Dynamical Modeling of Movement

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 time-derivative: 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 non-diagonal 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