Satellite Dynamics: Principal MOI, Inertia Matrix Analysis & Rotation

This assignment provides a comprehensive solution to analyzing satellite rotation, focusing on the principal moments of inertia (MOI) and the inertia matrix. It begins by establishing assumptions about the satellite as a rotating body and discusses the importance of principal axes. The solution details the computation of the inertia tensor, angular momentum, and rotational energy. MATLAB is utilized to compute eigenvalues and eigenvectors of the inertia matrix, and the results are interpreted to determine the Direction Cosine Matrix (DCM). The assignment further explores angular momentum values and applies gyro-sensor measurements to analyze angular velocity. Finally, similarity transformations are performed to validate the results, demonstrating the relationship between different coordinate systems and the inertia matrix. Desklib offers this solved assignment and many other resources to support students in their studies.

SOLUTION
Question 1
Assumption: The satellite is a rotating body about a fixed axis. From the z-axis, it is possible to
determine that,
I z = ( I x , I y , Iz )
Objects are not constrained to rotate about a fixed axis and a tumbling object in free fall.
The satellite has certain spin axis directions for which the angular momentum, L, and the rotation
axis, ω, are in the same direction, and they are referred to as the principal axes. Any symmetries
of the object make these principal axes and the calculations are easier when symmetries are
implemented. One can compute the single inertia tensor through a point that takes account of the
shape of the object independent of the axis of rotation, and the angular momentum about any
axis through that point can be determined. The arbitrary angular velocity vector is given as,
ω=(ωx , ω y ,ωz )
The angular momentum is given as,
L=∑ mα rα x v α=∑ mα rα x ( ω x rα )
The position is given in terms of r for the x, y, and z coordinate system,
r = ( x , y , z )
¿ r x ( ω x r )
r x ( ω x r )=
[ ( y2 + z2 ) ωx−xy ωy−xz ωz ,
− yx ωx+ ( z2 + x2 ) ωy − yz ωz ,
−zx ωx−zy ω y+ ( x2 + y2 ) ωz
]
The cross-product is written as,
( A x ( B x C ) =B ( A .C ) −C ( A . B ) )
The angular momentum is expressed using the following components,

Such that,
Using the Einstein summation notation, the function can be summarized as,
Li=I ik ωk
The matrix may be re-written in the form,
L=Iω
The inertia tensor is given as,
The elements of the moment of inertia tensor is given as,

The mass moment of inertia, ⃑
H=I⃑ Ω
The cross term products of the inertia are equal hence, I xy =I yx⃑
H=
[ H x
H y
H z ]=
[ I xx −I xy −I xz
−I yx I yy −I yz
−I zx −I zy Izz ][ Ωx
Ω y
Ωz ]
The mass moment of inertia for a rigid body,
I =∑ r2 dm=∫
m
❑
r2 dm
I =∫r2 ρdV
The rotational energy is obtained as,
I xx=∫( y2 +z2)dm
I yy=∫(x2+z2 )dm
I zz=∫( y2 + x2)dm
I xy =∫(xy )dm
I xz=∫(xz )dm

I yz=∫( yz )dm
E=1
2 I ij Ωi Ω j
Using the symmetric tensor, the MOI tensor can be reduced to diagonal form through the
appropriate choice of x, y, z coordinates, such that,
I =
[I xx 0 0
0 I yy 0
0 0 I zz ]
It gives the principle moments of inertia. The spin stabilization of the entire spacecraft
rotates about the vertical axis and the spinning sensors and payloads. It is cylindrical geometry
and solar arrays. For satellite precision, the satellite thruster is fired to change its spin axis and
the satellite is rotated by a small angle such that,
∆ ψ =2 FR ( Δ t )
Iω ; ω= Δϕ
Δt
∆ ψ =2 FR ( Δϕ )
I ω2
QUESTION 2
I A= [ 3700 −400 −150
−400 5800 200
−150 200 8800 ] ( kg m2 )
A – Defines the arbitrary coordinate system used when the satellite is being designed.
To calculate the principle MOI outputs,
From the output of the system based on the inertia matrix,
I xx=3700
I yy=5800
I zz=8800

To find the principle axes by obtaining the diagonalization of the inertia of the tensor for the
satellite,
A=UT IU
U = ( p1 , p2 , p3 ) −rotation ¿
I=∑
i=1
N
mi [ ( ri . ri ) ∑
α =1
3
e0∗eα−ri∗ri ]
Therefore, using MATLAB to perform the computation,
[ 3700 0 0
0 5800 0
0 0 8800 ]
QUESTION 3
Satellite rotation principal MOI and inertia matrix
I=[3700 -400 -150; -400 5800 200;-150 200 8800]
% Find eigenvalues and vectors of T
[V,D] = eig(I)
disp(D)
I =
3700 -400 -150
-400 5800 200
-150 200 8800
V =
0.9836 -0.1768 -0.0347
0.1788 0.9813 0.0706
0.0216 -0.0757 0.9969

D =
1.0e+03 *
3.6240 0 0
0 5.8566 0
0 0 8.8194
Extracting the first column
e1=V(:,1)
e1 =
0.9836
0.1788
0.0216
Extracting the Second column
e2=V(:,2)
e2 =
-0.1768
0.9813
-0.0757
Creating the cross product of the 1st two columns
… performing a cross product of the first two columns e3=V(:,3);
e3=cross(e1,e2)

e3 =
-0.0347
0.0706
0.9969
To detemrine the DCM using the eigenvectors as the columns
C_BA=[e1,e2,e3]
C_BA =
0.9836 -0.1768 -0.0347
0.1788 0.9813 0.0706
0.0216 -0.0757 0.9969
QUESTION 4

To find the values of angular momentum,
ω= L
I =rotation about the dia gonal
Using MATLAB, having obtained the eigenvectors and the inertia matrix,
% To obtain the angular values, w
L=D;
w=L\I;
ω= [ 1.021 −0.1104 −0.0414
−0.0683 0.9903 0.0341
−0.0170 0.0227 0.9978 ]
QUESTION 5
A gyro-sensor measures the angular velocity of the satellite when its aligned with the A axes.
The sensor is measuring,
ω A= [0.4
0.5
0.8 ] ( rad
s )
r x ( ω x r )=
[ ( y2 + z2 ) ωx−xy ωy−xz ωz ,
− yx ωx+ ( z2 + x2 ) ωy − yz ωz ,
−zx ωx−zy ω y+ ( x2 + y2 ) ωz
]
Using the values of the coordinates as obtained,
1.0e+03 *
3.6240 0 0
0 5.8566 0
0 0 8.8194
In matrix notation,
¿ [3624 0 0
0 5856.6 0
0 0 8819.4 ]

QUESTION 6
The similarity transformations are performed and output obtained as results with labels and units
I B=CAB I A CBA
C AB= [ CBA ]T
The similarity uses lemma notation s to prove that the input matrix is the same as the output
matrix after the transformation is performed,
CB A =
[ 0.9836 −0.1768 −0.0347
0.1788 0.9813 0.0706
0.0216 −0.0757 0.9969 ]
C AB= [ 0.9836 0.1788 0.0216
−0.1768 0.9813 −0.0757
−0.0347 0.0706 0.9969 ]
I A= [ 3700 −400 −150
−400 5800 200
−150 200 8800 ]
I B=CAB I A CBA
I B= [0.9836 −0.1768 −0.0347
0.1788 0.9813 0.0706
0.0216 −0.0757 0.9969 ]∗
[ 3700 −400 −150
−400 5800 200
−150 200 8800 ]∗
[ 0.9836 0.1788 0.0216
−0.1768 0.9813 −0.0757
−0.0347 0.0706 0.9969 ]
I B= [3624 0 0
0 5856.6 0
0 0 8819.4 ]