Question 10.4: The inertial components of the angular momentum of a torque-...

Substituting the Euler angles from (b) into Equation 9.117, we obtain the matrix of the transformation from the inertial frame to the body frame,

[Q]_{Xx}=\left[\begin{matrix} 0.03086 &0.6720& 0.7399\\−0.9646& −0.1740& 0.1983\\0.2620& −0.7198& 0.6428\end{matrix} \right]                    (d)

We use this to obtain the components of H_{G} in the body frame,
\left\{H_{G}\right\}_{x}= [Q]_{Xx}\left\{H_{G}\right\}_{X}=\left[\begin{matrix} 0.03086 &0.6720& 0.7399\\−0.9646& −0.1740& 0.1983\\0.2620& −0.7198& 0.6428\end{matrix} \right]\left\{\begin{matrix} 320\\−375\\450 \end{matrix} \right\}

=\left\{\begin{matrix} 90.86\\−154.2\\643.0 \end{matrix} \right\}(kg · m^{2}/s)                                                     (e)

In the body frame \left\{{HG}\right\}_{x} = [I_{G}]\left\{{ω}\right\}_{x}, where \left\{{ω}\right\}_{x}are the components of angular
velocity in the body frame. Thus
\left\{\begin{matrix} 90.86\\−154.2\\643.0 \end{matrix} \right\}=\left[\begin{matrix} 1000 & 0 & 0 \\ 0 & 2000 &0 \\ 0& 0 & 3000 \end{matrix} \right]\left\{\omega \right\}_{x}

or, solving for \left\{ω\right\}_{x},
\left\{\omega \right\}_{x}=\left[\begin{matrix} 1000 & 0 & 0 \\ 0 & 2000 &0 \\ 0& 0 & 3000 \end{matrix} \right]^{-1}\left\{\begin{matrix} 90.86\\−154.2\\643.0 \end{matrix} \right\}=\left\{\begin{matrix}0.09086\\−0.07709\\0.2144 \end{matrix} \right\}(rad/s)                   (f)

Euler’s equations of motion (Equation 9.72a) may be written for the case at hand as
M_{net}=\dot{H})_{rel}+\omega \times H                                        (9.72a)

[I_{G}]\left\{\alpha \right\}_{x}+\left\{\omega \right\}_{x}\times ([I_{G}]\left\{\omega \right\}_{x} )=\left\{0\right\}                  (g)
where \left\{\alpha \right\}_{x} is the absolute acceleration in body frame components. Substituting (c) and (f) into this expression, we get
\left[\begin{matrix}1000&0&0\\0&2000&0\\0&0&3000 \end{matrix} \right]\left\{\alpha \right\}_{x}+\left\{\begin{matrix} 0.09086\\−0.07709\\0.2144 \end{matrix} \right\}

so that, finally
\left\{\alpha \right\}_{x}=-\left[\begin{matrix}1000&0&0\\0&2000&0\\0&0&3000 \end{matrix} \right]^{-1} \left\{\begin{matrix}-16.52\\−38.95\\−7.005 \end{matrix} \right\}=\left\{\begin{matrix}0.01652\\0.01948\\0.002335 \end{matrix} \right\}(rad/s^{2})                         (h)

These are the components of the angular acceleration in the body frame. To transform them into the inertial frame we use

That is,
\underline{\alpha =−0.01766\hat{I} + 0.006033\hat{J}+ 0.01759\hat{K}(rad/s^{2})}