Momenta and inertia matrices Basilio Bona DAUIN – Politecnico di Torino Semester 1, 2015-16 B. Bona (DAUIN) Kinematics Semester 1, 2015-16 1 / 22
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 .
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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.
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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)
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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)
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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
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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 =
Γxx Γxy ΓxzΓyx Γyy ΓyzΓzx Γzy Γzz
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.
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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
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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.
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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)
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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.
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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)
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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)
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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
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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)
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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.
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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.
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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.
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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
=
Γxx Γxy ΓxzΓyx Γyy ΓyzΓzx Γzy Γzz
(16)
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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.
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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)
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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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