Momenta and inertia matrices - polito.it · Momenta and inertia matrices Basilio Bona DAUIN–PolitecnicodiTorino Semester 1, ... If the system rotates around a generic axis with

Momenta and inertia matrices

Basilio Bona

DAUIN – Politecnico di Torino

Semester 1, 2015-16

B. Bona (DAUIN) Kinematics Semester 1, 2015-16 1 / 22

Introduction

Consider a discrete system with N point masses mi , i = 1, · · · ,N, as inFigure.

These systems are also called multipoint systems.

Given a generic base RF Rb, mass position is represented by thegeometrical vector ri , and the velocity by the physical vector vi .


If the system rotates around a generic axis with total angular velocity ω,every mass will acquire a linear (tangential) velocity vi due to the bodyrotation:

vi (t) = ω(t)× ri (t) (1)

The linear momentum (or translational momentum) pi(t) is the product ofthe mass mi times the velocity vi(t)

pi(t) = mivi (t)

In this context the symbol p indicates the linear momentum and NOT thepose of a rigid body in space.


Angular momentum

Given a point O in space and a point mass mi with position ri , we definethe moment of the linear momentum or angular momentum hi , the crossproduct between the mass position and its linear momentum

hi (t) = ri(t)× pi (t) = ri (t)× (mivi (t))

Replacing the last term with that in (1) and omitting for simplicity thetime dependency, we have

hi = mi (ri × (ω × ri ))

Summing up all contributions of the N point masses, we obtain the totalangular momentum of the system

h(t) =

N∑

i=1

hi =∑

i

mi(ri × (ω × ri)) (2)


Recalling the triple cross product property

a× (b× c) = (aTc)b− (aTb)c,

where aTc and aTb are scalar products, the angular moment becomes

h(t) =∑

i

mi

(

(rTi ri)ω − (rTi ω)ri

)

We use the symbol r2i for the norm ‖ri‖2 = rTi ri = (x2i + y2i + z2i ), and

computing rTi ω = (xiωx + yiωy + ziωz), one obtains

h =∑

i

mi

r2i

ωx

ωy

ωz

− (xiωx + yiωy + ziωz)

xiyizi

that is

h =

hxhyhz

=∑

i

mi (r2i − x2i )ωx −mixiyiωy −mixiziωz

−mixiyiωx +mi(r2i − y2i )ωy −miyiziωz

−mixiziωx −miyiziωy +mi (r2i − z2i )ωz

(3)


The relation (3) can be written in matrix form as

h =∑

i

Γxx ,i Γxy ,i Γxz ,iΓyx ,i Γyy ,i Γyz ,iΓzx ,i Γzy ,i Γzz ,i

ωx

ωy

ωz

=∑

i

Γiω (4)

where

Γxx ,i = mi (r2i − x2i ) = mi (y

2i + z2i )

Γyy ,i = mi (r2i − y2i ) = mi (x

2i + z2i )

Γzz ,i = mi (r2i − z2i ) = mi(x

2i + y2i )

Γxy ,i = Γyx ,i = −mixiyi

Γxz ,i = Γzx ,i = −mixizi

Γyz ,i = Γzy ,i = −miyizi


Defining a matrix Γ as

Γ =

Γxx Γxy ΓxzΓyx Γyy ΓyzΓzx Γzy Γzz

=∑

i

Γxx ,i Γxy ,i Γxz ,iΓyx ,i Γyy ,i Γyz ,iΓzx ,i Γzy ,i Γzz ,i

=∑

i

Γi

from (4) we can write

h =


ωx

ωy

ωz

= Γω (5)

Now, indicating the time dependency, we have

h(t) = Γ(t)ω(t)

Γ(t) is the inertia matrix or inertia tensor.


The diagonal terms of Γ are the inertia moments

Γxx =∑

i

mi (r2i − x2i ) =

∑

i

mi (y2i + z2i )

Γyy =∑

i

mi (r2i − y2i ) =

∑

i

mi (x2i + z2i )

Γzz =∑

i

mi (r2i − z2i ) =

∑

i

mi(x2i + y2i )

the other terms are the inertia products

Γxy = Γyx = −∑

i

mixiyi

Γxz = Γzx = −∑

i

mixizi

Γyz = Γzy = −∑

i

miyizi


Inertia moments

We observe that the inertia moments are the sum of the products of eachmass times the Euclidean distance with respect to the x , y and z axesrespectively.

Inertia products

When the inertia products are all zero the inertia matrix is diagonal

Γ =

Γxx 0 00 Γyy 00 0 Γzz

In this case the RF axes result to be aligned with the body principalinertia axes and Γ is called principal inertia matrix.


Another mathematical representation of h is obtained considering (2) andthe properties of skew-symmetrical matrices:

h =∑

i

mi (ri × (ω × ri)) =∑

i

miS(ri )S(ω)ri (6)

Now, recalling the relation S(ω)ri = −S(ri)ω one can write

h =∑

i

miS(ri )S(ω)ri =∑

i

−miS(ri )S(ri )ω (7)

and since ω is one for the body, at the end we have

h =

(

−∑

i

miS(ri)S(ri)

)

ω = Γω (8)

conΓ = −

∑

i

miS(ri)S(ri ) (9)


Γ represent the inertial properties of a rigid body with respect to arotation, in the same way as the mass of a body represents the inertialproperties with respect to translations.

It is important to notice that Γ depends on the geometrical vectors ri thatdescribe the positions of each mass mi in a given reference frame,assuming a fixed point O (see slide 4); this point may be fixed or changing(as the center of mass C of a moving body); in the first case we write Γo ,in the second case we write Γc .

Moreover, if one chooses two different RF with the same origin but withrotated axes, two different inertia matrices are obtained. Their mutualrelations will be defined in a successive slide.


Linear and angular momenta

Relation (5) is similar to the linear momentum of a rigid body, that weknow to be the product of the total mass for the linear velocity, i.e.,p(t) = mv(t).

We can therefore write p(t) = Mv(t), where M = mI.

In conclusion

p(t) = Mv(t)

h(t) = Γ(t)ω(t)


Distributed mass bodies

Assuming now that the body is a continuous distribution of infinitesimalmasses dm contained in a finite space region V having volume V , withpoint density ρ(r) = ρ(x , y , z) function of the mass position, we can writedm = ρ(x , y , z)dV .

The total volume is V =∫

VdV and the total mass is therefore

mtot =

∫

V

dm =

∫

V

ρ(x , y , z)dV

To keep things simple, we can apply the derivation use for point massesreplacing the sum operator with the integral operator defined in thevolume V; in this way we can write

Γ =

∫

V

ρ(x , y , z)

y2 + z2 −xy −xz

−xy x2 + z2 −yz

−xz −yz x2 + y2

dV (10)


obtaining the inertia moments

Γxx =

∫

V

ρ(r)(y2 + z2) dV

Γyy =

∫

V

ρ(r)(x2 + z2) dV (11)

Γzz =

∫

V

ρ(r)(x2 + y2) dV

and the inertia products

Γxy = Γyx = −

∫

V

ρ(r)xy dV

Γxz = Γzx = −

∫

V

ρ(r)xz dV (12)

Γyz = Γzy = −

∫

V

ρ(r)yz dV


Since ρ(r)dV is equal to the infinitesimal mass dm(r), one can write (10)as follows

Γ =

∫

V

y2 + z2 −xy −xz

−xy x2 + z2 −yz

−xz −yz x2 + y2

dm(r) (13)


Some observation

The inertia matrix is defined starting from the angular moment h; thisone, as noticed before, is computed with reference to an arbitrarychosen point O.

The cartesian components of h are represented in some RF, thatoften, but not always, has the origin coincident with; so, the inertiatensor describes the way in which the mass is distributed with respectto the axes of the reference frame R0(O; x , y , z). If this referenceframe is fixed to the body (i.e., is a body frame) the inertia tensor istime invariant..

The above observations clarify why sometimes one says that theinertia moment is with respect to a point, and sometimes that theinertia moments are with respect to the axes. This shall not createproblems, since both sentences are based on meaningful definitions.


If one chooses a different body frame in the same origin O, the hcomponents change, and therefore also the matrix Γ, changes, but itremains time invariant.

If one translates the origin in a different point O ′, Γ changesaccording to the parallel axes theorem.

If the body rotates with respect to the reference frame, and thereforethis is NOT a body frame, the components of Γ vary with time.


Body center-of-mass

Given a rigid body B, and its body reference frame Rb, the position of thecenter-of-mass (CoM) C of the body is defined by the vector rc thatsatisfies the following relation:

∫

B

rdm = rc

∫

B

dm = rc mtot (14)

where rb represents the position in Rb of the generic point mass dmbelonging to the body B; mtot =

∫

Bdm is the total mass of the body B.

If a force is applied to the CoM, the body moves along the force directionwithout rotation.


Inertia matrix and body center-of-mass

If the body frame Rc has its origin in the center of mass C , we have

rc = 0 =

∫

B

rdm = 0 (15)

and the inertia matrix Γc will be defined as:

Γc = −

∫

B

S2(r)dm =

∫

B

[

‖r‖2 I− r(r)T]

dm

=


(16)


Rotation of the reference system

Take two rotated reference frames with the same origin O, Ra and Rb,with matrix Ra

b representing the rotation from Ra to Rb as well asrepresenting Rb in Ra.

Call Γao and Γb

o the two inertia matrices; the relation between them is

Γao = Ra

bΓboR

ba = Ra

bΓbo(R

ab)

T (17)

orΓaoR

ab = Ra

bΓbo .

using the usual notation Rba = (Ra

b)T.

If both RF are body frames, the matrix Rab is constant, otherwise is

time-variant; in this last case also the inertia matrix is time-variant.


Parallel axes theorem

Let us assume to know the inertia matrix Γc with respect to a referenceframe Rc (with origin in the center of mass C ), and willing to computethe inertia matrix with respect to another frame Ro , with different originO, only translated with respect to Rc .

We express the relation between the infinitesimal mass ρ dV = dm wrt thetwo points O and C . If rc is the position of dm in Rc , then

ro = toc + rc

where toc =[

tx ty tz]T

is the translation of the reference frame from Oto C .

It is now possible to compute the new inertia matrix Γo , taking intoaccount (16), as

Γo = Γc −mtotS2(toc ) = Γc +mtot

[

‖toc‖2 I− toc (t

oc )

T]

(18)


Parallel axes theorem

Hence

Γo = Γc +mtot

(t2y + t2z ) −txty −txtz−txty (t2x + t2z ) −ty tz−txtz −ty tz (t2x + t2y )

(19)

that givesΓo,xx = Γc,xx +mtot

(

t2y + t2z)

Γo,yy = Γc,yy +mtot

(

t2x + t2z)

Γo,zz = Γc,zz +mtot

(

t2x + t2y)

.

(20)

The inertia products can be written as

Γo,xy = Γc,xy −mtottxty

Γo,xz = Γc,xz −mtottx tz

Γo,yz = Γc,yz −mtotty tz

(21)


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Momenta and inertia matrices - polito.it · Momenta and inertia matrices Basilio Bona DAUIN–PolitecnicodiTorino Semester 1, ... If the system rotates around a generic axis with