0
where represent the independent components of . If that means
→S
S
1 2
that there is no contact, so no reaction forces and momentums and the rigid
body has full freedom in the space (adequately descripted by six independent
paramters). The following tables describes the principal ideal joint mechanisms.
55
4.2. JOINT MODELING 56
Figure 4.6
Figure 4.7
Figure 4.8
56
4.2. JOINT MODELING 57
Figure 4.9
Figure 4.10
57
Chapter 5
Lagrange’s equation of motion
analytical mechanics
In mathematical physics, is a collection of closely
related alternative formulations of classical mechanics. It was developed by
many scientists and mathematicians during the 18th century, after Newtonian
mechanics. Since Newtonian mechanics considers vector quantities of motion,
particularly accelerations, momenta, forces, of the constituents of the system.
By contrast, analytical mechanics uses scalar properties of motion represent
ing the system in terms of its total kinetic energy and its potential energy, not
using Newton’s vectorial forces of individual particles. A scalar is a quantity,
whereas a vector is represented by quantity and direction. The equations of
motion are derived from the scalar quantity by some underlying principle about
the scalar’s variation. Analytical mechanics takes advantage of a system’s con
straints to solve problems. The constraints limit the degrees of freedom the
system can have, and can be used to reduce the number of coordinates needed
to solve for the motion. The formalism is well suited to arbitrary choices of
coordinates, known in the context as generalized coordinates. The kinetic and
potential energies of the system are expressed using these generalized coordi
nates or momenta, and the equations of motion can be readily set up, thus
analytical mechanics allows numerous mechanical problems to be solved with
greater efficiency than fully vectorial methods. It does not always work for
nonconservative forces or dissipative forces like friction, in which case one may
revert to Newtonian mechanics. Analytical mechanics does not introduce new
physics and is not more general than Newtonian mechanics. Rather it is a
collection of equivalent formalisms which have broad application. In fact the
same principles and formalisms can be used in relativistic mechanics and gen
eral relativity, and with some modification, quantum mechanics and quantum
field theory also. The methods of analytical mechanics apply to discrete par
ticles, each with a finite number of degrees of freedom. They can be modified
58
I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher Meliuk di informazioni apprese con la frequenza delle lezioni di Dynamical Modeling of Movement e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università Politecnico delle Marche  Univpm o del prof Cheze Laurence.
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