The inertial components of the angular momentum of a torque-free rigid body are
$H_{G} = 320\hat{I}− 375\hat{J}+ 450\hat{K}(kg·m^{2}/s)$ (a)
The Euler angles are
$\begin{matrix} \phi =20° & \theta =50° & \psi =75°\end{matrix}$ (b)
If the inertia tensor in the body-fixed principl frame is
$[I_{G}]=\left[\begin{matrix} 1000&0&0\\0&2000&0\\0&0&3000 \end{matrix}\right] (kg · m^{2})$ (c)

Question

The inertial components of the angular momentum of a torque-free rigid body are
[latex]H_{G} = 320\hat{I}− 375\hat{J}+ 450\hat{K}(kg·m^{2}/s)[/latex] (a)
The Euler angles are
[latex]\begin{matrix} \phi =20° & heta =50° & \psi =75°\end{matrix}[/latex] (b)
If the inertia tensor in the body-fixed principl frame is
[latex][I_{G}]=\left[\begin{matrix} 1000&0&0\0&2000&0\0&0&3000 \end{matrix} ight] (kg · m^{2})[/latex] (c)

calculate the inertial components of the (absolute) angular acceleration.

Accepted Answer

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,

[latex][Q]_{Xx}=\begin{bmatrix}\cos \phi \cos \psi − \sin\phi \sin \psi \cos heta & \sin \phi \cos\psi +\cos \phi \cos heta \sin \psi & \sin heta \sin\psi \ -\cos \phi \sin \psi - \sin\phi \cos heta \cos \psi &−\sin \phi \sin \psi +\cos \phi \cos heta \cos \psi & \sin heta \cos \psi \\sin \phi \sin heta & -\cos \phi \sin heta & \cos heta \end{bmatrix}[/latex] (9.117)

[latex][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} ight][/latex] (d)

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

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

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

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

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

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

[latex] imes \left(\left[\begin{matrix}1000&0&0\0&2000&0\0&0&3000 \end{matrix} ight]\left\{\begin{matrix} 0.09086\−0.07709\0.2144 \end{matrix} ight\} ight)=\left\{\begin{matrix}0\0\0 \end{matrix} ight\}[/latex]

[latex]\left[\begin{matrix}1000&0&0\0&2000&0\0&0&3000 \end{matrix} ight]\left\{\alpha ight\}_{x}+ \left\{\begin{matrix}-16.52\−38.95\−7.005 \end{matrix} ight\}=\left\{\begin{matrix}0\0\0 \end{matrix} ight\}[/latex]

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

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

[latex]\left\{\alpha ight\}_{X}=[Q]_{xX}\left\{\alpha ight\}_{x}=([Q]_{Xx})^{T}\left\{\alpha ight\}_{x}[/latex]

[latex]=\left[\begin{matrix} 0.03086& −0.9646& 0.2620\0.6720& −0.1740& −0.7198\0.7399 &0.1983& 0.6428\end{matrix} ight]\left\{\begin{matrix} 0.01652\0.01948\0.002335 \end{matrix} ight\} =\left\{\begin{matrix} −0.01766\0.006033\0.01759 \end{matrix} ight\}(rad/s^{2})[/latex]

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

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

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