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Inertial components of the angular acceleration

Example 10.4, Orbital Mechanics for Engineering Students, 2nd Edition.

The inertial components of the angular momentum of a torque-free rigid body are

Hg = [320; -375; 450 ];       %[kg*m^2/s] IJK
% the Euler angles [deg] are
fi      =  20;
theta   =  50;
psi     =  75;
% The inertia tensor in the body-fixed principal frame is
Ig = [1000, 0,    0;
         0, 2000, 0;
         0, 0,    3000]; %[kg*m^2]

Obtain the inertial components of the (absolute) angular acceleration Matrix of the transformation from body-fixed frame to inertial frame

QxX = [-sind(fi)*cosd(theta)*sind(psi) + cosd(fi)*cosd(psi), ...
 -sind(fi)*cosd(theta)*cosd(psi) - cosd(fi)*sind(psi),sind(fi)*sind(theta);
 cosd(fi)*cosd(theta)*sind(psi) + sind(fi)*cosd(psi),...
  cosd(fi)*cosd(theta)*cosd(psi) - sind(fi)*sind(psi),-cosd(fi)*sind(theta);
    sind(theta)*sind(psi),   sind(theta)*cosd(psi),    cosd(theta)
       ]
QxX =

    0.0309   -0.9646    0.2620
    0.6720   -0.1740   -0.7198
    0.7399    0.1983    0.6428

Matrix of the transformation from inertial frame to body-fixed frame

QXx = QxX'
QXx =

    0.0309    0.6720    0.7399
   -0.9646   -0.1740    0.1983
    0.2620   -0.7198    0.6428

Obtain the components of HG in the body frame

Hgx = QXx*Hg            % [kg*m^2/s]
Hgx =

   90.8616
 -154.1810
  643.0376

The components of angular velocity in the body frame

Ig_inv = inv(Ig);
wx = Ig_inv*Hgx         % [rad/s]
wx =

    0.0909
   -0.0771
    0.2143

From Euler ’s equations of motion we calculate angular acceleration in the body frame.

alfa_x = - Ig_inv*cross(wx,Ig*wx)               % [rad/s^2]
alfa_x =

    0.0165
    0.0195
    0.0023

Angular acceleration in the inertial frame

alfa_X = QxX*alfa_x                             % [rad/s^2] IJK
alfa_X =

   -0.0177
    0.0060
    0.0176

Published with MATLAB® 7.10

 

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